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ELEMENTARY
COMMUTATIVE ALGEBRA
LECTURE NOTES
H.A. NIELSEN
DEPARTMENT OF MATHEMATICAL SCIENCES
UNIVERSITY OF AARHUS
2005
Elementary Commutative Algebra
H.A. Nielsen
Contents
7
Prerequisites
1.
A dictionary on rings and ideals
1.1. Rings
1.2. Ideals
1.3. Prime ideals
1.4. Chinese remainders
1.5. Unique factorization
1.6. Polynomials
1.7. Roots
1.8. Fields
1.9. Power series
9
9
11
13
14
15
16
18
19
20
2.
Modules
2.1. Modules and homomorphisms
2.2. Submodules and factor modules
2.3. Kernel and cokernel
2.4. Sum and product
2.5. Homomorphism modules
2.6. Tensor product modules
2.7. Change of rings
21
21
23
25
28
30
34
37
3.
Exact sequences of modules
3.1. Exact sequences
3.2. The snake lemma
3.3. Exactness of Hom
3.4. Exactness of tensor
3.5. Projective modules
3.6. Injective modules
3.7. Flat modules
41
41
45
50
53
55
56
60
4.
Fraction constructions
4.1. Rings of fractions
4.2. Modules of fractions
4.3. Exactness of fractions
4.4. Tensor modules of fractions
4.5. Homomorphism modules of fractions
4.6. The polynomial ring is factorial
63
63
65
67
69
70
71
5.
Localization
5.1. Prime ideals
5.2. Localization of rings
73
73
75
5
6
CONTENTS
5.3.
5.4.
5.5.
5.6.
Localization of modules
The local-global principle
Flat ring homomorphisms
Faithfully flat ring homomorphisms
77
79
81
83
6.
Finite modules
6.1. Finite modules
6.2. Free modules
6.3. Cayley-Hamilton’s theorem
6.4. Nakayama’s lemma
6.5. Finite presented modules
6.6. Finite ring homomorphisms
87
87
89
91
93
94
99
7.
Modules of finite length
7.1. Simple modules
7.2. The length
7.3. Artinian modules
7.4. Artinian rings
7.5. Localization
7.6. Local artinian ring
101
101
102
105
107
109
110
8.
Noetherian rings
8.1. Noetherian modules
8.2. Noetherian rings
8.3. Finite type rings
8.4. Power series rings
8.5. Localization of noetherian rings
8.6. Prime filtrations of modules
113
113
115
117
118
120
121
9.
Primary decomposition
9.1. Zariski topology
9.2. Support of modules
9.3. Ass of modules
9.4. Primary modules
9.5. Decomposition of modules
9.6. Decomposition of ideals
123
123
126
128
132
134
136
10. Dedekind rings
10.1. Principal ideal domains
10.2. Discrete valuation rings
10.3. Dedekind domains
139
139
141
143
Bibliography
145
Index
147
Prerequisites
The basic notions from algebra, such as groups, rings, fields and their homomorphisms together with some linear algebra, bilinear forms, matrices and determinants.
Linear algebra: Fraleigh & Beauregard, Linear algebra, New York 1995.
Algebra: Niels Lauritzen, Concrete abstract algebra, Cambridge 2003.
Also recommended: Jens Carsten Jantzen, Algebra 2, Aarhus 2004.
The propositions are stated complete and precise, while the proofs are quite short.
No specific references to the literature are given. But lacking details may all be
found at appropriate places in the books listed in the bibliography.
A proposition being important when working in commutative algebra or a proposition containing a final result is named “Theorem”.
Nielsen, University of Aarhus, Spring 2005
7
1
A dictionary on rings and ideals
1.1. Rings
1.1.1. Definition. An abelian group is a set A with an addition A×A → A, (a, b) 7→
a + b and a zero 0 ∈ A satisfying
(1)
(2)
(3)
(4)
associative: (a + b) + c = a + (b + c)
zero: a + 0 = a = 0 + a
negative: a + (−a) = 0
commutative: a + b = b + a
for all a, b, c ∈ A. A subset B ⊂ A is a subgroup if 0 ∈ B and a − b ∈ B for
all a, b ∈ B. The factor group A/B is the abelian group whose elements are the
cosets a + B = {a + b|b ∈ B} with addition (a + B) + (b + B) = (a + b) + B. A
homomorphism of groups φ : A → C respects addition φ(a + b) = φ(a) + φ(b).
The projection π : A → A/B, a 7→ a + B is a homomorphism. If φ(b) = 0 for all
b ∈ B, then there is a unique homomorphism φ0 : A/B → C such that φ = φ0 ◦ π.
1.1.2. Definition. A ring is an abelian group R, addition (a, b) 7→ a + b and zero
0, together with a multiplication R × R → R, (a, b) 7→ ab and an identity 1 ∈ R
satisfying
(1)
(2)
(3)
(4)
associative: (ab)c = a(bc)
distributive: a(b + c) = ab + ac, (a + b)c = ac + bc
identity : 1a = a = a1
commutative : ab = ba
for all a, b, c ∈ R. If (4) is not satisfied then R is a noncommutative ring. A
subring R0 ⊂ R is an additive subgroup such that 1 ∈ R0 and ab ∈ R0 for all
a, b ∈ R0 . The inclusion R0 ⊂ R is a ring extension. A homomorphism of rings
φ : R → S is an additive group homomorphism respecting multiplication and
identity
φ(a + b) = φ(a) + φ(b), φ(ab) = φ(a)φ(b), φ(1) = 1
An isomorphism is a homomorphism φ : R → S having an inverse map φ−1 :
S → R which is also a homomorphism. The identity isomorphism is denoted
1R : R → R.
1.1.3. Remark. (1) A bijective ring homomorphism is an isomorphism.
(2) Recall the usual formulas: a + (−b) = a − b, 0a = 0, (−1)a = −a.
(3) The identity 1 is unique.
(4) A ring R is nonzero if and only if the elements 0 6= 1.
(5) If φ : R → S is a ring homomorphism, then φ(0) = 0 and φ(R) is a subring
of S.
(6) The unique additive group homomorphism Z → R, 1 7→ 1 is a ring homomorphism.
9
10
1. A DICTIONARY ON RINGS AND IDEALS
1.1.4. Proposition. Let R1 , R2 be rings. The product ring is the product of additive groups R1 ×R2 , ((a1 , a2 ), (b1 , b2 )) 7→ (a1 +b1 , a2 +b2 ), with coordinate multiplication ((a1 , a2 ), (b1 , b2 )) 7→ (a1 b1 , a2 b2 ). The element (1, 1) is the identity. The
projections R1 × R2 → R1 , (a1 , a2 ) 7→ a1 and R1 × R2 → R2 , (a1 , a2 ) 7→ a2 are
ring homomorphisms.
1.1.5. Lemma. In a ring R the binomial formula is true
n X
n n−k k
n
(a + b) =
a
b
k
k=0
a, b ∈ R and n a positive integer.
Proof. The multiplication is commutative, so the usual proof for numbers works.
Use the binomial identity
n
n
n+1
+
=
k−1
k
k
together with induction on n.
1.1.6. Definition. a ∈ R is a nonzero divisor if ab 6= 0 for all b 6= 0 otherwise a
zero divisor. a is a unit if there is a b such that ab = 1.
1.1.7. Remark. (1) A unit is a nonzero divisor.
(2) If ab = 1 then b is uniquely determined by a and denoted b = a−1 .
1.1.8. Definition. A nonzero ring R is a domain if every nonzero element is a
nonzero divisor and a field if every nonzero element is a unit. Clearly a field is a
domain.
1.1.9. Example. The integers Z is a domain. The units in Z are {±1}. The rational
numbers Q, the real numbers R and the complex numbers C are fields. The
natural numbers N is not a ring.
1.1.10. Example. The set of n × n-matrices with entries from a commutative ring
is an important normally noncommutative ring.
1.1.11. Exercise.
(1) Show that the product of two domains is never a domain.
(2) Let R be a ring. Show that the set of matrices
a b U2 =
a, b ∈ R
0 a
with matrix addition and matrix multiplication is a ring.
(3) Show that the set of matrices with real number entries
a −b a, b ∈ R
b a
with matrix addition and multiplication is a field isomorphic to C.
(4) Show that the composition of two ring homomorphisms is again a ring homomorphism.
(5) Show the claim 1.1.3 that a bijective ring homomorphism is a ring isomorphism.
(6) Let φ : 0 → R be a ring homomorphism from the zero ring. Show that R is itself the
zero ring.
1.2. IDEALS
11
1.2. Ideals
1.2.1. Definition. Let R be a ring. An ideal I is an additive subgroup of R such
that ab ∈ I for all a ∈ R, b ∈ I. A proper ideal is an ideal I 6= R.
1.2.2.
P Lemma.
T Let {Iα } be a family of ideals in R. Then the additive subgroups
I
and
α
α
α Iα are ideals.
P
P
Proof.
The claim for
is clear. Use the formulas
bα + cα =
P
P the intersection
P
(bα + cα ) and a bα = abα to conclude the claim for the sum.
1.2.3. Definition. The intersection, 1.2.2, of all ideals containing a subset B ⊂ R
is the ideal generated by B and denoted (B) = RB = BR. It is the smallest ideal
containing B. A principal ideal (b) = Rb is an ideal generated by one element. A
finite ideal (b1 , . . . , bn ) is an ideal generated by finitely many elements. The zero
ideal is (0) = {0}. The ring itself is a principal ideal, (1) = R. The ideal product
of two ideals I, J is denoted IJ and is the ideal generated by all ab, a ∈ I, b ∈ J.
This generalizes to the product of finitely many ideals. The power of an ideal is
denoted I n . The colon ideal I : J is the ideal of elements a ∈ R such that aJ ⊂ I.
1.2.4. Example. (1) Every ideal in Z is principal.
(2) In a field (0), (1) are the only ideals.
(3) A subring is normally not an ideal.
(4) Let K be a field. In K × K there are four ideals (0), (1), ((1, 0)), ((0, 1)).
P
1.2.5. Proposition. Let R be a ring and B a subset, then RB = b∈B Rb
RB = {a1 b1 + · · · + an bn |ai ∈ R, bi ∈ B}
A principal ideal is
(b) = Rb = {ab|a ∈ R}
A finite ideal is
(b1 , . . . , bn ) = Rb1 + · · · + Rbn
Proof. The righthand side is contained in the ideal RB. Moreover the righthand
side is an ideal containing B, so equality.
1.2.6. Definition. Let φ : R → S be a ring homomorphism. For an ideal J ⊂ S
the contracted ideal is φ−1 (J) ⊂ R and denoted J ∩ R. The kernel is the ideal
Ker φ = φ−1 (0). For an ideal I ⊂ R the extended ideal is the ideal φ(I)S ⊂ S
and denoted IS. Note that (J ∩ R)S ⊂ J and I ⊂ (IS) ∩ R.
1.2.7. Lemma. Let I ⊂ R be an ideal and let R/I be the additive factor group.
The multiplication
R/I × R/I → R/I, (a + I, b + I) 7→ ab + I
is well defined. Together with the addition the conditions of 1.1.2 are satisfied.
Proof. If a + I = a0 + I and b + I = b0 + I then a − a0 , b − b0 ∈ I and so
ab − a0 b0 = a(b − b0 ) + b0 (a − a0 ) ∈ I. Therefore ab + I = a0 b0 + I and the
multiplication is well defined. Clearly 1.1.2 are satisfied.
1.2.8. Definition. Let R be a ring and I an ideal, then the factor ring is the additive
factor group R/I with addition (a + I, b + I) 7→ (a + b) + I and multiplication,
1.2.7, (a + I, b + I) 7→ ab + I. The projection π : R → R/I, a 7→ a + I is a ring
homomorphism.
12
1. A DICTIONARY ON RINGS AND IDEALS
1.2.9. Theorem. Let φ : R → S be a ring homomorphism.
(1) Let I ⊂ Ker φ be an ideal. Then there is a unique ring homomorphism
φ0 : R/I → S such that φ = φ0 ◦ π.
φ
R
π
/
=S
φ0
R/I
(2) The homomorphism φ0 : R/ Ker φ → S is a ring isomorphism onto the
subring φ(R) of S.
φ
R
π
/ φ(R)
:
φ0
R/ Ker φ
(3) For any ideal J ⊂ S, I = φ−1 (J) ⊂ R is an ideal and the map φ0 : R/I →
S/J is an injective ring homomorphism.
Proof. The statements are clear for the addition and the factor map φ0 (a + I) =
φ(a) is clearly a ring homomorphism.
1.2.10. Corollary. Let π : R → R/I be the projection. The map I 0 7→ J =
π −1 (I 0 ) gives a bijective correspondence between ideals I 0 in R/I and ideals J in
R containing I. Also I 0 = π(J) = J/I. This correspondence preserves inclusions,
sums and intersections of ideals.
1.2.11. Corollary. Let I ⊂ J ⊂ R be ideals. Then there is a canonical isomorphism
R/J → (R/I)/(J/I)
Proof. The kernel of the surjective east-south composite
/ R/I
R
/ (R/I)/(J/I)
R/J
is J. By 1.2.9 the horizontal lower factor map gives the isomorphism.
1.2.12. Example. For any integer n the ideals in the factor ring Z/(n) correspond
to ideals (m) ⊂ Z where m divides n.
1.2.13. Definition. Let R be a ring. The additive kernel of the unique ring homomorphism Z → R is a principal ideal generated by a natural number char(R), the
characteristic of R. Z/(char(R)) is isomorphic to the smallest subring of R.
1.2.14. Proposition. If the characteristic char(R) = p is a prime number, then
the Frobenius homomorphism R → R, a 7→ ap is a ring homomorphism.
Proof. By the binomial formula 1.1.4
p X
p p−k k
p
(a + b) =
a b = ap + bp
k
k=0
1.3. PRIME IDEALS
since a prime number p divides
1.2.15. Exercise.
p
k
13
, 0 < k < p. Clearly (ab)p = ap bp .
(1) Let I, J be ideals in R. Show that the ideal product
IJ = {a1 b1 + · · · + an bn |ai ∈ I, bi ∈ J}
Let I ⊂ R be an ideal. Show that I = I : R.
Show that a ∈ R is a unit if and only if (a) = R.
Show that a ring is a field if and only if (0) 6= (1) are the only two ideals.
Show that a nonzero ring K is a field if and only if any nonzero ring homomorphism
φ : K → R is injective.
(6) Let m, n be natural numbers. Determine the ideals in Z
(2)
(3)
(4)
(5)
(m, n), (m) + (n), (m) ∩ (n), (m)(n)
as principal ideals.
(7) Show that a additive cyclic group has a unique ring structure.
(8) Let p be a prime number. What is the Frobenius homomorphism on the ring Z/(p)?
1.3. Prime ideals
1.3.1. Definition. Let R be a ring and P 6= R a proper ideal.
(1) P is a prime ideal if for any product ab ∈ P either a ∈ P or b ∈ P . This
amounts to: if a, b ∈
/ P then ab ∈
/ P.
(2) P is a maximal ideal if no proper ideal 6= P contains P .
1.3.2. Proposition. Let P be a prime ideal and I1 , . . . In ideals such that I1 . . . In ⊂
P , then some Ik ⊂ P .
Proof. If there exist ak ∈ Ik \P for all k, then since P is prime a1 . . . an ∈
I1 . . . In \P contradicting the inclusion I1 . . . In ⊂ P .
1.3.3. Proposition. Let R be a ring and P an ideal.
(1) P is a prime ideal if and only if R/P is a domain.
(2) P is a maximal ideal if and only if R/P is a field.
Proof. Remark P 6= R ⇔ R/P 6= 0. (1) Assume a + P, b + P are nonzero in
R/P . Then a, b ∈
/ P . So if P is prime then by 1.3.1 ab ∈
/ P and ab + P is nonzero
in R/P . It follows that R/P is a domain. The converse is similar.
(2) Assume R/P is a field and a ∈
/ P . Then a + P is a unit in R/P and there is
b such that ab − 1 ∈ P . It follows that the ideal (a) + P = R and therefore P is
maximal. The converse is similar.
1.3.4. Corollary. (1) A maximal ideal is a prime ideal.
(2) A ring is an domain if and only if the zero ideal is a prime ideal.
(3) A ring a field if and only if the zero ideal is a maximal ideal.
1.3.5. Corollary. (1) If φ : R → S is a ring homomorphism and Q ⊂ S a prime
ideal then φ−1 (Q) is a prime ideal of R.
(2) Let I ⊂ R be an ideal. An ideal I ⊂ P is a prime ideal in R if and only if
P/I is a prime ideal in R/I.
(3) Let I ⊂ R be an ideal. An ideal I ⊂ P is a maximal ideal in R if and only if
P/I is a maximal ideal in R/I.
Proof. (1) By 1.2.9 R/φ−1 (Q) is a subring of the domain S/Q. (2) (3) By 1.2.11
R/P ' (R/I)/(P/I).
14
1. A DICTIONARY ON RINGS AND IDEALS
1.3.6. Example. An ideal in Z is a prime ideal if it is generated by 0 or a prime
number. Any nonzero prime ideal is a maximal ideal.
1.3.7. Definition. For an ideal I in a ring R the radical is
√
I = {a ∈ R|an ∈ I for some n}
n
a is nilpotent
√ is a = 0 for some positive integer n. A ring is reduced if the
nilradical 0 = 0, that is if 0 is the only nilpotent element.
1.3.8. Proposition. (1) The radical of an ideal is an ideal.
(2) The nilradical is contained in any prime ideal.
(3) A domain is reduced.
Proof. (1) If am , bn ∈ I then by the binomial formula
m+n
X m + n m+n
(a + b)
=
am+n−k bk ∈ I
k
k=0
and the radical is an ideal. (2) (3) Clearly a nilpotent element is contained in any
prime ideal.
1.3.9. Exercise.
(2)
(3)
(4)
(5)
(6)
(7)
(1) Show that the characteristic of a domain is either 0 or a prime
number.
Let m, n be a natural numbers. Show that n + (m) ∈ Z/(m) is a unit if and only if
m, n are relative prime.
Let m be a natural number. Show that Z/(m) is reduced if m is square free.
Show that a product of reduced rings is reduced.
Let a be nilpotent. Show that √
1 − a is√
a unit.
√
√
Let I, J be ideals. Show that IJ = I ∩ J = I ∩ J. √
Assume an ideal I is contained in a prime ideal P . Show that I ⊂ P .
1.4. Chinese remainders
1.4.1. Definition. Ideals I, J ⊂ R are comaximal ideals if I + J = R.
1.4.2. Theorem (Chinese remainder theorem). Let I1 , . . . , Ik be pairwise comaximal ideals in a ring R. Then
(1) For a1 , . . . , ak ∈ R there is a a ∈ R, such that a−am ∈ Im for m = 1, . . . , k
(2)
I1 · · · Ik = I1 ∩ · · · ∩ Ik
(3) The product of projections
R/I1 · · · Ik → R/I1 × · · · × R/Ik
is an isomorphism.
Proof. (1) For each m
R=
Y
(Im + In ) = Im +
n6=m
Y
In
n6=m
Q
So choose um ∈ Im and vm ∈ n6=m In with um + vm = 1. Put a = a1 v1 + · · · +
ak vk . Then a − am = · · · + am um + · · · ∈ Im . (2) For a in the intersection assume
by induction that a ∈ I2 · · · Ik . From the proof of (1) a = u1 a + av1 ∈ I1 · · · Ik .
(3) Surjectivity follows from (1). The kernel is the intersection which by (2) is the
product. 1.2.9 gives the isomorphism.
1.5. UNIQUE FACTORIZATION
15
1.4.3. Corollary. Let P1 , . . . , Pk be pairwise different maximal ideals in a ring R.
Then
P1n1 · · · Pknk = P1n1 ∩ · · · ∩ Pknk
and
R/P1n1 · · · Pknk → R/P1n1 × · · · × R/Pknk
is an isomorphism.
1.4.4. Definition. An element e in a ring R is idempotent if e = e2 . A nontrivial
idempotent is an idempotent e 6= 0, 1.
1.4.5. Proposition. A ring R is a product of two nonzero rings if and only if it
contains a nontrivial idempotent e.
Proof. Use that the ideals Re and R(1 − e) are proper and comaximal.
(1) Show that for a prime number p the rings Z/(p2 ) and Z/(p) ×
Z/(p) are not isomorphic.
(2) Let n = pn1 1 . . . pnk k be a factorization into different primes. Show that
1.4.6. Exercise.
Z/(pn1 1 · · · pknk ) → Z/(pn1 1 ) × · · · × Z/(pnk k )
is an isomorphism.
(3) Let elements e1 + e2 = 1 with e1 e2 = 0 be given in a ring R. Show that R '
R/(e1 ) × R/(e2 ).
√ √
(4) Let I, J be ideals such that I, J are comaximal. Show that I, J are comaximal.
1.5. Unique factorization
1.5.1. Lemma. Let R be a domain and (a) = (b) principal ideals. Then there is a
unit u ∈ R such that b = ua.
Proof. b = ua and a = vb giving b = uvb. If b 6= 0 then uv = 1 showing that u is
a unit.
1.5.2. Definition. Let R be a domain and P the set of principal ideals different
from (0) and R. By 1.5.1 multiplication of generators gives a well defined multiplication of principal ideals on P. Inclusion in P, (a) ⊂ (b) is division and written
b|a. An element in P maximal for inclusion is an irreducible principal ideal. A
generator of an irreducible element is an irreducible element in R. A principal
prime ideal is a prime divisor. A generator of a prime divisor is a prime element
in R. Clearly a prime divisor is irreducible.
1.5.3. Definition. A domain R is a unique factorization domain if
(1) Every irreducible ideal in P is a prime divisor.
(2) Every element in P is a product of irreducible elements.
1.5.4. Proposition. In a unique factorization domain the factorization of elements
in P into prime divisors is unique up to order.
Proof. Proceed by induction on the shortest factorization of an element in P. Let
(a1 ) . . . (am ) = (b1 ) . . . (bn ) be factorizations into irreducibles. (a1 ) is a prime
ideal, so by 1.3.2 and reordering (b1 ) = (a1 ). By cancellation m − 1 = n − 1 and
(ai ) = (bi ) after a reordering.
1.5.5. Definition. A domain R is a principal ideal domain if every ideal is principal.
16
1. A DICTIONARY ON RINGS AND IDEALS
1.5.6. Theorem. A principal ideal domain R is a unique factorization domain.
Proof. Let (a) be irreducible and x, y ∈
/ (a). Then (a, x), (a, y) are principal ideals
properly containing (a) giving (a, x) = (a, y) = R. Let ba + cx = da + ey = 1
and look at (ba+cx)(da+ey) = 1 to see that xy ∈
/ (a). It follows that (a) is prime,
part (1) of 1.5.3. If (b) is not irreducible, then (b) = (a1 )(b1 ) for some irreducible
(a1 ) and (b) ⊂ (b1 ). Continue this process to get (b) = (a1 ) . . . (an )(bn ) for some
irreducibles
(a1 ), . . . (an ). The chain of ideals (b) ⊂ (b1 ) ⊂ · · · ⊂ (bn ) has a
S
union n (bn ) which is a principal ideal. The generator of the union must be in
some (bn ). Therefore (bn ) = (bn+1 ) giving that (bn ) is irreducible. This gives a
factorization required in 1.5.3 part (2).
1.5.7. Example. The integers Z is a principal ideal domain and therefore a unique
factorization domain.
1.5.8. Definition. The supremum in the set of all principal ideals of a set of elements in P is the greatest common divisor and an infimum is the least common
multiple.
1.5.9. Corollary. In a unique factorization domain the greatest common divisor
and the least common multiple of a finite set of elements exist.
If (a) = (p1 )m1 . . . (pk )mk and (b) = (p1 )n1 . . . (pk )nk with mi , ni ≥ 0 then greatest common divisor is (p1 )min(m1 ,n1 ) . . . (pk )min(mk ,nk ) and least common multiple
is (p1 )max(m1 ,n1 ) . . . (pk )max(mk ,nk ) .
1.5.10. Exercise.
(1) Show that an irreducible element in a principal ideal domain generates a maximal ideal.
(2) Show that
√ there are infinitely many prime numbers. √
√
(3) Let Z[ −1] be the smallest subring of C containing −1. Show that Z[ −1] is a
principal
√ ideal domain.
√
√
(4) Let Z[ −5] be the smallest subring of C containing −5. Show that Z[ −5] is not
a unique factorization domain.
1.6. Polynomials
1.6.1. Definition. Let R be a L
ring. The polynomial ring R[X] is the additive
group given by the direct sum n RX n , n = 0, 1, 2, . . . consisting of all finite
sums f = a0 + a1 X + . . . am X m , a polynomial with an ∈ R being the n0 th
coefficient. Multiplication is given by X i X j = X i+j extended by linearity. If
g = b0 + b1 X + . . . bn X n is an other polynomial, then
f + g = (a0 + b0 ) + (a1 + b1 )X + · · · + (ak + bk )X k + . . .
f g = a0 b0 + (a0 b1 + a1 b0 )X + · · · + (a0 bk + a1 bk−1 + · · · + ak b0 )X k + . . .
A monomial is polynomial of form aX n . The construction may be repeated to
give the polynomial ring in n-variables R[X1 , . . . , Xn ] or even in infinitely many
variables.
1.6.2. Definition. The degree, deg(f ), of a polynomial 0 6= f ∈ R[X] is the
index of the highest nonzero coefficient, the leading coefficient. A polynomial
with leading coefficient the identity is a monic polynomial.
1.6.3. Remark. (1) R is identified with the subring of constants in the polynomial ring R[X1 , . . . , Xn ].
(2) The nonzero constants are the polynomials of degree 0.
1.6. POLYNOMIALS
17
(3) The constant polynomial 1 is the unique monic polynomial of degree 0 and
the identity in the polynomial ring.
1.6.4. Proposition. Let 0 6= f, g ∈ R[X].
(1) If f g 6= 0 then deg(f g) ≤ deg(f ) + deg(g).
(2) If the leading coefficient of f or g is a nonzero divisor in R, then f g 6= 0 and
deg(f g) = deg(f ) + deg(g)
Proof. (1) This is clear. (2)Clearly the leading coefficient of the product is the
product of the leading coefficients.
1.6.5. Corollary. Let R be a domain.
(1) The polynomial ring R[X] is a domain.
(2) The units in R[X] are the constants, which are units in R.
1.6.6. Theorem. Let R be a domain, 0 6= f, d ∈ R[X] polynomials with d monic.
Then there are a unique q, r ∈ R[X] such that
f = qd + r,
r = 0 or deg(r) < deg(d)
Proof. Induction on deg(f ). If deg(f ) < deg(d) then q = 0, r = f . Otherwise
if a is the leading coefficient of f , then f − adX deg(f )−deg(d) has degree less than
deg(f ). By induction f − adX deg(f )−deg(d) = qd + r giving the claim.
1.6.7. Proposition. Let φ : R → S be a ring homomorphism. For any element
b ∈ S there is a unique ring homomorphism R[X] → S extending φ and mapping
X 7→ b.
Proof.
a0 + a1 X + . . . am X m 7→ φ(a0 ) + φ(a1 )b + . . . φ(am )bm
is clearly the one and only choice.
1.6.8. Definition. The homomorphism in 1.6.7 is the evaluation map at b in S.
The image of a polynomial f ∈ R[X] is denoted f (b) ∈ S.
1.6.9. Proposition. Let I ⊂ R be an ideal. Then there is a canonical isomorphism.
(R/I)[X] ' R[X]/IR[X]
Proof. There is an obvious pair of inverse homomorphisms constructed by 1.2.9
and 1.6.7.
1.6.10. Corollary. If P ⊂ R is a prime ideal, then P R[X] ⊂ R[X] is a prime
ideal.
1.6.11. Definition. Let φ : R → S be a ring homomorphism and B ⊂ S a subset.
The ring generated over R by B is
R[B] = φ(R)[B] ⊂ S
the smallest subring of S containing φ(R) ∪ B. If there is a finite subset B such
that R[B] = S then S is a finite type ring or a finitely generated ring over R.
1.6.12. Corollary. Let φ : R → S be a ring homomorphism.
18
1. A DICTIONARY ON RINGS AND IDEALS
(1) If bα ∈ S and Xα is a family of variables, then there is a surjective ring
homomorphism
R[Xα ] → R[bα ], Xα 7→ bα
making R[bα ] a factor ring of the polynomial ring R[Xα ].
(2) If S is a finite type ring over R, then S is a factor ring of a polynomial ring
in finitely many variables over R.
1.6.13. Exercise.
(2)
(3)
(4)
(5)
(1) Let K be a field. Show that there are infinitely many prime ideals
in K[X].
What are the units in the ring Z[X]/(1 − 2X)?
Determine the prime ideals in Q[X]/(X − X 2 ).
Show that the ring Z[X] is not a principal ideal domain.
Show that the ring Q[X, Y ] is not a principal ideal domain.
1.7. Roots
1.7.1. Definition. Let φ : R → S be a ring homomorphism and f ∈ R[X] a
polynomial. An element b ∈ S is a root of f (in S) if the evaluation f (b) = 0.
1.7.2. Proposition. Let R be a domain. An element a ∈ R is a root of the polynomial f ∈ R[X] if and only if there is a q ∈ R[X] such that
f = q(X − a)
Proof. By 1.6.6 f = q(X −a)+r. It follows that a is a root if and only if r = 0.
1.7.3. Corollary. Let R is a domain. There are at most deg(f ) roots in a nonzero
polynomial f ∈ R[X].
1.7.4. Definition. The multiplicity of a root a of a nonzero polynomial f ∈ R[X]
is highest m such that
f = q(X − a)m
A root of multiplicity m = 1 is a simple root.
1.7.5. Corollary. Let R is a domain. If m1 , . . . , mk are the multiplicities of the
roots of a nonzero polynomial f ∈ R[X], then m1 + · · · + mk ≤ deg(f ).
P
1.7.6. Definition. The derivative of a polynomial f = an X n ∈ R[X] is
X
f0 =
nan X n−1
1.7.7. Lemma. The derivative satisfies
(1) (f + g)0 = f 0 + g 0 .
(2) (f g)0 = f 0 g + f g 0
(3) If f is constant, then f 0 = 0.
1.7.8. Proposition. Let R is a domain. An element a ∈ R is a root of multiplicity
m > 1 of a nonzero f ∈ R[X] if and only if a is a root of f and f 0 .
Proof. By 1.6.6 f = q(X − a)2 + cX + d and by 1.7.7 f 0 = q 0 (X − a)2 + 2q(X −
a) + c. I follows that a is a root of multiplicity m > 1 if and only if c = d = 0.
1.7.9. Exercise.
(1) Let a1 , . . . ak be roots with multiplicities m1 , . . . , mk in a polynomial f . Show that m1 + · · · + mk ≤ deg(f ).
(2) Let K be a field and let a1 , . . . , an ∈ K. Show that the ideal (X1 −a1 , . . . , Xn −an )
is maximal in K[X1 , . . . , Xn ].
(3) Let the characteristic char(R) = n > 0. What is (X n )0 in R[X].
1.8. FIELDS
19
1.8. Fields
1.8.1. Definition. Let p ∈ Z be a prime number. The factor ring Fp = Z/(p) is a
field with p elements. Together with Q they constitute the prime fields.
1.8.2. Theorem. Let K be a field then the polynomial ring K[X] is a principal
ideal domain.
Proof. Let d 6= 0 be a polynomial of lowest degree in an ideal I. Given f ∈ I then
by 1.6.6 f = qd + r with r = f − qd ∈ I. By degree considerations r = 0 and
I = (d).
1.8.3. Corollary. Let K be a field then the polynomial ring K[X] is a unique
factorization domain.
Proof. Follows from 1.5.6.
1.8.4. Definition. A subfield is a subring, which is a field. A field extension is the
inclusion of a subfield K ⊂ L in a field. A finite field extension K ⊂ L is an
extension, where L is finite dimensional as a vector space over K.
1.8.5. Example. (1) Let K be a field and f an irreducible polynomial in K[X].
Then K ⊂ K[X]/(f ) is a finite field extension.
(2) Let K ⊂ R ⊂ L be a subring in a finite field extension. Then R is a field.
Namely multiplication on R with a nonzero element of R is a K-linear map
on the finite dimensional K-vector space R and therefore an isomorphism.
1.8.6. Proposition. (1) Let K be a field and f a polynomial in K[X]. Then there
is a finite field extension K ⊂ L such that f factors in linear factors in L[X].
(2) If K ⊂ L1 and K ⊂ L2 are finite field extensions then there is a finite field
extension K ⊂ L such that L1 ∪ L2 ⊂ L.
Proof. (1) Assume f irreducible. In L = K[X]/(f ) the class X + (f ) is a root of
f . In general proceed adjoining roots of irreducible factors of f . (2) An element x
in a finite field extension K ⊂ K 0 is the root of the irreducible monic polynomial
f generating the kernel of the evaluation homomorphism K[X] → K 0 , X 7→ x.
Now proceed by (1) adjoining elements in L2 to L1 .
1.8.7. Proposition. Let p ∈ Z be a prime number. For any power q = pn there is
a field Fq with q elements, unique up to isomorphism.
Proof. Let Fp ⊂ K be a field extension, where X q − X factors into linear factors, 1.8.6 (1). The subset of roots is the set of elements fixed under n-times the
Frobenius and therefore a subring being a subfield by 1.8.5 (2). The derivative
(X q − X)0 = −1 so by 1.7.8 there are q elements in this subfield. Uniqueness
follows from 1.8.6 (2).
(1) Show that the ring R[X]/(X 2 + 1) is isomorphic to the field of
complex numbers.
Show that the ring F2 [X]/(X 2 + X + 1) is a field with 4 elements.
Show that the ring F2 [X]/(X 3 + X + 1) is a field with 8 elements.
Let K ⊂ L be a field extension of fields of characteristic 0 and let a ∈ L be a root of
an irreducible polynomial f ∈ K[X]. Show that a is a simple root.
Let p be a prime number. Show that Fp is the only ring with p elements.
Let p be a prime number. Show that a ring with p2 elements is isomorphic to one of
four non isomorphic Z/(p2 ), Fp × Fp , Fp [X]/(X 2 ), Fp2 .
1.8.8. Exercise.
(2)
(3)
(4)
(5)
(6)
20
1. A DICTIONARY ON RINGS AND IDEALS
1.9. Power series
1.9.1. Q
Definition. Let R be a ring. The power series
P ring R[[X]] is the additive
group n RX n , n = 0, 1, 2, . . . of all power series an X n with n0 th coefficient
an ∈ R. Multiplication
by X i X j = X i+j extended by linearity. For
P is given
n
another power series bn X the sum and product are
X
X
X
an X n +
bn X n =
(an + bn )X n
X
X
XX
an X n ·
bn X n =
(
an−k bk )X n
k
The construction may be repeated to give the power series ring in n-variables
R[[X1 , . . . , Xn ]] or even in infinitely many variables.
1.9.2. Remark. The polynomial ring is identified as a subring R[X] ⊂ R[[X]] of
power series with only finitely many nonzero terms.
1.9.3. Definition. The order, o(f ), of a power series 0 6= f ∈ R[[X]] is the index
of the least nonzero coefficient.
1.9.4. Proposition. If R is a domain, then R[[X]] is a domain and for 0 6= f, g ∈
R[X]
o(f g) = o(f ) + o(g)
Proof. Clearly the lowest nonzero coefficient in the product is the product of the
two lowest nonzero coefficients.
P
1.9.5. Proposition. A power series f =
an X n is a unit if and only if a0 is a
unit.
Proof. It suffices to look at a power series f = 1 − gX. Then the power series
1/f = 1 + gX + g 2 X 2 + · · · + g n X n + . . . is well defined and f · 1/f = 1.
1.9.6. Theorem. If K is a field, then K[[X]] is a principal ideal domain. and (X)
is the only nonzero prime ideal.
Proof. If the lowest order of an element in an ideal I is n. Then clearly I =
(X n ).
1.9.7. Corollary. If K is a field, then K[[X]] is a unique factorization domain.
1.9.8. Proposition. Let I ⊂ R be an ideal. Then there is a canonical surjective
homomorphism
R[[X]]/IR[[X]] → R/I[[X]]
1.9.9. Corollary. If Q ⊂ R[[X]] is a maximal ideal, then P = Q ∩ R ⊂ R is a
maximal ideal and Q = (P, X).
Proof. X ∈ Q so R/Q ∩ R ' R[[X]]/Q.
1.9.10. Exercise.
(1) Show that the ring Z[[X]] is not a principal ideal domain.
(2) Show that the ring Q[[X, Y ]] is not a principal ideal domain.
(3) Let K be a field. Show that (X1 , . . . , Xn ) is the unique maximal ideal in the power
series ring K[[X1 , . . . , Xn ]].
(4) Let a ∈ R be nilpotent. Show that the ring R[[X]]/(X − a) is isomorphic to R.
(5) What is R[[X]]/(X − a) if a ∈ R is a unit?
(6) Let I ⊂ R be an ideal. Show that IR[[X]] ⊂ R[[X]] is not a maximal ideal.
2
Modules
2.1. Modules and homomorphisms
2.1.1. Definition. Let R be a ring. A module (R-module) is an abelian group
M , addition (x, y) 7→ x + y and zero 0, together with a scalar multiplication
R × M → M, (a, x) 7→ ax satisfying
(1) associative : (ab)x = a(bx)
(2) bilinear : a(x + y) = ax + ay, (a + b)x = ax + bx
(3) identity: 1x = x
for all a, b ∈ R, x, y ∈ M . A submodule M 0 ⊂ M is an additive subgroup such
that ax ∈ M 0 for all a ∈ R, x ∈ M 0 . A homomorphism is an additive group
homomorphism f : M → N respecting scalar multiplication
f (x + y) = f (x) + f (y), f (ax) = af (y)
for all a ∈ R, x, y ∈ M . An isomorphism is a homomorphism f : M → N
having an inverse map f −1 : N → M which is also a homomorphism. The
identity isomorphism is denoted 1M : M → M .
2.1.2. Lemma. Let R be a ring and M a module.
(1) a0 = 0 = 0x
(2) (−1)x = −x
(3) (−a)x = −(ax) = a(−x)
for all a ∈ R, x ∈ M .
Proof. (1) Calculate a0 = a(0 + 0) = a0 + a0 and cancel to get a0 = 0. Similarly
0x = 0. (2) By (1) 0 = 0x = (1 + (−1))x = x + (−1)x, so conclude −x =
(−1)x. (3) Calculate (−a)x = ((−1)a)x = (−1)(ax) = −(ax). Similarly
a(−x) = −(ax).
2.1.3. Lemma. Let R be a ring and f : M → N a homomorphism of modules.
(1) f (0) = 0.
(2) f (ax + by) = af (x) + bf (y).
(3) f (−x) = −f (x) for all x ∈ M .
for all a, b ∈ R, x, y ∈ M .
Proof. (1) Calculate f (0) = f (0 + 0) = f (0) + f (0) and conclude f (0) = 0.
(2) Calculate f (ax + by) = f (ax) + f (by) = af (x) + bf (y). (3) By 2.1.2
f (−x) = f ((−1)x) = (−1)f (x) = −f (x).
2.1.4. Example. (1) The zero group is the zero module.
(2) Over the zero ring the zero module is the only module.
(3) The zero subgroup of a module is the zero submodule.
(4) The ring R is a module under multiplication. An ideal is a submodule.
21
22
2. MODULES
(5) If R is a field, a module is a vector space and a homomorphism is a linear
map.
(6) A module over Z is an abelian group and an additive map of abelian groups
is a homomorphism.
2.1.5. Proposition. A bijective homomorphism is an isomorphism.
Proof. Let f : M → N be a bijective homomorphism of R-modules and let g :
N → M be the inverse map. For x, y ∈ N write x = f (g(x)), y = f (g(y)) and
get additivity of g, g(x + y) = g(f (g(x)) + f (g(y))) = g(f (g(x) + g(y))) =
g(x) + g(y). Similarly for a ∈ R g(ax) = g(af (g(x))) = g(f (ag(x))) = ag(x),
so g respects scalar multiplication and is a homomorphism.
2.1.6. Lemma. Let a ∈ R and M be a module. The map M → M, x 7→ ax is a
homomorphism.
Proof. Let f (x) = ax and calculate f (x+y) = a(x+y) = ax+ay = f (x)+f (y)
and f (bx) = a(bx) = (ab)x = (ba)x = b(ax) = bf (x) to get that f is a
homomorphism. Remark that the last calculation uses that R is commutative 1.1.2
(4).
2.1.7. Definition. Let a ∈ R and M be a module.
(1) The scalar multiplication with a is the homomorphism, 2.1.6,
aM : M → M, x 7→ ax
(2) a ∈ R is a nonzero divisor on M if scalar multiplication aM is injective, i.e.
ax 6= 0 for all 0 6= x ∈ M . Otherwise a is a zero divisor.
2.1.8. Remark. The two notions of nonzero divisor on R : 1.1.6 as element in the
ring and 2.1.4 as scalar multiplication on the ring coincide.
2.1.9. Example. If R is a field, then scalar multiplication on a vector space is either
zero or an isomorphism.
2.1.10. Lemma. Let φ : R → S be a ring homomorphism and N an S-module.
The map
R × N → N, (a, x) 7→ φ(a)x
is an R-scalar multiplication on N , 2.1.1.
Proof. Let a, b ∈ R, x, y ∈ N and µ(a, x) = φ(a)x. Calculate µ(a + b, x) =
φ(a + b)x = φ(a)x + φ(b)x = µ(a, x) + µ(b, x), µ(a, x + y) = φ(a)(x + y) =
φ(a)x + φ(a)y = µ(a, x) + µ(a, y), µ(1, x) = φ(1)x = 1x = x and µ(ab, x) =
φ(ab)x = φ(a)φ(b)x = µ(a, µ(bx)) showing the conditions 2.1.1.
2.1.11. Definition. Let φ : R → S be a ring homomorphism. The restriction of
scalars of an S-module N is the same additive group N viewed as an R-module
through φ. The scalar multiplication is 2.1.10
R × N → N, (a, x) 7→ ax = φ(a)x
An S-module homomorphism g : N → N 0 is also an R-module homomorphism.
2.1.12. Example. (1) The scalar multiplication with a Restriction of scalars for
the unique ring homomorphism Z → R give just the underlying abelian group
of a module, 2.1.4 (6).
2.2. SUBMODULES AND FACTOR MODULES
23
(2) Let I ⊂ R be an ideal. Restriction of scalars along the projection R → R/I
gives any R/I-module as an R-module.
2.1.13. Proposition. Let R be a ring. There is a dictionary:
(1) To an R[X]-module N associate the pair (N, f ) consisting of N as R-module
through restriction of scalars and f = XN : N → N, f (y) = Xy scalar
multiplication with X as an R-module homomorphism. An R[X]-homomorphism
g : N → N 0 gives an R-homomorphism such that g ◦ f = f 0 ◦ g.
(2) To a pair (N, f ) of an R-module and a homomorphism f : N → N associate
the R[X]-module with abelian group N and scalar multiplication determined
by Xy = f (y) for y ∈ N . Note
X
X
(
an X n )y =
an f ◦n (y)
An R-homomorphism g : N → N 0 such that g ◦ f = f 0 ◦ g is an R[X]homomorphism.
Proof. The statement is an algorithm to follow.
2.1.14. Proposition. Let R be a ring and M a module. The abelian group R ⊕ M
with multiplication
(a + x)(b + y) = ab + (ay + bx)
is a ring. R is a subring and M is an ideal.
Proof. Simple calculations show that the conditions 1.1.2 are satisfied.
2.1.15. Exercise.
(1) Show that a composition of homomorphisms is a homomorphism.
(2) Show that composition of scalar multiplications with a, b ∈ R on a module M is a
scalar multiplication with the product, aM ◦ bM = (ab)M .
(3) Let φ : R → S be a ring homomorphism. Show that φ is an R-module homomorphism, when S is viewed as R-module through restriction of scalars 2.1.11.
(4) Fill out the dictionary 2.1.13.
2.2. Submodules and factor modules
2.2.1. Lemma. Let R be a ring and
T Let Nα be a family of submodPM a module.
ules. Then the additive subgroups α Nα and α Nα are submodules.
P
P
P
P
P
Proof. Use thePformulas xα + yα = (xα + yα ) and a xα =
axα to
conclude thatT Nα is a submodule. If x, y ∈ Nα for all α, then x + y, ax ∈ Nα
for all α, so Nα is a submodule.
2.2.2. Definition. Let R be a ring and M a module. The intersection of all submodules containing a subset Y ⊂ M is the submodule generated by Y and denoted
RY . This is the smallest submodule, 2.2.1, of M containing Y . The module M is
generated by Y if RY = M . Let I be an ideal. The submodule generated by all
products ax, a ∈ I, x ∈ M is denoted IM .
2.2.3.
Proposition. Let R be a ring and M a module. If Y ⊂ M , then RY =
P
y∈Y Ry,
RY = {a1 y1 + · · · + an yn |ai ∈ R, yi ∈ Y }
Proof. The righthand side is contained in the submodule RY . Moreover the righthand side is a submodule containing Y , so equality.
2.2.4. Corollary. Let I ⊂ R be an ideal and M a module.
24
2. MODULES
(1)
IM = {a1 y1 + · · · + an yn |ai ∈ I, yi ∈ M }
(2) If I = (a) is principal, then
aM = (a)M = {ay|y ∈ M }
P
P
Proof.
bi ayi = a bi yi = ay for
P(1) is clear. (2) By (1) an element in aM is
y = bi yi ∈ M as claimed.
2.2.5. Lemma. Let R be a ring, M a module and N ⊂ M a submodule. Let M/N
be the abelian factor group, then the map
R × M/N → M/N,
(a, x + N ) 7→ ax + N
is well defined and a scalar multiplication, 2.1.1.
Proof. If x+N = y+N then x−y ∈ N and so a(x−y) = ax−ay ∈ N . Therefore
ax + N = ay + N and the multiplication is well defined. Since representatives
may be chosen such that (x + N ) + (y + N ) = x + y + N, a(x + N ) = ax + N ,
the laws for scalar multiplication are satisfied.
2.2.6. Definition. Let R be a ring, M a module and N ⊂ M a submodule, then the
factor module is the additive factor group M/N with, 2.2.5, scalar multiplication
a(x + N ) = ax + N . The projection p : M → M/N, x 7→ x + N is a surjective
homomorphism.
2.2.7. Lemma. Let R be a ring, N ⊂ M a submodule and p : M → M/N
the projection. p is surjective and if Y ⊂ M generates M , then p(Y ) ⊂ M/N
generates the factor module.
Proof. Clearly if RY = M then Rp(Y ) = p(RY ) = M/N .
2.2.8. Example. (1) A submodule of R is the same as an ideal.
(2) Both an ideal I ⊂ R and a factor ring R/I are modules.
(3) The module structure on R/I as a factor module and the structure by restriction of scalars through the projection R → R/I are identical.
2.2.9. Proposition. Let R = R1 ×R2 be the product ring 1.1.4. There is a bijective
(up to natural isomorphism) correspondence.
(1) If M1 is an R1 -module and M2 is an R2 -module, then M = M1 × M2 is an
R-module with coordinate scalar multiplication. A pair of homomorphisms
induce a homomorphism on the product.
(2) If M is an R-module then M1 = (1, 0)M is an R1 -module and M2 =
(0, 1)M is an R2 -module. A homomorphism induces a pair of homomorphisms.
2.2.10. Remark. The correspondence 2.2.9 indicates that the structure of modules
and homomorphisms over a product ring is identified with the structure of pairs of
modules and homomorphisms over each component ring in the product.
2.2.11. Exercise.
(1) Give an example of two submodules N, L ⊂ M such that the
union N ∪ L is not a submodule.
(2) Let R be a ring and a ∈ R. Show that the R-module R[X]/(X − a) is isomorphic
to R.
(3) Show that the projection p 2.2.6 is a homomorphism.
(4) Fill in the details in the dictionary 2.2.9
2.3. KERNEL AND COKERNEL
25
2.3. Kernel and cokernel
2.3.1. Lemma. Let R be a ring and f : M → N a homomorphism of modules.
Given submodules M 0 ⊂ M, N 0 ⊂ N , then f −1 (N 0 ) ⊂ M and f (M 0 ) ⊂ N are
submodules.
Proof. If x, y ∈ f −1 (N 0 ) then f (x + y) = f (x) + f (y) ∈ N 0 and for a ∈ R
f (ax) = af (x) ∈ N 0 so x+y, ax ∈ f −1 (N 0 ) proving f −1 (N 0 ) to be a submodule.
The same equations prove that f (M 0 ) is a submodule.
2.3.2. Definition. Let f : M → N be a homomorphism of modules. Then there
are submodules, 2.3.1.
(1) The kernel Ker f = f −1 (0).
(2) The image Im f = f (M ).
(3) The cokernel Cok f = N/ Im f .
2.3.3. Proposition. Let f : M → N be a homomorphism of modules.
(1) f is injective if and only if Ker f = 0.
(2) f is surjective if and only if Cok f = 0.
(3) f is an isomorphism if and only if Ker f = 0 and Cok f = 0.
Proof. (1) If f is injective and x ∈ Ker f then f (x) = 0 = f (0) so x = 0.
Conversely if Ker f = 0 and f (x) = f (y) then f (x − y) = 0 so x = y. (2) The
factor module N/ Im f = 0 if and only if Im f = N . (3) This follows from (1)
and (2).
2.3.4. Example. Let a ∈ R give scalar multiplication aM : M → M, x 7→ ax.
(1) Im aM = aM = {ax ∈ M |x ∈ M }.
(2) Ker aM = {x ∈ M |ax = 0}.
(3) Cok aM = M/aM .
2.3.5. Theorem. Let f : M → N be a homomorphism of modules.
(1) Let L ⊂ Ker f be a submodule. Then there is a unique homomorphism f 0 :
M/L → N such that f = f 0 ◦ p.
M
p
f
/
<N
f0
M/L
(2) The homomorphism f 0 : M/ Ker f → N is a module isomorphism onto the
submodule Im f of N .
M
p
f
/ Im f
9
f0
M/ Ker f
Proof. (1) If x + L = x0 + L then x − x0 ∈ L giving f (x) = f (x0 ). The factor map
f 0 : M/L → N, x + L 7→ f (x) is therefore well defined and f = f 0 ◦ p. Since
f, p are homomorphisms and p is surjective it follows that f 0 is a homomorphism.
(2) The kernel of f 0 is Ker f / Ker f = 0 so by 2.3.3 it is an isomorphism.
26
2. MODULES
2.3.6. Corollary. Let p : M → M/N be the projection onto a factor module. The
map L0 7→ L = p−1 (L0 ) gives a bijective correspondence between submodules
in M/N and submodules in M containing N . Also L0 = p(L) = L/N . This
correspondence preserves inclusions, additions and intersections of submodules.
Proof. If L0 is a submodule of M/N then clearly p(p−1 (L0 )) = L0 . If N ⊂ L
is a submodule of M then clearly L ⊂ p−1 (p(L)). Moreover if x ∈ p−1 (p(L))
then p(x) = p(y) for some y ∈ L and therefore y − x ∈ N . It follows that
L = p−1 (p(L)) and the correspondence is bijective. Inclusions are easily seen to
be preserved. Also easily p(L1 + L2 ) = p(L1 ) + p(L2 ) and p−1 (L01 ∩ L02 ) =
p−1 (L01 ) ∩ p−1 (L02 ) hold. The two resting equalities are consequences of this and
bijectivity of the correspondence.
2.3.7. Corollary. Let L ⊂ N ⊂ M be submodules. Then there is a canonical
isomorphism
M/N → (M/L)/(N/L)
Proof. The kernel of the surjective east-south composite
/ M/L
M
/ (M/L)/(N/L)
M/N
is N . By 2.3.5 the horizontal lower factor map gives the isomorphism.
2.3.8. Corollary. Let L, N ⊂ M be submodules. Then there is a canonical isomorphism
N/N ∩ L → N + L/L
given by x + N ∩ L 7→ x + L.
Proof. The kernel of the east-south composite
/ N +L
N
N/N ∩ L
/ N + L/L
is N ∩ L. Since x + y + L = x + L for x ∈ N, y ∈ L this composite is also
surjective. By 2.3.5 the horizontal lower factor map gives the isomorphism.
2.3.9. Proposition. Let f : M → N and g : N → L be homomorphisms such that
Im f ⊂ Ker g. Then there is a unique homomorphism g 0 : Cok f → L such that
g = g 0 ◦ p.
M
f
g
/N
/L
FF
O
FF p
FF
0
FF g
"
Cok f
Proof. This follows from 2.3.5.
2.3.10. Lemma. Let R be a ring and M a module. For x ∈ M the map R →
M, a 7→ ax is the unique homomorphism such that 1 7→ x.
2.3. KERNEL AND COKERNEL
27
Proof. Let f (a) = ax. f (ab) = (ab)x = a(bx) = af (b) shows that f is a
homomorphism. This argument does not use the commutativity 1.1.2 of R.
2.3.11. Definition. Denote the homomorphism 2.3.10
1x : R → M, a 7→ ax
The annihilator of x is the ideal
Ann(x) = Ker 1x = {a ∈ R|ax = 0}
T
For a subset Y ⊂ M the annihilator Ann(Y ) = y∈Y Ann(y) is the ideal of
elements
Ann(Y ) = {a ∈ R|ay = 0, for all y ∈ Y }
2.3.12. Proposition. Let R be a ring and Y a subset of a module M .
(1) Ann(Y ) = Ann(RY ).
(2) a ∈ Ann(M ) if and only if aM = 0.
(3) If modules M ' M 0 then Ann(M ) = Ann(M 0 ).
(4) The induced homomorphism
10x : R/ Ann(x) → Rx, a + Ann(x) 7→ ax
is an isomorphism.
P
P
Proof. (1) If a ∈ Ann(Y ) then a bi yi = bi ayi = 0 giving the not so obvious
Ann(Y ) ⊂ Ann(RY ). (2) Clear since aM (x) = ax, 2.1.7. (3) Let f : M → M 0
be an isomorphism and a ∈ R. Then af (x) = f (ax) expresses that f ◦ aM =
aM 0 ◦ f . Since f is bijective, aM = 0 if and only if aM 0 = 0. By (2) Ann(M ) =
Ann(M 0 ). (4) This follows from 2.3.5.
2.3.13. Corollary. Let I, J ⊂ R be ideals.
(1) Ann(R/I) = I
(2) If R/I ' R/J then I = J.
2.3.14. Lemma. Let I ⊂ R be an ideal and M an R-module. If I ⊂ Ann(M )
then M is an R/I-module with the scalar multiplication
R/I × M → M, (a + I, x) 7→ ax
That is, M is an R/ Ann(M )-module.
2.3.15. Example. Let a ∈ R give scalar multiplication aM : M → M, x 7→ ax.
(1) a ∈ Ann(Ker aM ).
(2) a ∈ Ann(Cok aM ).
(3) Ker aM and Cok aM are modules over the factor ring R/(a), 2.3.14.
2.3.16. Definition. Let R be a ring and L, N ⊂ M submodules. The colon ideal
N : L is the ideal of elements a ∈ R such that aL ⊂ N .
2.3.17. Proposition. The colon ideal of submodules L, N ⊂ M is
N : L = Ann(N + L/N ) = Ann(N/N ∩ L)
Proof. If a ∈ N : L then aL ⊂ N . Therefore a(N + L) ⊂ N and a ∈ Ann(N +
L/N ). Conversely if a ∈ Ann(N + L/N ) then aL ⊂ N and therefore a ∈ N : L.
The last equality follows from 2.3.8.
2.3.18. Exercise.
(1) Give an example of a homomorphism f : M → N submodules
M1 , M2 ⊂ M such that f (M1 ∩ M2 ) 6= f (M1 ) ∩ f (M2 ).
28
2. MODULES
(2) Give an example of a homomorphism f : M → N submodules N1 , N2 ⊂ N such
that f −1 (N1 + N2 ) 6= f −1 (M1 ) + f −1 (M2 ).
(3) Let R be a ring and M a module. Show that M may be regarded as an R/ Ann(M )module in a natural way.
(4) Let L, N ⊂ M be submodules. Show that Ann(L + N ) = Ann(L) ∩ Ann(N ).
(5) Let f : M → N be a surjective homomorphism. Show that Ann(M ) ⊂ Ann(N ).
(6) Let f : M → N be an injective homomorphism. Show that Ann(N ) ⊂ Ann(M ).
2.4. Sum and product
2.4.1.
Lemma. Let R be a ring and (Mα ) a family of modules. The product
Q
M
α is the abelian group of all families (xα ), xα ∈ Mα with term wise adα
dition. The setting
r(xα ) = (rxα )
Q
L
is aQscalar multiplication on α Mα . The direct sum α Mα is the subgroup
of α Mα consisting of families with only finitely many nonzero terms. This is a
submodule.
Proof. The laws in 2.1.1 are true for each factor and therefore trivially verified for
the product and sum.
2.4.2. Definition. Let R be a ring and Mα a family of modules. By 2.4.1 there are
modules and homomorphisms
Q
(1) The direct product is Qα Mα .
(2) The projections pβ : α Mα → Mβ are the homomorphisms pβ ((xα )) =
xβ .
L
L
(3) The
P direct sum is α Mα . Elements in α Mα are written as finite sums
xα .
L
(4) The injections iβ : Mβ → α Mα are the homomorphisms given by iβ (x) =
(xα ), where xβ = x and xα = 0, α 6= β.
2.4.3. Theorem. Let R be a ring and Mα a family of modules.
(1) Given a family of homomorphisms
fα : L → Mα , then there exists a unique
Q
homomorphism f : L → Mα such that fα = pα ◦ f .
L@
f
@@
@@
@
fα @@
Mα
/
Q
v
vv
vvp
v
v
v{ v α
α Mα
(2) Given a family of homomorphisms
gα : Mα → L, then there exists a unique
L
homomorphism g :
Mα → L such that gα = g ◦ iα .
L
g
/
α Mα
>L
dH
HH
HH
HH
iα HH
Mα
~~
~~
~
~ g
~~ α
P
P
Proof. (1) f (y) = (fα (y)) is the unique homomorphism. (2) g( xα ) = gα (xα )
is well defined since only finitely many xα 6= 0 and a homomorphism.
2.4. SUM AND PRODUCT
29
2.4.4. Definition.PA family of submodules Mα ⊂ M constitutes a direct sum if
any element x ∈ α Mα has a unique finite representation
X
x=
xα , xα ∈ Mα
α
2.4.5. Proposition. The following conditions are equivalent:
(1) The family Mα ⊂ M constitutes a direct sum.
(2) The natural homomorphism
M
X
Mα →
Mα
α
α
is an isomorphism.
(3) For all β
Mβ ∩
X
Mα = 0
α6=β
P
P
Proof. (1) ⇔ (2): This is clear. (1) ⇒ (3): If xβ = α6=β xα ∈ Mβ ∩ α6=β Mα ,
P
P
then xβ − α6=β xα = 0. Therefore by uniqueness xβ = 0. (3) ⇒ (1): If α xα =
P
P
0, then xβ = − α6=β xα ∈ Mβ ∩ α6=β Mα = 0. for any β. This shows
uniqueness.
L
2.4.6. Definition. Let R be a ring. A module isomorphic to a direct sum α R of
copies of the ring R is a free module.
A basis of a module,
is a subset Y such that any element admits a unique finite
P
representation α aα yαL
, where aα ∈ R, yα ∈ Y .
The standard basis of α R consists of the elements eα , where each is a family
with 0 for β 6= α and exactly 1 at index α.
2.4.7. Proposition. A module is free if and only if it admits a basis. If yα is a basis
of F then F = Ryα as direct sum and there is an isomorphism
M
f:
R→F
α
P
P
where f ( aα ) = aα yα .
L
Proof. Given a free module f :
α R → F then yα = f (eα ) is a basis. Conversely given a basis yα ∈ F then 1L
yα : R → F 2.3.11 is a family
P of homomorP
phisms giving a homomorphism f : α R → F by 2.4.3. As f ( aα ) = aα yα
it follows that f is bijective and therefore by 2.1.5 an isomorphism.
2.4.8. Remark. The polynomial ring R[X1 , . . . , Xn ] is free as R-module.
2.4.9. Example. (1) A nonzero ideal is a free module if and only if it has a basis
consisting of a nonzero divisor. Namely if x1 6= x2 where in a basis then the
product x1 x2 has two different representations.
(2) Let I ⊂ R be an ideal. The module R/I is free if and only if I = 0 or I = R.
2.4.10. Proposition. Any module over a field is free. Conversely if any module
over a nonzero ring is free, then the ring is a field
Proof. By Zorn’s lemma any vector space admits a basis. If I ⊂ R is an ideal and
R/I is free, then I = Ann(R/I) is either 0 or R. So R is a field.
30
2. MODULES
2.4.11. Proposition. Let F be a free module with basis yα . For a module M and
a family of elements xα ∈ MPthere is a unique
homomorphism g : F → M such
P
that g(yα ) = xα given by g( aα yα ) = aα xα .
L
Proof. The basis yα ∈ F gives the isomorphism 2.4.7
f :
α R → F . The
L
R
→
M
by 2.4.3. Then
family 1xα : R → M gives a homomorphism g 0 :
α
g = g 0 ◦ f −1 .
L
2.4.12. Corollary. Let M be an R-module and M R the free module with basis
ex indexed by x ∈ M . The homomorphism
M
X
X
R → M,
ax ex 7→
ax x
M
is surjective identifying M as a factor module of a free module in a natural way.
2.4.13. Definition. A module is indecomposable if it is not isomorphic to a direct
sum of two nonzero submodules, otherwise decomposable.
1 m2
2.4.14. Example. Q is an indecomposable Z-module. Namely if m
n1 , n2 are
m2
1
nonzero numbers in two submodules, then n1 m2 m
n1 = n2 m1 n2 is a nonzero number in the intersection.
2.4.15. Exercise.
(1) Show that if a ring is decomposable as a module, then it is the
product of two nonzero rings.
L
Q
(2) Let Mα be a finite family of modules. Show that
Mα = Mα ,
(3) Let Nα ⊂ Mα be a family of submodules modules. Show that
M
M
M
Mα /
Nα '
Mα /Nα
and that
Y
Mα /
Y
Nα '
Y
Mα /Nα
2.5. Homomorphism modules
2.5.1. Lemma. Let R be a ring and f, g : M → N homomorphisms.
(1) (f + g)(x) = f (x) + g(x) is a homomorphism.
(2) If a ∈ R, then (af )(x) = af (x) is a homomorphism.
Proof. Calculate according to 2.1.1. (1) (f + g)(x + y) = f (x + y) + g(x + y) =
f (x) + f (y) + g(x) + g(y) = (f + g)(x) + (f + g)(y) and (f + g)(ax) =
f (ax) + g(ax) = a(f (x) + g(x)) = a(f + g)(x). (2) (af )(x + y) = af (x + y) =
af (x) + af (y) = (af )(x) + (af )(y) and (af )(bx) = af (bx) = abf (x) =
baf (x) = b(af )(x). The last calculation uses that R is commutative 1.1.2 (4).
2.5.2. Definition. Let R be a ring and M, N modules. By 2.5.1, the homomorphism module HomR (M, N ) is the additive group of all homomorphism with
scalar multiplication
R × HomR (M, N ) → HomR (M, N ), (a, f ) 7→ af = [x 7→ af (x)]
2.5.3. Definition. Let a ∈ R be a ring and f : M → M 0 , g : N → N 0 , h, k :
M 0 → N homomorphisms of modules. By 2.5.1
(1) (h + k) ◦ f = h ◦ f + k ◦ f .
(2) (ah) ◦ f = a(h ◦ f ).
(3) g ◦ (h + k) = g ◦ h + g ◦ k.
(4) g ◦ (ah) = a(g ◦ h).
2.5. HOMOMORPHISM MODULES
31
There is induced a homomorphism
Hom(f, g) : HomR (M 0 , N ) → HomR (M, N 0 )
(h : M 0 → N ) 7→ (g ◦ h ◦ f : M → N 0 )
of R-modules.
2.5.4. Definition. Let R, S be rings. A functor is a construction T , which to an
R-module M associates S-modules T (M )
M 7→ T (M )
and for two R-modules M, N there is an additive homomorphism
HomR (M, N ) → HomS (T (M ), T (N ))
f 7→ T (f )
such that
(1) T (1M ) = 1T (M )
(2) T (g ◦ f ) = T (g) ◦ T (f )
In case the homomorphism changes domain and target
HomR (M, N ) → HomS (T (N ), T (M ))
f 7→ T (f )
and
(1) T (1M ) = 1T (M )
(2) T (g ◦ f ) = T (f ) ◦ T (g)
the functor is contravariant.
Clearly (contravariant) functors transform isomorphisms into isomorphisms.
Given functors T, T 0 a natural homomorphism is a family νM : T (M ) → T 0 (M )
of homomorphisms, such that for each f : M → N the following diagram commutes
νM
/ T 0 (M )
T (M )
T (f )
T (N )
νN
T 0 (f )
/ T 0 (N )
In the contravariant case the diagram is
T (M )
νM
O
T 0 (f )
T (f )
T (N )
/ T 0 (M )
O
νN
/ T 0 (N )
A natural isomorphism is a natural homomorphism such that each νM is an isomorphism.
2.5.5. Proposition. Let R be a ring.
(1) The construction
N 7→ HomR (M, N )
g 7→ Hom(1M , g)
32
2. MODULES
is a functor and
HomR (N, N 0 ) → HomR (HomR (M, N ), HomR (M, N 0 ))
is an R-homorphism.
(2) The construction
M 7→ HomR (M, N )
f 7→ Hom(f, 1N )
is a contravariant functor and
HomR (M, M 0 ) → HomR (HomR (M 0 , N ), HomR (M, N ))
is an R-homorphism.
Proof. By 2.5.3 the construction on homomorphisms are homomorphisms. Given
also homomorphisms f 0 : M 0 → M 00 and g 0 : N 0 → N 00 . Then
Hom(1M , g 0 ◦ g) = Hom(1M , g 0 ) ◦ Hom(1M , g)
and
Hom(f 0 ◦ f, 1N ) = Hom(f, 1N ) ◦ Hom(f 0 , 1N )
showing the conditions on compositions. Clearly also
Hom(1, g + g 0 ) = Hom(1, g) + Hom(1, g 0 )
Hom(1, ag) = a Hom(1, g)
and similarly for the resting properties.
2.5.6. Corollary. Let a ∈ R give scalar multiplications aM , aN .
(1) The three constructions
Hom(aM , 1N ) = Hom(1M , aN ) = aHomR (M,N )
are the same homomorphism
HomR (M, N ) → HomR (M, N ) f 7→ af
(2) The map R → HomR (M, M ), a 7→ aM is a homomorphism.
2.5.7. Example. Let R = R1 × R2 be the product ring. The constructions in 2.2.9
is: (1) A functor which to a pair of an R1 -module and an R2 -module associates an
R-module.(2) A functor which to an R-module associates a pair of an R1 -module
and an R2 -module.
2.5.8. Proposition. Let R be a ring and Mα a family of modules. For any module
N there are natural isomorphisms
L
Q
(1) HomR ( αQMα , N ) ' Q α HomR (Mα , N )
(2) HomR (N, α Mα ) ' α HomR (N, Mα )
L
Proof. This is 2.4.3 reformulated. (1) A homomorphism g :
α Mα → N is
uniquely determined
by
the
family
g
=
g
◦
i
:
M
→
N
.
(2)
A
homomorphism
α
α
α
Q
f : N → α Mα is uniquely determined by the family fα = pα ◦ f : N → Mα .
2.5.9. Lemma. Let R be a ring and M, N modules. For x ∈ M there is a homomorphism HomR (M, N ) → N, f 7→ f (x).
Proof. Calculate according to 2.1.1 (f + g) 7→ (f + g)(x) = f (x) + g(x) and
(af ) 7→ (af )(x) = af (x).
2.5. HOMOMORPHISM MODULES
33
2.5.10. Definition. The natural homomorphism 2.5.9
evx : HomR (M, N ) → N, f 7→ f (x)
is the evaluation at x.
2.5.11. Lemma. There is a natural homomorphism
M → HomR (HomR (M, N ), N ), x 7→ evx
Proof. Calculate according to 2.1.1 evx+y (f ) = f (x + y) = f (x) + f (y) =
evx (f ) + evy (f ). and evax (f ) = f (ax) = af (x) = aevx (f ) to see that the map
is a homomorphism.
2.5.12. Proposition. Let R be a ring and M a module. The evaluation
ev1 : HomR (R, M ) ' M, f 7→ f (1)
is a natural isomorphism. x 7→ 1x 2.3.11 is the inverse.
Proof. Calculate the composite ev1 (x 7→ 1x ) = 1x (1) = x and 1f (1) (a) =
af (1) = f (a) proving the claims.
2.5.13. Definition. Let R be a ring and M a module. The dual module is
M ∨ = HomR (M, R)
If f : M → N is a homomorphism, then the dual homomorphism is
f ∨ = Hom(f, 1R ) : N ∨ → M ∨
This construction is a contravariant functor.
2.5.14. Lemma. There is a natural homomorphism
M → M ∨∨ = HomR (HomR (M, R), R), x 7→ evx
where evx (f ) = f (x) 2.5.10.
Proof. This is a special case of 2.5.11
2.5.15. Definition. A module M is a reflexive module if the homomorphism 2.1.14,
M → M ∨∨
is an isomorphism.
2.5.16. Example. Let R be a ring.
(1) The module R is reflexive.
(2) If (a) =
6 R, (0) in a domain, then R/(a) is not reflexive.
2.5.17. Exercise.
(1) Show that HomZ (Q, Z) = 0.
(2) Calculate HomZ (Z/(m), Z/(n)) = 0 for integers m, n.
(3) Let I ⊂ R be an ideal and M a module. Show that HomR (R/I, M ) = {x ∈ M |I ⊂
Ann(x)}.
(4) Let R be a ring. Show that a free module with a finite basis is a reflexive module.
(5) If (n) ⊂ (m) ⊂ Z, then show that (m)/(n) is a reflexive Z/(n)-module.
34
2. MODULES
2.6. Tensor product modules
2.6.1. Definition. Let R be a ring and M, N modules. The tensor product module
M ⊗R N = F/F 0
is the factor module F/F 0 , where F = ⊕M ×N R is the free module with basis
(x, y) = e(x,y) 2.4.6 and F 0 is the submodule generated by all elements of form
(x1 + x2 , y) − (x1 , y) − (x2 , y), (x, y1 + y2 ) − (x, y1 ) − (x, y2 )
(ax, y) − a(x, y), (x, ay) − a(x, y)
The projection of the basis element (x, y) onto M ⊗R N is x ⊗ y = (x, y) + F 0 .
2.6.2. Remark. The relations are interpreted.
(1) There are identities in M ⊗R N
(x1 + x2 ) ⊗ y = x1 ⊗ y + x2 ⊗ y, x ⊗ (y1 + y2 ) = x ⊗ y1 + x ⊗ y2
ax ⊗ y = a(x ⊗ y) = x ⊗ ay
(2) The map
⊗ : M × N → M ⊗R N, (x, y) 7→ x ⊗ y
has partial maps x 7→ x⊗y : M → M ⊗R N and y 7→ x⊗y : N → M ⊗R N
that are all homomorphisms.
(3) The formation of partial homomorphism are again homomorphisms.
N → HomR (M, M ⊗R N )), y 7→ (x 7→ x ⊗ y)
and
M → HomR (N, M ⊗R N )), x 7→ (y 7→ x ⊗ y)
2.6.3. Theorem. Given a map µ : M × N → L such that the partial maps x 7→
µ(x, y) : M → L and y 7→ µ(x, y) : N → L are homomorphisms. Then there
exists a unique homomorphism u : M ⊗R N → L such that u(x ⊗ y) = µ(x, y).
⊗
/ M ⊗R N
M × NM
MMM
MMM
u
µ MMM
MM& L
On a general element u satisfies
X
X
X
u(
xα ⊗ yα ) =
u(xα ⊗ yα ) =
µ(xα , yα )
Proof. By 2.6.1 M ⊗R N = F/F 0 . The homomorphism 2.4.11 F → L, (x, y) 7→
µ(x, y) has F 0 in the kernel. 2.3.5 gives the homomorphism u.
2.6.4. Remark. Assume xα generate M and yβ generate N .
(1) xα ⊗ yβ generate M ⊗R N .
(2) Two homomorphisms u, v : M ⊗R N → L are equal if u(xα ⊗ yβ ) =
v(xα ⊗ yβ ).
2.6.5. Proposition. Let R be a ring and f : M → M 0 , g : N → N 0 homomorphisms of modules. Then there is induced a homomorphism
M ⊗R N → M 0 ⊗R N 0 , x ⊗ y 7→ f (x) ⊗ g(y)
2.6. TENSOR PRODUCT MODULES
35
Proof. The map south-east
M ×N
f ×g
M0 × N0
⊗
⊗
/ M ⊗R N
/ M0 ⊗ N0
R
satisfies the assumptions in 2.6.3 to induce the right vertical map satisfying x⊗y 7→
f (x) ⊗ g(y).
2.6.6. Definition. f ⊗ g : M ⊗R N → M 0 ⊗R N 0 is the induced homomorphism
2.6.5. On a general element
X
X
f ⊗ g(
xα ⊗ yα ) =
f (xα ) ⊗ g(yα )
2.6.7. Proposition. Let R be a ring. The constructions
(1)
M 7→ M ⊗R N, f 7→ f ⊗ 1N
(2)
N 7→ M ⊗R N, g 7→ 1M ⊗ g
are functors and
HomR (M, M 0 ) → HomR (M ⊗R N, M 0 ⊗R N )
HomR (N, N 0 ) → HomR (M ⊗R N, M ⊗R N 0 )
are R-homomorphism.
Proof. Given also homomorphisms f 0 : M 0 → M 00 and g 0 : N 0 → N 00 . Then by
2.6.4
f 0 ◦ f ⊗ g0 ◦ g = f 0 ⊗ g0 ◦ f ⊗ g
The rest follows directly from 2.6.5. E.g.
(f + g) ⊗ 1 = f ⊗ 1 + g ⊗ 1, (af ) ⊗ 1 = a(f ⊗ 1)
and similarly.
2.6.8. Corollary. Let a ∈ R give scalar multiplications aM , aN . The homomorphisms
aM ⊗ 1N = 1M ⊗ aN = aM ⊗R N
are the same homomorphism
M ⊗R N → M ⊗R N, x ⊗ y 7→ a(x ⊗ y)
2.6.9. Example. Let R be a ring.
(1) Then there is an isomorphism
R ⊗R R ' R, a ⊗ b 7→ ab
(2) Let M, N, L be modules. Composition of maps gives a homomorphism
HomR (N, L) ⊗R HomR (M, N ) → HomR (M, L), g ⊗ f 7→ g ◦ f
by 2.5.3. This is a natural homomorphism in each variable.
36
2. MODULES
(3) For a module M composition
HomR (M, M ) ⊗R HomR (M, M ) → HomR (M, M ), g ⊗ f 7→ g ◦ f
gives HomR (M, M ) a structure of a normally noncommutative ring. The
map R → HomR (M, M ), a 7→ aM is a ring homomorphism.
2.6.10. Proposition. Let R be a ring and M, N, L modules. Then there are natural
isomorphisms
(1) (unit:) M ⊗R R ' M, x ⊗ a 7→ ax
(2) (commutative:) (M ⊗R N ) ' N ⊗R M, x ⊗ y 7→ y ⊗ x
(3) (associative:) (M ⊗R N )⊗R L ' M ⊗R (N ⊗R L), (x⊗y)⊗z 7→ x⊗(y ⊗z)
Proof. (1) M × R → M, (x, a) 7→ ax induces the homomorphism M ⊗R R →
M, x ⊗ a 7→ ax by 2.6.3. The map M → R ⊗R M, x 7→ 1 ⊗ x is the inverse. (2)
M × N → N ⊗R M, (x, y) 7→ y ⊗ x induces the homomorphism (M ⊗R N ) '
N ⊗R M, x ⊗ y 7→ y ⊗ x by 2.6.3. The inverse is constructed similarly and the
composites are the identities by the uniqueness statement in 2.6.3. (3) For a fixed
z ∈ L the map M × N → M ⊗R (N ⊗R L), (x, y) 7→ x ⊗ (y ⊗ z) induces the
homomorphism µz : M ⊗R N → M ⊗R (N ⊗R L), x ⊗ y 7→ x ⊗ (y ⊗ z) by 2.6.3.
Finally the map (M ⊗R N )×L → M ⊗R (N ⊗R L), (x⊗y, z) 7→ µz (x⊗y). induces
the homomorphism (M ⊗R N )⊗R L ' M ⊗R (N ⊗R L), (x⊗y)⊗z 7→ x⊗(y⊗z)
by 2.6.3. The inverse is constructed similarly and the composites are the identities
by the uniqueness statement in 2.6.3.
2.6.11. Proposition. Let R be a ring and Mα a family of modules. For any module
N there is a natural isomorphism
M
M
(
Mα ) ⊗R N '
(Mα ⊗R N )
α
α
P
P
giving the identification ( xα ) ⊗ y = (xα ⊗ y).
Proof. 2.4.3 and 2.6.3
L give a pair of inverse homomorphisms. Fix y ∈ N . The
family gα : L
Mα → (M
induces by 2.4.3 aL
homomorLα ⊗R N ), xα 7→ xα ⊗ y L
phismP
gy :
Mα → P (Mα ⊗R N ). The map ( Mα ) × N → L (Mα ⊗R
N ), (L xα , y) 7→ gy (P xα ) inducesPby 2.6.3 the homomorphism ( Mα ) ⊗R
NL→ (Mα ⊗R N ), ( xα )⊗y 7→
xα ⊗y. The family iα ⊗1N : Mα ⊗R N →
( Mα ) ⊗R N induces by 2.4.3 the inverse.
2.6.12. Example. Let R be a ring, F a free module with basis yα and G a free
module with basis zβ . Then F ⊗R G is a free module with basis yα ⊗ zβ .
2.6.13. Theorem. Let R be a ring and M, N, L modules. Then there is a natural
isomorphism
HomR (M ⊗R N, L) ' HomR (M, HomR (N, L))
f 7→ (x 7→ [y 7→ f (x ⊗ y)])
(x ⊗ y 7→ g(x)(y)) ←[ g
Proof. A given f : M ⊗R N → L is mapped to the composite homomorphism
M → HomR (N, M ⊗R N ) → HomR (N, L), 2.5.4 and 2.6.2. This is a homomorphism as map of f by 2.5.4. Given g : M → HomR (N, L) the map M × N →
L, (x, y) 7→ g(x)(y) induces a homomorphism M ⊗R N → L, x ⊗ y 7→ g(x)(y)
2.7. CHANGE OF RINGS
37
by 2.6.3. Clearly the maps are inverse to each other and therefore giving an isomorphism by 2.1.5.
2.6.14. Exercise.
(1) Show that Q ⊗Z Q/Z ' 0.
(2) Show that Z/(m) ⊗Z Z/(n) = 0 if (m, n) = Z.
(3) Let P, Q ⊂ R be different maximal ideals and M a module. Show that M/P M ⊗R
M/QM = 0.
2.7. Change of rings
2.7.1. Proposition. (1) Let φ : R → S be a ring homomorphism and N an S
module. The restriction scalars 2.1.11 viewing N as an R-module through
φ, R × N → N, (a, x) 7→ ax = φ(a)x, is a functor from S-modules to
R-modules.
(2) Let I ⊂ R be an ideal. Restriction of scalars along R → R/I identifies
R/I-modules M with R-modules such that I ⊂ Ann(M ). For R/I-modules
M, N there is a natural isomorphism
HomR (M, N ) ' HomR/I (M, N )
Proof. This is a restatement of 2.1.11 using 2.5.4.
2.7.2. Lemma. Let R → S be a ring homomorphism, M an R-module and N an
S-module. Then
S × M ⊗R N → M ⊗R N, (b, x ⊗ y) 7→ x ⊗ by
is an S-scalar multiplication.
Proof. For fixed b ∈ S the map M × N → M ⊗R N, (x, y) 7→ x ⊗ by induces
the homomorphism µb : M ⊗R N → M ⊗R N, x ⊗ y 7→ x ⊗ by by 2.6.3. This
gives a well defined scalar multiplication S × M ⊗R N → M ⊗R N, (b, x ⊗ y) 7→
µb (x ⊗ y).
2.7.3. Definition. Let R → S be a ring homomorphism and M an R-module. The
change of ring S-module is M ⊗R S with S-scalar multiplication 2.7.2
S × M ⊗R S → M ⊗R S, (b, x ⊗ c) 7→ x ⊗ bc
2.7.4. Proposition. The construction
M 7→ M ⊗R S
and
f : M → M 0 7→ f ⊗ 1S : M ⊗R S → M 0 ⊗R S
is a functor from R-modules to S-modules.
HomR (M, M 0 ) → HomS (M ⊗R S, M 0 ⊗R S)
is an R-homomorphism.
Proof. By 2.6.7 this is a functor to R-modules and by 2.7.2 f ⊗ 1S is an Shomomorphism.
2.7.5. Proposition. Let R → S be a ring homomorphism, M, M 0 two R-modules
and N an S-modules.
38
2. MODULES
(1) There is a natural isomorphism of S-modules.
M ⊗R S ⊗S N ' M ⊗R N
x ⊗ b ⊗ y 7→ x ⊗ by
(2) Then there is a natural isomorphism of S-modules.
(M ⊗R S) ⊗S (M 0 ⊗R S) ' (M ⊗R M 0 ) ⊗R S
x ⊗ b ⊗ y ⊗ c 7→ x ⊗ y ⊗ bc
Proof. (1) The homomorphism v : S ⊗S N → N, b ⊗ y 7→ by is an isomorphism,
2.6.10. The homomorphism 1M ⊗ v : M ⊗R S ⊗S N ' M ⊗R N is an R-module
isomorphism, 2.6.7. The identity x ⊗ bc ⊗ y = x ⊗ b ⊗ cy proves this to be an
S-module homomorphism. (2) Use (1) on N = M 0 ⊗R S and conclude by the
canonical isomorphisms 2.6.10.
2.7.6. Theorem. Let R → S be a ring homomorphism, M, M 0 two R-modules
and N an S-modules.
(1) For any R-homomorphism f : M → N , then there exists a unique Shomomorphism f 0 : M ⊗R S → N, x ⊗ b 7→ bf (x) such that f (x) =
f 0 (x ⊗ 1).
f
M II
II
II
II
I$
M ⊗R S
/N
:
f0
(2) There is a natural isomorphism
HomR (M, N ) ' HomS (M ⊗R S, N )
f 7→ (x ⊗ b 7→ bf (x))
(3) There is a natural homomorphism
HomR (M, N ) ⊗R S → HomS (M ⊗R S, M 0 ⊗R S)
f ⊗ c 7→ (x ⊗ b 7→ f (x) ⊗ bc)
Proof. (1) From 2.6.3 f 0 is the unique R-homomorphism and by 2.7.3 this is an
S-homomorphism. (2) This is a restatement of (1). A given f is mapped to the
composite M ⊗R S → N ⊗R S → N which is an S-homomorphism. Given a
homomorphism g : M ⊗R S → N then the composite M → M ⊗R S → N is an
R-homomorphism and an inverse to the first given map. (3) This follows from (1)
applied to HomR (M, N ) → HomS (M ⊗R S, M 0 ⊗R S).
2.7.7. Lemma. Let R → S be a ring homomorphism, M an R-module and N an
S-module. Then
S × HomR (N, M ) → HomR (N, M )
(b, f : N → M ) 7→ (y 7→ f (by))
is an S-scalar multiplication.
Proof. The map (b, f ) 7→ f ◦ bN satisfies the laws 2.1.1.
2.7. CHANGE OF RINGS
39
2.7.8. Definition. Let R → S be a ring homomorphism and M an R-module. The
induced module is the S-module HomR (S, M ) with S-scalar multiplication 2.7.7
S × HomR (S, M ) → HomR (S, M )
(b, f : S → M ) 7→ (c 7→ f (bc))
2.7.9. Proposition. The induced module
M 7→ HomR (S, M )
and
f : M → M 0 7→ Hom(1S , f ) : HomR (S, M ) → HomR (S, M 0 )
is a functor from R-modules to S-modules.
HomR (M, M 0 ) → HomS (HomR (S, M ), HomR (S, M 0 ))
is an R-homomorphism.
Proof. By 2.5.5 this is a functor to R-modules and by 2.7.7 Hom(1, f ) is an Shomomorphism.
2.7.10. Theorem. Let R → S be a ring homomorphism and M an R-module and
N an S-modules. Then there is a natural isomorphism
HomR (N, M ) ' HomS (N, HomR (S, M ))
f 7→ (y 7→ [b 7→ f (by)])
Proof. g 7→ (y 7→ g(y)(1)) is an inverse.
2.7.11. Example. Let I ⊂ R be an ideal and R → R/I the projection.
(1) The change of ring functor maps an R-module M to the R/I-module M/IM .
The natural isomorphism 2.7.6 is
HomR (M, N ) ' HomR/I (M/IM, N )
for any R/I-module N .
(2) The induced module functor maps an R-module M to the R/I-module {x ∈
M |I ⊂ Ann(x)}. The natural isomorphism 2.7.10 is
HomR (N, M ) ' HomR/I (N, HomR (R/I, M ))
for any R/I-module N .
2.7.12. Definition. Let R → S, S 0 be ring homomorphisms. The tensor product
ring over R is S ⊗R S 0 with multiplication given by (b ⊗ b0 )(c ⊗ c0 ) = bc ⊗ b0 c0
extended by linearity. R → S ⊗R S 0 , r 7→ r ⊗ 1 = 1 ⊗ r is the natural ring
homomorphism.
2.7.13. Proposition. Let φ, φ0 : R → S, S 0 and ψ, ψ 0 : S, S 0 → T give a commutative diagram of ring homomorphisms, ψ ◦ φ = ψ 0 ◦ φ0 . Then b ⊗ b0 7→ ψ(b)ψ 0 (b0 )
40
2. MODULES
is the unique homomorphism making the following diagram commutative.
/ S0
55
55
55
55
S SSSS / S ⊗R S 0 555
SSS
55
SSS
SSS
SSS 555
SSS # S)
R
T
Proof. This is clear by 2.6.3.
2.7.14. Example. Let R → S be a ring homomorphism. Then
R[X] ⊗R S ' S[X]
is an isomorphism.
2.7.15. Exercise.
(1) Show that the change of rings of a free R-module is a free Smodule.
(2) Let φ : R → S be a ring homomorphism. Show that the change of rings of a
scalar multiplication a : M → M on an R-module is a scalar multiplication φ(a) :
M ⊗R S → M ⊗R S.
(3) Show that the change of rings of the composition of two homomorphisms is the
composition of the change of rings of each homomorphism.
(4) Show the isomorphism
R[X] ⊗R R[Y ] ' R[X, Y ]
3
Exact sequences of modules
3.1. Exact sequences
3.1.1. Definition. Let f : M → N and g : N → L be homomorphisms of
modules. The sequence
M
f
/N
g
/L
of homomorphisms is a
(1) 0-sequence: g ◦ f = 0 or equivalently Im f ⊂ Ker g
(2) exact sequence: Im f = Ker g
For a sequence of more homomorphisms the conditions should be satisfied for
every consecutive composition. E.g. The sequence
M
f
g
/N
/L
h
/K
is a 0-sequence if g ◦ f = 0 and h ◦ g = 0. The sequence is exact if Im f = Ker g
and Im g = Ker h.
3.1.2. Remark. An interpretation of 2.3.3 gives:
(1) The sequence
/M
0
f
/N
is exact if and only if f is injective.
(2) The sequence
M
f
/0
/N
is exact if and only if f is surjective.
(3) The sequence
0
/M
f
/N
/0
is exact if and only if f is an isomorphism.
3.1.3. Proposition.
0
(1) For a homomorphism f : M → N the sequence
/ Ker f
/M
f
/N
/ Cok f
/0
is exact.
(2) For scalar multiplication with a ∈ R on M the sequence
0
/ Ker aM
/M
aM
is exact.
41
/M
/ M/aM
/0
42
3. EXACT SEQUENCES OF MODULES
3.1.4. Proposition. Given a 0-sequence
f
/M
0
/N
g
/L
then the following statements are equivalent:
(1) The sequence is exact.
(2) f is an isomorphism onto Ker g.
(3) Given a homomorphism h : K → N such that g ◦ h = 0 then there is a
unique h0 : K → M such that h = f ◦ h0 .
f
/N
|>
|
h ||
||
|
|
/M
O
0
h0
K
g
/L
Proof. (1) ⇔ (2): This is clear. (1) ⇒ (3): Im h ⊂ Ker g = Im f, so by (2) put
h0 = f −1 ◦ h. (3) ⇒ (2): Apply it to Ker g → M to see that (2) is satisfied.
3.1.5. Proposition. Given a 0-sequence
M
f
g
/N
/L
/0
then the following statements are equivalent:
(1) The sequence is exact.
(2) The factor homomorphism 2.3.9 g 0 : Cok f → L induced by g is an isomorphism.
(3) Given a homomorphism k : N → K such that k ◦ f = 0 then there is a
unique k 0 : L → K such that k = k 0 ◦ g.
f
M
/N g /L
AA
AA k
AA
0
AA k
/0
K
Proof. (1) ⇔ (2): This follows from 2.3.9. (1) ⇒ (3): By 2.3.5 there is k 00 :
Cok f → K such that k 00 ◦ p = k. By (1) put k 0 = k 00 ◦ g −1 . (3) ⇒ (2) satisfied
and apply it to N → Cok f to see that (2) is satisfied.
3.1.6. Proposition. Let
/ Nα
Mα
/ Lα
be a family of exact sequences. Then there are exact sequences:
(1) The sum
L
Mα
Q
Mα
/
L
Nα
Q
Nα
/
L
Lα
(2) The product
/
/
Q
Lα
Proof. Clear since kernel and image are calculated componentwise.
3.1. EXACT SEQUENCES
43
3.1.7. Definition. An exact sequence
/M
0
f
/N
g
/L
/0
is a short exact sequence. That is f is injective, Im f = Ker g and g is surjective.
(1) Let I ⊂ R be an ideal, then there is a short exact se-
3.1.8. Proposition.
quence
0
/I
/R
/ R/I
/0
(2) Let M ⊂ N be a submodule, then there is a short exact sequence
0
/M
/N
/ N/M
/0
(3) For scalar multiplication with nonzero divisor a ∈ R on M the sequence
0
aM
/M
/ M/aM
/M
/0
is a short exact sequence.
(4) Given a homomorphism f : M → N there are associated two short exact
sequences.
0
/ Ker f
0
/ Im f
/M
f
/ Im f
/0
/ Cok f
/0
and
/N
(5) For scalar multiplication with any a ∈ R on M there are associated two short
exact sequences.
0
/ Ker aM
0
/ aM
/M
aM
/ aM
/0
/ M/aM
/0
and
/M
3.1.9. Definition. Let f : M → N be a homomorphism.
(1) f has a retraction if there is a homomorphism u : N → M such that u ◦ f =
1M .
(2) f has a section if there is a homomorphism v : N → M such that f ◦v = 1N .
3.1.10. Proposition. Let f : M → N be a homomorphism.
(1) If f has a retraction u : N → M then f is injective, u is surjective and
N = Im f ⊕ Ker u
(2) If f has a section v : N → M then f is surjective, v is injective and
M = Ker f ⊕ Im v
Proof. (1) u(f (x)) = x so f is injective and u is surjective. If y ∈ N then
y = f (u(y)) + (y − f (u(y)) and u(y − f (u(y))) = 0, so N = Im f + Ker u. Let
y ∈ Im f ∩ Ker u. Then y = f (x) gives x = u(f (x)) = u(y) = 0, so y = 0.
Conclude by 2.4.5 that the sum is direct. (2) y = f (v(y)) so f is a retraction of v.
Finish by (1).
44
3. EXACT SEQUENCES OF MODULES
3.1.11. Lemma. For a short exact sequence
0
f
/M
/N
g
/L
/0
the following statements are equivalent:
(1) f has a retraction.
(2) g has a section.
For any retraction u there is a unique section v and wise-verse such that
1N = f ◦ u + v ◦ g
Proof. If u is a retraction of f , then Ker g = Im f ⊂ Ker(1N − f ◦ u). By 3.1.5
there is a homomorphism v : L → N such that v ◦ g = 1N − f ◦ u. This is a
section of g. Conversely if v is a section of g then Im(1N − v ◦ g) ⊂ Ker g, so
there is a homomorphism u : N → M such that f ◦ u = 1N − v ◦ g, 3.1.4. u is a
retraction of f . The equation is clearly satisfied.
3.1.12. Definition. Let R be a ring and f : M → N, g : N → L homomorphisms.
A short exact sequence
0
f
/M
/N
g
/L
/0
is a split exact sequence if equivalently 3.1.11 f has a retraction or g has a section.
3.1.13. Theorem. A sequence
0
f
/M
/N
g
/L
/0
is a split exact sequence if and only if there are homomorphism u : N → M, v :
L → N satisfying
u ◦ f = 1M , g ◦ v = 1L , f ◦ u + v ◦ g = 1N
If the sequence is split exact then
0
/L
/N
v
/M
u
/0
is split exact and (x, y) 7→ f (x)+v(y) and z 7→ u(z)+g(z) gives the isomorphism
M ⊕L'N
Proof. The sequence is a 0-sequence f is injective and g is surjective. From f ◦
u + v ◦ g = 1N follows that z ∈ Ker g ⊂ Im f , so the sequence is short exact. The
rest is contained in 3.1.10.
3.1.14. Corollary. A (contravariant) functor preserves split exact sequences. If
0
/M
f
/N
g
/L
/0
is split exact and T a functor, then
0
/ T (M )
T (f )
/ T (N )
T (g)
/ T (L)
/0
/ T (M )
/0
is split exact. If T is contravariant, then
0
is split exact.
/ T (L)
T (g)
/ T (N )
T (f )
3.2. THE SNAKE LEMMA
45
Proof. By 3.1.13 a split exact sequence is characterized by a set of equations.
These are preserved by the functor, 2.5.4.
3.1.15. Example. A short exact sequence
f
/M
0
g
/N
/F
/0
where F is a free module is a split exact sequence. Namely let xα ∈ F be a
basis and choose yα ∈ N with g(yα ) = xα . The define v : L → N by setting
v(xα ) = yα , 2.4.11.
3.1.16. Example. Let Zi be the family of modules each a copy of Z indexed by
the natural numbers. Then the short exact sequence
L
Q
Q
L
/
/
/
/0
0
i Zi
i Zi
i Zi /
i Zi
is not split exact.
L
The element f = (1, 2, 22 , . . . , 2n , . . . ) + i Zi is divisible by 2k for all k. If
L
Q
L
fk = (0, . . . , 0, 2n−k , . . . ) + i Zi for n ≥ k, then 2k fk = f in i Zi / i Zi .
Q
But in i Zi the only element divisible with all 2k is 0, so no section exists.
3.1.17. Exercise.
(1) Show that the sequence
/Q
/ Q/Z
/0
/Z
/ Z/(n)
/0
/Z
0
is short exact, but not split exact.
(2) Show that the sequence
/Z
0
n
is exact, but not split exact for n 6= 0, 1.
(3) Show that the sequence
/ Z/(2)
0
17→2
/ Z/(4)
/ Z/(2)
/0
17→3
/ Z/(6)
/ Z/(3)
/0
is exact, but not split exact.
(4) Show that the sequence
/ Z/(2)
0
is split exact.
3.2. The snake lemma
3.2.1. Example. Given a commutative diagram of homomorphisms
M
f
/N
u
M0
v
f0
/ N0
f
/N
there is induced a commutative diagram
0
/ Ker f
0
/M
u
/ Ker f 0
u
/ M0
where the rows are exact sequences.
v
f0
/ N0
/ Cok f
/0
v
/ Cok f 0
/0
46
3. EXACT SEQUENCES OF MODULES
The diagram splits into two diagrams
/ Ker f
0
f
/M
u
/ Im f
u
0
/ Ker f 0
0
/ Im f
/ M0
v
f0
/0
/ Im f 0
/0
/ Cok f
/0
and
/N
v
v
/ Im f 0
0
v
/ Cok f 0
/ N0
/0
where the rows are short exact sequences.
3.2.2. Lemma. Given a commutative diagram of homomorphisms
f
M
u
/ M0
0
f0
g
/N
/L
v
/ N0
g0
/0
w
/ L0
where the rows exact sequences. The snake homomorphism δ : Ker w → Cok u
is well defined by: For z ∈ Ker w choose y ∈ N such that g(y) = z. The
element v(y) ∈ Ker g 0 so there is x0 ∈ M 0 such that f 0 (x0 ) = v(y). Then δ(z) =
x0 + Im u ∈ Cok u.
Proof. Assume g(y 0 ) = z and f 0 (x00 ) = v(y 0 ). There is x ∈ M with f (x) = y−y 0 .
Now f 0 (u(x)) = v(f (x)) = v(y − y 0 ) = f 0 (x0 − x00 ) so u(x) = x0 − x00 since f 0
is injective. Then x0 + Im u = x00 + Im u as wanted. The choices made respect
addition and scalar multiplication showing that δ is a homomorphism.
3.2.3. Remark. The snake is
Ker u
f
/ Ker v
g
/ Ker w
mk
j
l
g
/
/
/0
M
N
L
_________________________________
!on
!! 0
!hi
f
u
/ M0
^^. Cok u
f0
f0
v
/ N0
/ Cok v
g0
g0
w
/ L0
/ Cok w
3.2. THE SNAKE LEMMA
47
The construction of δ is schematically
z_
Ker w
M0
f0
/L
g
N
y_ v
/ N0
x_0
/z
/ v(y)
Cok u
δ(z)
3.2.4. Theorem (snake lemma). Given a commutative diagram of homomorphisms
f
M
u
f0
/ M0
0
g
/N
/L
v
g0
/ N0
/0
w
/ L0
where the rows exact sequences. There is induced a six term long exact sequence
f
/.
Ker u
g
/ Ker v
/ Ker w
*+
δ
f0
Cok u
g0
/ Cok v
/ Cok w
Proof. By construction of δ it is clear that the sequence is a 0-sequence: If y ∈
Ker v then to calculate δ(g(y)) the choice v(y) = 0 gives δ ◦ g = 0. Also
f 0 (δ(z)) = v(y) + Im v shows that f 0 ◦ δ = 0. Exactness at Ker v and Cok v
are clear. Given z ∈ Ker w such that δ(z) = 0. By 3.2.2 choose y, g(y) = z
and x0 , f 0 (x0 ) = v(y). Then δ(z) = x0 + Im u = 0, so choose x, u(x) = x0 .
Now v(f (x)) = f 0 (u(x)) = v(y) so y − f (x) ∈ Ker v and g(y − f (x)) =
g(y) = z. Therefore exactness at Ker w. Given x0 + Im u ∈ Cok u such that
f 0 (x0 ) + Im v = 0 ∈ Cok v. Choose y, v(y) = f 0 (x0 ) and put z = g(y). Then
w(g(y)) = g 0 (v(y)) = g 0 (f 0 (x0 )) = 0. Now z ∈ Ker w and δ(z) = x0 + Im u.
Therefore exactness at Cok u.
3.2.5. Corollary. If f is injective then the f : Ker u → Ker v is injective and the
long exact sequence is
f
/ Ker u
/.
0
/ Ker v
g
/ Ker w
*+
δ
f0
Cok u
/ Cok v
g0
/ Cok w
If g 0 is surjective then g 0 : Cok v → Cok w is surjective and the long exact sequence is
/.
Ker u
Cok u
f
/ Ker v
g
/ Ker w
*+
δ
f0
/ Cok v
g0
/ Cok w
/0
48
3. EXACT SEQUENCES OF MODULES
3.2.6. Corollary. (1) If v is injective and u is surjective, then w is injective.
(2) If v is surjective and w is injective, than u is surjective.
(3) If v is an isomorphism, then w is injective if and only if u is surjective.
3.2.7. Proposition. Given submodules N, L ⊂ M , then there is a short exact
sequence
0
x7→(x,x)
/ M/N ∩ L
(x,y)7→x−y
/ M/N ⊕ M/L
/ M/N + L
/0
Proof. The commutative diagram
0
/ N ∩L
0
/M
/ N ⊕L
/ N +L
/0
(x,y)7→x−y /M
/ M ⊕M
x7→(x,x)
/0
where the rows are short exact sequences, gives by 3.2.4 a five term long exact
sequence
0
/ M/N ⊕ M/L
/ M/N ∩ L
/ M/N + L
/0
3.2.8. Proposition (five lemma). Given a commutative diagram of homomorphisms
M1
/ M2
u1
M10
u2
/ M0
2
/ M3
/ M4
u3
/ M0
3
/ M5
u4
/ M0
4
u5
/ M0
5
where the rows are exact sequences. If u1 is surjective, u2 , u4 are isomorphism
and u5 is injective, then u3 is an isomorphism.
0 . Split the given diagram in three as
Proof. Let fi : Mi → Mi+1 , fi0 : Mi0 → Mi+1
follows
M1
u1
0
/ Ker f 0
2
0
/ Cok f2
u00
3
0
/ Ker f 0
4
0
/ Im f2
u03
0
/ Im f 0
2
/ M2
u2
/ M0
2
/ M4
u4
/ M0
4
/ M3
u3
/ M0
3
/ Cok f1
/0
u03
/ Im f 0
2
/0
/ Im f4
/0
u5
/ M0
5
/ Cok f2
/0
u00
3
/ Cok f 0
2
/0
Note that Cok f1 ' Im f2 and Cok f20 ' Ker f40 . Now use 2.3.3 and the snake
lemma to conclude that Ker u3 = 0 and Cok u3 = 0 and u3 is therefore an isomorphism.
3.2. THE SNAKE LEMMA
3.2.9. Corollary.
49
(1) Given a commutative diagram of homomorphisms
f
M
/N
u
f0
M0
v
/ N0
If u, v are isomorphism, then the induced homomorphisms u : Ker f →
Ker f 0 and v : Cok f → Cok f 0 are isomorphisms.
(2) Given a commutative diagram of homomorphisms
/N
M
u
/L
v
/ N0
M0
w
/ L0
If u, v, w are isomorphism, then the upper row is exact if and only if the lower
row is exact.
3.2.10. Proposition (windmill lemma). Given homomorphism M
There is induced an eight term long exact sequence
0
/ Ker f
/.
g
Cok f
/ Cok g ◦ f
/ Cok g
Proof. Look at the two diagrams
0
/M
/ Ker g
0
M
f
/L
0
By the snake lemma the sequences
/.
/M
g
f
/N
/N
/0
g◦f
/L
/ Cok f
1
/0
/0
/L
/ Ker f
0
1
g
g◦f
/ Ker g ◦ f
*+
δ=f
Ker g
Ker g ◦ f
/.
Cok g ◦ f
/N
/ Ker g
*+
f
/ Ker g ◦ f
f
/ Cok f
/ Ker g
/ Cok g ◦ f
/ Cok f
*+
δ=g
/ Cok g
are exact and overlap to give the windmill sequence.
/0
/0
g
/L.
50
3. EXACT SEQUENCES OF MODULES
3.2.11. Remark. The windmill is
/ Ker g
LLL
v
LLL
vv
v
LLL
vv
v
v
L%
zvv
f
/ Cok f
/N
KerO f
@@

@@

@@


 g
g◦f @@

0 eLLL
ll L RRRRR
LLL
lll
RRR
l
l
l
RRR
LLL
ll
RRR
LL
lll
R)
ulll
Cok g o
Cok g ◦ f
Ker
g ◦If
9
s
sss
s
s
s
sss
3.2.12. Exercise.
II
II
II
II
$
/M
(1) Given a short exact sequence
/M
0
/N
/L
/0
/L
/0
Show that Ann(N ) ⊂ Ann(M ) ∩ Ann(L).
(2) Give a short exact sequence
/M
0
/N
such that Ann(N ) 6= Ann(M ) ∩ Ann(L).
(3) Given ideals I, J ⊂ R. Show that there is a short exact sequence.
0
/ R/I ⊕ R/J
/ R/I ∩ J
/ R/I + J
/0
3.3. Exactness of Hom
3.3.1. Definition. A functor T is a left exact functor if given an exact sequence
/M
0
f
g
/N
/L
the following sequence is exact
/ T (M )
0
/ T (N )
/ T (L)
T is a right exact functor if given an exact sequence
M
f
g
/N
/L
/0
the following sequence is exact
/ T (N )
T (M )
/ T (L)
/0
A functor is an exact functor if it is both left and right exact.
A contravariant functor T is a left exact contravariant functor if given an exact
sequence
/M
0
the following sequence is exact
f
/N
/ T (N )
T (L)
g
/L
/ T (M )
/0
T is a right exact contravariant functor if given an exact sequence
M
f
/N
g
/L
/0
3.3. EXACTNESS OF HOM
51
the following sequence is exact
/ T (N )
/ T (L)
0
/ T (M )
A contravariant functor is an exact functor if it is both left and right exact.
3.3.2. Proposition. Let T be a functor. The following conditions are equivalent:
(1) T is a exact.
(2) Given an exact sequence
M
f
/N
g
/L
then the following sequence is exact
/ T (N )
T (M )
/ T (L)
(3) Given a short exact sequence
0
/M
f
/N
g
/L
/0
Then the following sequence is short exact
/ T (M )
0
/ T (N )
/ T (L)
/0
Let T be a contravariant functor. The following conditions are equivalent:
(1) T is a exact.
(2) Given an exact sequence
M
f
/N
g
/L
then the following sequence is exact
T (L)
/ T (N )
/ T (M )
(3) Given a short exact sequence
0
/M
f
/N
g
/L
/0
Then the following sequence is short exact
/ T (L)
0
/ T (N )
/ T (M )
/0
Proof. Let T be a functor. (1) ⇒ (2): The exact sequence M → Im f → 0 gives
T (M ) → T (Im f ) → 0 exact and the exact sequence 0 → Im f → N gives
0 → T (Im f ) → T (N ) exact. It follows that Im T (f ) ' T (Im f ). The sequence
0 → Im f → N → L is exact so 0 → T (Im f ) → T (N ) → T (L) is exact. Now
conclude that Im T (f ) = Ker T (g). (2) ⇒ (3): This is clear. (3) ⇒ (1): Given
f
g
/N
/ L exact, then the sequences 0 → T (M ) → T (N ) →
0 → M
T (Im g) → 0 and 0 → T (Im g) → T (L) are exact, so Im T (f ) = ker T (g) and T
is left exact. Similarly T is rigth exact.
3.3.3. Corollary. Let T be an exact functor. For a homomorphism f : M → N
there are natural isomorphisms.
(1) Ker T (f ) ' T (Ker f ).
(2) Im T (f ) ' T (Im f ).
(3) Cok T (f ) ' T (Cok f ).
52
3. EXACT SEQUENCES OF MODULES
Let T be a contravariant functor. For a homomorphism f : M → N there are
natural isomorphisms.
(1) Ker T (f ) ' T (Cok f ).
(2) Im T (f ) ' T (Im f ).
(3) Cok T (f ) ' T (Ker f ).
Proof. Represent the statements using short exact sequences. (1) The kernel is
determined by the exact sequence 0 → Ker f → M → N , 3.1.4. (3) The cokernel
is determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2) The
image is determined by the exact sequence 0 → Ker f → M → Im f → 0, 2.3.5,
3.1.4.
3.3.4. Corollary. Let T be an exact functor. For submodules N, L ⊂ M there are
natural identifications
(1) T (M/N ) = T (M )/T (N ).
(2) T (N + L) = T (N ) + T (L).
(3) T (N ∩ L) = T (N ) ∩ T (L).
Proof. Represent the statements using short exact sequences. (1) 0 → N → M →
M/N → 0 is short exact giving 0 → T (N ) → T (M ) → T (M/N ) → 0. The
wanted isomorphism follows form 3.1.5. (2) N + L is the image of N ⊕ L → M .
(3) N ∩ L is the kernel of N ⊕ L → M .
3.3.5. Proposition. The functor HomR (K, −) is left exact. Given an exact sequence
f
g
/N
/L
/M
0
and an R-module K. Then the following sequence is exact
0
/ HomR (K, M )
/ HomR (K, N )
/ HomR (K, L)
Proof. Given h : K → M such that f ◦ h = 0 then h = 0 since f is injective.
Given k : K → N such that g ◦ k = 0 then by 3.1.4 there is h : K → M such that
f ◦ h = k. So the sequence is exact.
3.3.6. Proposition. The contravariant functor HomR (−, K) is right exact. Given
an exact sequence
f
g
/N
/L
/0
M
and a module K. Then the following sequence is exact
0
/ HomR (L, K)
/ HomR (N, K)
/ HomR (M, K)
Proof. Given h : L → K such that h ◦ g = 0 then h = 0 since g is surjective.
Given k : N → K such that k ◦ f = 0 then by 3.1.5 there is h : L → K such that
h ◦ g = k. So the sequence is exact.
3.3.7. Proposition. Given a sequence
M
f
/N
g
/L
/0
such that for any module K, the following sequence is exact
0
/ HomR (L, K)
Then the original sequence is exact.
/ HomR (N, K)
/ HomR (M, K)
3.4. EXACTNESS OF TENSOR
53
Proof. For K = L the identity 1L is mapped to 1L ◦ g ◦ f = 0 so it is a 0-sequence.
Take K = Cok g then pg : L → Cok g has pg ◦ g = 0, but by exactness 0 is the
unique homomorphism satisfying this, so pg = 0. Therefore Cok g = 0 and g is
surjective. Take K = Cok f , p : N → Cok f the projection. p ◦ f = 0 so by
exactness there exists a unique q : L → Cok f such that q ◦ g = p. It follows that
Ker g ⊂ Ker p = Im f . All together the original sequence is exact.
3.3.8. Proposition. Given a sequence
/M
0
f
/N
g
/L
/0
The following conditions are equivalent:
(1) The sequence is split exact.
(2) For any K the following sequence is exact
0
/ HomR (K, N )
/ HomR (K, M )
/ HomR (K, L)
/0
/ HomR (M, K)
/0
(3) For any K the following sequence is exact
0
/ HomR (N, K)
/ HomR (L, K)
If the conditions are true, then the sequences (2) and (3) are split exact.
Proof. (1) ⇒ (2), (1) ⇒ (3) are clear by 3.1.14 giving that the sequences (2), (3)
are split exact. (2) ⇒ (1): Let K = L, then there is a section to g. By 3.1.11 the
original sequence is split exact. (3) ⇒ (1): Let K = M , then there is a retraction
to f . By 3.3.7and 3.1.11 the original sequence is split exact.
3.3.9. Exercise.
0
(1) Show that the sequence
/ HomZ (Q/Z, Q)
/ HomZ (Q/Z, Z)
/ HomZ (Q/Z, Q/Z)
is exact, but the rightmost map is not surjective.
(2) Show that the sequence
0
/ HomZ (Z/(n), Z)
/ HomZ (Z/(n), Z)
n
/ HomZ (Z/(n), Z/(n))
is exact, but the rightmost map is not surjective.
3.4. Exactness of tensor
3.4.1. Theorem. The functor − ⊗R K is right exact. Given an exact sequence
M
f
/N
g
/L
/0
and an R-module K. Then the following sequence is exact
M ⊗R K
/ N ⊗R K
/ L ⊗R K
/0
Proof. Let K 0 be any module. By 3.3.7 it is enough to see that the sequence
0
/ HomR (L ⊗R K, K 0 )
/ HomR (N ⊗R K, K 0 )
HomR (M ⊗R K, K 0 )
54
3. EXACT SEQUENCES OF MODULES
is exact. By 2.6.13 it amounts to see that the sequence
/ HomR (M, HomR (K, K 0 ))
/ HomR (L, HomR (K, K 0 ))
0
HomR (N, HomR (K, K 0 ))
is exact. This follows from 3.3.6
3.4.2. Proposition. Given a split exact sequence
0
/M
f
/N
g
/L
/0
and a module K. Then the following sequence is split exact
/ K ⊗R M
0
/ K ⊗R N
/ K ⊗R L
/0
Proof. This follows from the functor properties 3.1.14.
3.4.3. Proposition. Let I ⊂ R be an ideal. For any module M , the homomorphism
M ⊗R R/I → M/IM, x ⊗ a + I 7→ ax + IM
is an isomorphism.
Proof. There is a commutative diagrm
M ⊗R I
/ M ⊗R R
/ M ⊗R (R/I)
/0
/ IM
/M
/ M/IM
/0
0
The conclusion follows from the snake lemma, 3.2.4.
3.4.4. Corollary. Let I, J ⊂ R be ideals. Then
R/I ⊗R R/J → R/(I + J), a + I ⊗ b + J 7→ ab + I + J
is an isomorphism.
3.4.5. Proposition. Let I1 , . . . , Ik ⊂ R be pairwise comaximal ideals and M a
module. Then the product of projections
M/I1 · · · Ik M → M/I1 M × · · · × M/Ik M
is an isomorphism.
Proof. By Chinese remainders 1.4.2
R/I1 · · · Ik → R/I1 × · · · × R/Ik
is an isomorphism. Tensor with M and use 3.4.3 to get the isomorphism.
3.4.6. Exercise.
(1) Calculate Z/(m) ⊗Z Z/(n) for all integers m, n.
(2) Let I ⊂ R be an ideal. Show that R/I ⊗R R/I ' R/I.
(3) Let I ⊂ R be an ideal. Show that I ⊗R R/I ' I/I 2 .
(4) Let 2Z be scalar multiplication. Show that 2Z ⊗ 1Z/(2) : Z ⊗Z Z/(2) → Z ⊗Z Z/(2)
is not injective.
3.5. PROJECTIVE MODULES
55
3.5. Projective modules
3.5.1. Definition. An R-module F is a projective module if for any surjectice
homomorphism N → L the homomorphism
HomR (F, N ) → HomR (F, L)
is surjective.
3.5.2. Proposition. Let F be an R-module. The following conditions are equivalent:
(1) F is projective.
(2) The functor HomR (F, −) is exact.
(3) Given an exact sequence
/L
/N
M
then the following sequence is exact
/ HomR (F, N )
HomR (F, M )
/ HomR (F, L)
(4) Given a short exact sequence
0
f
/M
g
/N
/L
/0
Then the following sequence is short exact
0
/ HomR (F, M )
/ HomR (F, N )
/ HomR (F, L)
/0
Proof. This is clear from 3.3.2. (1) ⇔ (2): This follows since HomR (F, −) is
allways left exact. (1) ⇔ (3) ⇔ (4): This is true for any exact functor.
3.5.3. Proposition. A module F is projective if and only if any surjective homomorphism M → F → 0 has a section.
Proof. Assume F projective and g : M → F surjective. Then HomR (F, M ) →
HomR (F, F ) → 0 is exact. So there exists a v : F → M such that g ◦ v = 1F .
v is then a section. Conversely given g : N → L surjective and h : F → L. Let
M = Ker N ⊕F → L, (y, z) 7→ g(y)−h(z) and pN : M → N, pF : M → F the
projections, then g ◦ pN = h ◦ pF . Now pF is surjective since g is. Let v : F → M
be a section of pF , then h0 = pN ◦ v satisfies h = g ◦ h0 .
g
NO `
h0
pN
M
v
t
pF
/L
O
/0
h
/F
3.5.4. Corollary. A short exact sequence
0
/M
f
/N
g
/F
/0
where F is a projective module is a split exact sequence.
3.5.5. Corollary. A free module is projective. Over a field every module is projective.
3.5.6. Example. Let I ⊂ R be an ideal. If R/I is projective, then there is a ring
decomposition R/I × R0 ' R.
56
3. EXACT SEQUENCES OF MODULES
3.5.7. Proposition. A direct summand in a projective module is projective.
Proof. Let F ⊕ F 0 be projective and g : M → F surjective. By 3.5.3 there is a
section v 0 : F ⊕ F 0 → M ⊕ F 0 to (g, 1F 0 ). Then v(y) = pM ◦ v 0 (y, 0) is a section
to g and F is projective.
3.5.8. Proposition. A module is projective if and only if it is a direct summand in
a free module.
Proof. By 2.4.12 any module is a factor module of a free module. By 3.5.3 a
projective factor module has a section, and is therefore by 3.1.10 a direct summand.
3.5.9.
L Proposition. Let Fα be a family of projective modules, then the direct sum
α Fα is a projective module.
Proof. Let N → L be surjective. Then by 2.5.8
M
M
HomR (
Fα , N ) → HomR (
Fα , L)
is the product
Y
Y
HomR (Fα , N ) →
HomR (Fα , L)
L
which is surjective by 3.1.6. So
Fα is projective.
3.5.10. Proposition. Let F, F 0 be projective modules. Then F ⊗R F 0 is projective.
Proof. F ⊗R F 0 is clearly a direct summand in a free module.
3.5.11. Proposition. Let R → S be a ring homomorphism and F a projective
module. The change of ring module F ⊗R S is a projective S-module.
Proof. A direct summand of a free R-module is changed to a direct summand of a
free S-module.
3.5.12. Proposition. Any module M admits an exact sequence
F →M →0
where F is a projective module. That is, any module is a factor module of a projective module.
Proof. Take F free, 2.4.12.
3.5.13. Exercise.
(1) Let R = R1 × R2 . Show that R1 , R2 are projective ideals in R.
(2) Show that the ideal (2)/(6) in the ring Z/(6) is projective.
(3) Show that the ideal (2)/(4) in the ring Z/(4) is not projective.
3.6. Injective modules
3.6.1. Definition. An R-module E is an injective module if for any injective homomorphism M → N the homomorphism
HomR (N, E) → HomR (M, E)
is surjective.
3.6.2. Proposition. Let E be an R-module. The following conditions are equivalent:
(1) E is injective.
3.6. INJECTIVE MODULES
57
(2) The contravariant functor HomR (−, E) is exact.
(3) Given an exact sequence
/L
/N
M
then the following sequence is exact
/ HomR (N, E)
HomR (L, E)
/ HomR (M, E)
(4) Given a short exact sequence
0
/M
f
g
/N
/L
/0
Then the following sequence is short exact
0
/ HomR (N, E)
/ HomR (L, E)
/ HomR (M, E)
/0
Proof. This is clear from 3.3.2. (1) ⇔ (2): This follows since the contravariant
functor HomR (−, E) is allways right exact. (1) ⇔ (3) ⇔ (4): This is true for any
exact contravariant functor.
3.6.3. Proposition. A module E is an injective module if and only if any injective
homomorphism 0 → E → L has a retraction.
Proof. Assume E injective and f : E → L injective. Then HomR (L, E) →
HomR (E, E) → 0 is exact. So there exists a u : L → E such that u ◦ f = 1E .
Then u is a retraction. Conversely given f : M → N injective and h : M → E.
Let L = Cok M → E ⊕ N, x 7→ h(x) − f (x) and iE : E → L, iN : N → L the
injections, then iN ◦ f = iE ◦ h. Now iE is injective since f is. Let u : L → E be
a retraction of iE , then h0 = u ◦ iN satisfies h = h0 ◦ f .
f
/M
0
/N
h0
h
~
Ei
iE
iN
/L
u
3.6.4. Corollary. A short exact sequence
0
/E
f
/N
g
/L
/0
where E is an injective module is a split exact sequence.
3.6.5. Example. Let I ⊂ R be an ideal. If I is injective, then there is a ring
decomposition R/I × R0 ' R.
3.6.6. Proposition. A direct summand in an injective module is injective.
Proof. Let E ⊕ E 0 be an injective module and f : E → L an injective homomorphism. By 3.6.3 there is a retraction u0 : L ⊕ E 0 → E ⊕ E 0 to (f, 1E 0 ). Then
u(y) = pE ◦ u0 (y, 0) is a retraction to f and E is injective.
3.6.7.
Proposition. Let Eα be a family of injective modules, then the direct product
Q
E
α α is an injective module
58
3. EXACT SEQUENCES OF MODULES
Proof. Let M → N be injective. Then by 2.5.8
Y
Y
HomR (N,
Eα ) → HomR (M,
Eα )
is the product
Y
Y
HomR (N, Eα ) →
HomR (M, Eα )
Q
which is surjective by 3.1.6. So Eα is injective.
3.6.8. Proposition. A module E is injective, if for any ideal I ⊂ R
HomR (R, E) → HomR (I, E) → 0
is exact.
Proof. Let f : E → L be an injective homomorphism. The set of submodules
f (E) ⊂ L0 ⊂ L and retractions u0 : L0 → E is nonempty and inductively ordered.
By Zorn’s lemma a maximal L0 , u0 exists. If L/L0 6= 0 choose a y ∈ L\L0 . The
homomorphism Ann(y + L0 ) → E, a 7→ u0 (ay) extends to u00 : R → E by
hypothesis. The setting L0 + Ry → E, x + ay 7→ u0 (x) + u00 (a) is a well defined
retraction. This contradicts maximality.
3.6.9. Definition. Let R be a domain. M is a divisible module if scalar multiplication with a nonzero a ∈ R is surjective.
3.6.10. Proposition. (1) An injective module over a domain is divisible.
(2) A divisible module over a principal ideal domain is injective.
(3) Over a field any module is injective.
Proof. (1) Let E be an injective module and a 6= 0. Choose x ∈ E and look
at 1x : R → E. Scalar multiplication aR is injective, so there is an extension
1y : R → E such that 1x = 1y ◦ aR . Then ay = x. (2) Let E be divisible. To
extend f : (a) → E, choose x ∈ E such that ax = f (a), then 1x : R → E, 1 → x
extends f . (3) This is clear.
3.6.11. Proposition. Let R → S be a ring homomorphism and E an injective
R-module. The induced module HomR (S, E) is an injective S-module.
Proof. Let M → N be an injective homomorphism of S-modules. Then by
2.7.10 HomS (N, HomR (S, E)) → HomS (M, HomR (S, E)) is HomR (N, E) →
HomR (M, E) being surjective since E is an injective R-module.
3.6.12. Lemma. (1) Q and Q/Z are divisible and therefore injective Z modules.
(2) The homomorphism 2.5.11
x 7→ evx : M → HomZ (HomZ (M, Q/Z), Q/Z)
is injective.
Proof. (1) Clear by 3.6.10. (2) Let x 6= 0 and choose h : Z/ Ann(x) → Q/Z
nonzero. Now Z/ Ann(x) ' Zx ⊂ M so extend h to h0 : M → Q/Z. Then
evx (h0 ) = h(1) 6= 0.
3.6.13. Theorem. Any module M admits an exact sequence
0→M →E
where E is an injective module. That is, any module is a submodule of an injective
module.
3.6. INJECTIVE MODULES
59
Proof. By 2.4.12 choose a surjection F → HomZ (M, Q/Z) where F is a free
R-module. Then
0 → M → HomZ (F, Q/Z)
L
is exact by 3.6.12. Since F ' α R the module
Y
HomZ (F, Q/Z) '
HomZ (R, Q/Z)
α
is injective, 3.6.7 and 3.6.11.
3.6.14. Proposition. A homomorphism f : M → N is injective if and only if
HomR (N, E) → HomR (M, E) → 0
is surjective for any injective module E.
Proof. Assume that Ker f 6= 0 and choose 0 → Ker f → E, 3.6.13. The sequence
HomR (N, E) → HomR (M, E) → HomR (Ker f, E) → 0
is exact. So HomR (N, E) → HomR (M, E) is not surjective.
/ M , then the fol3.6.15. Lemma. Given an injective homomorphism 0 / N
lowing conditions are equivalent:
(1) Any nonzero submodule L ⊂ M has nonzero retraction i−1 (L) 6= 0.
(2) Given a homomorphism f : M → K such that the restriction f ◦i is injective,
then f is injective.
i
i /
3.6.16. Definition. An injective homomorphism 0 / N
M satisfying the
equivalent conditions 3.6.15 is an essential extension. An essential extension
0
/M
i
/ E is an injective envelope of M .
3.6.17. Proposition. Any module M has an injective envelope. If M ⊂ E, E 0 are
two injective envelopes, then there is an isomorphism f : E → E 0 fixing M .
Proof. By 3.6.13 choose M ⊂ E 0 with E 0 injective. By Zorn’s lemma choose
M ⊂ E ⊂ E 0 maximal among the essential extensions of M . If E 6= E 0 then the
set of modules 0 6= N 0 ⊂ E 0 such that E ∩ N 0 is nonempty by maximality of E.
Let N be maximal among these by Zorn’s lemma. Since E 0 is injective there is a
commutative diagram
/ E0
/ E 0 /N
E
mm
mmm
m
m
mmm
mv mmmm
0
E
The composite f :
→ E 0 /N → E 0 has E ⊂ Im f . If Im f 6= E then by
maximality of E there is a submodule 0 6= L ⊂ Im f with E ∩ L = 0. Now
N
f −1 (L) and E ∩ f −1 (L) = E ∩ L = 0 gives a contradiction. Therefore
Im f = E and f is a retraction making E a direct summand in E 0 . By 3.6.6 E is
an injective module. Now uniqueness: given two envelopes, let f : E → E 0 be
any homomorphism fixing M . Then f is injective, since M ⊂ E is essential. If f
is not surjective, then E 0 ' Im f ⊕ E 00 contradicting that M ⊂ E 0 is essential.
E0
3.6.18. Exercise.
module.
(1) Let R be a domain. Show that the fraction field is an injective
60
3. EXACT SEQUENCES OF MODULES
(2) Let R be a domain. Show that the torsion free divisible module injective.
(3) Show that for a ring that all modules are projective if and only all modules are injective.
3.7. Flat modules
3.7.1. Definition. An R-module F is a flat module if for any injective homomorphism M → N the homomorphism
M ⊗R F → N ⊗R F
is injective.
3.7.2. Proposition. Let F be an R-module. The following conditions are equivalent:
(1) F is a flat.
(2) The functor − ⊗R F is exact.
(3) Given an exact sequence of R-modules
/N
M
/L
Then the sequence of change of rings modules is exact
/ L ⊗R F
/ N ⊗R F
M ⊗R F
(4) Given a short exact sequence
0
/M
f
g
/N
/L
/0
Then the following sequence is exact
0
/ M ⊗R F
/ N ⊗R F
/ L ⊗R F
/0
Proof. This is follows from 3.3.2. (1) ⇔ (2): This follows since −⊗R is allways
right exact. (1) ⇔ (3) ⇔ (4): This is true for any exact functor.
3.7.3. Proposition. A direct summand in a flat module is flat.
Proof. Let F ⊕ F 0 be flat and M → N injective. Then M ⊗R (F ⊕ F 0 ) →
N ⊗R (F ⊕ F 0 ) is injective. Conclusion by 2.6.11.
L
3.7.4. Proposition. Let Fα be family of flat modules, then the direct sum α Fα
is a flat module.
L
L
Proof. Let M → N be injective. Then M ⊗R ( Fα ) → N ⊗R ( Fα ) is
injective by 2.6.11 and 3.1.6, so the product is flat.
3.7.5. Corollary. A free module is flat.
Proof. R is flat since M ⊗R R ' M , so
L
αR
is flat.
3.7.6. Corollary. A projective module is flat.
Proof. By 3.5.8 a projective module is a direct summand in a free. Conclusion by
3.7.4.
3.7.7. Proposition. Let F, F 0 be flat modules. Then F ⊗R F 0 is flat.
Proof. Let M → N be injective. Then by 2.6.10 M ⊗R (F ⊗R F 0 ) → N ⊗R
(F ⊗R F 0 ) is (M ⊗R F ) ⊗R F 0 → (N ⊗R F ) ⊗R F 0 being injective by using the
definition twice.
3.7. FLAT MODULES
61
3.7.8. Proposition. Let R → S be a ring homomorphism and F a flat R-module.
The change of ring module F ⊗R S is a flat S-module.
Proof. Let M → N be an injective homomorphism of S-modules. Then by 2.7.5
M ⊗S (F ⊗R S) → N ⊗S (F ⊗R S) is M ⊗R F → N ⊗R F being injective since
F is a flat R-module.
3.7.9. Theorem. Let R be a ring and F a module. The following conditions are :
(1) F is a flat module.
(2) HomR (F, E) is an injective module for any injective module E.
Proof. Let M → N be injective. By 3.6.14 M ⊗R F → N ⊗R F is injective if and
only if HomR (N ⊗R F, E) → HomR (M ⊗R F, E) is surjective for any injective
module E. By 2.6.13 this is HomR (N, HomR (F, E)) → HomR (M, HomR (F, E)).
3.7.10. Corollary. Given a short exact sequence
0
/N
/M
/F
/0
where F is a flat module. Then M is flat if and only if N is flat.
Proof. Let E be an injective module. By 3.7.9 and 3.6.4 the sequence
/ HomR (F, E)
0
/ HomR (N, E)
/ HomR (M, E)
/0
is split exact. By 3.6.6 and 3.6.7 HomR (N, E) is injective if and only if HomR (M, E)
is so. Conclusion by 3.7.9.
3.7.11. Corollary. Given a short exact sequence
0
/M
/N
/F
/0
where F is a flat module. For any module L there is a short exact sequence
0
/ L ⊗R M
/ L ⊗R N
/ L ⊗R F
/0
Proof. Let E be an injective module. By 3.7.9 and 3.6.4 the sequence
/ HomR (F, E)
0
/ HomR (N, E)
/ HomR (M, E)
/0
is split exact. So also the sequence
/ HomR (L, HomR (F, E))
0
/ HomR (L, HomR (N, E))
/0
HomR (L, HomR (M, E))
is split exact. By 2.6.13 this is natural isomorphic to the sequence
0
/ HomR (L ⊗R F, E)
/ HomR (L ⊗R N, E)
HomR (L ⊗R M, E)
Conclusion by 3.6.14.
/0
62
3. EXACT SEQUENCES OF MODULES
3.7.12. Theorem. A module F is flat, if for any ideal I ⊂ R
0 → I ⊗R F → R ⊗R F
is exact.
Proof. By 3.7.9 it suffices to see that HomR (F, E) is injective for any injective E.
By 3.6.8 this amounts to HomR (R, HomR (F, E)) → HomR (I, HomR (F, E)) being surjective. By 2.6.13 this homomorphism is HomR (R⊗R F, E) → HomR (I⊗R
F, E) which is surjective since E is injective.
3.7.13. Corollary. A module F is flat, if for any finite ideal J ⊂ R
0 → J ⊗R F → R ⊗R F
is exact.
P
P
Proof.
Let I be any ideal. Given bi ⊗ xi ∈ I ⊗R F such that bi xi = 0, then
P
bi ⊗ xi = 0 ∈ (bi ) ⊗R M and therefore also in I ⊗R M .
3.7.14. Exercise.
(1) Show that if a ∈ R is a nonzero divisor and M is a flat modules,
then aM is injective and a is a nonzero divisor on M .
(2) Show that Z/(n) is not a flat Z-module for n 6= 0, 1.
(3) Show that Z/(2) is a flat Z/(6)-module.
(4) Show that Z/(2) is not a flat Z/(4)-module.
4
Fraction constructions
4.1. Rings of fractions
4.1.1. Lemma. Let R be a ring and U ⊂ R such that 1 ∈ U and for u, v ∈ U the
product uv ∈ U . On U × R is defined a relation
(u, a) ∼ (u0 , a0 ) ⇔ there is v ∈ U such that vu0 a = vua0
(1) The relation is an equivalence relation.
(2) If (u, a) ∼ (u0 , a0 ) and (v, b) ∼ (v 0 , b0 ) then (uv, va + ub) ∼ (u0 v 0 , v 0 a0 +
u0 b0 ).
(3) If (u, a) ∼ (u0 , a0 ) and (v, b) ∼ (v 0 , b0 ) then (uv, ab) ∼ (u0 v 0 , a0 b0 ).
Proof. The claims are proved by simple calculations. (1) Symmetry is clear. Reflexive follows as 1 ∈ U . Transitive: if (u, a) ∼ (u0 , a0 ), (u0 , a0 ) ∼ (u00 , a00 ) then
vu0 a = vua0 , v 0 u00 a0 = v 0 u0 a00 . Since multiplication is commutative (v 0 vu0 )u00 a =
v 0 u00 vua0 = (v 0 vu0 )ua00 give (u, a) ∼ (u00 , a00 ). (2) From wu0 a = wua0 , w0 v 0 b =
w0 vb0 follow that w0 vv 0 wu0 a = w0 vv 0 wua0 , wuu0 w0 v 0 b = wuu0 w0 vb0 . So now
ww0 (u0 v 0 )(va + ub) = ww0 (uv)(v 0 a0 + u0 b0 ) as needed. (3) This is similar.
4.1.2. Definition. Let R be a ring. U ⊂ R is a multiplicative subset if 1 ∈ U
and for u, v ∈ U the product uv ∈ U . The ring of fractions U −1 R is given by
equivalence classes
a
= {(u0 , v 0 )|(u0 , v 0 ) ∼ (u, v)}
u
on U × R under the relation 4.1.1
(u, a) ∼ (u0 , a0 ) ⇔ there is v ∈ U such that vu0 a = vua0
The addition is
a
b
va + ub
+ =
u v
uv
and the multiplication is
a b
ab
· =
u v
uv
This is well defined and U −1 R is a ring, verified analog to the rational numbers Q
being a ring. The canonical ring homomorphism is
a
1
4.1.3. Theorem. Let φ : R → S be a ring homomorphism and U ⊂ R a multiplicative subset. If all elements in φ(U ) ⊂ S are units, then there exists a unique
ι : R → U −1 R, a 7→
63
64
4. FRACTION CONSTRUCTIONS
ring homomorphism φ0 : U −1 R → S such that φ = φ0 ◦ ι.
R FF
φ
FF
FF
ι FFF
"
/
<S
φ0
U −1 R
Proof. φ0 ( ua ) = φ(a)φ(u)−1 is the well defined unique ring homomorphism. Observe that the elements of form bv −1 satisfy the rules for fractions.
4.1.4. Proposition. Let U ⊂ R be a multiplicative subset.
(1) a1 6= 0 in U −1 R if and only if Ann(a) ∩ U = ∅.
(2) The canonical ring homomorphism R → U −1 R is injective if and only if U
consists only of nonzero divisors.
Proof.
a
1
= 0 if and only if va = 0 for some v ∈ U .
4.1.5. Corollary. If R is a domain and U is the nonzero elements then K = U −1 R
is a field and the canonical homomorphism identifies R as a subring.
4.1.6. Definition. The field K in 4.1.5 is the fraction field of R.
4.1.7. Corollary. If R is a domain with fraction field K and φ : R → L is an
injective ring homomorphism into a field, then there is a unique homomorphism
K → L extending φ.
4.1.8. Definition. The total ring of fractions of R is U −1 R, where U is the set of
nonzero divisors of R.
4.1.9. Corollary. A ring is a subring of its total ring of fractions.
4.1.10. Proposition. Let U = {un } the powers of an element u ∈ R. There is an
isomorphism
1
R[X]/(uX − 1) → U −1 R, X 7→
u
The ring of fractions with only one denominator is of finite type.
Proof. A pair of inverse homomorphism are constructed by 1.6.7 and 4.1.3.
4.1.11. Definition. Let U = {un } the powers of an element u ∈ R. The ring
U −1 R, 4.1.10, is denoted
1
R[ ]
u
4.1.12. Proposition. If U ⊂ V ⊂ R are multiplicative, then there is a canonical
identification
a
a/1
V −1 R ' (V /1)−1 U −1 R, 7→
v
v/1
4.1.13. Proposition. Let U ⊂ R1 × R2 be a multiplicative subset of a product of
rings and let Ui ⊂ Ri be the projections of U . Then there is an isomorphism
U −1 (R1 × R2 ) ' U1−1 R1 × U2−1 R2
a1 a2
(a1 , a2 )
7→
,
(u1 , u2 )
u1 u2
4.2. MODULES OF FRACTIONS
65
Proof. Injective: If ua11 = 0 and ua22 = 0 then there is (v1 , v2 ), (w1 , w2 ) ∈ U
such that v1 a1 = 0and w2a2 = 0. It follows that (v1 , v2 )(w1 , w2 )(a1 , a2 ) = 0.
Surjective: Given ua11 , ua22 then there is (u1 , v2 ), (v1 , u2 ) ∈ U . It follows that
(a1 v1 ,a2 v2 )
a1 a2
→
7
,
u1 u 2 .
(u1 v1 ,u2 v2 )
4.1.14. Exercise.
(1) Show that if U contains a nilpotent element, then U −1 R = 0.
(2) Show that
Ker R → U −1 R = {a ∈ R|ua = 0, for some u ∈ U }
(3) Let U = {u1 , . . . , um } and u = u1 · · · um . Then show that
U −1 R = {un }−1 R
(4) Let R1 , R2 be domains with fraction fields K1 , K2 . Show that the total ring of fractions of R1 × R2 is K1 × K2 .
(5) Show that the ring
a
{ n ∈ Q|a ∈ Z, n ∈ N}
2
is of finite type over Z.
4.2. Modules of fractions
4.2.1. Lemma. Let R be a ring and U ⊂ R multiplicative. On U × M is defined
a relation
(u, x) ∼ (u0 , x0 ) ⇔ there is v ∈ U such that vu0 x = vux0
(1) The relation is an equivalence relation.
(2) If (u, x) ∼ (u0 , x0 ) and (v, y) ∼ (v 0 , y 0 ) then (uv, vx + uy) ∼ (u0 v 0 , v 0 x0 +
u0 y 0 ).
(3) If (u, a) ∼ (u0 , a0 ) in U × R 4.1.1 and (v, x) ∼ (v 0 , x0 ) then (uv, ax) ∼
(u0 v 0 , a0 x0 ).
Proof. The claims are proved by simple calculations. See the proof 4.1.1.
4.2.2. Definition. Let R be a ring and U a multiplicative subset. The module of
fractions U −1 M is given by equivalence classes ux on U × M under the relation
4.2.1
(u, x) ∼ (u0 , x0 ) ⇔ there is v ∈ U such that vu0 x = vux0
The addition is
x y
vx + uy
+ =
u v
uv
−1
and the U R-scalar multiplication is
a y
ay
· =
u v
uv
The canonical homomorphism is
M → U −1 M, x 7→
x
1
66
4. FRACTION CONSTRUCTIONS
4.2.3. Theorem. Let U ⊂ R be a multiplicative subset. Given an R-module M , a
U −1 R-module N and a R-homomorphism f : M → N , then there exists a unique
U −1 R-homomorphism f 0 : U −1 M → N, ux 7→ u1 f (x) such that f = f 0 ◦ i.
f
/N
;
M G
GG
GG
GG
i GG#
U −1 M
f0
Proof. The claims are proved by simple calculations. If (u, x) ∼ (u0 , x0 ) then
vu0 x = vux0 and therefore vu0 f (x) = vuf (x0 ) showing u1 f (x) = u10 f (x0 ). So the
map is well defined. The rest is similar.
4.2.4. Corollary. Let f : M → N a homomorphism of R-modules. Then there is
−1 R-modules.
a homomorphism U −1 f : U −1 M → U −1 N, ux 7→ f (x)
u of U
4.2.5. Proposition. The construction 4.2.2
M 7→ U −1 M
and 4.2.4
f : M → N 7→ U −1 f : U −1 M → U −1 N
is a functor from R-modules to U −1 R-modules.
Proof. Follows from the definitions by simple calculations as in the proof of 4.2.3.
= f (x)
For example, U −1 (f + g) = U −1 f + U −1 g, follows from f (x)+g(x)
u
u +
g(x)
u .
4.2.6. Remark. The induced homomorphism relates to the canonical homomorphism such that the diagram is commutative.
M
U −1 M
f
U −1 f
/N
/ U −1 N
That is, the canonical homomorphism is a natural homomorphism.
4.2.7. Proposition. Let U ⊂ R be a multiplicative subset and M a module.
(1) x1 6= 0 in U −1 M if and only if Ann(x) ∩ U = ∅.
(2) The canonical homomorphism M → U −1 M is injective if and only if U
consists only of nonzero divisors on M .
Proof.
x
1
= 0 if and only if vx = 0 for some v ∈ U .
4.2.8. Proposition. If Mα is a family of modules, then the homomorphism
M
M
U −1 (
Mα ) →
U −1 Mα
α
is a natural isomorphism of U −1 R-modules.
α
4.3. EXACTNESS OF FRACTIONS
67
L
L
−1 M which
Proof. By 2.4.3 there is an R-homomorphism α Mα →
α
αU
gives the U −1 R-homomorphism in in question by 4.2.3. An inverse is given by
2.4.3. This is also the method of common denominators in a finite sum.
X xi
1 X
=
(Πj6=i uj )xi
ui
Πi ui
i
i
4.2.9. Definition. Let U = {un } the powers of an element u ∈ R. The R[ u1 ]module, 4.1.10, U −1 M is denoted
1
M[ ]
u
4.2.10. Proposition. If U ⊂ V ⊂ R are multiplicative, then there is a canonical
identification
x
x/1
V −1 M ' (V /1)−1 U −1 M, 7→
v
v/1
4.2.11. Exercise.
(1) Show that if U contains a nilpotent element, then U −1 M = 0.
(2) Show that
Ker M → U −1 M = {x ∈ M |ux = 0, for some u ∈ U }
(3) Let U = {u1 , . . . , um } and u = u1 · · · um . Then show that
U −1 M = {un }−1 M
(4) Show that U −1 M = 0 if and only if U ∩ Ann(x) 6= ∅ for all x ∈ M .
(5) Show that the fraction homomorphism of a composition is the composition of the
respective fraction homomorphisms.
(6) Let M be a free R-module. Show that U −1 M is a free U −1 R-module
(7) Show that the homomorphism
Y
Y 1
1
( Z)[ ] →
Z[ ]
2
2
N
N
is not surjective.
4.3. Exactness of fractions
4.3.1. Theorem. Let R be a ring and U a multiplicative subset. The functor
U −1 (−) is exact. Given an exact sequence of R-modules
M
f
/N
g
/L
Then the following sequence is exact
/ U −1 N
U −1 M
/ U −1 L
Proof. If uy ∈ U −1 N maps to g(y)
u = 0 then there is v ∈ U such that 0 = vg(y) =
(x)
x
g(vy). Choose x ∈ M such that f (x) = vy. Then vu
maps to fvu
= uy proving
exactness.
4.3.2. Corollary. Given a short exact sequence
0
/M
f
/N
g
/L
/0
Then the following sequence is exact
0
/ U −1 M
/ U −1 N
/ U −1 L
/0
68
4. FRACTION CONSTRUCTIONS
If the first sequence is split exact, also the second sequence is split exact.
4.3.3. Corollary. For a homomorphism f : M → N there are natural isomorphisms of U −1 R-modules.
(1) U −1 Ker f ' Ker U −1 f .
(2) U −1 Im f ' Im U −1 f .
(3) U −1 Cok f ' Cok U −1 f .
Proof. Represent the statements using short exact sequences. (1) The kernel is
determined by the exact sequence 0 → Ker f → M → N , 3.1.4. (3) The cokernel
is determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2) The
image is determined by the exact sequence 0 → Ker f → M → Im f → 0, 2.3.5,
3.1.4.
4.3.4. Corollary. For submodules N, L ⊂ M there are natural identifications of
U −1 R-submodules and factor modules.
(1) U −1 (M/N ) = U −1 M/U −1 N .
(2) U −1 (N + L) = U −1 N + U −1 L.
(3) U −1 (N ∩ L) = U −1 N ∩ U −1 L.
Proof. Represent the statements using short exact sequences. (1) 0 → N → M →
M/N → 0 is short exact giving 0 → U −1 N → U −1 M → U −1 (M/N ) → 0. The
wanted isomorphism follows form 3.1.5. (2) N + L is the image of N ⊕ L → M
so conclude by 4.3.3. (3) N ∩ L is the kernel of N ⊕ L → M so conclude by
4.3.3.
4.3.5. Corollary. For ideals I, J ⊂ R there are natural identifications in U −1 R.
(1)
(2)
(3)
(4)
U −1 (R/I) = U −1 R/U −1 I.
U −1 (I + J) = U −1 I + U −1 J.
U −1 (I ∩ J) = U −1 I ∩ U −1 J.
U −1 (IJ) = U −1 IU −1 J.
Proof. (1) (2) (3) These are special cases of 4.3.4. (4) Both sides have the same
generators ab
u , a ∈ I, b ∈ J, u ∈ U .
4.3.6. Proposition. Let R → U −1 R be the canonical homomorphism.
(1) For an ideal I ⊂ R the extended ideal
IU −1 R = U −1 I
(2) For an ideal J ⊂ U −1 R the extended contracted ideal
U −1 (J ∩ R) = J
(3) For an ideal I ⊂ R the contracted extended ideal
I ⊂ IU −1 R ∩ R
Proof. (1) This is clear. (2) U −1 (J ∩ R) ⊂ J is true for any ring homomorphism
1.2.6. If ub ∈ J then 1b ∈ J giving b ∈ J ∩ R and u1 b = ub ∈ U −1 (J ∩ R). (3) This
is true for any ring homomorphism 1.2.6.
4.4. TENSOR MODULES OF FRACTIONS
69
4.3.7. Proposition. Let R → U −1 R be the canonical homomorphism. For an
ideal I ⊂ R the contracted extended ideal
I = IU −1 R ∩ R
if and only if each u ∈ U is a nonzero divisor on R/I.
Proof. Apply the snake lemma 3.2.4 to
0
/I
/R
/ R/I
0
/0
/ U −1 R/U −1 I
U −1 R/U −1 I
/0
/0
and get the exact sequence
0 → I → IU −1 R ∩ R → Ker(R/I → U −1 (R/I)) → 0
Conclusion from 4.2.7.
4.3.8. Corollary. Let P ⊂ R be a prime ideal. Then U −1 P ⊂ U −1 R is either a
prime ideal or the whole ring.
Proof. By 4.1.5 and 4.3.5 U −1 R/U −1 P = U −1 (R/P ) is either 0 or a domain.
4.3.9. Corollary. Let R be a principal ideal domain. Then U −1 R is a principal
ideal domain.
Proof. A restricted ideal is principal by hypothesis and the extension of a principal
ideal is principal. Conclude by 4.3.6.
4.3.10. Theorem. Let R be a unique factorization domain. Then U −1 R is a unique
factorization domain.
Proof. By 4.3.7 the extension of an irreducible element is either a unit or an irreducible element. Since a principal ideal ( ua ) = ( a1 ), a factorization into irreducibles
in R gives a factorization in U −1 R. Now if (a) = (p1 ) . . . (pn ) is a factorization
in R and ( a1 ) is irreducible in U −1 R. Then all but one p1i is a unit in U −1 R so
( a1 ) = ( p1i ) is a prime ideal 4.3.9. The conditions 1.5.3 are satisfied.
4.3.11. Exercise.
(1) Let U ⊂ R be multiplicative. Show that
U −1 (R[X]) = (U −1 R)[X]
(2) Let φ : R → S be a ring homomorphism and U ⊂ R a multiplicative subset. Show
that
U −1 S = φ(U )−1 S
4.4. Tensor modules of fractions
4.4.1. Theorem. Let R be a ring and U a multiplicative subset. For any module
M , the homomorphism
a
ax
M ⊗R U −1 R → U −1 M, x ⊗ 7→
u
u
is a natural isomorphism of U −1 R-modules.
Proof. By 2.6.3 there is an R-module homomorphism x ⊗ ua → ax
u . By definition
a
ba
bax
b ax
b
−1
R-homomorphism. The
v (x ⊗ u ) = x ⊗ vu 7→ vu = v u , so the this is a U
map U −1 M → M ⊗R U −1 R, ux 7→ x ⊗ u1 is an inverse.
70
4. FRACTION CONSTRUCTIONS
4.4.2. Corollary. The two constructions, module change of ring to a fraction ring
and fraction module are natural isomorphic functors from modules to modules over
the fraction ring.
4.4.3. Corollary. Let R be a ring and U a multiplicative subset. Then U −1 R is a
flat R-module.
Proof. This follows from 4.4.1 and 4.3.1.
4.4.4. Corollary. Let R be a ring and U a multiplicative subset and M, N modules. Then there is a natural isomorphism
U −1 (M ⊗R N ) ' U −1 M ⊗U −1 R U −1 N
Proof. This follows from 2.7.5 and 4.4.1.
4.4.5. Corollary. Let I ⊂ R be an ideal and U a multiplicative subset For a
module M module
U −1 (IM ) ' U −1 IU −1 M
Proof. Use that IM = Im(I ⊗R M → M ) and 4.4.3,4.4.4.
4.4.6. Proposition. Let U ⊂ R be a multiplicative subset and M an R-module.
Let i : M → U −1 M be the canonical homomorphism and N ⊂ U −1 M any
U −1 R-submodule. Then N is extended
U −1 (i−1 (N )) ' N
Proof. The inclusion U −1 (i−1 (N )) ⊂ N is clear. If
x ∈ i−1 (N ) and u1 x = ux ∈ U −1 (i−1 (N )).
4.4.7. Exercise.
x
u
∈ N then
x
1
∈ N giving
(1) Let F be a flat R-module. Show that U −1 F is a flat U −1 R-
module.
(2) Let M be a projective R-module. Show that U −1 M is a projective U −1 R-module.
(3) Show that the homomorphism
Y
Y
1
1
Z ⊗Z Z[ ]
( Z) ⊗Z Z[ ] →
2
2
N
N
is not surjective.
4.5. Homomorphism modules of fractions
4.5.1. Proposition. Let R be a ring and U a multiplicative subset. For any modules
M, N there is a natural homomorphism
U −1 HomR (M, N ) → HomU −1 R (U −1 M, U −1 N )
of U −1 R-modules.
Proof. Given f : M → N and u ∈ U the setting
homomorphism U −1 M → U −1 N .
x
v
7→
f (x)
uv
is a U −1 R-
4.5.2. Proposition. Let R be a ring and U a multiplicative subset. For any Rmodule M and any U −1 R-modules N there is a natural isomorphism
HomR (M, N ) → HomU −1 R (U −1 M, N )
of U −1 R-modules.
4.6. THE POLYNOMIAL RING IS FACTORIAL
71
Proof. This is the change of rings isomorphism 2.7.6 interpreted according to
4.4.2. Also this is a reinterpretation of 4.2.3.
4.5.3. Example. Let R be a ring and U a multiplicative subset. R → U −1 R the
canonical homomorphism. The induced module functor maps an R-module M to
the U −1 R-module HomR (U −1 R, M ). The natural isomorphism 2.7.10 is
HomR (N, M ) ' HomU −1 R (N, HomR (U −1 R, M ))
for any U −1 R-module N .
4.5.4. Example. The homomorphism 4.5.1 is in general neither injective nor surjective.
(1) Not surjective:
0 = HomZ (Q, Z) → HomQ (Q, Q) = Q
(2) Not injective:
0 6= HomZ (Q, Q/Z) ⊗Z Q → HomQ (Q, Q/Z ⊗Z Q) = 0
4.5.5. Exercise.
(1) Let M be a Z-module and N a Q-module. Show that there is an
isomorphism HomZ (M, N ) ' HomQ (M ⊗Z Q, N ).
4.6. The polynomial ring is factorial
4.6.1. Definition. Let R be a unique factorization domain and let f = an X n +
· · · + a0 be a polynomial over R. Then the content of polynomial f , c(f ), is the
greatest common divisor of the coefficients a0 , . . . , an .
4.6.2. Lemma (Gauss’ lemma). Let R be a unique factorization domain. For polynomials f, g ∈ R[X]
c(f g) = c(f )c(g)
Proof. Assume by cancellation that c(f ), c(g) are units in R. For any irreducible
p ∈ R the projections of f, g in R/(p)[X] are nonzero. Since R has unique factorization the ideal (p) is a prime ideal. It follows that the projection of the product
f g in R/(p)[X] is also nonzero and therefore p is not a common divisor of the
coefficients of the product f g.
4.6.3. Theorem. Let R be a unique factorization domain. Then the ring of polynomials R[X] is a unique factorization domain.
Proof. Let K be the fraction field of R, then the polynomial ring K[X] is a principal ideal domain. Let f ∈ R[X] and use unique factorization in K[X] to get
0 6= a ∈ R and p1 , . . . , pn ∈ R[X], irreducible in K[X], such that
af = p1 . . . pn
Assume by 4.6.2 that a = 1 and c(p1 ), . . . , c(pn ) are units in R. Apply 4.6.2 and
1.6.5 to see that p1 , . . . , pn are irreducible in R[X]. An irreducible p ∈ R generates
a prime ideal (p) ⊂ R[X]. A non constant irreducible p ∈ R[X] generates a prime
ideal (p) ⊂ K[X] and therefore also a prime ideal (p) ⊂ R[X]. So conditions
1.5.3 are satisfied.
4.6.4. Corollary. Let K be a field. Then the polynomial ring K[X1 , . . . , Xn ] is a
unique factorization domain.
Proof. Follows by induction from 4.6.3.
72
4.6.5. Exercise.
4. FRACTION CONSTRUCTIONS
(1) Let f ∈ Z[X] be monic and assume f = gh where g, h ∈ Q[X]
are monic. Show that g, h ∈ Z[X].
(2) Let f ∈ Z[X] be monic and irreducible in Z/(n)[X]. Show that f is irreducible
Q[X].
(3) Let K be a field. Show that the polynomial ring K[X1 , X2 , . . . ] in countable many
variables is a unique factorization domain
5
Localization
5.1. Prime ideals
5.1.1. Theorem (Krull). A nonzero ring contains a maximal ideal.
Proof. The nonempty set of ideals different from R is ordered by inclusion. Given
an increasing chain Iα then ∪Iα is an ideal different from R which is a maximum
for the chain. Conclusion by Zorn’s lemma.
5.1.2. Corollary. Any proper ideal in a ring is contained in a maximal ideal.
Proof. If I 6= R then the factor ring R/I 6= 0 and contains a maximal ideal.
Conclusion by 1.2.10.
5.1.3. Theorem. Let P1 , . . . , Pn ⊂ R be ideals with at most 2 not being prime
ideals. If an ideal
I ⊂ P1 ∪ · · · ∪ Pn
then I ⊆ Pi for some i.
Proof. Assume n > 1 and I is not contained in any sub union. Moreover assume
the numbering such that P3 , . . . , Pn are prime ideals. Then for each i there is
ai ∈ (I ∩ Pi )\ ∪j6=i Pj
The element
an + a1 . . . an−1
is in I but not in any Pi , giving a contradiction. So n = 1.
5.1.4. Proposition. Let P1 , . . . , Pn ⊂ R be prime ideals and I any ideal. If for
some a ∈ R
a + I ⊂ P1 ∪ · · · ∪ Pn
then I ⊆ Pi for some i.
Proof. If a ∈ ∩i Pi then conclusion by 5.1.3. On the contrary after renumbering
there exists j with 1 ≤ j < n such that
a ∈ P1 ∩ · · · ∩ Pj \Pj+1 ∪ · · · ∪ Pn
Assume no inclusions between the prime ideals and choose by 5.1.3
b ∈ I ∩ Pj+1 ∩ · · · ∩ Pn \P1 ∪ · · · ∪ Pj
Then a + b ∈
/ ∪i Pi contradicts the hypothesis.
5.1.5. Theorem. Let R → U −1 R be the canonical homomorphism. Extension and
contraction gives a bijective correspondence between prime ideals in R disjoint
from U and all prime ideals in U −1 R.
(1) For a prime ideal P ⊂ R\U the extended ideal P U −1 R is a prime ideal in
U −1 R and the contracted P U −1 R ∩ R = P .
73
74
5. LOCALIZATION
(2) For a prime ideal Q ⊂ U −1 R the contracted ideal Q ∩ R is a prime ideal
and the extended (Q ∩ R)U −1 R = Q
R → U −1 R
P 7→ P U −1 R
Q ∩ R ←[ Q
Proof. Conclusion by 4.3.6, 4.3.7, 4.3.8.
5.1.6. Corollary. Let R be a ring and U a multiplicative subset.
(1) An ideal P ⊂ R is maximal among the ideals disjoint from U if and only if
P U −1 R is a maximal ideal in U −1 R.
(2) An ideal maximal among the ideals disjoint from U is a prime ideal.
(3) Any ideal disjoint from U is contained in a prime ideal disjoint form U .
Proof. The prime ideals disjoint from U are the prime ideals in U −1 R.
5.1.7. Theorem. The nilradical of a ring R is the intersection of all prime ideals
P.
\
√
0=
P
P
Proof. By 1.3.8 the nilradical is contained in any prime ideal. Suppose u ∈ R is
not nilpotent. Then {un }−1 R is nonzero. Then contraction of a maximal ideal,
5.1.1, is a prime ideal in R not containing u.
5.1.8. Corollary. Let R be a ring.
(1) The radical of an ideal I is the intersection of all prime idealsP containing I
\
√
I=
P
√
√
I⊂P
√
(2) For ideals I, J ⊂ R, I ∩ J = I ∩ √
J.
√
(3) If U is a multiplicative subset, then U −1 0 = 0 in U −1 R.
If R is reduced, then U −1 R is reduced.
Proof.
√ (1) Use
√ 5.1.7 on the factor ring R/I. (2) This follows from (1). (3) Clearly
U −1 0 ⊂ 0. If ua is nilpotent, then van = 0 for some v ∈ U . So (va)n = 0 and
√
va
a
−1 0.
u = vu ∈ U
5.1.9. Definition. A prime ideal minimal for inclusion among prime ideals is a
minimal prime ideal.
5.1.10. Theorem. Any prime ideal of Q ⊂ R contains a minimal prime ideal
P ⊂ Q.
Proof. The set of prime ideals in R is ordered by inclusion. Given a decreasing
chain Pα then ∩Pα is a prime ideal. Conclusion by Zorn’s lemma.
5.1.11. Corollary. Let R ⊂ S be a subring and P ⊂ R a minimal prime ideal.
Then there is a minimal prime ideal Q ⊂ S contracting to P = Q ∩ R.
5.1.12. Proposition. Let R be a domain. The following conditions are equivalent:
(1) R is a unique factorization domain.
(2) Any nonzero prime ideal contains a nonzero principal prime ideal.
5.2. LOCALIZATION OF RINGS
75
Proof. (1) ⇒ (2): A prime ideal P 6= 0 contains an irreducible element generating
a prime ideal. (2) ⇒ (1): Let U be the multiplicative subset generated by generators
of principal prime ideals. Given a ∈ R nonzero and not a unit. If a ∈
/ U then by
5.1.6 there is a prime ideal (a) ⊂ P such that P ∩ U = ∅. Such a P contains no
principal prime ideal, so a ∈ U and 1.5.3 are satisfied.
5.1.13. Exercise.
(1) Let K ⊂ R be an infinite subfield and I, P1 , . . . , Pn any ideals.
Show that if I ⊂ P1 ∪ · · · ∪ Pn then I ⊂ Pi for some i.
(2) Let P, P1 , P2 be proper ideals. Show that if P is a maximal ideal and P n ⊂ P1 ∪ P2
then P = P1 of P = P2 .
5.2. Localization of rings
5.2.1. Definition. A ring R which contains precisely one maximal ideal P is a
local ring and denoted (R, P ). The residue field of R is R/P denoted by k(P ).
A ring homomorphism φ : R → S of local rings (R, P ), (S, Q) is a local ring
homomorphism if φ(P ) ⊂ Q.
5.2.2. Proposition. Let R be a ring.
(1) If (R, P ) is a local ring, then R\P is the set of units of R.
(2) If the subset of non units in R is an ideal P , then (R, P ) is a local ring.
Proof. (1) If u ∈
/ P then by 5.1.1 (u) = R and u is a unit. (2) Any ideal I 6= R
contains only non units, so I ⊂ P .
5.2.3. Proposition. A ring homomorphism φ : R → S of local rings (R, P ), (S, Q)
is a local ring homomorphism if the extended ideal P S ⊂ Q or the contracted ideal
Q ∩ R = P . The residue homomorphism k(P ) → k(Q) is a field extension.
Proof. The contraction Q ∩ R is a prime ideal containing P . The rest is clear.
5.2.4. Lemma. Let R be a ring and P a prime ideal. Then U = R\P is a multiplicative subset. The ring of fractions U −1 R is a local ring. The maximal ideal is
the extended ideal P U −1 R. The residue field is U −1 R/P U −1 R which is canonical isomorphic to the fraction field of R/P .
5.2.5. Definition. Let R be a ring, P a prime ideal and U = R\P . The localized
ring at P is the local ring RP = U −1 R. The residue field is denoted k(P ) =
RP /P RP .
R
R/P
/ RP
/ k(P )
Note that
k(P ) = RP /P RP = (R/P )P = (R/P )(0)
5.2.6. Proposition. Given a ring homomorphism R → S and a prime ideal Q ⊂
S. Then P = Q ∩ R is a prime ideal and there is a local ring homomorphism
76
5. LOCALIZATION
(RP , P RP ) → (SQ , QSQ ) fitting into the following commutative diagram.
/5 SQ
k5 S
kkk
kkk
k
k
k
k
k
k
kkk
kkk
kkk
kkk
k
k
k
k
kk
kkk
kkk
/ RP k
R
/ k(Q)
5
ll
lll
lll
lll
l
l
l
l
l
l
lll
lll
lllll
lllll
/ k(P )
R/P
l5 S/Q
Proof. This is clear from the constructions.
5.2.7. Example. Let the ring be Z.
(1) The local ring at (0) is the fraction field Q = Z(0) .
(2) The local ring Z(p) for a prime number p is identified with a subring of Q
m
|p not dividing n}
n
The residue field Fp = Z(p) /(p).
(3) Any nonzero ideal in Z(p) is principal of the form (pn ) for some n.
Z(p) = {
5.2.8. Proposition. Let (R, P ) be a local ring. One of the following conditions is
satisfied:
(1) The characteristic char(R) = 0. P ∩ Z = (0) and Q ⊂ R is a subfield.
Q → k(P ) is a field extension.
(2) The characteristic char(R) = 0. P ∩ Z = (p), p a prime number. Z(p) ⊂ R
is a local subring. Fp → k(P ) is a field extension.
(3) The characteristic char(R) = pn , a power of a prime number. Z/(pn ) ⊂ R
is a local subring. Fp → k(P ) is a field extension.
Proof. (1) (2) are clear by 5.2.3 and 5.2.7. (3) If the characteristic is nonzero then
any prime ideal contracts Q ∩ Z = (p). So a prime number q 6= p gives a unit in
R. There is a local homomorphism Z(p) → R, 4.1.3. The nontrivial kernel is (pn )
by 5.2.7.
5.2.9. Example. (1) A field K is a local ring with maximal ideal (0). The power
series ring K[[X]] is a local ring with maximal ideal (X) and residue field K.
(2) Let (R, P ) be a local ring. The power series ring R[[X]] is a local ring with
maximal ideal (P, X) and residue field k(P, X) = k(P ).
5.2.10. Proposition. Let R be a domain.
(1) The local ring at (0) is the fraction field K = R(0) .
(2) Any local ring RP is identified with a subring of the fraction field K.
(3) The intersection
\
R=
RP , P a maximal ideal
P
5.2.11. Proposition. Let R × S be a product of rings.
5.3. LOCALIZATION OF MODULES
77
(1) A prime ideal is of the form P × S or R × Q for uniquely determined prime
ideals P ⊂ R or Q ⊂ S.
(2) The local ring at P × S is identified with RP through the projection R × S →
R.
(3) The local ring at R × Q is identified with SQ through the projection R × S →
S.
5.2.12. Theorem. Let P be a prime ideal and R → RP the canonical homomorphism. Extension and contraction gives a bijective correspondence between prime
ideals in R contained in P and all prime ideals in RP .
(1) For a prime ideal Q ⊂ P the extended ideal QRP is a prime ideal in RP and
the contracted QRP ∩ R = Q.
(2) For a prime ideal Q0 ⊂ RP the contracted ideal Q0 ∩ R ⊂ P is a prime ideal
and the extended (Q0 ∩ R)RP = Q0
R → RP
P → P RP
Q → QRP
Q0 ∩ R ←[ Q0
Proof. This is a special case of 5.1.5.
5.2.13. Proposition. Let U ⊂ R be a multilicative subset and P ∩ U = ∅ a prime
ideal. Then P U −1 R is a prime ideal in U −1 R and canonically
RP = (U −1 R)P U −1 R
Proof. This follows from 4.1.12 and reflects the fraction rule ua / wv =
av
uw .
5.2.14. Corollary. Let Q ⊂ P ⊂ R be prime ideals. Then QRP is a prime ideal
in RP and canonically
RQ = (RP )QRP
5.2.15. Definition. The intersection of all maximal ideals in a ring is the Jacobson
radical.
5.2.16. Remark. The Jacobson radical contains the nilradical. In a local ring the
Jacobson radical is the maximal ideal.
5.2.17. Exercise.
(2)
(3)
(4)
(5)
(1) Show that a local ring is never a product of two nonzero rings.
Show that a ∈ R is in the Jacobson radical if and only if 1 + ab is a unit for all b ∈ R.
Let p be a prime number. Describe the prime ideals in the ring Z(p) .
Let P be a prime ideal. Show that k(P ) is the fraction field of R/P .
Let (R, P ) be a local ring. Show that (R[[X]], (P, X) is a local ring and the canonical
homomorphism R → R[[X]] is a local homomorphism.
5.3. Localization of modules
5.3.1. Definition. Let R be a ring, P a prime ideal and U = R\P . For a module
M , the localized module at P is the module MP = U −1 M over the local ring RP .
For a homomorphism f : M → N the localized homomorphism is fP : MP →
NP and the residue homomorphism is f (P ) : M ⊗R k(P ) → N ⊗R k(P ).
78
5. LOCALIZATION
The constructions are functors, 2.7.4, 4.2.5. If f : M → N is a homomorphism,
then the following diagram is commutative.
/
ii4 NP
iii4 N
i
i
i
i
i
i
iiii
iiii
iiii
iiii
i
i
i
i
i
i
ii
iiii
iiii
/ MP i
M
/ N (P )
jj4
jjjj
jjjj
j
j
j
j
j
j
jj
jj
jjjj jjjj
jjjj
jjjj
/ M (P )
M/P M
N/P N
j4
5.3.2. Proposition. Let R be a ring, P a prime ideal and M a module.
(1) MP ' M ⊗R RP are natural isomorphic exact functors.
(2) MP /P RP MP ' M ⊗R k(P ) are natural isomorphic functors.
Proof. See 4.4.1.
5.3.3. Corollary. For a homomorphism f : M → N
(1) (Ker f )P ' Ker fP .
(2) (Im f )P ' Im fP .
(3) (Cok f )P ' Cok fP .
Proof. See 4.3.3.
5.3.4. Corollary. Let R be a ring and P a prime ideal. For submodules N, L ⊂ M
(1) (M/N )P ' MP /NP .
(2) (N + L)P ' NP + LP .
(3) (N ∩ L)P ' NP ∩ LP .
Proof. See 4.3.4.
5.3.5. Proposition. Let R be a ring and P a prime ideal. If Mα is a family of
modules, then the homomorphism
M
M
(
Mα )P →
(Mα )P
α
α
is an isomorphism of RP -modules.
Proof. See 4.2.8.
5.3.6. Proposition. Let R be a ring, P a prime ideal.
(1) For an R module M and an RP -module N there is a natural isomorphism
M ⊗R RP ⊗RP N ' M ⊗R N
(2) For an R module M, L there is a natural isomorphism
(M ⊗R L)P ' MP ⊗RP LP
(3) For an R module M, L there is a natural isomorphism
(M ⊗R L)(P ) ' M (P ) ⊗k(P ) L(P )
Proof. See 4.4.4 and 2.7.4.
5.4. THE LOCAL-GLOBAL PRINCIPLE
79
5.3.7. Proposition. Let U ⊂ R be a multilicative subset and P ∩ U = ∅ a prime
ideal. Then P U −1 R is a prime ideal in U −1 R and canonically for any R-module
M
MP = (U −1 M )P U −1 R
Proof. This follows from 4.2.10 extending the set of denominators.
5.3.8. Corollary. Let Q ⊂ P ⊂ R be prime ideals and M an R-module Then
QRP is a prime ideal in RP and canonically
MQ = (MP )QRP
5.3.9. Definition. Let R be a ring. F is a locally free module is FP is a free
RP -module for all prime ideals P .
5.3.10. Proposition. Let R be a ring. F is a locally free module if FQ is a free
RQ -module for all maximal ideals Q.
Proof. A prime ideal P ⊂ Q is contained in a maximal ideal. By 5.3.8 FP '
(FQ )PQ is free.
5.3.11. Example. A free module is a locally free module.
5.3.12. Exercise.
(1) Let P ⊂ R be a prime ideal and R → S a ring homomorphism.
Show that RP → SP is a ring homomorphism.
(2) Let Q ⊂ S be a prime ideal and R → S a ring homomorphism. Show that RQ∩R →
SQ is a local ring homomorphism.
(3) Let R = K × L be a product of fields. Show that ideal K × {0} is locally free but
not free.
5.4. The local-global principle
5.4.1. Theorem. Let R be a ring and M a module. The following conditions are
equivalent:
(1) M = 0.
(2) MP = 0 for all prime ideals P .
(3) MP = 0 for all maximal ideals P .
Proof. (1) ⇒ (2) ⇒ (3) is clear. (3) ⇒ (1): Let 0 6= x ∈ M be given. Then
6 x1 ∈ MP
Ann(x) ⊂ P is contained in a maximal ideal, 5.1.2. Clearly 0 =
contradicts (3).
5.4.2. Corollary. Let R be a ring and f : M → N a homomorphism. The following conditions are equivalent:
(1) f is injective.
(2) fP is injective for all prime ideals P .
(3) fP is injective for all maximal ideals P .
Proof. Use 5.4.1 on Ker f .
5.4.3. Corollary. Let R be a ring and f : M → N a homomorphism. The following conditions are equivalent:
(1) f is surjective.
(2) fP is surjective for all prime ideals P .
(3) fP is surjective for all maximal ideals P .
80
5. LOCALIZATION
Proof. Use 5.4.1 on Cok f .
5.4.4. Corollary. Let R be a ring and f : M → N a homomorphism. The following conditions are equivalent:
(1) f is an isomorphism.
(2) fP is an isomorphism for all prime ideals P .
(3) fP is an isomorphism for all maximal ideals P .
5.4.5. Corollary. Let R be a ring and
f
M
/N
g
/L
a sequence of homomorphisms. The following conditions are equivalent:
(1) The sequence is exact.
(2) The sequence
fP
/ NP
gP
/ LP
/ NP
MP
is exact for all maximal ideals P .
gP
/ LP
g
/L
MP
is exact for all prime ideals P .
(3) The sequence
fP
5.4.6. Corollary. Let R be a ring and
0
/M
f
/N
/0
a sequence of homomorphisms. The following conditions are equivalent:
(1) The sequence is short exact.
(2) The sequence
0
/ MP
fP
/ NP
gP
/ LP
/0
gP
/ LP
/0
is short exact for all prime ideals P .
(3) The sequence
0
/ MP
fP
/ NP
is short exact for all maximal ideals P .
5.4.7. Corollary. Let R be a ring and F a module. The following conditions are
equivalent:
(1) F is flat.
(2) FP is flat for all prime ideals P .
(3) FP is flat for all maximal ideals P .
Proof. Let 0 → M → N . Use 5.3.6 and 5.4.2 on M ⊗R F → N ⊗R F .
5.4.8. Proposition. Let R be a ring and M a module. Then there is an exact
sequence
Y
0→M →
MP
P maximal
Proof. Let 0 6= x ∈ M be given. Then Ann(x) ⊂ P is contained in a maximal
ideal, 5.1.2. Clearly 0 6= x1 ∈ MP .
5.5. FLAT RING HOMOMORPHISMS
81
5.4.9. Corollary. Let R be a ring. Then there is an injective ring homomorphism
Y
R→
RP
P maximal
5.4.10. Corollary. Let R be a ring. The following conditions are equivalent:
(1) R is reduced.
(2) RP is reduced for all prime ideals P .
(3) RP is reduced for all maximal ideals P .
Proof. Use 5.1.8 and 5.4.9.
5.4.11. Exercise.
(1) Let R be a ring and
/M
0
f
/N
g
/L
/0
a split exact sequence. Show that the localized sequence is split exact for all prime
ideals P .
5.5. Flat ring homomorphisms
5.5.1. Definition. A ring homomorphism R → S is a flat ring homomorphism if
S is a flat R-module, 3.7.1.
5.5.2. Proposition. Let R → S be a ring homomorphism. The following conditions are equivalent:
(1) R → S is a flat ring homomorphism.
(2) If M → N is injective, then the change of rings homomorphism M ⊗R S →
N ⊗R S is injective.
(3) The change of rings functor − ⊗R S is an exact functor.
(4) Given an exact sequence of R-modules
/N
M
/L
Then the sequence of change of rings modules is exact
M ⊗R S
/ N ⊗R S
/ L ⊗R S
(5) Given a short exact sequence
0
/M
f
/N
g
/L
/0
Then the following sequence is exact
0
/ M ⊗R S
/ N ⊗R S
/ L ⊗R S
/0
Proof. This follows from the definition 3.7.1 and 3.7.2.
5.5.3. Corollary. (1) Let U ⊂ R be a multiplicative subset. Then the canonical
ring homomorphism R → U −1 R is flat.
(2) Let P ⊂ R be a prime ideal. Then the canonical ring homomorphism R →
RP is flat.
5.5.4. Corollary. Let R → S be a flat ring homomorphism. For a homomorphism
f : M → N of R-modules there are natural isomorphisms of S-modules.
(1) Ker f ⊗R S ' Ker f ⊗ 1.
(2) Im f ⊗R S ' Im f ⊗ 1.
(3) Cok f ⊗R S ' Cok f ⊗ 1.
82
5. LOCALIZATION
Proof. See 3.3.3. Represent the statements using short exact sequences. (1) The
kernel is determined by the exact sequence 0 → Ker f → M → N , 3.1.4. (3) The
cokernel is determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2)
The image is determined by the exact sequence 0 → Ker f → M → Im f → 0,
2.3.5, 3.1.4.
5.5.5. Corollary. Let R → S be a flat ring homomorphism. For submodules
N, L ⊂ M there are natural identifications of S-submodules and factor modules.
(1) (M/N ) ⊗R S = M ⊗R S/N ⊗R S.
(2) (N + L) ⊗R S = N ⊗R S + L ⊗R S.
(3) (N ∩ L) ⊗R S = N ⊗R S ∩ L ⊗R S.
Proof. See 3.3.4. Represent the statements using short exact sequences. (1) 0 →
N → M → M/N → 0 is short exact giving 0 → N ⊗R S → M ⊗R S →
(M/N ) ⊗R S → 0. The wanted isomorphism follows form 3.1.5. (2) N + L is the
image of N ⊕ L → M . (3) N ∩ L is the kernel of N ⊕ L → M .
5.5.6. Corollary. Let R → S be a flat ring homomorphism. For ideals I, J ⊂ R
there are natural identifications in S.
(1) I ⊗R S = IS.
(2) (I ∩ J)S = IS ∩ JS.
Proof. (1) The sequence 0 → I ⊗R S → R ⊗R S → (R/I) ⊗R S → 0. is exact.
(2) This follows from (1) and 5.5.4.
5.5.7. Proposition. Let φ : R → S be a flat ring homomorphism.
(1) If F is a flat S-module, then the restriction of scalars makes F a flat Rmodule.
(2) If S → T is a flat ring homomorphism, then the composite R → T is a flat
ring homomorphism.
Proof. (1) There is a natural isomorphism M ⊗ RF ' (M ⊗R S) ⊗S F and clearly
the composition of two exact functors is exact. (2) This follows from (1).
5.5.8. Proposition. Let φ : R → S be a ring homomorphism. The following
conditions are equivalent:
(1) φ is flat.
(2) φP : RP → SP is flat for all prime ideals P ⊂ R.
(3) φP : RP → SP is flat for all maximal ideals P ⊂ R.
Proof. Use 5.4.7.
5.5.9. Proposition. Let R → S be a ring homomorphism. The following conditions are equivalent:
(1) R → S is flat.
(2) RP → SQ is flat for all prime ideals Q ⊂ S and P = Q ∩ R.
(3) RP → SQ is flat for all maximal ideals Q ⊂ S and P = Q ∩ R.
Proof. RP → SQ is the composite RP → SP → SP QSP = SQ , 5.3.8. (1) ⇒
(2): The composite RP → SP → SQ is flat by 5.5.8 and 5.5.3. (3) ⇒ (1): If
0 → M → N is exact, then 0 → M ⊗R RP → N ⊗R RP is exact, 5.5.3. By the
hypothesis 0 → M ⊗R RP ⊗RP SQ → N ⊗R RP ⊗RP SQ is exact, so cancelation
and canonical isomorphism, 2.7.5 and 5.3.2, give 0 → (M ⊗R S)Q → (N ⊗R S)Q
5.6. FAITHFULLY FLAT RING HOMOMORPHISMS
83
exact for all maximal ideals Q. By 5.4.2 0 → M ⊗R S → N ⊗R S exact and
R → S is flat.
5.5.10. Exercise.
(1) Let R → S and S → T be flat homomorphisms. Show that the
composite R → S is flat.
(2) Show that Q is a flat but not
√faithfully flat Z-module.
(3) Let R be a ring and I = 0 the nilradical. Show that IR[X] is the nilradical of
R[X].
5.6. Faithfully flat ring homomorphisms
5.6.1. Definition. A flat R-module F is faithfully flat if for any R-module M 6= 0
the tensor product M ⊗R F 6= 0.
5.6.2. Proposition. (1) A nonzero free module is faithfully flat.
(2) Let F, F 0 be faithfully flat modules, then F ⊗R F 0 is faihfully flat.
(3) Let R → S be a ring homomorphism and F a faithfully flat R-module. The
change of ring module F ⊗R S is a faithfully flat S-module.
Proof. (1) This is clear from properties of tensor product. (2) F ⊗R F 0 is flat by
3.7.7 and M 6= 0 gives M ⊗R F 6= 0 so M ⊗R F ⊗ RF 0 6= 0. (3) F ⊗R S is flat
by 3.5.8 and if N 6= 0 is an S-module (3) F ⊗R S is flat by 3.5.8 and if N 6= 0 is
an S-module then N ⊗S (F ⊗R S) ' N ⊗R F 6= 0 by 2.7.5.
5.6.3. Proposition. Let F be an R-module. The following conditions are equivalent:
(1) F is faithfully flat.
(2) A homomorphism M → N is injective if and only if M ⊗R F → N ⊗R F is
injective.
(3) A sequence of R-modules
M
f
/N
g
/L
is exact if and only if the sequence
M ⊗R F
/ N ⊗R F
/ L ⊗R F
is exact.
(4) A sequence
f
/N
/M
0
is short exact if and only if the sequence
0
/ M ⊗R F
g
/ N ⊗R F
/L
/0
/ L ⊗R F
/0
is short exact.
Proof. (1) ⇔ (2): f : M → N is injective if and only if Kerf = 0 if and only if
Ker f ⊗ 1F = (Ker f ) ⊗R F = 0. (1) ⇔ (3) ⇔ (4): M → N → L is exact if and
only if (Im f + ker g)/ Ker g = 0 = (Im f + ker g)/ Im f .
5.6.4. Proposition. Let F be a flat R-module. The following conditions are equivalent:
(1) F is faithfully flat.
(2) P F 6= F for all maximal ideals P ⊂ R.
(3) F ⊗ k(P ) 6= 0 for all maximal ideals P ⊂ R.
84
5. LOCALIZATION
Proof. (1) ⇒ (2) ⇔ (3): k(P ) ⊗R R ' F/P F 6= 0. (2) ⇒ (1): For 0 6= x ∈ M
let Ann(x) = I ⊂ P be a maximal ideal. Then R/I ⊗R F → R/P ⊗R F → 0
and 0 → R/I ⊗R F → M ⊗R F → 0 are exact, so M ⊗R F 6= 0.
5.6.5. Proposition. Let R be a ring and F a module. The following conditions are
equivalent:
(1) F is faithfully flat.
(2) FP is faithfully flat for all prime ideals P .
(3) FP is faithfully flat for all maximal ideals P .
Proof. (1) ⇒ (2) ⇒ (3): 5.6.2. (3) ⇒ (1): F is flat by 5.4.7. If M 6= 0, then
MP 6= 0 for some P . Therefore (M ⊗R F )P ' MP ⊗RP FP 6= 0.
5.6.6. Definition. A ring homomorphism R → S is a faithfully flat ring homomorphism if S is a faithfully flat R-module.
5.6.7. Proposition. Let R → S be a ring homomorphism. The following conditions are equivalent:
(1) R → S is faithfully flat.
(2) A homomorphism M → N is injective if and only if the change of rings
homomorphism M ⊗R F → N ⊗R S is injective.
(3) A sequence of R-modules
M
f
/N
g
/L
is exact if and only if the sequence
M ⊗R S
/ N ⊗R S
/ L ⊗R S
is exact.
(4) A sequence
f
/M
/N
0
is short exact if and only if the sequence
0
/ M ⊗R S
g
/ N ⊗R S
/L
/0
/ L ⊗R S
/0
is short exact.
Proof. See 5.6.3.
5.6.8. Proposition. Let R → S be a faithfully flat ring homomorphism and F an
R-module.
(1) If F ⊗R S is a flat S-module, then F is a flat R-module.
(2) If F ⊗R S is a faithfully flat S-module, then F is a faithfully flat R-module.
Proof. (1) If 0 → M → N is exact, then 0 → M ⊗R S ⊗S F ⊗R S → N ⊗R
S ⊗S F ⊗R S is exact. This is natural isomorphic to 0 → M ⊗R F ⊗R S →
N ⊗R F ⊗R S, so by 5.6.3 0 → M ⊗R F → N ⊗R F is exact. (2) If M 6= 0 then
M ⊗R S ⊗S F ⊗R S ' M ⊗R F ⊗R S 6= 0 and therefore M ⊗R F 6= 0.
5.6.9. Proposition. Let R → S be a faithfully flat ring homomorphism.
(1) If F is a faithfully flat S-module, then the restriction of scalars makes F a
faithfully flat R-module.
(2) If S → T is a faithfully flat ring homomorphism, then the composite R → T
is a faithfully flat ring homomorphism.
5.6. FAITHFULLY FLAT RING HOMOMORPHISMS
85
Proof. (1) F is a flat R-module by 5.5.7. If M 6= 0 is an R-module, then M ⊗R
F ' M ⊗R S ⊗S F 6= 0. (2) This is a special case of (1).
5.6.10. Proposition. Let R → S be a ring homomorphism.
(1) If F is a faithfully flat S-module such that the restriction of scalars makes F
a flat R-module. Then R → S is a flat ring homomorphism.
(2) If F is a faithfully flat S-module such that the restriction of scalars makes F a
faithfully flat R-module. Then R → S is a faithfully flat ring homomorphism.
(3) If S → T is a faithfully flat ring homomorphism and the composite R → T
is a faithfully flat ring homomorphism. Then R → S is a faithfully flat ring
homomorphism.
(4) If S → T is a faithfully flat ring homomorphism and the composite R → T
is a flat ring homomorphism. Then R → S is a flat ring homomorphism.
Proof. Let M → N be an R-homomorphism. Then M ⊗R S⊗S F → N ⊗R S⊗S F
is natural isomorphic to M ⊗R F → N ⊗R F . (1) Since F is flat as R-module
and faithfully flat as S-module, it follows by 5.6.3 that if M → N is injective
then M ⊗R S → N ⊗R S is injective. (2) Since F is faithfully flat both as Rmodule and S-module, it follows by 5.6.3 that M → N is injective if and only if
M ⊗R S → N ⊗R S is injective. (3),(4) These are special cases of (1),(2).
5.6.11. Proposition. Let R → S be a faithfully flat ring homomorphism. for any
ideal I ⊂ R the extended contracted returns I, i.e.
I = IS ∩ R
Proof. Tensor the homomorphism R/I → R/IS∩R with S. The induced R/I ⊗R
S → R/IS ∩ R ⊗R S is canonically isomorphic to the identity S/IS → S/IS.
By 5.6.3 R/I → R/IS ∩ R is injective giving I = IS ∩ R.
5.6.12. Corollary. A faithfully flat ring homomorphism R → S is injective.
5.6.13. Proposition. A flat local homomorphism (R, P ) → (S, Q) is faithfully
flat.
Proof. P S ⊂ Q 6= S so S is faithfully flat by 5.6.4.
5.6.14. Theorem (going-down). Let φ : R → S be a flat ring homomorphism and
Q ⊂ S a prime ideal. For any prime ideal P 0 ⊂ P = Q ∩ R there is a prime ideal
Q0 ⊂ Q contracting to P 0 = Q0 ∩ R.
R
P
nS
nnn
nnn
Q
ooo
o
o
o
Q0
P0
Proof. The local homomorphism RP → SQ is faithfully flat. The ring k(P 0 ) ⊗R
SQ is nonzero and therefor contains a maximal ideal Q00 . The contraction to S,
Q0 = Q00 ∩ S contracts to P 0 = R ∩ Q0 .
86
5. LOCALIZATION
5.6.15. Proposition. Let R → S be a flat homomorphism. The following conditions are equivalent:
(1) R → S is faithfully flat.
(2) Any prime ideal P ⊂ R is the contraction P = Q ∩ R of a prime ideal
Q ⊂ S.
oo S
R
o
ooo
Q
P
Proof. (1) ⇒ (2): The ring SP /P SP ' k(P ) ⊗R S 6= 0, so let Q0 be a maximal
ideal. The contraction Q = Q0 ∩ S is a prime ideal such that P = Q ∩ R. (2) ⇒
(1): Let M 6= 0 be an R-module. Then MP 6= 0 for some P . Let P = Q ∩ R, then
(M ⊗R S)Q ' MP ⊗RP SQ 6= 0 as RP → SQ is faithfully flat.
5.6.16. Proposition. The inclusion R → R[X1 , . . . , Xn ] is a faithfully flat homomorphism
Proof. The R-module R[X1 , . . . , Xn ] is free.
5.6.17. Exercise.
(1) Show that a free module is faithfully flat.
(2) Let F be a faithfully flat module and G a flat module. Show that F ⊕ G is faithfully
flat.
(3) Show that Q is a flat but not
√faithfully flat Z-module.
(4) Let R be a ring and I = 0 the nilradical. Show that IR[X] is the nilradical of
R[X].
L
(5) Show that
RP over all maximal ideals P is a faithfully flat R-module.
6
Finite modules
6.1. Finite modules
6.1.1. Definition. Let R be a ring. A finite module is generated by finitely many
elements. The finite free module with standard basis e1 , . . . , en is denoted Rn .
6.1.2. Lemma. Let R be a ring and M a module. The following conditions are
equivalent:
(1) M is generated by n elements x1 , . . . , xn .
(2) There is a surjective homomorphism Rn → M → 0, ei 7→ xi .
Proof. See 2.4.12.
6.1.3. Proposition. Let R → S be a ring homomorphism. If an R-module M is
generated by n elements x1 , . . . , xn . Then the change of rings S-module M ⊗R S
is generated by x1 ⊗ 1, . . . , xn ⊗ 1 over S.
Proof. Follows from 6.1.2 and 3.4.1
6.1.4. Corollary. Let R be a ring and U a multiplicative subset. If M is a finite
R-module, then U −1 M is a finite U −1 R-module.
6.1.5. Proposition. For a short exact sequence
f
/M
0
/N
g
/L
/0
the following hold
(1) If N is finite, then L is finite.
(2) If M, L are finite, then N is finite.
Proof. (1) If y1 , . . . , yn generates N , then g(y1 ), . . . , g(yn ) generates L. (2) Choose
u : Rn → M → 0 and v : Rm → L → 0 exact. By 3.5.5 there is w : Rm → N
such that g ◦ w = v. There is a diagram
/ Rn ⊕ Rm
/ Rn
0
u
/M
0
f
/ Rm
f ◦u+w
g
/N
/0
v
/L
/0
Conclusion by the snake lemma 3.2.4.
6.1.6. Corollary. Let
0
/M
f
/N
g
/L
/0
be a split exact sequence. Then N is finite if and only if M, L are finite.
Proof. Let u be a retraction of f . By 6.1.5 Im u = M is finite. The rest is contained
in 6.1.5.
87
88
6. FINITE MODULES
6.1.7. Corollary. Let f : M → N be a homomorphism.
(1) If M is finite, then Im f is finite.
(2) If Ker f, Im f are finite, then M is finite.
(3) If N is finite, then Cok f is finite.
(4) If Im f, Cok f are finite, then N is finite.
Proof. Use 6.1.5 on the exact sequences 3.1.8.
6.1.8. Corollary. Let M, N be modules. Then M ⊕ N is finite if and only if M
and N are finite.
Proof. Use 6.1.6.
6.1.9. Proposition. Let R be a ring and M, N finite modules. Then M ⊗R N is
finite.
Proof. Use 6.1.2 and 6.1.5. Let Rm → M and Rn → N be surjective. Then
Rm ⊗R Rn → M ⊗R N is surjective, 3.4.1.
6.1.10. Proposition. Let R be a ring and F a module. The following conditions
are equivalent:
(1) F is finite and projective.
(2) F is a direct summand in a finite free module.
Proof. Use 6.1.2 and 6.1.8.
6.1.11. Proposition. Let R be a ring and F a finite projective module.
(1) If N is finite, then HomR (F, N ) is finite.
(2) If N is projective, then HomR (F, N ) is projective.
Proof. Use 6.1.8.
6.1.12. Proposition. Let F be a finite projective module and E an injective module.
(1) F ⊗R E is injective.
(2) HomR (F, E) is injective.
Proof. (1) (2) Both modules become summands in injective modules.
6.1.13. Proposition. Let R be a ring and U a multiplicative subset. For a finite
module M the following hold:
(1) U −1 M = 0 if and only if there is a u ∈ U such that uM = 0.
(2) Ann(U −1 M ) = U −1 Ann(M ) = Ann(M )U −1 R in the ring U −1 R.
Proof. Let x1 , . . . , xn generate M . (1) U −1 M = 0 if and only if u1 x1 = · · · =
un xn = 0. Put u = u1 . . . un . (2) Ann(M ) = Ann(x1 )∩· · ·∩Ann(xn ). The exact
sequence 0 → Ann(x1 ) → R → Rx1 → 0 localizes to 0 → U −1 Ann(x1 ) →
U −1 R → U −1 Rx1 → 0 giving that U −1 Ann(x1 ) = Ann(U −1 Rx1 ). Conclude
by 4.3.4.
6.1.14. Proposition. Let R be a ring and Mα a family of modules. For any finite
module N there is a natural isomorphism
M
M
HomR (N,
Mα ) '
HomR (N, Mα )
α
α
6.2. FREE MODULES
Proof. A homomorphism f : N →
L
89
Mα has an image in a finite sum.
L
6.1.15. Exercise.
(1) Show that n Z/(n) is not a finite Z-module.
(2) Let K be a field and R = K[X1 , X2 , . . . ] the polynomial ring in countable many
variables. Show that R is a finite module, but the ideal (X1 , X2 , . . . ) is not a finite
module.
6.2. Free modules
6.2.1. Definition. Let R be a ring and let Rn be the free module with standard
basis e1 , . . . , en .
(1) Let A = (aij ) be a m × n-matrix with m rows and n columns, where the
entry aij ∈ R. Identify Rn with n-columns. Then matrix multiplication
X
x = (xj ) 7→ y = Ax, yi =
aij xj
j
Rn
Rm .
defines a homomorphism
→
(2) Let f : Rn → Rm be a homomorphism. Then define a m × n-matrix A =
(aij ) by
X
f (ej ) =
aij ei
i
6.2.2. Proposition. (1) The dictionary defined in 6.2.1 gives a canonical isomorphism between the module of m × n-matrices and HomR (Rn , Rm ).
(2) Matrix multiplication corresponds to composition of homomorphisms and the
identity matrix corresponds to the identity homomorphism.
(3) Invertible matrices correspond to isomorphisms.
Proof. Do linear algebra homework.
6.2.3. Definition. Let R be a ring.
(1) Let A = (aij ) be a m × n-matrix. The (m − 1) × (n − 1) matrix derived
from A be deleting i-row and j-column is Aij .
(2) For a square matrix A the determinant is defined by row expansion and induction:
det(a11 ) = a11
X
det A =
(−1)1+j a1j det A1j
j
(3) The determinant of a k × k-matrix derived from A by choosing entries from
k rows and columns is a k-minor of A.
(4) If A is a n × n-matrix, then the cofactor matrix A0 = (a0ij ) has entries
a0ij = (−1)i+j det Aji
given by (n − 1)-minors.
6.2.4. Proposition. (1) For a fixed a ∈ R, the scaled determinant a det is the
unique map from n × n-matrices to R, which satisfies
(a) multilinear: For n − 1 fixed columns a•1 , . . . , a•j−1 , a•j+1 , . . . , a•n , the
map
Rn → R, x 7→ a det(a•1 , . . . , a•j−1 , x, a•j+1 , . . . , a•n )
90
6. FINITE MODULES
is a homomorphism:
a det(..bx + cy..) = ba det(..x..) + ca det(..y..)
(b) alternating: For two equal columns
a det(..x..x..) = 0
(c) normed:
a det(1n ) = a
(2) The determinant is calculated by any expansion
X
X
det A =
(−1)i+j aij det Aij =
(−1)i+j aij det Aij
i
j
(3) If A, B are n × n-matrices then the product rule holds
det AB = det A det B
(4) Let A be an n × n-matrix with cofactor matrix A0 . Then the matrix Cramer’s
rule holds
AA0 = A0 A = det A(1n )
Where (1n ) is the n × n identity matrix.
(5) A square matrix A is invertible if and only if det A is a unit in R.
Proof. More linear algebra homework. (1) First prove that any map satisfying (a)(c) must satisfy the rule for interchanging columns and the rule for a modification
of a column. Then continue proof by induction. (2) Use the uniqueness from (1).
(3) To get the product rule det AB = det A det B, you fix A. Then both sides
satisfy (1) with norm det A. (4) Writeout (2) and use alternating from (1).
6.2.5. Theorem. Let f : Rn → Rn be a homomorphism represented by an n × nmatrix A.
(1) f is a surjective if and only if det A is a unit.
(2) f is injective if and only if det A is a nonzero divisor.
Proof. (1) If f is surjective, then a section is represented by a matrix B such that
AB = (1n ). Then det A is a unit. Conversely by 6.2.4. (2) If det A is a nonzero
divisor, then by 6.2.4 A0 A = det A(1n ) gives an injective homomorphism, so f is
injective. If det A is a zero divisor, then there is number k < n such that Ann(1 −
minors) = · · · = Ann(k − minors) = 0 and 0 6= b ∈ Ann((k + 1) − minors).
Assume that the k −minor from first k rows and columns ck+1 has ck+1 b 6= 0. Let
cj (−1)k+1+j be the k-minor from first k rows and first k + 1 columns excluding
number j and put cj = 0, j > k + 1. Then A(cj ) is a column with entries being
(k + 1) − minors so f ((bcj )) = A(bcj ) = bA(cj ) = 0 and f is not injective.
6.2.6. Proposition. Let R be a ring and f : Rn → Rm homomorphism.
(1) If f is surjective, then n ≥ m.
(2) If f is injective, then n ≤ m.
(3) f is an isomorphism if and only if m = n and f is surjective.
Proof. (1) If n < m let p : Rm → Rn be the projection onto first n coordinates.
Then f ◦p is surjective and represented by an m×m-matrix A with m-column zero.
A section to f ◦ p is represented by an m × m-matrix B such that BA = (1m ). But
the product AB must have a zero m-column, so the contradiction gives n ≥ m.
(2) If n > m let i : Rm → Rn be injection into first m coordinates. Then i ◦ f is
6.3. CAYLEY-HAMILTON’S THEOREM
91
injective and represented by an n × n-matrix A with n-row zero. Then det A = 0
contradicting 6.2.5. So n ≤ m. (3) If f is an isomorphism then m = n by (1), (2).
If f is surjective, then f is an isomorphism by 6.2.5 (1) and 6.2.4 (5).
6.2.7. Theorem. (1) A finite free module has a finite basis.
(2) The number of elements in a basis for a finite free module is independent of
the basis.
Proof. Let F be finite free generated by n elements. If y1 , . . . , ym is part of a
basis, then by projection F → Rm there is a surjective homomorphism Rn → Rm .
Conclusion by 6.2.6.
6.2.8. Definition. The number of elements 6.2.7 in a basis for a finite free module
F is the rank, rankR F .
6.2.9. Proposition. If x1 , . . . , xn generates a free module F of rank n, then they
constitutes a basis.
Proof. Choose a basis and an isomorphism f : Rn → F . The homomorphism
g : Rn → F, ei 7→ xi is surjective. The composite f −1 ◦ g : Rn → Rn is
surjective and therefore by 6.2.6 an isomorphism. Then g is an isomorphism and
xi a basis.
6.2.10. Proposition. Let F, F 0 be finite free modules. Then
(1) F ⊕ F 0 is free and rankR F ⊕ F 0 = rankR F + rankR F 0 .
(2) F ⊗R F 0 is free and rankR F ⊗R F 0 = rankR F · rankR F 0 .
(3) HomR (F, F 0 ) is free and rankR HomR (F, F 0 ) = rankR F · rankR F 0 .
6.2.11. Exercise.
(1) Let Rn → Rn be a surjective homomorphism. Show that it is an
isomorphism.
6.3. Cayley-Hamilton’s theorem
6.3.1. Remark. Let R be a ring and f : M → M a homomorphism. By 2.1.13
view M as an R[X]-module, where Xx = f (x) for x ∈ M . The homomorphism
1.6.7, 2.6.9, R[X] → HomR (M, M ), a 7→ aM , X 7→ f is a ring homomorphism.
The image is R[f ] the smallest subring containing 1M , f . M is naturally a R[f ]module and the R[X]-module above is the restriction of scalars.
6.3.2. Proposition. Let A be an n × n-matrix and I the ideal generated by the
entries aij . The polynomial
det(X1n − A) = a0 + a1 X + . . . an−1 X n−1 + X n
has a0 , . . . , an−1 ∈ I and gives the relation
a0 (1n ) + a1 A + . . . an−1 An−1 + An = 0
as n × n-matrix.
Proof. View Rn as a module over the ring R[X], 6.3.1, with scalar multiplication
Xx = Ax, x ∈ Rn
Let p : R[X]n → Rn be the R[X]-homomorphism determined by p(ej ) = ej
and denote U = X(1n ) − A the n × n-matrix with entries from R[X]. Then
p(U ej ) = p(Xej − Aej ) = 0 in Rn , so p(U y) = 0 for all y ∈ R[X]n . Let U
have cofactor matrix U 0 over R[X]. Then p(U U 0 ej ) = 0. By Cramer’s rule 6.2.4,
92
6. FINITE MODULES
U U 0 ej = det U ej in R[X]n . Therefore p(det U ej ) = det U p(ej ) = det U ej =
0 in Rn . By calculation
det U = a0 + a1 X + · · · + an−1 X n−1 + X n
in R[X], with ai ∈ I and for all x ∈ Rn
det U x = (a0 (1n ) + a1 A + . . . an−1 An−1 + An )x = 0
6.3.3. Theorem. Let I ⊂ R be an ideal and f : M → M a homomorphism on
a finite module generated by n elements. Suppose Im f ⊂ IM , then there exist
a0 , . . . , an−1 ∈ I such that
a0 1M + a1 f + . . . an−1 f n−1 + f n = 0
in HomR (M, M ).
Proof. Let x1 , . . . , xn generate M and write
X
f (xj ) =
aij xi
i
for an n × n-matrix A with entries aij ∈ I. View Rn , x 7→ Ax and M, x 7→ f (x)
as modules over the ring R[X], 6.3.1. Then the R-homomorphism p : Rn →
M, ej 7→ xj is an R[X]-homomorphism. By 6.3.2
det(X1n − A) = a0 + a1 X + . . . an−1 X n−1 + X n
gives 0 = p(det(X(1n ) − A) ej ) = det(X(1n ) − A) xj . It follows that
(a0 1M + a1 f + . . . an−1 f n−1 + f n )xj = 0
for all j.
6.3.4. Proposition. Let I ⊂ R be an ideal and M a finite module. If IM = M
then I + Ann(M ) = R. That is there is a ∈ I such that (1 + a)M = 0.
Proof. By 6.3.3
a0 1M + · · · + 1M = (a0 + · · · + an−1 )1M + 1M = 0
Put a = a0 + · · · + an−1 .
6.3.5. Corollary. Let I ⊂ R be an ideal and M a finite module. If IM = M and
all elements 1 + I are nonzero divisors on M , then M = 0.
6.3.6. Corollary. Let I ⊂ R be an ideal and N ⊂ M a submodule. Suppose M/N
is a finite module and M = N + IM . Then I + (N : M ) = R.
6.3.7. Proposition. Let R be a ring and M a finite module. If a homomorphism
f : M → M is surjective, then it is an isomorphism.
Proof. Regard M, f as a module over R[X] 6.3.1. Then (X)M = M , so by 6.3.4
there is p ∈ R[X] such that 1 + pX ∈ Ann(M ). For any u ∈ Ker f , calculate
u = u + p(f ) ◦ f (u) = (1 + pX)u = 0. So f is an isomorphism.
6.3.8. Exercise.
(1) Let
√
0M = M . Show that M = 0.
6.4. NAKAYAMA’S LEMMA
93
6.4. Nakayama’s lemma
6.4.1. Theorem (Nakayama). Let (R, P ) be a local ring and M a finite module.
The following conditions are equivalent:
(1) M = 0.
(2) P M = M .
(3) M ⊗R k(P ) = 0.
Proof. (1) ⇒ (2) ⇔ (3) is clear. (2) ⇒ (1): Elements in 1 + P are units in R, so
by 6.3.5 M = 0.
6.4.2. Corollary. Let (R, P ) be a local ring and N ⊂ M a submodule. Suppose
M/N is a finite module and M = N + P M . Then N = M .
6.4.3. Corollary. Let (R, P ) be a local ring and M, N finite modules. If M ⊗R
N = 0, then M = 0 or N = 0.
Proof. If M, N 6= 0 then M ⊗R k(P ), N ⊗R k(P ) 6= 0 are vector spaces over
k(P ). Now M ⊗R N ⊗R k(P ) ' M ⊗R k(P )⊗k(P ) , N ⊗R k(P ) 6= 0, giving the
statement.
6.4.4. Corollary. Let R be a ring and M a finite module. The following conditions
are equivalent:
(1) M = 0.
(2) P MP = MP for all prime ideals P .
(3) P MP = MP for all maximal ideals P .
(4) M ⊗R k(P ) = 0 for all prime ideals P .
(5) M ⊗R k(P ) = 0 for all maximal ideals P .
Proof. Combine 6.4.1 with 5.4.1.
6.4.5. Corollary. Let (R, P ) be a local ring and f : M → N a homomorphism.
Assume N is finite. The following conditions are equivalent:
(1) f is surjective.
(2) f (P ) is surjective.
Proof. f is surjective if and only if Cok f = 0. Cok f is finite, so it is zero if and
only if Cok f ⊗R k(P ) = Cok(f (P )) = 0, 6.4.1.
6.4.6. Corollary. Let R be a ring and f : M → N a homomorphism. Assume N
is finite. The following conditions are equivalent:
(1) f is surjective.
(2) f (P ) is surjective for all prime ideals P .
(3) f (P ) is surjective for all maximal ideals P .
6.4.7. Corollary. Let (R, P ) be a local ring and M a finite module. Let x1 , . . . , xn ∈
M . The following conditions are equivalent:
(1) x1 , . . . , xn generate M .
(2) x1 ⊗ 1, . . . , xn ⊗ 1 generate M ⊗R k(P ).
The minimal number of generators of M is rankk(P ) M ⊗R k(P ).
6.4.8. Corollary. Let R be a ring and M a finite module. Let x1 , . . . , xn ∈ M .
The following conditions are equivalent:
(1) x1 , . . . , xn generate M .
94
6. FINITE MODULES
(2) x1 ⊗ 1, . . . , xn ⊗ 1 generate M ⊗R k(P ) for all prime ideals P .
(3) x1 ⊗ 1, . . . , xn ⊗ 1 generate M ⊗R k(P ) for all maximal ideals P .
6.4.9. Proposition. Let f : Rn → Rm be a homomorphism represented by an
m × n-matrix A. The following statements are equivalent:
(1) f is surjective.
(2) n ≥ m and the ideal (m − minors) = R.
Proof. (1) ⇒ (2): If f is surjective, then for any maximal ideal f (P ) is surjective
linear map. So n ≥ m and some m-minor is nonzero in k(P ). Therefore (m −
minors) is not contained in P , so (m − minors) = R. (2) ⇒ (1): f (P ) is
surjective for all maximal ideals, so f surjective by 6.4.6.
6.4.10. Proposition. Let f : Rn → Rm be a homomorphism represented by an
m × n-matrix A. The following statements are equivalent:
(1) f is injective.
(2) n ≤ m and the ideal Ann(n − minors) = 0.
Proof. (1) ⇒ (2): See the proof 6.2.5 (2) and note that the construction only uses
n ≤ m. (2) ⇒ (1): Assume f (x) = Ax = 0 and choose n rows in A to give a
n × n-matrix B. Then Bx = 0 and by Cramer’s rule det B x = 0. Since det B
run through all n-minors it follows that the coordinates of x belong to Ann(n −
minors) = 0.
(1) Let R be a domain and f : Rn → Rm a homomorphism represented by an m × n-matrix. Show that f is injective if and only if n ≤ m and some
n − minor 6= 0.
6.4.11. Exercise.
6.5. Finite presented modules
6.5.1. Definition. Let R be a ring. A finite presented module is a module M
having an exact sequence
Rn → Rm → M → 0
or equivalently there is a short exact sequence
0 → N → Rm → M → 0
with N finite.
6.5.2. Proposition. (1) Let R → S be a ring homomorphism and M a finite
presented R-module Then the change of rings S-module M ⊗R S is finite
presented.
(2) Let U ⊂ R be a multiplicative subset and M a finite presented R-module,
then U −1 M is a finite presented U −1 R-module.
6.5.3. Proposition. For a short exact sequence
0
/M
f
/N
g
/L
/0
the following hold:
(1) If M, L are finite presented, then N is finite presented.
(2) If L is finite presented and N is finite , then M is finite.
(3) If N is finite presented and M is finite , then L is finite presented.
6.5. FINITE PRESENTED MODULES
95
Proof. (1) Choose u : Rn → M → 0 and v : Rm → L → 0 exact with finite
kernels. By 3.5.5 there is w : Rm → N such that g ◦ w = v. There is a diagram
/ Rn
0
/ Rn ⊕ Rm
u
/M
0
f ◦u+w
f
/ Rm
g
/N
/0
v
/L
/0
By the snake lemma 3.2.4 the sequence 0 → Ker u → Ker f ◦u+w → Ker v → 0
is exact. By 6.1.5 Ker f ◦ u + w is finite. (2) Choose v : Rm → L → 0 exact with
finite kernel and w : Rm → N such that g ◦ w = v. There is a diagram
0
/0
/ Rm
/M
0
f
Rm
w
/N
g
/0
v
/L
/0
By the snake lemma 3.2.4 the sequence 0 → Ker w → Ker v → M → Cok w →
0 is exact. By 6.1.5 M is finite. (3) Choose w : Rm → N → 0 exact with finite
kernel. Then v = g ◦ w : Rm → L → 0 is exact and there is a diagram
/ Rm
/0
0
/M
0
f
Rm
w
/N
g
/0
v
/L
/0
By the snake lemma 3.2.4 the sequence 0 → Ker w → Ker v → M → 0 is exact.
By 6.1.5 Ker v is finite and L is finite presented.
6.5.4. Corollary. Let
0
f
/M
/N
g
/L
/0
be a split exact sequence. Then N is finite presented if and only if M, L are finite
presented.
Proof. By 3.1.13 there is a split exact sequence
0
/L
v
/N
u
/M
/0
so the statement follows from 6.5.3.
6.5.5. Corollary. Let f : M → N be a homomorphism.
(1)
(2)
(3)
(4)
If M is finite and Im f finite presented, then Ker f is finite.
If Ker f, Im f are finite presented, then M is finite presented.
If N is finite presented and Im f finite, then Cok f is finite presented.
If Im f, Cok f are finite presented, then N is finite presented.
Proof. Use the sequences 3.1.8.
6.5.6. Proposition. Let R be a ring and M, N finite presented modules.
(1) M ⊕ N is finite presented.
(2) M ⊗R N is finite presented.
96
6. FINITE MODULES
Proof. (1) This is clear from 6.5.4. (2) If M = Rn then M ⊗R N is finite presented
by (1). In general chose u : Rn → M → 0 exact with finite kernel. The sequence
Ker u ⊗R N → Rn ⊗R N → M ⊗R N → 0 is exact. So Ker u ⊗ 1N is finite.
Conclusion by 6.5.5.
6.5.7. Proposition. Given submodules N, L ⊂ M . Then
(1) If M/N, M/L are finite and M/N + L is finite presented, then M/N ∩ L is
finite.
(2) If M/N, M/L are finite presented and M/N ∩ L is finite, then M/N + L is
finite presented.
Proof. Use the sequence 3.2.7
/ M/N ∩ L
0
/ M/N ⊕ M/L
/ M/N + L
/0
together with 6.5.3.
6.5.8. Theorem. Let M be a finite presented module and Nα a family of modules.
Then there is a natural isomorphism
Y
Y
M ⊗R
Nα →
(M ⊗R Nα )
α
α
Proof. If M = R this is an isomorphism. Then this is also an isomorphism for
M = Rn since both functors respect finite direct sums. In general choose Rn →
Rm → M → 0 exact. There is a diagram
Q
Q
Q
/ M ⊗R
/0
/ Rm ⊗R
Nα
Nα
Rn ⊗R Nα
Q
Rn ⊗R Nα
/
Q
Rm ⊗R Nα
/
Q
/0
M ⊗R Nα
giving isomorphism by the five lemma 3.2.8.
6.5.9. Theorem. Let R → S be a flat ring homomorphism.
(1) For a finite module M and any module N the natural homomorphism
HomR (M, N ) ⊗R S → HomS (M ⊗R S, N ⊗R S)
is injective.
(2) For a finite presented module M and any module N the homomorphism
HomR (M, N ) ⊗R S → HomS (M ⊗R S, N ⊗R S)
is a natural isomorphism.
Proof. (1) If M = R this is an isomorphism. Then this is also an isomorphism
for M = Rn since both functors respect finite direct sums. In general choose
0 → K → Rn → M → 0 exact. There is a diagram
/
0
0
/
/
HomR (M, N ) ⊗R S
HomS (M ⊗R S, N ⊗R S)
/
/
HomR (Rn , N ) ⊗R S
HomS (Rn ⊗R S, N ⊗R S)
/
HomR (K, N ) ⊗R S
HomS (K ⊗R S, N ⊗R S)
giving injectivity. (2) In (1) K is finite, so the last vertical map is injective. Conclusion by the five lemma 3.2.8.
6.5.10. Corollary. Let R be a ring and U a multiplicative subset.
6.5. FINITE PRESENTED MODULES
97
(1) For a finite module M and any module N the natural homomorphism
U −1 HomR (M, N ) → HomU −1 R (U −1 M, U −1 N )
is injective.
(2) For a finite presented module M and any module N the homomorphism
U −1 HomR (M, N ) → HomU −1 R (U −1 M, U −1 N )
is a natural isomorphism.
Proof. Use that R → U −1 R is flat.
6.5.11. Corollary. Let P ⊂ R be a prime ideal and M a finite presented module.
For any module N there is a natural isomorphism
HomR (M, N )P ' HomRP (MP , NP )
6.5.12. Corollary. Let R be a ring and
M
f
/N
g
/L
a sequence of homomorphisms with L a finite presented module. The following
conditions are equivalent:
(1) The sequence is split exact.
(2) The sequence
fP
gP
/ LP
fP
gP
/ LP
/ NP
MP
is split exact for all prime ideals P .
(3) The sequence
/ NP
MP
is split exact for all maximal ideals P .
Proof. The statement for exactness is true in general 5.4.5. The sequence splits if
HomR (L, N ) → homR (N, N ) is surjective. By 5.4.3 this holds if HomR (L, N )P →
homR (N, N )P is surjective for all maximal ideals. Now L is finite presented, so
the localized is HomRP (LP , NP ) → HomRP (NP , NP ) giving the conclusion.
6.5.13. Theorem. Let R be a ring and F a module. The following conditions are
equivalent:
(1) F is flat.
P
(2) For any module N and a relation P
0 = i yi ⊗ xi ∈ N ⊗P
R F , there exist
zj ∈ F and aij ∈ R such that 0 = i aij yi ∈ N and xi = j aij zj ∈ F .
P
(3) For any
relation
0
=
∈ F , there exist zj ∈ F and aij ∈ R such that
i bi xiP
P
0 = i aij bi ∈ R and xi = j aij zj ∈ F .
n
Proof. (1) ⇒ (2): Let f : Rn → N, ei 7→ yj , then 0 → Ker f ⊗R F
P→ F →
N ⊗R F is exact. By assumption (xi ) ∈ Ker f ⊗R F , so (xi ) = j aij ⊗ zj
with
be an ideal and
P (aij ) ∈ Ker f . (2) ⇒ (3) is clear. (3) ⇒
P(1): Let I ⊂ R P
bi ⊗ xi ∈ Ker(I ⊗R F → F ). Then 0 = i aij bi and xi = j aij zj . Now
P
P P
calculate bi ⊗ xi = j i aij bi ⊗ xi = 0. By 3.7.12 F is flat.
6.5.14. Theorem. Let (R, P ) be a local ring and F a finite presented module. The
following conditions are equivalent:
(1) F is free.
98
6. FINITE MODULES
(2) F is projective.
(3) F is flat.
(4) P ⊗R F → F is injective.
Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) are clear. (4) ⇒ (1): Choose xi ∈ F such that
xi ⊗ 1 give a basis for F ⊗R k(P ). The homomorphism f : Rn → F, ei 7→ xi
is surjective by 6.4.5 and 0 → K → Rn → F → 0 is exact with K finite, 6.5.5.
There is a diagram
0
P ⊗R K
/ P ⊗R Rn
/ P ⊗R F
/0
/K
/ Rn
/F
/0
/ k(P ) ⊗R Rn
/ k(P ) ⊗R F
/0
k(P ) ⊗R K
∼
By the snake lemma 3.2.4 k(P ) ⊗R K = 0 and therefore K = 0 by 6.4.1, so F is
free.
6.5.15. Corollary. Let (R, P ) be a local ring and f : F → F 0 a homomorphism
of finite free modules. The following conditions are equivalent:
(1) f has a retraction u : F 0 → F .
(2) f is injective and Cok f is free.
(3) f (P ) is injective.
Proof. (1) ⇒ (2): Cok f is projective, so free by 6.5.14. (2) ⇒ (3) is clear. (3) ⇒
(1): Let f ∨ : F 0∨ → F ∨ be the dual homomorphism. f ∨ (P ) is surjective, so f ∨ is
surjective, 6.4.5. A section v of f ∨ gives a retraction u = v ∨ .
6.5.16. Corollary. Let R be a ring and F a finite presented module. The following
conditions are equivalent:
(1)
(2)
(3)
(4)
F is projective.
F is flat.
P ⊗R F → F is injective for all maximal ideals P .
FP is free for all maximal ideals P .
Proof. (1) ⇒ (2) ⇒ (3) are clear by 6.5.14. (3) ⇒ (4): P RP ⊗RP FP → FP is
(P ⊗R F )P → FP and therefore injective. Conclusion by 6.5.14. (4) ⇒ (1): Let
N → L → 0 be exact. By hypothesis HomRP (FP , NP ) → HomRP (FP , LP ) →
0 is exact for all maximal ideals. By 6.5.11 HomR (F, N )P → HomR (F, L)P → 0
is exact for all maximal ideals. By 5.4.3 HomR (F, N ) → HomR (F, L) → 0 is
exact and F is projective.
6.5.17. Exercise.
(1) Show that a finite projective module is finite presented.
(2) Let I ⊂ R be an ideal. Show that R/I is a finite presented R-module if and only if
I is a finite ideal.
(3) Show that Q is a flat, but not projective Z-module.
6.6. FINITE RING HOMOMORPHISMS
99
6.6. Finite ring homomorphisms
6.6.1. Definition. A ring homomorphism φ : R → S is a finite ring homomorphism if S is a finite R-module. If R ⊂ S is a subring, then a finite ring homomorphism is a finite ring extension.
6.6.2. Proposition. Let R be a ring.
(1) Let f ∈ R[X] be a monic polynomial. Then the homomorphism R →
R[X]/(f ) is finite.
(2) Let f : M → M be a homomorphism of a finite R-module. Then the homomorphism 6.3.1, R → R[f ] is finite.
Proof. (2) Follows from (1) and 6.3.3.
6.6.3. Lemma. Let φ : R → S be a finite ring homomorphism.
(1) If N is a finite S-module, then by restriction along φ the R-module N is finite.
(2) Let U ⊂ R be a multiplicative subset, then U −1 R → U −1 S is a finite ring
homomorphism.
6.6.4. Proposition. Let R ⊂ S be a finite ring extension of domains. Then R is a
field if and only if S is a field.
Proof. Let R be a field, a minimal equation 6.3.3 for scalar multiplication by a
nonzero b ∈ S, bS as R-module homomorphism
bn + · · · + a0 = 0
gives
n−2
+ · · · + a1 ) ∈ S
b−1 = −a−1
0 (an−1 b
Let S be a field and 0 6= a ∈ R. An equation 6.3.3 for scalar multiplication a−1
S as
R-homomorphism
a−n + · · · + a0 = 0
gives
a−1 = −(a0 an−1 + · · · + an−1 ) ∈ R
6.6.5. Corollary. Let R → S be a finite ring homomorphism. A prime ideal Q ⊂ S
is maximal if and only if the contraction Q ∩ R is maximal.
Proof. R/Q ∩ R ⊂ S/Q is a finite extension of domains.
6.6.6. Theorem. Let R ⊂ S be a finite ring extension and P ⊂ R a prime ideal.
Then there is a prime ideal Q ⊂ S contracting P = Q ∩ R.
R
S
ooo
o
o
o
Q
P
Proof. RP ⊂ SP is a finite ring extension. Since SP 6= 0 there is a maximal
ideal in SP contracting to P RP by 6.6.5. The corresponding prime ideal Q ⊂ S
contracts to P .
100
6. FINITE MODULES
6.6.7. Corollary (going-up). Let R → S be a finite ring homomorphism, Q ⊂ S
a prime ideal and P = Q ∩ R the contraction. For any prime ideal P ⊂ P 0 in R
there is a prime ideal Q ⊂ Q0 in S contracting P 0 = Q0 ∩ R.
R
nS
nnn
n
n
n
Q0
P0
P
Q
ooo
o
o
o
Proof. R/P ⊂ S/Q is a finite ring extension. So there is a prime ideal in S/Q
contracting to P 0 /P , 6.6.6. The corresponding prime ideal Q ⊂ Q0 ⊂ S contracts
to P 0 .
6.6.8. Theorem. Let R ⊂ S be a finite ring extension and E an R-module. The
following conditions are equivalent:
(1) E is an injective R-module.
(2) HomR (S, E) is an injective S-module.
Proof. (1) ⇒ (2): This is 3.6.11. (2) ⇒ (1): Let E ⊂ E 0 be an injective envelope,
3.6.17. HomR (S, E) → HomR (S, E 0 ) is injective and identifies HomR (S, E) as
an essential submodule. For this, let f : S → E 0 be nonzero and S = Rb1 +
· · · + Rbn . Then use E ⊂ E 0 essential to construct with renumbering a maximal
sequence a1 , . . . , am ∈ R such that
0 6= a1 f (b1 ) ∈ E, . . . , 0 6= a1 · · · am f (bm ) ∈ E
Then a1 · · · am f : S → E is nonzero, so the extension of Hom’s is essential. Next,
by (1) ⇒ (2) HomR (S, E 0 ) is an injective S module, so the extension of Hom’s
above is trivial, 3.6.17. It follows that
HomR (S, E)
HomR (S, E 0 )
HomR (R, E)
/ HomR (R, E 0 )
E
/ E0
with the right down map being surjective. So E = E 0 and E is injective.
√
(1) Show that Z → Z[ −5] is finite.
(2) Let p be a prime number. Show that Z → Z(p) is not finite.
6.6.9. Exercise.
7
Modules of finite length
7.1. Simple modules
7.1.1. Definition. A nonzero module M is a simple module if 0 and M are the
only submodules.
7.1.2. Lemma. Let f : M → M 0 be a homomorphism
(1) If M is simple, then f is either zero or injective.
(2) If M 0 is simple, then f is either zero or surjective.
(3) If M, M 0 are both simple, then f is either zero or an isomorphism.
Proof. This follows from 2.3.3. (1) Ker f ⊂ M is either 0 or M . (2) Im f ⊂ N is
either 0 or N .
7.1.3. Theorem. Let M be a simple R-module.
(1) M is isomorphic to the factor module R/P , where P = Ann(M ) is a maximal ideal.
(2) M = Rx is finite and generated by any one nonzero element x ∈ M .
Proof. A nonzero f : R → M must be surjective. If Ker f = P then and R/P '
M is simple exactly when P is maximal. Clearly P = Ann(M ).
7.1.4. Corollary. Any nonzero finite module has a simple factor module.
Proof. If M 6= 0 then MP 6= 0 for some maximal ideal 5.4.1. By 6.4.1 M ⊗R
k(P ) = M/P M 6= 0. Choose a nonzero linear projection M/P M → k(P ),
giving M → k(P ) → 0 exact.
7.1.5. Corollary. Let (R, P ) be a local ring and M a finite module. The following
conditions are equivalent:
(1) M = 0.
(2) HomR (M, k(P )) = 0.
7.1.6. Proposition. Let M, M 0 be simple modules. The following conditions are
equivalent:
(1) M ' M 0 .
(2) M ⊗R M 0 6= 0.
(3) HomR (M, M 0 ) 6= 0.
Moreover if M ' M 0 there are non natural isomorphisms
M ⊗R M 0 ' M ' HomR (M, M 0 )
Proof. This follows from 7.1.3 with M = R/P .
7.1.7. Proposition. Let I ⊂ R be an ideal. A simple R/I-module M is a simple
R-module such that I ⊂ Ann(M ).
101
102
7. MODULES OF FINITE LENGTH
Proof. Maximal ideals in R/I are of the form P/I and (R/I)/(P/I) ' R/P .
7.1.8. Proposition. Let U ⊂ R be a multiplicative subset and M is a simple Rmodule with Ann(M ) = P .
(1) If U ∩ P = ∅, then is U −1 M = M is a simple U −1 R-module.
(2) If U ∩ P 6= ∅, then U −1 M = 0.
Proof. Let P ⊂ R be a maximal ideal, then the fractions U −1 (R/P ) is either 0 or
R/P , 5.1.5.
7.1.9. Proposition. Let R = R1 × R2 be a product of rings. A simple R-module
is of the form M1 × 0 or 0 × M2 , where Mi is a simple Ri -module.
Proof. Clear by 2.2.9.
7.1.10. Example. (1) If K is a field, a one dimensional vector space is simple.
(2) If R is a principal ideal domain, then R/(p) is a simple module for all irreducible (p).
(3) Z/(p) are simple for all prime numbers p.
(4) Let K be a field. K[X]/(X − a) is a simple K[X]-module.
7.1.11. Exercise.
(1) Show that Z does not contain any simple modules.
(2) Show that Q does not have any simple factor Z-module.
(3) Let L, L0 ⊂ M be simple submodules. Show that either L ∩ L0 = 0 of L = L0 .
(4) Let L 6= L0 ⊂ M be simple submodules. Show that L ⊕ L0 = L + L0 .
7.2. The length
7.2.1. Definition. Let R be a ring. A module M has finite length if it admits a
composition series by submodules
0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn−1 ⊂ Mn = M
such that each factor Mi /Mi−1 is a simple module.
7.2.2. Lemma. Any two finite composition series have the same number of submodules.
Proof. Let l(M ) be the least length of a composition series. M is simple if and
only if l(M ) = 1. Let 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M be any composition
series. For a submodule N ⊂ M there is a filtration 0 = N ∩ M0 ⊂ N ∩ M1 ⊂
· · · ⊂ N ∩ Mn = N with factors N ∩ Mi /N ∩ Mi−1 ⊂ Mi /Mi−1 being either
simple or 0. It follows that l(N ) ≤ l(M ). If l(N ) = l(M ) then each factor is
nonzero, so N ∩ Mi /N ∩ Mi−1 = Mi /Mi−1 . This gives Mi ⊂ N and finally
N = M . Applying this to the composition series of M gives l(M0 ) < l(M1 ) <
· · · < l(Mn ) so n ≤ l(M ) as needed.
7.2.3. Definition. The number of nontrivial submodules in a filtration as above
will be denoted `R (M ) and is the length of M . Remark that a module of finite
length is simple if and only if the length is 1.
7.2.4. Theorem. Given an exact sequence
0
/M
f
/N
then the following statements are equivalent:
(1) N has finite length.
g
/L
/0
7.2. THE LENGTH
103
(2) M and L have finite length.
If N has finite length, then
`R (N ) = `R (M ) + `R (L)
Proof. (1) ⇒ (2): A finite filtration Ni with simple factors in N induces a filtration
f −1 (Ni ) in M and a filtration g(Ni ) in L with factors being simple or 0. (2) ⇒
(1): A finite composition series Mi in M and Lj in L may be spliced together
· · · ⊂ f (M ) = g −1 (0) ⊂ g −1 (L1 ) ⊂ . . .
to give a composition series of N . The length formula follows from 7.2.2.
7.2.5. Corollary. Let f : M → N be a homomorphism.
(1) M has finite length if and only if Ker f, Im f have finite length. If finite length
`R (M ) = `R (Ker f ) + `R (Im f )
(2) N has finite length if and only if Im f, Cok f have finite length. If finite length
`R (N ) = `R (Im f ) + `R (Cok f )
Proof. Use the sequences 3.1.8.
7.2.6. Corollary. Let f : M → M be a homomorphism on a module of finite
length. Then
`R (Ker f ) = `R (Cok f )
and the following conditions are equivalent:
(1) f is injective.
(2) f is surjective.
(3) f is an isomorphism.
Proof. Use the sequences 3.1.8.
7.2.7. Corollary. Let N ⊂ M be a submodule and suppose M has finite length.
(1) `R (N ) ≤ `R (M ).
(2) `R (N ) = `R (M ) if and only if N = M .
7.2.8. Corollary. Let N, L ⊂ M be submodules and suppose N, L has finite
length. Then N + L, N ∩ L have finite length and
`R (N + L) + `R (N ∩ L) = `R (N ) + `R (L)
Proof. By 2.3.8 N/N ∩ L ' N + L/L.
7.2.9. Proposition. Given submodules N, L ⊂ M . The following statements are
equivalent:
(1) M/N, M/L have finite length.
(2) M/N ∩ L has finite length.
If finite length
`R (M/N + L) + `R (M/N ∩ L) = `R (M/N ) + `R (M/L)
Proof. Use the exact sequence 3.2.7
0
/ M/N ∩ L
together with 7.2.4.
/ M/N ⊕ M/L
/ M/N + L
/0
104
7. MODULES OF FINITE LENGTH
7.2.10. Proposition. A R-module M of finite length is finite and generated by
`R (M ) or less elements.
f
/M g /L
/ 0 with L simple, 7.1.4.
Proof. There is an exact sequence 0 / N
If xi generate N and g(y) generate L, then f (xi ), y generate M . Conclusion by
induction.
7.2.11. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M has finite
length. Then M has finite length as R-module and
`R/I (M ) = `R (M )
Proof. From 7.1.7 follows that a composition series as R/I-module is also a composition series as R-module. and 7.2.2.
7.2.12. Proposition. Let U ⊂ R be a multiplicative subset. Suppose an R-module
M has finite length. Then U −1 M has finite length as U −1 R-module and
`U −1 R (U −1 M ) ≤ `R (M )
Proof. This follows from 7.1.8 and 7.2.2. The fraction construction of a composition series may be refined to a composition series.
7.2.13. Proposition. Let (R, P ) be a local ring and M a module of finite length.
The following conditions are equivalent:
(1) M = 0.
(2) HomR (k(P ), M ) = 0.
(3) HomR (M, k(P )) = 0.
Proof. Clear by a composition series as k(P ) is the only simple module.
7.2.14. Proposition. Let M, M 0 be modules of finite length.
(1) M ⊕ M 0 has finite length and `R (M ⊕ M 0 ) = `R (M ) + `R (M 0 ).
(2) M ⊗R M 0 has finite length and `R (M ⊗R M 0 ) ≤ `R (M ) · `R (M 0 ).
(3) HomR (M, M 0 ) has finite length and `R (HomR (M, M 0 )) ≤ `R (M )·`R (M 0 ).
Proof. (1) This is clear from 7.2.4. (2) If M = R/P is simple then M ⊗R M 0 '
M 0 /P M 0 is a factor module, so of finite length and the equality follows from 7.2.4.
In general there is an exact sequence 0 → N → M → L → 0 with L simple giving
an exact sequence N ⊗R M 0 → M ⊗R M 0 → L ⊗R M 0 → 0. Conclusion by
induction and 7.2.4. (2) If M = R/P is simple then HomR (M, M 0 ) ⊂ M 0 is a
submodule, so of finite length and the equality follows from 7.2.4. In general there
is an exact sequence 0 → N → M → L → 0 with L simple giving an exact
sequence 0 → HomR (L, M 0 ) → HomR (M, M 0 ) → HomR (N, M 0 ). Conclusion
by induction and 7.2.4.
7.2.15. Proposition. Let R = R1 × R2 be a product of rings. An R-module
M1 × M2 is of the finite length if and only if Mi is a finite length Ri -module.
Moreover
`R (M1 × M2 ) = `R1 (M1 ) + `R2 (M2 )
Proof. This follows from 7.1.9 and 7.2.11.
7.3. ARTINIAN MODULES
105
7.2.16. Example. Let K be a field. A module M is of finite length if and only if it
is a finite dimensional vector space. Then
`K (M ) = rankK (M )
7.2.17. Exercise.
(1) Compute
`Z (Z/(pn1 1 . . . pnk k )) = n1 + · · · + nk
(2) Compute
`K[X] (K[X]/((X − a1 )n1 . . . (X − ak )nk )) = n1 + · · · + nk
7.3. Artinian modules
7.3.1. Lemma. Let M be a module. The following conditions are equivalent:
(1) Any descending chain · · · ⊂ Mi+1 ⊂ Mi of submodules is stationary: there
is n such that Mi+1 = Mi for i > n.
(2) Any nonempty subset of submodules of M contains a minimal element.
Proof. (1) ⇒ (2): Suppose a nonempty subset of submodules do not contain a
minimal element. Then choose a non stationary descending chain. (2) ⇒ (1): A
descending chain containing a minimal element is stationary.
7.3.2. Definition. A module M which satisfies the conditions of 7.3.1 is an artinian module.
7.3.3. Example. (1) A module of finite length is artinian. Just note that a descending chain produces a decreasing sequence of lengths.
(2) A vector space is artinian if and only if it is finite.
7.3.4. Theorem. Let
0
/M
f
/N
g
/L
/0
be an exact sequence of modules over the ring R. The following statements are
equivalent:
(1) N is artinian.
(2) M and L are artinian.
Proof. (1) ⇒ (2): A chain Mi in M gives a chain f (Mi ) in N which becomes
stationary. So M is artinian. A chain in Li in L gives a chain in g −1 (Li ) in N ,
which becomes stationary. Then also the original chain Li = g(g −1 (Li )) becomes
stationary and L is artinian. (2) ⇒ (1): A chain Ni in N , induces chains f −1 (Ni )
in N and g(Ni ) in L which become stationary. By the snake lemma 3.2.4
f −1 (Ni )/f −1 (Ni+1 ) → Ni /Ni+1 → g(Ni )/g(Ni+1 )
is exact and the original chain is stationary.
7.3.5. Corollary. (1) Let N ⊂ M be a submodule. Then M is artinian if and
only if both N and M/N are artinian.
(2) Let f : M → N be a homomorphism.
(a) M is artinian if and only if Ker f, Im f are artinian.
(b) N is artinian if and only if Im f, Cok f are artinian
Proof. Use the sequences 3.1.8.
106
7. MODULES OF FINITE LENGTH
7.3.6. Proposition. Let f : M → M be a homomorphism on an artinian module.
The following conditions are equivalent:
(1) f is injective
(2) f is an isomorphism
Proof. There is a number n such that Im f ◦n = Im f ◦n+1 . For x ∈ M there is y
such that f ◦n (x) = f ◦n+1 (y). Then f ◦n (x − f (y)) = 0 so x = f (y) since f is
injective. It follows that f is surjective.
7.3.7. Proposition. A finite direct sum M1 ⊕ · · · ⊕ Mn of artinian modules Mi is
artinian.
Proof. Use the exact sequences
0 → M1 ⊕ · · · ⊕ Mn−1 → M1 ⊕ · · · ⊕ Mn → Mn → 0
together with induction and 7.3.4.
7.3.8. Proposition. Given submodules N, L ⊂ M , then the following statements
are equivalent:
(1) M/N, M/L are artinian
(2) M/N ∩ L is artinian
Proof. Use the exact sequences 3.2.7
0
/ M/N ∩ L
/ M/N ⊕ M/L
/ M/N + L
/0
together with 7.3.4.
7.3.9. Proposition. Let M be a finite and N an artinian R-module.
(1) M ⊗R N is artinian.
(2) HomR (M, N ) is artinian.
Proof. Choose Rn → M → 0 exact. (1) N n ' Rn ⊗R N → M ⊗R N → 0 is
exact, so conclusion by 7.3.4. (2) 0 → HomR (M, N ) → HomR (Rn , N ) ' N n is
exact, so conclusion by 7.3.4.
7.3.10. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M is artinian. Then M is an artinian R-module.
Proof. This is clear since a sequence of R-submodules is a sequence of R/Imodules.
7.3.11. Proposition. Let U ⊂ R be a multiplicative subset. Suppose an R-module
M is artinian. Then U −1 M is an artinian U −1 R-module.
Proof. Let i : M → U −1 M . By 4.4.6 any U −1 R-submodule N ⊂ U −1 M is
extended U −1 (i−1 (N )) ' N . So a chain is stationary.
7.3.12. Proposition. Let R = R1 × R2 be a product of rings. An R-module
M1 × M2 is artinian if and only if Mi is an artinian Ri -module.
Proof. This follows from the dictionary 2.2.9.
7.3.13. Exercise.
mensional.
(1) Show that a vector space is artinian if and only if it is finite di-
7.4. ARTINIAN RINGS
107
7.4. Artinian rings
7.4.1. Definition. A ring R is an artinian ring if R is an artinian module over
itself.
7.4.2. Example. (1) A field is artinian.
(2) A finite product of fields is artinian.
7.4.3. Proposition. (1) Let R be an artinian ring and I an ideal. Then the factor
ring R/I is artinian.
(2) A product ring R1 × R2 is artinian if and only if each Ri is artinian.
(3) Let R be an artinian ring and U a multiplicative subset. Then the fraction
ring U −1 R is artinian.
Proof. This follows from 7.3.10 and 7.3.11.
7.4.4. Proposition. Given ideals I, J ⊂ R, then the following statements are
equivalent:
(1) R/I, R/J are artinian
(2) R/I ∩ J is artinian
Proof. This follows from 7.3.8.
7.4.5. Proposition. An artinian domain is a field.
Proof. By 7.3.6 scalar multiplication with a nonzero element is an isomorphism.
7.4.6. Theorem. Let R be an artinian ring. Then all prime ideals are maximal and
there are only finitely many such.
Proof. By 7.4.5 primes are maximal. If Pi are different maximal ideals, then
P1 · · · Pn+1 ⊂ P1 · · · Pn is a strictly decreasing chain. So there are only finitely
many maximal ideals.
7.4.7. Proposition. Let R be an artinian ring.
√
(1) The factor ring R/ 0 is a finite product of fields.
√
√ k
(2) The nilradical 0 is nilpotent, 0 = 0 for some k.
Proof. (1) Let P1 , . . . , Pn be the prime and maximal ideals 7.4.6. The nilradical
√ k
√
0 = P1 ∩ · · · ∩ Pn by 5.1.7. Conclude by Chinese remainders 1.4.3. (2) 0 =
√ k+1
√ k
0
for some k. If 0 6= 0 let (a) be minimal among ideals I such that
√ k
√ k
√ k
I 0 6= 0. By minimality (a) = (a) 0 , so a = ab for some b ∈ 0 . But b is
√ k
nilpotent 1.3.8, so a = 0 gives a contradiction. It follows, that 0 = 0.
7.4.8. Theorem. A ring R is artinian if and only if it has finite length.
√ i √ i+1
√
Proof. The factor module 0 / 0
is an artinian module over R/ 0 which by
7.4.7 is a product of fields, so it has finite length. By the exact sequences
√ i √ i+1
√ i+1
√ i
0→ 0/ 0
→ R/ 0
→ R/ 0 → 0
√ k
√
and induction follows that R/ 0 is artinian for all k. Since 0 is nilpotent 7.4.7
it follows that R has finite length.
7.4.9. Corollary. Let R be an artinian ring and M a module. The following conditions are equivalent:
108
7. MODULES OF FINITE LENGTH
(1) M has finite length.
(2) M is finite.
(3) M is finite presented.
7.4.10. Corollary. Let R be artinian and R → S a finite ring homomorphism.
(1) S is artinian.
(2) A finite length S-module N is by restriction of scalars a finite length Rmodule.
(3) A finite length R-module M gives by change of rings M ⊗R S as finite length
S-module.
7.4.11. Proposition. Let M be an R-module of finite length.
(1) The ring R/ Ann(M ) is artinian.
(2) There are only finitely many prime ideals Ann(M ) ⊂ P .
(3) Any prime ideal Ann(M ) ⊂ P is maximal.
Proof. (1) Let x ∈ M , then R/ Ann(x) ' Rx ⊂ M is artinian. If x1 , . . . , xn
generate M , then Ann(M ) = Ann(x1 ) ∩ · · · ∩ Ann(xn ). By 7.4.4 R/ Ann(M )
is artinian. (2)(3) This follows by (1) and 7.4.6.
7.4.12. Proposition. Let M be a R-module. The following conditions are equivalent:
(1) M has finite length.
(2) M is finite and artinian.
Proof. (1) ⇒ (2): See 7.2.10 and 7.3.3. (2) ⇒ (1): Let x ∈ M , then R/ Ann(x) '
Rx ⊂ M is artinian. If x1 , . . . , xn generate M , then Ann(M ) = Ann(x1 ) ∩ · · · ∩
Ann(Xn ). By 7.4.8 R/ Ann(M ) is artinian and a finite module over an artinian
ring has finite length.
7.4.13. Corollary. Let (R, P ) be a local ring and M an artinian module. The
following conditions are equivalent:
(1) M = 0.
(2) HomR (k(P ), M ) = 0.
Proof. A finite submodule has finite length.
7.4.14. Example. Let K be a field. The ring R =
artinian.
(1) There are maximal ideals in R
Q
NK
= {a : N → K} is not
Pi = {a : N → K|a(i) = 0}
(2) Corresponding simple types are
R/Pi ' Ji = {a : N → K|a(j) = 0, i 6= j}
(3) Ji are simple ideals and the sum is an ideal
M
Ji = J 6= R
i
(4) J is an ideal which has no complement in R.
7.4.15. Exercise.
(1) Show that a vector space is artinian if and only if it is finite dimensional.
(2) Show that Z is not artinian.
7.5. LOCALIZATION
(3)
(4)
(5)
(6)
109
Show that R[X] is not artinian for a nonzero ring R.
Show that Q[X]/(X 2 − X) is artinian.
Show that a ring with finitely many ideals is artinian.
Let R be a principal ideal domain and a 6= 0. Show that R/(a) is artinian.
7.5. Localization
7.5.1. Theorem. Any artinian ring is the product of finitely many artinian local
rings. Let R be artinian with maximal ideals P1 , . . . , Pk . Then there is n such that
P1n . . . Pkn = 0 and
R ' R/P1n × · · · × R/Pkn
Each R/Pin is a local artinian ring.
Proof. This follows from 7.4.7, 7.4.8 and Chinese remainders 1.4.2.
7.5.2. Example. A reduced artinian ring is a finite product of fields.
7.5.3. Corollary. Let R be an artinian ring and P ⊂ R a prime ideal. Then the
localization RP is an artinian ring.
7.5.4. Lemma. Let P ⊂ R be a maximal ideal and M a module. Assume u ∈
/P
and n ∈ N.
(1) (u) + P n = R.
(2) Scalar multiplication uM/P n M : M/P n M → M/P n M is an isomorphism.
(3) The canonical map M/P n M → (M/P n M )P is an isomorphism.
(4) If M is finite then M/P n M has finite length.
7.5.5. Theorem. Let M be an R-module of finite length and Ann(M ) ⊂ P1 , . . . , Pk
the maximal ideals. Then the sequence
0 → Pin M → M → MPi → 0
is exact and M/Pin M ' MPi is a finite length RPi -module. There are isomorphisms
M
M
M'
MPi '
M/Pin M
i
i
and the length is
`R (M ) =
X
`RPi (MPi )
i
R/P1n
Proof. The ring R/ Ann(M ) '
+ Ann(M ) × · · · × R/Pkn + Ann(M ) by
7.4.11 and 7.5.1. Therefore M ' M/P1n M × · · · × M/Pkn M and by localization
MPi ' M/Pin M .
7.5.6. Corollary. Let R be artinian with maximal ideals P1 , . . . , Pk and M a finite
module. Then the sequence
0 → Pin M → M → MPi → 0
is exact and M/Pin M ' MPi is a finite RPi -module. There are isomorphisms
M
M
M'
MPi '
M/Pin M
i
i
and the length is
`R (M ) =
X
i
`RPi (MPi )
110
7. MODULES OF FINITE LENGTH
7.5.7. Proposition. Let (R, P ) → (S, Q) be a local ring homomorphism and assume that k(P ) → k(Q) is a finite extension. If N is a finite length S-module, then
N is a finite length R-module and
`R (N ) = rankk(P ) (k(Q)) · `S (N )
Proof. Reduce to N = k(Q).
7.5.8. Proposition. Let (R, P ) → (S, Q) be a local ring homomorphism and assume that S/P S is a finite length S-module. Let M be a finite length R-module.
(1) M ⊗R S is a finite length S-module.
(2) In general
`S (M ⊗R S) ≤ `S (S/P S) · `R (M )
(3) If R → S is flat then
`S (M ⊗R S) = `S (S/P S) · `R (M )
Proof. The case M = k(P ) is clear. Conclude by induction.
7.5.9. Exercise.
(1) Let K ⊂ L be a finite field extension and W a finite vector space
over L. Show that
rankK (W ) = rankK (L) · rankL (W )
(2) Let K ⊂ L be a finite field extension and V a finite vector space over K. Show that
rankL (V ⊗K L) = rankK (V )
7.6. Local artinian ring
7.6.1. Lemma. Let (R, P ) be a local artinian ring and M any module. The following conditions are equivalent:
(1) M = 0.
(2) HomR (k(P ), M ) = 0.
Proof. (2) ⇒ (1): A nonzero submodule Rx ⊂ M has finite length and contains a
simple submodule k(P ) ' L ⊂ M . So HomR (k(P ), M ) 6= 0.
7.6.2. Proposition. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope.
(1) There are isomorphisms
k(P ) ' HomR (k(P ), k(P )) ' HomR (k(P ), E)
(2) For a finite module M , the module HomR (M, E) has finite length and
`R (HomR (M, E)) = `R (M )
Proof. (1) A nonzero homomorphism f : k(P ) → E has Im f = k(P ) since
the extension is essential. (2) An exact sequence 0 → N → M → k(P ) → 0
7.1.4 gives an exact sequence with HomR (−, E). Conclude by (1) and induction
on `R (M ).
7.6.3. Corollary. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. Then E has finite length
`R (E) = `R (R)
7.6. LOCAL ARTINIAN RING
111
7.6.4. Theorem. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. There is a natural isomorphism for any finite module M
x 7→ evx : M → HomR (HomR (M, E), E)
Proof. The case M = k(P ) is clear by 7.6.2. Let 0 → N → M → L → 0 be a
short exact sequence. Then the following diagram has exact rows.
/
0
0
/
/
N
HomR (HomR (N, E), E)
/
/
M
HomR (HomR (M, E), E)
/
/
0
/
0
L
HomR (HomR (L, E), E)
Conclusion by the five lemma 3.2.8 and induction on the length.
7.6.5. Corollary. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. There is an isomorphism
11E : R → HomR (E, E)
7.6.6. Lemma. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. If E 0 is injective and HomR (k(P ), E 0 ) is finite, then E 0 ' E n , where
n = `R (HomR (k(P ), E 0 ).
Proof. If n = 0 then E 0 = 0 by 7.6.1. For n > 0 a nonzero k(P ) → E 0 gives
an injective extension 0 → E → E 0 . Since E is injective there is a splitting
E 0 ' E ⊕ E 00 . Conclusion by induction on n.
7.6.7. Theorem. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. Let M → E 0 be an injective envelope. Then
(1) HomR (k(P ), M ) ' HomR (k(P ), E 0 )
(2) If M is artinian, then E 0 ' E n , where n = `R (HomR (k(P ), M ).
(3) An artinian M module has finite length and
`R (M ) ≤ `R (HomR (k(P ), M )) `R (R)
Proof. (1) A homomorphism f : k(P ) → E 0 has Im f ⊂ M since the extension
is essential. (2) E 0 ' E n by 7.6.6 and n is determined by (1) and 7.6.2. (3) This
follows from (2) and 7.6.3.
7.6.8. Corollary. Let R be an artinian ring and M an R-module. The following
conditions are equivalent:
(1) M has finite length.
(2) M is finite.
(3) M is artinian.
Proof. Reduce to the case where R is local by 7.5.1.
7.6.9. Theorem. Let (R, P ) be a local artinian ring and k(P ) ⊂ E an injective
envelope. The following conditions are equivalent:
(1) R is injective.
(2) R ' E.
(3) HomR (k(P ), R) ' k(P ).
Proof. (1) ⇒ (2): By 7.6.6 R ' E n and by 7.6.3 n = 1. (2) ⇒ (3): Immediate
from 7.6.6. (3) ⇒ (1): Let R ⊂ E 0 be an injective envelope. By 7.6.7 R = E 0 .
112
7. MODULES OF FINITE LENGTH
7.6.10. Definition. A ring satisfying the conditions 7.6.9 is a local artinian Gorenstein ring.
7.6.11. Example. Let R be a principal ideal domain and p ∈ R an irreducible
element. Then HomR (R/(p), R/(pn )) ' (pn−1 )/(pn ) ' R/P and R/(pn ) is a
local artinian Gorenstein ring, 7.6.9.
(1) Let p be a prime number. Show that Z/(pk ) is a local artinian
Gorenstein ring.
7.6.12. Exercise.
8
Noetherian rings
8.1. Noetherian modules
8.1.1. Lemma. Let M be a module. The following conditions are equivalent:
(1) Any increasing chain Mi ⊂ Mi+1 ⊂ . . . of submodules is stationary: there
is n such that Mi = Mi+1 for i > n.
(2) Any nonempty subset of submodules of M contains a maximal element.
(3) Any submodule of M is finite.
Proof. (1) ⇒ (2): Suppose a nonempty subset of submodules do not contain a
maximal element. Then choose a non stationary ascending chain. (2) ⇒ (1): An
ascending chain containing a maximal element is stationary. (2) ⇒ (3): Let N be a
submodule of M and choose a maximal element N 0 in the set of finite submodules
of N . For y ∈ N , the module N 0 + Ry is finite, so N 0 = N 0 + Ry gives N = N 0
finite. (3) ⇒ (1): The union ∪j Mj is a submodule, generated by x1 , . . . , xm ∈ Mn ,
so Mi = ∪j Mj , i > n.
8.1.2. Definition. A module M which satisfies the conditions of 8.1.1 is a noetherian module.
8.1.3. Theorem. Let
0
/M
f
/N
g
/L
/0
be an exact sequence of modules over the ring R. The following statements are
equivalent:
(1) N is noetherian.
(2) M and L are noetherian.
Proof. (1) ⇒ (2): A chain Mi in M gives a chain f (Mi ) in N which becomes
stationary. So M is noetherian. A chain in Li in L gives a chain in g −1 (Li )
in N , which becomes stationary. Then also the original chain Li = g(g −1 (Li ))
becomes stationary and L is noetherian. (2) ⇒ (1): A chain Ni in N , induces
chains f −1 (Ni ) in N and g(Ni ) in L which become stationary. By the snake
lemma 3.2.4
f −1 (Ni )/f −1 (Ni−1 ) → Ni /Ni−1 → g(Ni )/g(Ni−1 )
is exact and the original chain becomes stationary.
8.1.4. Corollary. (1) Let M ⊂ N be a submodule. Then N is noetherian if and
only if both M and N/M are noetherian.
(2) Let f : M → N be a homomorphism.
(a) M is noetherian if and only if Ker f, Im f are noetherian.
(b) N is noetherian if and only if Im f, Cok f are noetherian
113
114
8. NOETHERIAN RINGS
Proof. Use the sequences 3.1.8.
8.1.5. Proposition. Let f : M → M be a homomorphism on a noetherian module.
The following conditions are equivalent:
(1) f is surjective
(2) f is an isomorphism
Proof. Analog of the proof of 7.3.4. There is a number n such that Ker f ◦n+1 =
Ker f ◦n . For x ∈ Ker f there is y such that x = f ◦n (y), since f is surjective.
Then f ◦n+1 (y) = f (x) = 0, so y ∈ Ker f ◦n+1 = Ker f ◦n . Then x = f ◦n (y) = 0
and f is injective.
8.1.6. Proposition. A finite direct sum M1 ⊕ · · · ⊕ Mn of noetherian modules Mi
is noetherian.
Proof. Use the exact sequences
0 → M1 ⊕ · · · ⊕ Mn−1 → M1 ⊕ · · · ⊕ Mn → Mn → 0
together with induction and 8.1.3.
8.1.7. Proposition. Given submodules N, L ⊂ M , then the following statements
are equivalent:
(1) M/N, M/L are noetherian
(2) M/N ∩ L is noetherian
Proof. Use the exact sequence 3.2.7
0
/ M/N ∩ L
/ M/N ⊕ M/L
/ M/N + L
/0
together with 8.1.3.
8.1.8. Proposition. Let M be a finite and N a noetherian R-module.
(1) M ⊗R N is noetherian.
(2) HomR (M, N ) is noetherian.
Proof. Choose Rn → M → 0 exact. (1) N n ' Rn ⊗R N → M ⊗R N → 0 is
exact, so conclusion by 8.1.3. (2) 0 → HomR (M, N ) → HomR (Rn , N ) ' N n is
exact, so conclusion by 8.1.3.
8.1.9. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M is noetherian. Then M is a noetherian R-module.
Proof. This is clear since a chain of R-submodules is a chain of R/I-modules.
8.1.10. Proposition. Let U ⊂ R be a multiplicative subset. Suppose an R-module
M is noetherian. Then U −1 M is noetherian U −1 R-module.
Proof. Let i : M → U −1 M . By 4.4.6 any U −1 R-submodule N ⊂ U −1 M is
extended U −1 (i−1 (N )) ' N . So a chain is stationary.
8.1.11. Proposition. Let R = R1 × R2 be a product of rings. An R-module
M1 × M2 is noetherian if and only if Mi is an noetherian Ri -module.
Proof. This follows from 8.1.9 and 8.1.3.
8.1.12. Proposition. Let M be a R-module. The following conditions are equivalent:
8.2. NOETHERIAN RINGS
115
(1) M has finite length.
(2) M is noetherian and artinian.
Proof. (1) ⇒ (2): A chain produces a monoton bounded sequence of lengths. (2)
⇒ (1): M is finite and artinian, so conclusion by 7.4.12.
8.1.13. Example.
(1) A vector space is noetherian if and only if it is finite.
L
(1) Show that N Z is not noetherian.
(2) Show that N Z is not noetherian.
(3) Show that Q is not a noetherian Z-module.
(4) Let p be a prime number. Show that the Z(p) -submodules of Q containing Z(p) is
of form p−n Z(p) and conclude that Q/Z(p) is an artinian but not noetherian Z(p) module.
8.1.14. Exercise.
Q
8.2. Noetherian rings
8.2.1. Definition. A ring R is a noetherian ring if R is a noetherian module.
8.2.2. Proposition. The following conditions are equivalent:
(1) R is noetherian.
(2) Any ideal is finite.
(3) Any increasing sequence of ideals is stationary.
(4) Any nonempty subset of ideals of contains an ideal maximal for inclusion.
(5) Any ideal is noetherian.
Proof. This follows from 8.1.1.
8.2.3. Proposition. Let R be a noetherian ring.
(1) The nilradical is nilpotent. For some n
√ n
0 =0
(2) Some power of the radical of an ideal is contained in the ideal. For some n
√ n
I ⊂I
√
P
Proof. (1) 0 = (b1 , . . . , bm ) such that bki = 0. Then ( ai bi )mk = 0. (2) Use
(1) on R/I.
8.2.4. Proposition. (1) Let I ⊂ R be an ideal in a noetherian ring. Then the
factor ring R/I is a noetherian ring.
(2) The product ring R1 × R2 is noetherian if and only if each Ri is noetherian.
(3) Let U ⊂ R be a multiplicative subset in a noetherian ring. Then the fraction
ring U −1 R is a noetherian ring.
Proof. Use 8.1.3 and 8.1.10.
8.2.5. Proposition. Given ideals I, J ⊂ R, then the following statements are
equivalent:
(1) R/I, R/J are noetherian
(2) R/I ∩ J is noetherian
Proof. This is a special case of 8.1.7.
8.2.6. Theorem. Let R be a noetherian ring. Then any finite R-module is noetherian.
116
8. NOETHERIAN RINGS
Proof. Let M be a finite R-module. There is a surjective homomorphism Rn →
M → 0, 6.1.2. Conclusion by 8.1.3.
8.2.7. Corollary. Let R be a noetherian ring. Any finite R-module is finite presented.
8.2.8. Proposition. Let M be an R-module. The following Statements are equivalent:
(1) M is a noetherian.
(2) R/ Ann(M ) is a noetherian ring and M is finite.
Proof. (1) ⇒ (2): Let x ∈ M , then R/ Ann(x) ' Rx ⊂ M is noetherian. If
x1 , . . . , xn generate M , then Ann(M ) = Ann(x1 ) ∩ · · · ∩ Ann(Xn ). By 8.2.4
R/ Ann(M ) is noetherian. (2) ⇒ (1): By 8.2.6 M is a finite R/ Ann(M )-module.
Conclusion by 8.1.9.
8.2.9. Proposition. Let R be a noetherian ring and M, N noetherian modules.
(1) M ⊗R N is noetherian.
(2) HomR (M, N ) is noetherian.
Proof. Conclusion by 6.1.8, 6.1.9, 8.2.6.
8.2.10. Theorem. Let R be a noetherian ring and U a multiplicative subset. For
a finite module M and any module N the homomorphism
U −1 HomR (M, N ) → HomU −1 R (U −1 M, U −1 N )
is an isomorphism.
Proof. Conclusion by 6.5.10, 8.2.7.
8.2.11. Proposition. Let R be a noetherian ring. If E is an injective R-module,
then U −1 E is an injective U −1 R-module.
Proof. Let I ⊂ R be an ideal. Then HomR (R, E) → HomR (I, E) → 0 is exact.
By 8.2.10 HomU −1 R (U −1 R, U −1 E) → HomU −1 R (U −1 I, U −1 E) → 0 is exact.
So U −1 E is injective 3.6.8, using that any ideal is extended 4.3.6.
8.2.12. Theorem. For a ring R, the following statements are equivalent:
(1) R is a noetherian ring.
L
(2) For any family Eα of injective modules, the sum α Eα is injective.
L
Proof. (1) ⇒ (2): Let I ⊂ R be an ideal. A homomorphism f : I →
Eα
has
Im
f
contained
in
a
finite
sum,
which
is
injective
3.6.7.
So
f
extends
to
R
→
L
Eα , giving injectivity. (2) ⇒ (1): Let In ⊂ R be an increasing chain of ideals.
Put I = ∪I
envelope
I/In ⊂ En . The homomorphism
Ln and choose an injective
L
f :I →
En extends to f 0 : R →
En . Since Im f is contained in a finite
sum, I/In = 0 for n >> 0.
8.2.13. Theorem.QLet R be a noetherian ring and Fα a family of flat modules.
Then the product α Fα is flat.
Proof.
AnQideal I ⊂ R is finite presented. By 6.5.8 theQhomomorphsim Q
I ⊗R
Q
Fα → Fα is a product of injective homomorphisms (I ⊗R Fα ) → Fα ,
which is injective giving the conclusion.
8.3. FINITE TYPE RINGS
117
8.2.14. Theorem. If R is a ring such that every prime ideal is finite, then it is
noetherian.
Proof. If R is not noetherian, then the set of not finite ideals is nonempty and by
Zorn’s lemma it has a maximal element I. Since I is not prime there is a, b ∈
/ I
and ab ∈ I. The ideals I + (a) and I : (a) are both greater that I and therefore
finite. Let I + (a) = (a1 , . . . , am , a), ai ∈ I and I : (a) = (b1 , . . . , bn ). Assume
c = c1 a1 + · · · + cm am + da ∈ I. Then d ∈ I : (a) and d = d1 b1 + · · · + dn bn . So
I = (a1 , . . . , am , ab1 , . . . , abn ) is finite. It follows that R must be noetherian.
8.2.15. Exercise.
(1) Show that a principal ideal domain is noetherian.
(2) Let I ⊂ R be an ideal in a noetherian ring. Show that R/I is noetherian.
(3) Let K be a field and R = K[X1 , X2 , . . . ] the polynomial ring in countable many
variables. Show that R is not noetherian. Q
(4) Let K be a field. Show that the ring R = N K is not noetherian.
8.3. Finite type rings
8.3.1. Theorem (Hilbert’s basis theorem). Let R be a noetherian ring. Then the
ring of polynomials R[X] is noetherian.
Proof. Assume I ⊂ R[X] to be a not finite ideal. Choose a sequence f1 , f2 , · · · ∈
I such that
fi = ai X di + terms of lower degree , ai 6= 0
and fi+1 has lowest degree in I\(f1 , . . . , fi ). The ideal of leading coefficients is
finitely generated by a1 , . . . , an . Then an+1 = b1 a1 + · · · + bn an and d1 ≤ · · · ≤
dn+1 = d gives
fn+1 − b1 X d−d1 f1 − · · · − bn X d−dn fn
in I\(f1 . . . , fn ) of degree less than d. By contradiction the ideal I is finite.
8.3.2. Corollary. Let R be a noetherian ring.
(1) A polynomial ring R[X1 , . . . , Xn ] in finitely many variables is noetherian.
(2) If R → S a finite type ring over R, then S is noetherian.
8.3.3. Corollary. Let R be a noetherian ring and M a finite module. Then the ring
R ⊕ M , 2.1.14, is noetherian.
Proof. Let x1 , . . . , xn generate M . Then R[X1 , . . . , Xn ] → R ⊕ M, Xi 7→ xi is
surjective. So R → R ⊕ M is of finite type.
8.3.4. Example. Let I ⊂ R be an ideal in a noetherian ring. Then there are noetherian rings.
(1) GI R = ⊕n≥0 I n /I n+1 .
(2) BI R = ⊕n≥0 I n .
8.3.5. Theorem (Krull’s intersection theorem). Let I be an ideal in a noetherian
ring R. Then there is a ∈ I such that
\
1 + a ∈ Ann( I n )
n
118
8. NOETHERIAN RINGS
Proof. Let I = (u1 , . . . , um ). If b ∈ I n then b = fn (u1 , . . . , um ) where fn ∈
R[X1 , . . . , Xm ] are homogeneous of degree n. By Hilbert’s basis theorem, 8.3.1
there is N such that
fN +1 = f1 g1 + · · · + fN gN
where gn is homogeneous of degree N − n + 1 > 0. By substitution b ∈ bI and
I(∩I n ) = ∩I n . Now ∩I n ⊂ R is finite, so conclusion by 6.3.4.
8.3.6. Corollary. Let I be an ideal in a noetherian ring R such that the elements
1 + a, a ∈ I are nonzero divisors. Then
\
In = 0
n
8.3.7. Corollary. Let I be an ideal in a noetherian ring R and M a finite module.
Then there is a ∈ I such that
\
1 + a ∈ Ann( I n M )
n
Proof. Use 8.3.5 on the ring R ⊕ M and the ideal I ⊕ M . Clearly (I + M )n =
I n + I n−1 M .
8.3.8. Corollary. Let I be an ideal in a noetherian ring R and M a finite module
such that the elements 1 + a, a ∈ I are nonzero divisors on M . Then
\
I nM = 0
n
8.3.9. Theorem. If R ⊂ S be a finite extension. Then R is noetherian if and only
if S is noetherian.
Proof.
Assume S is noetherian. Let Eα be a family of injective R-modules. Then
L
HomR (S, Eα ) is an
is finite over R, there
L injective S-module. Since S L
L is an
isomorphism 6.1.14
HomR (S, Eα ) ' HomR (S, Eα ). By 6.6.8
Eα is
injective and by 8.2.12 R is noetherian.
8.3.10. Example. Let K be a field and R = K[X1 , X2 , . . . ]/(X1 − x2 X3 , x2 −
x3 X4 , . . . ). Put P = (X1 , X2 , . . . ).
(1) P is maximal
(2) P 2 = P .
(3) P (∩n P n ) = ∩n P n .
8.3.11. Exercise.
(1) Show that if R[X] is noetherian, then R is noetherian.
(2) Show that the subring Z[2X, 2X 2 , . . . ] ⊂ Z[X] is not noetherian. Conclude that the
extension is not finite.
8.4. Power series rings
8.4.1. Theorem. Let R be a noetherian ring. Then the power series ring R[[X]] is
noetherian.
Proof. Let P ⊂ R[[X]] be a prime ideal. Then P + (X)/(X) = (a1 , . . . , an ) ⊂ R
is a finite ideal. If X ∈ P then P = (a1 , . . . , an , X) is finite. Suppose X ∈
/ P
and choose fi = ai + terms of positive degree ∈ P . If g ∈ P then write g =
b10 f1 + · · · + bn0 fn + Xg1 for some bi0 ∈ R. Since
P . Now
P P is prime, g1 ∈ P
g1 = b11 f1 +· · ·+bn1 fn +Xg2 and so on. Put hi = k bik X k , then g = i hi fi .
So P = (h1 , . . . , hn ) is finite and R[[X]] is noetherian by 8.2.14.
8.4. POWER SERIES RINGS
119
8.4.2. Corollary. Let R be a noetherian ring.
(1) A power series ring R[[X1 , . . . , Xn ]] in finitely many variables is noetherian.
(2) Let I ⊂ R[[X1 , . . . , Xn ]] be an ideal, then the factor ring R[[X1 , . . . , Xn ]]/I
is noetherian.
8.4.3. Proposition. Let R be a principal ideal domain. Then the power series ring
R[[X]] is a unique factorization domain.
Proof. By 5.1.12 it is enough to show that a nonzero prime contains a principal
prime. Let P ⊂ R[[X]] be nonzero prime ideal. If X ∈ P then (X) ⊂ P is
a principal prime. Suppose P 6= (X) and P + (X)/(X) = (a) ⊂ R. Choose
f = a + terms of positive degree ∈ P . If g ∈ P then g = b0 f + Xg1 for
someP
b0 ∈ R. Since P is prime, g1 ∈ P . Now g1 = b1 f + Xg2 and so on. Put
h = k bk X k , then g = hf . So P = (f ) is a principal prime itself.
8.4.4. Proposition. Let I ⊂ R be an ideal in a noetherian ring. Then there is a
canonical isomorphism
R[[X]]/IR[[X]] ' R/I[[X]]
Proof. The projection R[[X]] → R[[X]]/IR[[X]] factors over R/I[[X]] since I is
finitely generated. Then there is an inverse to the homomorphism 1.9.8.
8.4.5. Corollary. If P ⊂ R is a prime ideal in a noetherian ring, then P R[[X]] ⊂
R[[X]] is a prime ideal.
8.4.6. Proposition. Let R be a noetherian ring.
(1) The inclusion R[X] ⊂ R[[X]] is a flat homomorphism.
(2) The inclusion R ⊂ R[[X]] is a faithfully flat homomorphism.
P
Proof. (1) Let I ⊂ R[X]
ai ⊗ fi ∈ K = Ker(I ⊗R[X]
Pbe an ideal and let
R[[X]] → R[[X]]. Then ai fi = 0. Chase the diagram
I ⊗R[X] (X n )
I ⊗R[X] R[[X]]
/ R[[X]]
/ R[[X]]/(X n )
I ⊗R[X] R[[X]]/(X n )
P
P
Since R[X]/(X n ) = R[[X]]/(X nP
) it follows that ai ⊗ fi = ai fi ⊗ 1 = 0 ∈
n ). Therefore
I ⊗R[X]
ai ⊗ fi ∈ X n (I ⊗R[X] R[[X]]). It follows that
T R[[X]]/(X
K ⊂ n X n (I⊗R[X] R[[X]]). I⊗R[X] R[[X]] is a finite R[[X]]-module and 1+aX
is a unit, so conclusion by 8.3.8. (2) The homomorphism R → R[X] → R[[X]]
is a composition of flat homomorphisms by (1). For any maximal ideal P ⊂ R
the homomorphism RP → RP [[X]] is local. It follows that P = Q ∩ R for some
Q ⊂ R[[X]], so faithfully flat by 5.6.15.
8.4.7. Exercise.
(1) Show that if R[[X]] is noetherian, then R is noetherian.
120
8. NOETHERIAN RINGS
8.5. Localization of noetherian rings
8.5.1. Theorem. Let R be a noetherian ring and P a prime ideal. Then the local
ring RP is a noetherian ring.
Proof. Any ideal is extended 4.3.6 or just 8.2.4.
8.5.2. Theorem. Let R be a noetherian ring and P a prime ideal. For a finite
module M and any module N the homomorphism
HomR (M, N )P → HomRP (MP , NP )
is an isomorphism.
Proof. A special case of 8.2.10.
8.5.3. Proposition. Let R be a noetherian ring and M a finite module. Let P be a
prime ideal. Then RP is a noetherian ring and MP is a finite RP -module.
8.5.4. Theorem (Krull’s intersection theorem). Let (R, P ) be a noetherian local
ring and M a finite module. Then
\
P nM = 0
n
Proof. From 8.3.7, (1 + a)
a unit.
T
nP
nM
= 0 for some a ∈ P . Now use that 1 + a is
8.5.5. Corollary. Let (R, P ) be a noetherian local ring and I ⊂ P an ideal. Then
\
In = 0
n
8.5.6. Theorem. Let (R, P ) be a noetherian local ring and F a finite module. The
following conditions are equivalent:
(1)
(2)
(3)
(4)
F
F
F
P
is free.
is projective.
is flat.
⊗R F → F is injective.
Proof. This follows from 6.5.14 as finite modules are finite presented.
8.5.7. Proposition. Let R be a noetherian ring and F a finite module. The following conditions are equivalent:
(1)
(2)
(3)
(4)
F is projective.
F is flat.
P ⊗R F → F is injective for all maximal ideals P .
FP is free for all maximal ideals P .
Proof. This follows from 6.5.16 as finite modules are finite presented.
8.5.8. Exercise.
(1) Is it true that if U −1 R is noetherian, then R is noetherian?
8.6. PRIME FILTRATIONS OF MODULES
121
8.6. Prime filtrations of modules
8.6.1. Proposition. Let R be a ring and M 6= 0 a nonzero module. An ideal
Ann(x) maximal in the set of ideals {Ann(y)|0 6= y ∈ M } is a prime ideal.
Proof. Let Ann(x) be a maximal annihilator. Suppose a, b ∈ R such that ab ∈
Ann(x) and b ∈
/ Ann(x). Then
Ann(x) ⊆ Ann(bx) 6= R
Consequently Ann(x) = Ann(bx), in particular a ∈ Ann(x).
8.6.2. Corollary. Let R be a noetherian ring and M 6= 0 a nonzero module. Then
there is x ∈ M such that Ann(x) is a prime ideal.
8.6.3. Theorem. Let R be a noetherian ring and M 6= 0 a finite R-module. Then
there exists a finite filtration of M by submodules
0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn−1 ⊂ Mn = M
such that Mi /Mi−1 , i = 1, . . . , n is isomorphic to an R-module of the form R/Pi
where Pi is a prime ideal in R.
Proof. The set of submodules of M for which the theorem is true is nonempty by
8.2.6. Let N ⊂ M be maximal in this set. Suppose N 6= M . By 8.2.6 applied to
M/N there is a chain N ⊂ N 0 ⊂ M such that N 0 /N is isomorphic to an R-module
of the form R/P 0 where P 0 is a prime ideal. This contradicts the maximality of N .
So N = M .
8.6.4. Corollary. Let R be a nonzero noetherian ring. Then there exists a finite
filtration of ideals
0 = I0 ⊂ I1 ⊂ · · · ⊂ In−1 ⊂ In = R
such that Ii /Ii−1 , i = 1, . . . , n is isomorphic to an R-module of the form R/Pi
where Pi is a prime ideal in R.
8.6.5. Example. In Z there is a filtration
0 ⊂ (pn ) ⊂ (pn−1 ) ⊂ · · · ⊂ (p) ⊂ Z
of any length with factors (pn ) ' Z and (pi−1 )/(pi ) ' Z/(p) for any prime
number p.
8.6.6. Theorem. Let R be a ring. The following statements are equivalent:
(1) R is artinian.
(2) R is noetherian and all prime ideals are maximal.
Proof. (1) ⇒ (2): By 7.4.8 an artinian ring has finite length and therefore noetherian. Primes are maximal by 7.3.11. (2) ⇒ (1): By 8.6.4 there is a finite composition
series.
8.6.7. Corollary. Let R be a ring and M a module. The following statements are
equivalent:
(1) M has finite length.
(2) R/ Ann(M ) is artinian and M is finite.
8.6.8. Theorem. If R ⊂ S be a finite extension. Then R is artinian if and only if
S is artinian.
122
8. NOETHERIAN RINGS
Proof. Suppose S is artinian. By 8.3.9 R is noetherian. By 6.6.5 any prime ideal
in R is maximal. Conclusion by 8.6.6.
8.6.9. Proposition. Let R be a noetherian ring. The number of minimal prime
ideals is finite.
Proof. Choose a filtration 8.6.4, 0 = I0 ⊂ · · · ⊂ In = R with Ii /Ii−1 ' R/Pi ,
where Pi is a prime ideal. Let P be a minimal prime ideal in R. Then (Ii /Ii−1 )P '
(R/Pi )P 6= 0 if and only if Pi ⊂ P . Thus P = Pi for some i since RP 6= 0.
8.6.10. Proposition. Let R be a noetherian ring such that the local rings RP are
domains for all maximal ideals P . Then R is a finite product of domains.
Proof. Let P1 , . . . , Pn be the minimal primes 8.6.9. The intersection is 0 since the
ring is reduced 5.4.10. They are comaximal since a domain has a unique minimal
prime. Conclusion by Chinese remainders 1.4.2.
8.6.11. Exercise.
(1) Compute a filtration 8.6.3 of the Z-module Z/(36).
9
Primary decomposition
9.1. Zariski topology
9.1.1. Definition. Let R be a ring.
(1) The set of prime ideals is the spectrum and denoted X = Spec(R).
(2) For a ring homomorphism φ : R → S restriction defines the associated map
a
φ : Spec(S) → Spec(R)
Q 7→ φ−1 (Q)
(3) For a subset B ⊂ R
V (B) = {P ∈ Spec(R)|B ⊂ P }
is a subset of the spectrum.
(4) For a subset B ⊂ R
XB = Spec(R)\V (B)
is the complement.
9.1.2. Lemma. Let R be a ring.
(1) For a subset B ⊂ R
√
V (B) = V (RB) = V ( RB)
So V (B) depends only on the radical of the ideal generated by B.
(2) For a subset B ⊂ R
\
V (B) =
V (b)
b∈B
(3) For a subset B ⊂ R
XB =
[
Xb
b∈B
9.1.3. Lemma. Let R be a ring.
(1) Spec(R) = V (0), ∅ = V (1).
(2) V
T(B1 ) ∪ V (B2 ) =
S V (B1 B2 ).
(3) α V (Bα ) = V ( α Bα ).
Proof. (2) Clearly V (Bi ) ⊂ V (B1 B2 ). If P ∈
/ V (Bi ) then choose bi ∈ Bi \P .
The product b1 b2 ∈
/ P , so P ∈
/ V (B1 B2 ).
9.1.4. Definition. Let R be a ring.
(1) The subsets V (B), B ⊂ R are the closed sets in the Zariski topology on
Spec(R).
(2) The subsets XB , B ⊂ R are the open sets in the Zariski topology on Spec(R).
(3) Subsets Xb , b ∈ R are the principal open subsets.
123
124
9. PRIMARY DECOMPOSITION
9.1.5. Proposition. Let R be a ring.
(1) Spec(R) = X1 , ∅ = X0 .
(2) Xb1 ∩ Xb2 = Xb1 b2S
.
(3) Spec(R)\V (B) = b∈B Xb
(4) The subset Xb , b ∈ R is a basis for the open sets in the Zariski topology.
9.1.6. Lemma. Let R be a ring and I, J ideals. The following conditions are
equivalent:
(1) V (I)√⊂ V (J).
(2) J
√ ⊂ I.
√
(3) J ⊂ I.
√
T
Proof. This is clear from I = I⊂P P , 5.1.8.
9.1.7. Proposition. Let R be a ring and X = Spec(R).
√
√
(1) V (I) = V (J) if and only if √
J = I.
(2) V (I) = X if and only if I ⊂ 0.
(3) V (I) = ∅ if and only if I = R.
(4) Xb = X if and only if b is√a unit.
(5) Xb = ∅ if and only if b ∈ 0.
Proof. This follows from 9.1.6.
9.1.8. Proposition. Let φ : R → S be a ring homomorphism.
(1) The map a φ : Spec(S) → Spec(R) is continuous.
(2) If I ⊂ R is an ideal, then a φ−1 (V (I)) = V (IS).
(3) If b ∈ R and X = Spec(R) then a φ−1 (Xb ) = Xφ(b) .
(4) If J ⊂ S is an ideal, then the closure a φ(V (J)) = V (J ∩ R).
(5) If X = Spec(R), then the closure a φ(X) = V (Ker φ).
Proof. (1) This follows from (2). (2) Calculate a φ−1 (V (I)) = {Q|a φ(Q) ∈
V (I)} = {Q|I ⊂ Q ∩ R} = V (IS).
9.1.9. Proposition. Let R be a ring and X = Spec(R).
(1) Let I ⊂ R be an ideal. Then
Spec(R/I) ' V (I) ⊂ Spec(R)
is a homeomorphism onto the closed subset V (I).
(2) Let b ∈ R. Then
Spec({bn }−1 R) ' Xb ⊂ Spec(R)
is a homeomorphism onto the principal open subset Xb .
Proof. This is a restatement of 1.3.5 and 5.1.5.
9.1.10. Theorem. Let R be a ring. Then any S
principal open subset Xb is quasicompact. That is, any open covering Xb = α Uα may be refined to an finite
covering Xb = Uα1 ∪ · · · ∪ Uαn .
Proof. Assume by 9.1.9 Xb = X. By 9.1.6 Uα = ∪Xbαβ . So V ({bαβ }) = ∅
and therefore some (bα1 , . . . , bαn ) = R. Clearly X = Xbα1 ∪ · · · ∪ Xbαn ⊂
Uα1 ∪ · · · ∪ Uαn .
9.1. ZARISKI TOPOLOGY
125
9.1.11. Theorem. Let R be a ring. Spec(R) is connected if and only if R is not a
product of two nonzero rings.
Proof. If R is a nontrivial product, then the are proper ideals I1 +I2 = R, I1 ∩I2 =
0. So Spec(R) = V (I1 ) ∪ V (I2 ) is not connected. Conversely if Spec(R) =
V (I1 )∪V (I2 ) is not connected, then the ideals are proper and I1 +I2 = R, I1 ∩I2 ⊂
√
0. Choose ai ∈ Ii such that a1 + a2 = 1 and n so big that an1 an2 = 0. Now
V (an1 ) ∪ V (an2 ) = V (a1 ) ∪ V (a2 ) = Spec(R) so for some bi , b1 an1 + b2 an2 = 1
and (b1 an1 )(b2 an2 ) = 0. It follows that R ' R/(b1 an1 ) × R/(b2 an2 ) is a product of
nonzero rings.
9.1.12. Lemma. Let X 6= ∅ be a topological space. The following conditions are
equivalent:
(1) X is not a union of two proper closed subsets.
(2) Any two nonempty open subsets intersects nonempty.
(3) Any nonempty open subset is dense in X.
Proof. (1) ⇔ (2): This is clear. (2) ⇒ (3): If U is open and nonempty, then
X\Ū ∩U = ∅ gives Ū = X. (3) ⇒ (2): If two open U1 ∩U2 = ∅, then Ū1 ⊂ X\U2 .
So U1 and U2 cannot both be nonempty.
9.1.13. Definition. An irreducible space is a topological space satisfying the conditions in 9.1.10. A subset of a topological space is an irreducible subset if it is
an irrecucible space in the induced topology. A maximal irreducible subset is an
irreducible component.
9.1.14. Lemma. Let X 6= ∅ be a topological space.
(1) If X is irreducible, then any nonempty open subset is irreducible.
(2) If Y ⊂ X is an irreducible subset, then the closure Ȳ is irreducible.
(3) If X is irreducible, then X is connected.
(4) IfX is irreducible and f : X → Y is continuous. Then f (X) is an irreducible
subset.
(5) Any irreducible component in X is closed.
(6) Any irreducible subset in X is contained in an irreducible component.
(7) X is the union of irreducible components.
Proof. Do easy topology homework.
9.1.15. Theorem. Let R be a ring.
√
(1) Spec(R) is irreducible if and only if the nilradical 0 is a prime
√ ideal.
(2) A closed subset V (I) is irreducible if and only if the radical I is a prime
ideal.
(3) An irreducible component is of the form V (P ) where P is a minimal prime
ideal.
√
Proof. Let X = Spec(R). (1) If X √
is irreducible and a1√
, a2 ∈
/ 0, then V (a1 a2 ) =
V (a1 ) ∪√V (a2 ) 6= X. So a1 a2 ∈
/ 0. Conversely, if 0 is a prime, then replace
with R/ 0 and assume R is a domain. If V (I1 I2 ) = V (I1 ) ∪ V (I2 ) = X then
I1 I2 = 0. It follows that I1 = 0 or I2 = 0. (2) Use (1) on R/I. (3) From 5.1.10
any prime ideal contains a minimal prime ideal.
9.1.16. Proposition. Let R be a ring. The following conditions are equivalent:
126
9. PRIMARY DECOMPOSITION
(1) Spec(R)
is a discrete topological space.
√
(2) R/ 0 is a finite product of fields.
Proof. (1) ⇒ (2): By 9.1.10 Spec(R) is finite and consists of maximal ideals.
Conclusion by Chinese remainders 1.4.3.
9.1.17. Exercise.
(1) Let K be a field. Show that Spec(K[X, Y ]/(XY )) is connected,
but not irreducible.
9.2. Support of modules
9.2.1. Definition. Let R be a ring and M a module.
(1) The support of M is
Supp(M ) = {P ∈ Spec(R)|MP 6= 0}
(2) A minimal prime ideal in Supp(M ) is a minimal prime of M .
9.2.2. Proposition. Let I ⊂ R be an ideal. Then as R-module
Supp(R/I) = V (I)
Proof. This is a restatement of 1.3.5.
9.2.3. Proposition. Let
0→M →N →L→0
be a short exact sequence of modules. Then
Supp(N ) = Supp(M ) ∪ Supp(L)
Proof. From 5.4.6 follows that 0 → MP → NP → LP → 0 is exact.
9.2.4. Corollary.
(1) Let N ⊂ M be a submodule. Then
Supp(M ) = Supp(N ) ∪ Supp(M/N )
(2) Given submodules N, L ⊂ M . Then
(a) Supp(M/N ∩ L) = Supp(M/N ) ∪ Supp(M/L) .
(b) Supp(M/N + L) ⊂ Supp(M/N ) ∩ Supp(M/L) .
Proof. (2) Use the exact sequence 3.2.7
0
/ M/N ∩ L
/ M/N ⊕ M/L
/ M/N + L
/0
together with 9.2.3.
9.2.5. Theorem. Let R be a ring and M a module.
(1) M = 0 if and only if Supp(M ) = ∅.
(2) Let M be a module and P ∈ Supp(M ), then V (P ) ⊂ Supp(M ).
(3) For any module
Supp(M ) ⊂ V (Ann(M ))
(4) If M is finite, then the support is a closed subset in the Zariski topology
Supp(M ) = V (Ann(M ))
Proof. (1) See 5.4.1. (2) If P ⊂ Q, then MP = (MQ )P RQ . (3) If MP 6= 0 then
for u ∈
/ P there is x ∈ M such that ux 6= 0. So Ann(M ) ⊂ P . (4) Let x1 , . . . , xn
generate M . If Ann(M ) = ∩ Ann(xi ) ⊂ P then some Ann(xi ) ⊂ P , so x1i 6= 0
in MP .
9.2. SUPPORT OF MODULES
127
9.2.6. Proposition. Let N1 , . . . , Nk ⊂ M be submodules such that Supp(M/Ni )∩
Supp(M/Nj ) = ∅, i 6= j. Then the homomorphism
M
M/ ∩i Ni →
M/Ni
i
is an isomorphism.
Proof. By 3.2.7
0
/ M/ ∩k−1 N ⊕ M/N
i
k
1
/ M/ ∩k Ni
1
/ M/ ∩k−1 N + N
i
k
1
/0
is exact. By 9.2.4 and induction on k
Supp(M/ ∩k−1
Ni + Nk ) ⊂ (∪k−1
Supp(M/Ni )) ∩ Supp(M/Nk )
1
1
So the support of the cokernel is empty. Conclusion by 9.2.5.
9.2.7. Proposition. Let R be a ring and M, N modules.
(1)
Supp(M ⊗R N ) ⊂ Supp(M ) ∩ Supp(N )
(2) If M, N are finite, then
Supp(M ⊗R N ) = Supp(M ) ∩ Supp(N )
(3) If M is finite, then
Supp(HomR (M, N )) ⊂ Supp(M ) ∩ Supp(N )
Proof. There is an isomorphism (M ⊗R N )P ' MP ⊗RP NP . (1) This is clear.
(2) This follows from 6.4.3. (3) By 6.5.10 there is an injective homomorphism
0 → HomR (M, N )P → HomRP (MP , NP ).
9.2.8. Proposition. Let (R, P ) → (S, Q) be a local homomorphism and M a finite
R-module. If Supp(M ) 6= ∅ then Supp(M ⊗R S) 6= ∅.
Proof. Calculate
M ⊗R S ⊗S k(Q) ' M ⊗R k(P ) ⊗k(P ) k(Q)
and conclude by Nakayama’s lemma 6.4.1.
9.2.9. Corollary. Let φ : R → S be a ring homomorphism and M an R-module.
(1) For the change of rings S-module
Supp(M ⊗R S) ⊂ a φ−1 (Supp(M ))
(2) If M is finite, then
Supp(M ⊗R S) = a φ−1 (Supp(M ))
9.2.10. Corollary. Let R be a ring, I an ideal in R and M a finite R-module. Then
a π : Spec(R/I) → Spec(R) defines a bijective correspondence
Supp(M/IM ) → Supp(M ) ∩ V (I) = V (Ann(M ) + I)
Q 7→ P = Q ∩ R
128
9. PRIMARY DECOMPOSITION
9.2.11. Corollary. Let R be a ring, U a multiplicative subset and M an R-module.
Then ι : R → U −1 R, a ι : Spec(U −1 R) → Spec(R) defines a bijective correspondence
Supp(U −1 M ) → Supp(M ) ∩ {P ∈ Spec(R)|P ∩ U = ∅}
Q 7→ Q ∩ R
Proof. Let Q ⊂ U −1 R be a prime and P = Q ∩ R. Then U −1 MQ ' MP gives
the result.
9.2.12. Proposition. Let M be a finite R-module and P ∈ Supp(M ). Then there
is a nonzero homomorphism M → R/P , that is HomR (M, R/P ) 6= 0.
Proof. By 7.1.4 there is a surjective homomorphism f : MP → k(P ) → 0. Let
x1 , . . . , xn generate M and choose u ∈ R\P such that uf (xi ) ∈ R/P . The
composite of M → MP with uf is a nonzero homomorphism M → R/P .
9.2.13. Proposition. Let R be a noetherian ring and M a finite module. M has
finite length if and only if Supp(M ) consists only of maximal ideals.
Proof. By 9.2.5 Supp(M ) = V (Ann(M )), so conclusion by 8.6.7.
9.2.14. Definition. A ring R with only finitely many maximal ideals is a semilocal ring.
9.2.15. Proposition. Let R be a semi-local noetherian ring and F a finite module.
If FP is free of rank m for all maximal ideals, then F is free of rank m.
Proof. Let P1 , . . . , Pk be the maximal ideals. If R ' R/P1n × · · · × R/Pkn is
artinian, then by 7.5.5 F ' FP1 × · · · × FPk ' (R/P1n )m × · · · × (R/Pkn )m is free
of rank m. In general choose x1 , . . . , xm ∈ F giving a basis for F/P1 · · · Pk F .
as R/P1 · · · Pk -module. The homomorphism Rm → F, ei 7→ xi is surjective by
Nakayama’s lemma 6.4.6 and the support of the kernel is empty, so the homomorphism is injective by 9.2.5.
9.2.16. Exercise.
(1) Let R = Z, M = Q and N = Z/(p). Show that Supp(M ⊗R
N ) 6= Supp(M ) ∩ Supp(N ).
9.3. Ass of modules
9.3.1. Definition. Let M be an R-module. A prime ideal P ⊂ R is an associated
prime ideal of M if P = Ann(x) for some x ∈ M . The set of prime ideals
associated to M is Ass(M ).
9.3.2. Proposition. Let R be a ring.
(1) Let M be a module and P a prime. P ∈ Ass(M ) if and only if there is an
injective homomorphism R/P → M .
(2) For any module M
Ass(M ) ⊂ Supp(M )
(3) Let 0 6= N ⊂ R/P be a nonzero submodule, then
Ass(N ) = {P }
(4) Let I be an ideal and M an R/I-module. The inclusion Spec(R/I) →
Spec(R) identifies Ass(M ) over the rings R/I and R.
9.3. ASS OF MODULES
129
Proof. This is clear from the definition. (1) R/P ' Rx ⊂ M . (2) 0 → R/P →
M gives 0 → k(P ) → MP . (3) P = Ann(x) of any nonzero x ∈ R/P . (4)
R/P = (R/I)/(P/I) gives by (1) identified associated primes.
9.3.3. Proposition. Let
0→M →N →L→0
be a short exact sequence of modules. Then
Ass(M ) ⊂ Ass(N ) ⊂ Ass(M ) ∪ Ass(L)
Proof. The left inclusion is trivial. Assume M ⊂ N and L = N/M . Next let
P ∈ Ass(N ) such that P ∈
/ Ass(M ). Choose a submodule K ⊂ N such that
K ' R/P . Then Ass(K ∩ M ) ⊂ Ass(K) ∩ Ass(M ). It follows 9.3.2 that
K ∩ M = 0. Therefore K ' K/K ∩ M ' K + M/M ⊂ N/M gives P ∈
Ass(K) ⊂ Ass(L).
9.3.4. Corollary. Let 0 → M → N → L → 0 be a split exact sequence of
modules.
Ass(N ) = Ass(M ) ∪ Ass(L)
(1) Let N ⊂ M be a submodule. Then
9.3.5. Corollary.
Ass(N ) ⊂ Ass(M ) ⊂ Ass(N ) ∪ Ass(M/N )
(2) Let M, N be modules. Then
Ass(M ⊕ N ) = Ass(M ) ∪ Ass(N )
(3) Given submodules N, L ⊂ M . Then
Ass(M/N ∩ L) ⊂ Ass(M/N ) ∪ Ass(M/L) ⊂ Ass(M/N ∩ L) ∪ Ass(M/N + L)
Proof. (3) Use the exact sequence 3.2.7
0
/ M/N ∩ L
/ M/N ⊕ M/L
/ M/N + L
/0
together with 9.3.3.
9.3.6. Proposition. Let M be a module and P ⊂ Ass(M ). Then there is a submodule N ⊂ M such that
Ass(N ) = P , Ass(M/N ) = Ass(M )\P
Proof. Choose by Zorn’s lemma N maximal in the set of submodules N 0 ⊂ M
for which Ass(N 0 ) ⊂ P. Let Q ∈ Ass(M/N ) and choose N ⊂ L ⊂ M with
L/N ' R/Q. Then Ass(L) ⊂ P ∪ {Q}. By maximality of N follows that
Ass(L) 6⊂ P. So Q ∈
/ P and Q ∈ Ass(L) ⊂ Ass(M ).
9.3.7. Theorem. Let R be a noetherian ring and M a module. The following
conditions are equivalent:
(1) M = 0.
(2) Supp(M ) = ∅.
(3) Ass(M ) = ∅.
(4) MP = 0 for all prime P ∈ Ass(M ).
Proof. (4) ⇒ (1): If M 6= 0, then by 8.6.2 and 9.3.2 there is P ∈ Ass(M ) ⊂
Supp(M ).
130
9. PRIMARY DECOMPOSITION
9.3.8. Corollary. Let R be a noetherian ring and f : M → N a homomorphism.
The following conditions are equivalent:
(1) f is injective.
(2) fP is injective for all prime ideals P ∈ Ass(M ).
Proof. (2) ⇒ (1): Ker fP = 0 for all P ∈ Ass(Ker f ) ⊂ Ass(M ). Conclusion by
9.3.7.
9.3.9. Theorem. Let R be a noetherian ring and M a module.
(1) a ∈ R is a nonzero divisor on M if and only if a ∈
/ ∪P ∈Ass(M ) P .
(2) The set of zero divisors on M is
∪P ∈Ass(M ) P
Proof. aM is injective if and only if aMP is injective for all P ∈ Ass(M ), 9.3.8.
This happens when a ∈
/ ∪P ∈Ass(M ) P .
9.3.10. Proposition. Let U ⊂ R be a multiplicative subset and M a module. Denote ι : R → U −1 R and a ι : Spec(U −1 R) → Spec(R).
(1) The map a ι−1 defines an inclusion
Ass(M ) ∩ {P ∈ Spec(R)|P ∩ U = ∅} → Ass(U −1 M )
P 7→ P RP
(2) If R is noetherian, then the map (1) is a bijective correspondence.
Proof. Let P be a prime ideal. For any homomorphism R/P → M there is a
commutative diagram
/M
R/P
U −1 R/P
/ U −1 M
(1) If P ∈ Ass(M ) and P ∩ U = ∅ then U −1 P ∈ Ass(U −1 M ). (2) If U −1 P ∈
Ass(U −1 M ), then P ∩U = ∅. 8.2.9 gives an isomorphism U −1 HomR (R/P, M ) '
HomU −1 R (U −1 R/P, U −1 M ), so there is a diagram as above and R/P → M is
injective. It follows that P ∈ Ass(M ).
9.3.11. Proposition. Let R be a noetherian ring and M a module. Then any minimal prime P ∈ Supp(M ) is contained in Ass(M ).
Proof. Assume P ∈ Supp(M ) is minimal. Then the RP -module MP has support
exactly in the maximal ideal, so {P RP } = Ass(MP ). Conclusion by 9.3.10.
9.3.12. Definition. A non minimal prime ideal in Ass(M ) is an embedded prime
of M .
9.3.13. Theorem. Let R be a noetherian ring and M a finite module. Then the
associated primes Ass(M ) is a finite set.
Proof. Follows immediately from 9.3.4 and 8.1.6. Let 0 = M0 ⊂ · · · ⊂ Mn = M
be a filtration with factors Mi /Mi−1 ' R/Pi . Then Ass(M ) ⊂ {Pi }.
9.3.14. Corollary. Let R be a noetherian ring and M a finite module. The following conditions are equivalent:
(1) M has finite length.
9.3. ASS OF MODULES
131
(2) Supp(M ) consists of maximal ideals.
(3) Ass(M ) consists of maximal ideals.
(4) Ass(M ) = Supp(M ).
Proof. Most is just a restatement. (4) ⇒ (2): Use that the minimal ideals in the
support are maximal.
9.3.15. Lemma. Let (R, P ) be a local ring M a module. Then P ∈ Ass(M ) if
and only if HomR (k(P ), M ) 6= 0.
Proof. k(P ) is simple, so a homomorphism k(P ) → M is either 0 or injective.
9.3.16. Proposition. Let R be a noetherian ring and M a module. For a prime P
the following conditions are equivalent:
(1) P ∈ Ass(M ).
(2) P RP ∈ Ass(MP ).
(3) HomRP (k(P ), MP ) 6= 0.
Proof. (1) ⇔ (2): 9.3.10. (2) ⇔ (3): 9.3.15.
9.3.17. Theorem. Let R be a noetherian ring and M a finite module. For any
module N
Ass(HomR (M, N )) = Supp(M ) ∩ Ass(N )
Proof. By 8.2.9 HomR (M, N )P ' HomRP (MP , NP ). So reduce to the case
where (R,P) is local. Now
HomR (k(P ), HomR (M, N )) = HomR (M, HomR (k(P ), N ))
= Homk(P ) (M ⊗R k(P ), HomR (k(P ), N ))
Conclusion by Nakayama’s lemma 6.4.1 and 9.3.15.
9.3.18. Corollary. Let R be a noetherian ring and M a finite module. For a prime
P the following conditions are equivalent:
(1) P ∈ Supp(M ).
(2) P ∈ Ass(HomR (M, R/P )).
9.3.19. Proposition. Let R be a noetherian ring and F a finite module. Assume
rank F ⊗R k(P ) = n for all primes P . Then F is locally free (projective) if and
only if FP is free for all P ∈ Ass(R).
n → F → 0 exact. Ass(K) ⊂
Proof. Let Q be a maximal ideal and 0 → K → RQ
Q
Ass(R), so KP = 0 for all P ∈ Ass(K). By 9.3.7 K = 0.
9.3.20. Theorem. Let (R, P ) be a noetherian local ring. If there is a nonzero finite
injective module E, then R is artinian.
Proof. Let Q be a prime and f : R/Q → E. If a ∈ P \Q then aR/Q is injective,
so there is f 0 : R/Q → E such that f = f 0 ◦ aR/Q . That is P HomR (R/Q, E) =
HomR (R/Q, E), so by Nakayama’s lemma 6.4.1 HomR (R/Q, E) = 0 if Q 6= P .
By 9.3.7 0 6= HomR (R/P, E) ⊂ HomR (R/Q, E), so Q = P . R is artinian by
8.6.6.
9.3.21. Exercise.
(1) Show that
Ass(Z/(n)) = {(p)|p prime dividing n}
132
9. PRIMARY DECOMPOSITION
(2) Show that
Ass(K[X, Y ]/(X) ∩ (X 2 , Y 2 )) = {(X), (X, Y )}
and point out an embedded prime.√
(3) Let I ⊂ R be an ideal such that I = I. Show that R/I has no embedded prime
ideals.
(4) Let I, J ⊂ R be a ideals such that JRP ⊂ IRP for all P ∈ Ass(R/I). Show that
J ⊂ I.
9.4. Primary modules
9.4.1. Definition. A submodule N ⊂ M is a primary submodule or more precisely P -primary if Ass(M/N ) = {P }. An ideal is a primary ideal if it is a
primary submodule of the ring.
9.4.2. Proposition. A prime ideal P ⊂ R is a P -primary ideal.
Proof. By 9.3.2 Ass(R/P ) = {P }.
9.4.3. Proposition. Let R be a noetherian ring and M a finite module. For a
proper submodule N ⊂ M the following conditions are equivalent:
(1) N ⊂ M is primary for some prime
p
(2) The set of zero divisors on M/N is contained in the radical Ann(M/N ).
(3) For any zero divisor a on M/N there is a power an ∈ Ann(M/N ).
p
Proof. (1) ⇒ (2),(3): Supp(M/N ) = V (P ) by 9.3.11. So by 9.1.6 Ann(M/N ) =
P is the set of zero divisors by 9.3.9. (2),(3) ⇒ (1): By 9.3.7 Ass(M/N
) 6= ∅. If
p
P1 , P2 ∈ Ass(M/N ) then by 9.3.9 the zero divisors in P1 ∪P2 ⊂ Ann(M/N ) ⊂
P1 ∩ P2 , so P1 = P2 .
9.4.4. Corollary. Let R be a noetherian ring and I ⊂ R a proper ideal. The
following conditions are equivalent:
(1) I ⊂ R is primary for some prime.
(2) Any zero divisor in the ring R/I is nilpotent.
(3) If ab ∈ I for some b ∈
/ I then some power an ∈ I.
9.4.5. Corollary. Let R be a noetherian ring.
√
(1) If an ideal I ⊂√R is P -primary then I = P .
(2) If the radical I = P is a maximal ideal, then I ⊂ R is P -primary.
(3) A finite power P n ⊂ R of a maximal ideal is P -primary.
√
Proof. (1) By 9.3.11 Supp(R/I) = V (P ), so √I = P . (2) Supp R/I = V (I) =
{P }. By 9.3.2 and 9.3.7 Ass(R/I) = {P }. (3) P n = P .
9.4.6. Proposition. Let R be a noetherian ring and M a module.
(1) If N, N 0 ⊂ M are P -primary, then N ∩ N 0 is P -primary.
(2) If N ⊂ M ispP -primary and M/N is finite, then Ann(M/N ) ⊂ R is P primary and Ann(M/N ) = P .
Proof. (1) From 9.3.5 Ass(M/N ∩ N 0 ) ⊂ Ass(M/N ) ∪ Ass(M/N 0 ). (2) Let
x1 , . . . , xn be nonzero generators of M/N , then R/ Ann(xi ) ' Rxi ⊂ M/N
shows
that Ann(xi ) is P -primary. By (1) Ann(M/N ) = ∩i Ann(xi ) is P -primary.
p
Ann(M/N ) = P by 9.4.5.
9.4. PRIMARY MODULES
133
9.4.7. Definition. If i : M → U −1 M is the canonical homomorphism to the
module of fractions and N 0 ⊂ U −1 M is a submodule, then abuse the notation
N 0 ∩ M = i−1 (N 0 ).
9.4.8. Theorem. Let R be a noetherian ring and M a module. Suppose U ⊂ R is
a multiplicative subset and P is a prime ideal.
(1) If U ∩ P 6= ∅ and N ⊂ M is P -primary, then U −1 N = U −1 M .
(2) If U ∩ P = ∅ and N ⊂ M is P -primary, then U −1 N ⊂ U −1 M is P U −1 Rprimary and N = U −1 N ∩ M .
(3) If N 0 ⊂ U −1 M is P U −1 R-primary, then N 0 ∩ M ⊂ M is P -primary and
N 0 = U −1 (N 0 ∩ M ).
Proof. From 9.3.10: (1) Ass(U −1 M/N ) = ∅. (2) Ass(U −1 M/N ) = {P RP }.
Now Ass(U −1 N ∩ M/N ) ⊂ Ass(M/N ) = {P }. But (U −1 N ∩ M/N )P =
NP ∩ MP /NP = 0, so N = U −1 N ∩ M by 9.3.7. (3) P U −1 R ∩ R = P ∈
Ass(M/N 0 ∩ M ) is the only associated prime disjoint from U . Let N 0 ∩ M ⊂
N ⊂ M be such that N/N 0 ∩ M ' R/Q. If Q ∩ U 6= ∅, then U −1 N = N 0 and
therefore N = N 0 ∩ M is a contradiction, so N 0 ∩ M ⊂ M is P -primary.
9.4.9. Corollary. Let R be a noetherian ring. Suppose U ⊂ R is a multiplicative
subset and P is a prime ideal.
(1) If U ∩ P 6= ∅ and I ⊂ R is P -primary, then IU −1 R = U −1 R.
(2) If U ∩ P = ∅ and I ⊂ R is P -primary, then IU −1 R ⊂ U −1 R is P U −1 Rprimary and I = IU −1 R ∩ R.
(3) If I 0 ⊂ U −1 R is P U −1 R-primary, then I 0 ∩ R ⊂ R is P -primary and I 0 =
(I 0 ∩ R)U −1 R.
9.4.10. Corollary. Let R be a noetherian ring and P a prime ideal.
(1) If P ∈ Supp(M ) of a finite module, then P n MP ∩ M ⊂ M is a P -primary
submodule.
(2) P n RP ∩ R ⊂ R is a P -primary ideal.
Proof. (1) By Nakayama’s lemma 6.4.1 Supp(MP /P n MP ) = {P RP }. Then
P n MP ⊂ MP is P RP -primary. Conclusion by 9.4.8.
9.4.11. Example. Let R = Z[X] and P = (p, X) for some prime number p. Then
R/P ' Z/(p), so P is a maximal
ideal. The ideal Q = (p2 , X) satisfies strict
√
inclusions P 2 ⊂ Q ⊂ P . So Q = P gives that Q is P -primary, but Q is not a
power of P .
9.4.12. Example. Define the subring R = Z[pX, X 2 ] ⊂ Z[X] for some prime
number p.
P
(1) R = { ai X|p|a2i+1 }.
P
(2) The ideal P = (pX, X 2 ) = { ai X|p|a2i+1 , a0 = 0} has R/P ' Z, so P
is a non maximal prime ideal. P
(3) The ideal Q = (q, pX, X 2 ) = { ai X|p|a2i+1 , q|a0 } has R/Q ' Z/(q), so
for q a prime number P ⊂ Q is a maximal
P ideal.
2
2
2
3
4
(4) The ideal P = (p X , pX , X ) = { ai X|p|a2i+1 , a0 = a1 = 0, p2 |a2 }
has p · pX 2 = p2 X 2 ∈ P 2 , pX 2 ∈
/ P 2 and pn ∈
/ P 2 , so by 9.4.4 P 2 is not
primary.
9.4.13. Exercise.
primary.
(1) Let K be a field. Show that (X 2 , Y ) ⊂ K[X, Y ] is (X, Y )-
134
9. PRIMARY DECOMPOSITION
(2) Let p be a prime number. Show that (pk ) ⊂ Z is a primary ideal.
9.5. Decomposition of modules
9.5.1. Definition. A submodule L ⊂ M has a primary decomposition if there
exist a family Ni ⊂ M of Pi -primary submodules, such that
L = N1 ∩ · · · ∩ Nk
A primary decomposition is a reduced primary decomposition if Pi 6= Pj for i 6= j
and no Ni can be excluded.
9.5.2. Lemma. Let R be a noetherian ring and L ⊂ M submodule with a primary
decomposition. Then by intersection there is a reduced primary decomposition.
Proof. By 9.4.6 intersection of P -primary submodules is P -primary, so replace
more P -primary factors by their intersection.
9.5.3. Lemma. Let R be a noetherian ring and M a finite module.
(1) For each Pi ∈ Ass(M ) there is a submodule Ni ⊆ M such that Ass(Ni ) =
Ass(M ) − {Pi } and Ass(M/Ni )) = {Pi }.
(2) The intersection
\
Ni = 0
i
(3) The module M injects
0→M →
M
M/Ni
i
Proof. The submodule Ni is given by 9.3.6. Ass(∩Ni ) = ∅, so conclusion by
9.3.7.
9.5.4. Theorem. Let R be a noetherian ring and M a finite module. A proper
submodule L ⊂ M has a reduced primary decomposition
L = N1 ∩ · · · ∩ Nk
with Ni Pi -primary. For any such
(1) The primes are determined
Ass(M/L) = {P1 , . . . , Pk }
(2) There is an inclusion
0 → M/L →
M
M/Ni
i
Proof. Apply 9.5.3 to M/L to get a reduced primary decomposition. By (2)
Ass(M/L) ⊂ {P1 , . . . , Pk }. For i there is an inclusion ∩j6=i Nj /L ' ∩j6=i Nj +
Ni /Ni ⊂ M/Ni . It follows since the decomposition is reduced that {Pi } ∈
Ass(∩j6=i Nj /L) ⊂ Ass(M/L). So {P1 , . . . , Pk } ⊂ Ass(M/L).
9.5.5. Proposition. Let R be a noetherian ring and M a finite module. If
L = N1 ∩ · · · ∩ Nk
is a reduced primary decomposition of L ⊂ M and Pi is one of the minimal primes
in Ass(M/L), then
Ni = LPi ∩ M
9.5. DECOMPOSITION OF MODULES
135
and therefore uniquely determined.
Proof. Since Pi is minimal, Nj Pi = MPi for j 6= i. By 9.4.8 follows that Ni =
NPi ∩ M = LPi ∩ M .
9.5.6. Proposition. Let R be a noetherian ring and M a finite module. Let L ⊂ M
such that M/L 6= 0 has finite length. If Ass(M/L) = {P1 , . . . , Pk }, then there is
a reduced primary decomposition
L = N1 ∩ · · · ∩ Nk
where
Ni = LPi ∩ M
and an isomorphism
M/L '
M
M/Ni
i
Proof. This follows from 9.5.4, 9.5.5 and 9.2.6. Since M/Ni have finite length
Supp(M/Ni ) = {Pi } and the conditions in 9.2.6 are satisfied.
9.5.7. Proposition. Let R be a noetherian ring and M a finite length module. If
Ass(M ) = {P1 , . . . , Pk }, then there is a reduced primary decomposition
0 = P1 n M ∩ · · · ∩ Pk n M
and exact sequences
0 → Pin M → M → MPi → 0
There are isomorphisms
M'
M
MPi '
i
M
M/Pi n M
i
Proof. This follows from 7.5.5.
9.5.8. Proposition. Let R be a noetherian ring and M a finite module. Let L ⊂ M
have a reduced primary decomposition
L = N1 ∩ · · · ∩ Nk
where Ni is Pi -primary. Assume U to be a multiplicative subset disjoint from
exactly P1 , . . . , Pm . Then
U −1 L = U −1 N1 ∩ · · · ∩ U −1 Nm
is a reduced primary decomposition.
Proof. This follows from 9.4.8 and 9.5.4.
9.5.9. Exercise.
(1) Describe the primary decomposition over a field.
136
9. PRIMARY DECOMPOSITION
9.6. Decomposition of ideals
9.6.1. Theorem. Let R be a noetherian ring. A proper ideal I has a reduced
primary decomposition
I = Q1 ∩ · · · ∩ Qk
with Qi Pi -primary. For any such
(1) The primes are determined
Ass(R/I) = {P1 , . . . , Pk }
(2) There is an inclusion
0 → R/I →
M
R/Qi
i
Proof. This is a case of 9.5.4.
9.6.2. Proposition. Let R be a noetherian ring. If
I = Q1 ∩ · · · ∩ Qk
is a reduced primary decomposition and Pi one of the minimal primes in Ass(R/I),
then
Qi = IRPi ∩ R
and therefore uniquely determined.
Proof. This is a case of 9.5.5.
9.6.3. Definition. Let P be a prime ideal. The symbolic power of P is
P (n) = R ∩ P n RP
9.6.4. Lemma. Let P ⊂ R be a prime ideal in a noetherian ring.
(1) If P is a maximal ideal then P (n) = P n is P -primary.
(2) P n RP ⊂ RP is P RP -primary.
(3) P (n) ⊂ R is P -primary.
Proof. (1) By 9.4.5 P n is primary and by 9.4.9 P n = P (n) . (2) This follows from
(1). (3) This follws from 9.4.10.
9.6.5. Proposition. Let R be a noetherian ring and let
P n = Q1 ∩ · · · ∩ Qk
be a reduced primary decomposition of a power of a prime ideal P . If Q1 is P primary, then
Q1 = P (n)
Proof. This is a case of 9.6.2.
9.6.6. Proposition. Let I ⊂ R be a proper ideal in a noetherian ring such that
R/I is artinian. If Ass(R/I) = {P1 , . . . , Pk }, then there is a reduced primary
decomposition
(n)
(n)
I = P1 ∩ · · · ∩ Pk
and an isomorphism
M
(n)
R/I '
R/Pi
i
Proof. This is a case of 9.5.6.
9.6. DECOMPOSITION OF IDEALS
137
9.6.7. Proposition. Let R be an artinian ring. If Ass(R) = {P1 , . . . , Pk }, then
there is a reduced primary decomposition
0 = P1n ∩ · · · ∩ Pkn
and exact sequences
0 → Pin → R → RPi → 0
There are isomorphisms
R'
M
i
RPi '
M
R/Pin
i
Proof. This is a case of 9.5.7.
9.6.8. Proposition. Let a proper ideal I ⊂ R in a noetherian ring have a reduced
primary decomposition
I = Q1 ∩ · · · ∩ Qk
where Qi is Pi -primary. Assume U to be a multiplicative subset disjoint from
exactly P1 , . . . , Pm . Then
IU −1 R = Q1 U −1 R ∩ · · · ∩ Qm U −1 R
is a reduced primary decomposition.
Proof. This is a case of 9.5.8.
9.6.9. Proposition. Let R be a noetherian ring and I an ideal. Then
√
I = P1 ∩ · · · ∩ Pk
√
is the primary decomposition and the elements in Ass(R/ I) = {P1 , . . . , Pk } are
minimal primes over I.
√
Proof. By 9.3.13 there are only finitely many minimal primes over I. By 5.1.8 I
is the intersection of these and
√ by 9.6.1 this is a reduced primary decomposition
determining the set Ass(R/ I).
9.6.10. Corollary. Let R be a reduced noetherian ring. Then all elements in
Ass(R) are minimal primes. That is, there are no embedded primes.
9.6.11. Corollary. Let R be a noetherian ring. The following statements are equivalent:
(1) R is reduced.
(2) RP is a field for all P ∈ Ass(R).
(3) RP is a domain for all P ∈ Ass(R).
Proof. (1) ⇒ (2): 0 = P1 ∩ · · · ∩ Pk where Pi are minimal. Then 0 = Pi√
RPi is
the decomposition
by√9.6.8, so the maximal ideal is zero.
√
√ (3) ⇒ (1): Ass( 0) ⊂
Ass(R) and 0P = 0 ⊂ RP for all P ∈ Ass(R). So 0 = 0 by 9.3.7.
9.6.12. Example. Let R be a unique factorization domain. A factorization into
powers of different irreducible primes is a reduced primary decomposition of a
principal ideal.
(1) Let (p) be a prime divisor. The homomorphism 1pn−1 : R → R/(pn ) fits to
an exact sequence 0 → R/(p) → R/(pn ) → R/(pn−1 ) → 0. By 9.3.3 and
induction Ass(R/(pn )) ⊂ Ass(R/(p)) ∪ Ass(R/(pn−1 )) = {(p)} gives that
any power (pn ) is a (p)-primary ideal.
138
9. PRIMARY DECOMPOSITION
(2) If (a) = (pn1 1 ) . . . (pnk k ) is a prime factorization then
(a) = (pn1 1 ) ∩ · · · ∩ (pnk k )
0 → R/(a) → R/(pn1 1 ) × · · · × R/(pnk k )
is a primary decomposition and Ass(R/(a)) = {(p1 ), . . . , (pk )}.
√
(1) Let I ⊂ R be an ideal. Show that if P = I is a maximal ideal,
then I is a P -primary ideal.
(2) Let I ⊂ R be an ideal. Show that if I contains a power of a maximal ideal P , then I
is a P -primary ideal.
√
(3) Let K be a field and I = (X 2 , XY ) ⊂ K[X, Y ]. Show that I = (X), but I is not
(X)-primary.
9.6.13. Exercise.
10
Dedekind rings
10.1. Principal ideal domains
10.1.1. Lemma. Let R be a domain. The set of elements x ∈ M in a module with
Ann(x) 6= 0 is a submodule.
Proof. If ax = by = 0 then ab(x + y) = 0. Now use that R is a domain.
10.1.2. Definition. Let R be a domain. An element x ∈ M in a module is a torsion
element if Ann(x) 6= 0. By 10.1.1 the set of torsion elements is a submodule
T (M ) = {x ∈ M | Ann(x) 6= 0}
of M , the torsion submodule. If T (M ) = 0 then M is a torsion free module. If
T (M ) = M then M is a torsion module.
10.1.3. Lemma. Let R be a domain and M a module.
(1) M is torsion free if and only if any nonzero a ∈ R is a nonzero divisor on M .
(2) The factor M/T (M ) is torsion free.
(3) Let K be the fraction field of R.
T (M ) = Ker(M → M(0) ) = Ker(M → M ⊗R K)
(4) If U ⊂ R is multiplicative, then U −1 T (M ) = T (U −1 M ).
(5) T is a left exact functor: if 0 → M → N → L is exact, then 0 → T (M ) →
T (N ) → T (L) is exact.
Proof. (1) x ∈ T (M ) ⇔ Ann(x) 6= 0. (2) Let x + T (M ) ∈ M/T (M ) and
0 6= a ∈ R such that ax ∈ T (M ). Then bax = 0 for some b 6= 0. It follows that
x + T (M ) = 0. (3) From 4.4.1 M → M ⊗R K is the canonical homomorphism
M → M(0) to the fractions by all nonzero denominators, so the kernel is T (M ) by
4.2.7. (4) By (3) U −1 T (M ) = Ker(U −1 M → U −1 M ⊗ K) = T (U −1 M ). (5)
If f : M → N is injective and f (x) ∈ T (N ), then x ∈ T (M ), so f (T (M )) =
Ker(T (N ) → T (L)).
10.1.4. Corollary. Let R be a domain and F a flat module. Then F is torsion free.
Proof. If 0 6= a ∈ R and F flat, then aF = aR ⊗ 1F is injective.
10.1.5. Corollary. Let R be a domain and M a module. The following conditions
are equivalent:
(1) M is torsion free.
(2) MP is torsion free for all prime ideals P .
(3) MP is torsion free for all maximal ideals P .
Proof. By 10.1.3 T (M )P = T (MP ). Conclusion by the local-global principle
5.4.1.
139
140
10. DEDEKIND RINGS
10.1.6. Corollary. Let R be a noetherian domain and M a module. The following
conditions are equivalent:
(1) M is a torsion module.
(2) (0) ∈
/ Supp(M ).
(3) (0) ∈
/ Ass(M ).
Proof. (1) ⇔ (2): By 10.1.3 T (M ) = Ker(M → M(0) ). (2) ⇔ (3): By 9.3.11 the
set of primes Ass(M ) and Supp(M ) have the same minimal elements.
10.1.7. Proposition. Let R be a principal ideal domain. A submodule of a finite
free module is free.
Proof. Let F ⊂ Rn be a submodule. If n = 1 F is a principal ideal and free.
Let n > 1 and p : Rn → R be the last projection. Then p(F ) is a principal ideal
and free. By induction F ∩ Ker p ⊂ Rn−1 is free, so the split exact sequence
0 → F ∩ Ker p → F → p(F ) → 0 gives F ' F ∩ Ker p ⊕ p(F ) is free.
10.1.8. Proposition. Let R be a principal ideal domain.
(1) A torsion free module is flat.
(2) A finite torsion free module is free.
(3) A finite torsion module has finite length.
Proof. (1) For a nonzero ideal (a) ⊂ R the composite R ' (a) → R is aM .
For a torsion free F the homomorphism aF : F ' (a) ⊗R F → F is injective.
So F is flat by 3.7.12. (2) By (1) a finite torsion free module F is flat. By 8.5.7
F is projective and therefore isomorphic to a submodule of a finite free module.
Conclusion by 10.1.7. (3) By 10.1.6 there are only maximal ideals in the support.
Conclusion by 9.2.13.
10.1.9. Proposition. Let R be a principal ideal domain. A finite module M decomposes
M = T (M ) ⊕ F
as a direct sum of the torsion submodule and a finite free submodule F .
Proof. By 10.1.8 M/T (M ) is free, so 0 → T (M ) → M → M/T (M ) → 0 is
split exact. This gives M/T (M ) ' F ⊂ M .
10.1.10. Proposition. Let R be a principal ideal domain. A finite torsion module
M has a decomposition
M
M=
Mp
(p)∈Ass(M )
where (p) is a prime divisor and
Mp = {x ∈ M |pn x = 0, for some n}
Proof. This comes from the primary decomposition 9.5.7. M ' ⊕p M/(pn )M .
If x ∈ Mp and q 6= p then R = (p, q n ) ⊂ Ann(x + (q n )M ). Therefore Mp =
Ker(M → ⊕q6=p M/(q n )M ) ' M/(pn )M .
10.1.11. Proposition. Let R be a principal ideal domain and (p) a prime divisor.
A finite torsion module M such that M = Mp has decomposition
M = R/(pn1 ) ⊕ · · · ⊕ R/(pnk )
where n1 ≥ · · · ≥ nk .
10.2. DISCRETE VALUATION RINGS
141
Proof. Let n1 = n be such that pn ∈ Ann(M ), but pn−1 x 6= 0 for some x ∈ M .
The short exact sequence 0 → Rx → M → M/Rx → 0 of R/(pn )-modules
is split exact, since R/(pn ) ' Rx is an injective module 7.6.11. Conclusion by
induction on `R (M ).
10.1.12. Theorem. Let R be a principal ideal domain and M a finite module.
Then M has decomposition
M = R/(pn1 1 ) ⊕ · · · ⊕ R/(pnk k ) ⊕ Rn
where (p1 ), . . . , (pk ) are not necessarily distinct prime divisors.
10.1.13. Example.
(1) A finite abelian group is isomorphic to
Z/(pn1 1 ) ⊕ · · · ⊕ Z/(pnk k )
where p1 , . . . , pk are not necessarily distinct prime numbers.
(2) A finite dimensional C vector space with a linear map to itself is by 2.1.13 a
C[X] module
C[X]/(X − λ1 )n1 ⊕ · · · ⊕ C[X]/(X − λk )nk
where λi are the not necessarily distinct eigenvalues. Choosing C-basis 1, (X−
λ), . . . , (X − λ)n−1 in C[X]/(X − λ)n , then the matrix of multiplication by
X is a lower Jordan block
λ
B1
B
B
@
0
1
λ
..
.
1
C
C
C
A
λ
Just use the formulas X(X − λ)i = (X − λ)i+1 + λ(X − λ)i . All together
this gives the Jordan block decomposition.
10.1.14. Exercise.
(1) Show that a nonzero finite Z-submodule of Q is a free module
of rank 1.
(2) Show that a finite torsion Z-module is a finite group.
(3) Let K be a field. Show that a finite torsion K[X]-module is a finite K vector space.
(4) Let R be a noetherian domain such that every nonzero prime ideal is maximal. Let
M be a finite module. Show that the torsion submodule T (M ) has finite length.
10.2. Discrete valuation rings
10.2.1. Definition. A local principal ideal domain, which is not a field, is a discrete valuation ring. A generator of the maximal ideal is a local parameter or a
uniformizing parameter.
10.2.2. Proposition. Let (R, (p)) be a discrete valuation ring. Then
\
(pn ) = 0
n
Proof. See Krull’s intersection theorem 8.5.5.
10.2.3. Proposition. Let (R, P ) be a noetherian local domain. The following
statements are equivalent:
(1) R is a discrete valuation ring.
(2) P is a principal ideal.
(3) `R (P/P 2 ) = 1.
142
10. DEDEKIND RINGS
Proof. (1) ⇒ (2) ⇔ (3): Nakayama’s lemma 6.4.1. (2) ⇒ (1): Let P = (p) and
0 6= a R, by 10.2.2 there is n such that a ∈ (pn ) − (pn+1 ). Since any ideal is
finitely generated it follows that a nonzero ideal is of the form (pn ).
10.2.4. Proposition. Let (R, (p)) be a discrete valuation ring. Any nonzero ideal
is of the form (pn ) for a unique n = 0, 1, 2, . . . .
Proof. (p) is the only prime divisor. Conclusion by unique factorization 1.5.6.
10.2.5. Corollary. Let (R, (p)) be a discrete valuation ring. Any nonzero element
in the fraction field K of R has a unique representation upn where u is a unit and
n ∈ Z.
10.2.6. Definition. Let K be a field. A surjective map v : K\{0} → Z satisfying
(1) v(xy) = v(x) + v(y)).
(2) If x + y 6= 0, then v(x + y) ≥ min(v(x), v(y)).
is a valuation on K.
10.2.7. Proposition. Let v be a valuation on a field K.
(1) The subset R = {x|v(x) ≥ 0} ∪ {0} is subring a of K.
(2) The units in R are {x|v(x) = 0}.
(3) Any element p, v(p) = 1 generates the same prime divisor (p).
(4) (R, (p)) is a discrete valuation ring.
Proof. (1) Calculate v(±1) = 0, so ±1 ∈ R. If v(x), v(y) ≥ 0 then v(xy) ≥ 0
and v(x + y) ≥ 0, so x, y ∈ R ⇒ xy, x − y ∈ R. Therefore R is a subring. (2)
v(x) = 0 ⇒ v(x−1 ) = 0. (3), (4) If v(p) = v(q) = 1 then pq −1 is a unit, so
(p) = (q). If 0 6= x ∈ R with v(x) = n > 0 then x = (xp−1 )p ∈ (p), so (p) is the
unique maximal ideal.
10.2.8. Definition. The ring in 10.2.7 is the discrete valuation ring of v.
10.2.9. Proposition. Let (R, (p)) be a discrete valuation ring with fraction field
K. The map
v : K\{0} → Z, upn 7→ n
is a valuation and R is the discrete valuation ring of v.
Proof. By 10.2.4 v is well defined. Calculate upm vpn = uvpm+n and if m ≤ n,
upm +vpn = (u+vpn−m )pm to see that v is a valuation. Clearly R is the valuation
ring of v.
10.2.10. Proposition. Let (R, (p)) be a discrete valuation ring. A finite module M
has decomposition
M = R/(pn1 ) ⊕ · · · ⊕ R/(pnk ) ⊕ Rn
where n1 ≥ · · · ≥ nk > 0.
Proof. This is a case of 10.1.12.
10.2.11. Example. Let (k[[X]], (X)) be the power series ring over a field. A finite
module is isomorphic to
k[[X]]/(X n1 ) ⊕ · · · ⊕ k[[X]]/(X nk ) ⊕ k[[X]]n
where n1 ≥ · · · ≥ nk > 0.
10.3. DEDEKIND DOMAINS
143
(1) Let K be a field. Show that the subring K[[X 2 , X 3 ]] ⊂ K[[X]]
is not a discrete valuation ring.
10.2.12. Exercise.
10.3. Dedekind domains
10.3.1. Definition. A noetherian domain R, which is not a field, is a Dedekind
domain if all local rings RP at nonzero prime ideals are discrete valuation rings.
10.3.2. Proposition. Let R be a Dedekind domain.
(1) Any nonzero prime ideal is maximal.
(2) If U ⊂ R is multiplicative, then U −1 R is a field or a Dedekind domain.
Proof. (1) Prime ideals in a nonzero prime ideal P correspond to prime ideals in
the discrete valuation ring RP . So (0), P are the only primes and P is maximal.
(2) This is clear from 5.2.13.
10.3.3. Proposition. Let R be a noetherian domain which is not a field. The following conditions are equivalent:
(1) R is a Dedekind domain.
(2) Every nonzero proper ideal in R is a product of finitely many maximal ideals.
(3) `R (P/P 2 ) = 1 for all maximal ideals.
Proof. (1) ⇒ (2): By 9.6.4 the primary ideals P (n) = P n . Conclusion by 9.6.5
and Chinese remainders 1.4.2. (2) ⇒ (1): Assume (R, P ) is local. By Nakayama’s
lemma 6.4.1 there is p ∈ P \P 2 . Since (p) = P n it follows that (p) = P and
R is a discrete valuation ring. (1) ⇔ (3): 10.2.3 and the isomorphism P/P 2 '
P RP /P 2 RP .
10.3.4. Proposition. Let R be Dedekind domain.
(1) If R is a unique factorization domain then it is a principal ideal domain.
(2) If R has only finitely many maximal ideals then it is a principal ideal domain.
Proof. (1) Any nonzero prime ideal is principal. Conclusion by 10.3.3. (2) Let
P, P2 . . . , Pk be the finitely many maximal ideals. Choose a ∈ P \P 2 ∪P2 · · ·∪Pk ,
5.1.3. Then (a) is P primary. By 10.3.3 (a) = P n , so (a) = P . As all maximal
ideals are principal, conclusion by 10.3.3.
10.3.5. Proposition. Let R be Dedekind domain. An ideal I is generated by at
most two elements.
Proof. Let Ass(R/I) = {P1 , . . . , Pk } and U = R\P1 ∪ · · · ∪ Pk , then by 10.3.4
U −1 R is a principal ideal domain. By 10.1.6 U −1 R ' U −1 I, so choose by 8.2.9
a homomorphism f : R → I such that U −1 f is an isomorphism. Then f is
injective and the ideal f (R) = (a) ⊂ I satisfies: Pi ∈
/ Supp(I/(a)) for any i.
Therefore Ass(I/(a)) ∩ V (I) = ∅. Let Q1 , . . . , Qm ∈ Ass(I/(a)) and choose
b ∈ I\Q1 ∪ · · · ∪ Qm . By 9.3.9 b is a nonzero divisor on I/(a) and therefore bI/(a)
is an isomorphism as I/(a) has finite length, 7.2.6. It follows that I = (a, b).
10.3.6. Theorem. Let R be Dedekind domain.
(1) A torsion free module is flat.
(2) A finite torsion free module is projective.
(3) Any ideal is projective.
144
10. DEDEKIND RINGS
(4) Let F be a finite torsion free module. Then there is a number n and an ideal
I such that
F ' Rn ⊕ I
(5) A finite torsion module has finite length.
Proof. (1), (2), (3), (5) These follow from 10.1.6, 6.5.16. (4) Let R have fraction
field K and assume rankK F ⊗R K = n + 1. Choose a nonzero homomorphism
F → R and get by induction on n, F ' I1 ⊕ · · · ⊕ In+1 for nonzero ideals Ij
in R. It suffices to treat the case n = 1. Let Ass(R/I1 ) = {P1 , . . . , Pk } and
U = R\P1 ∪ · · · ∪ Pk , then by 10.3.4 U −1 R is a principal ideal domain. By 10.1.8
U −1 I2 ' U −1 R, so choose by 8.2.9 a homomorphism f : I2 → R such that
U −1 f is an isomorphism. Then f is injective and the ideal f (I2 ) ⊂ R satisfies:
Pi ∈
/ Ass(R/f (I2 )) for any i. It follows that I1 + f (I2 ) = R. Conclusion by a
split exact sequence
0 → I → I1 ⊕ I2 → R → 0
10.3.7. Proposition. Let R be a Dedekind domain. A finite module M decomposes
M = T (M ) ⊕ F
as a direct sum of the torsion submodule and a finite torsion free submodule F .
Proof. By 10.3.6 the projection M → M/T (M ) splits.
10.3.8. Proposition. Let R be a Dedekind domain. A finite torsion module M has
a decomposition
M ' R/P1n1 ⊕ · · · ⊕ R/Pknk
where P1 , . . . , Pk are not necessarily distinct maximal ideals.
Proof. From 9.5.7 M ' ⊕P MP and by 10.2.10 the RP torsion module MP '
⊕i RP /P ni RP ' ⊕i R/P ni .
10.3.9. Theorem. Let R be a Dedekind domain and M a finite module. Then M
has a decomposition
M = R/P1n1 ⊕ · · · ⊕ R/Pknk ⊕ Rn ⊕ Q1 · · · Ql
where P1 , . . . , Pk , Q1 , . . . , Ql are not necessarily distinct maximal ideals.
10.3.10. Example. R = Z[X]/(X 2 −5) is a noetherian domain. The ideal (2, X +
1) is a prime ideal in Z[X] and X 2 −5 = (X +1)2 −2(X +1)−4 ∈ (2, X +1), so
this corresponds to a prime ideal P ⊂ R. Calculate P/P 2 = (2, X +1)/(4, 2(X +
1), (X +1)2 , X 2 −5) = (2, X +1)/(4, 2(X +1), (X +1)2 ) which has a nontrivial
submodule (2, (X +1)2 )/(4, 2(X +1), (X +1)2 ). P/P 2 is not simple, so by 10.3.3
R is not a Dedekind domain.
√
(1) √
Show that the ring Z[ −5] is a Dedekind domain.
(2) Show that the ring Z[ 5] is not a Dedekind domain.
10.3.11. Exercise.
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Index
(B), 11
(R, P ), 75
(aij ), 89
(xi ), 89
1M , 21
1x , 27
A/B, 9
Aij , 89
I : J, 11
IJ, 11
IM , 23
IS, 11
J ∩ R, 11
M (P ), 77
M/N , 24
M [ u1 ], 67
M ∨ , 33
MP , 77
N 0 ∩ M , 133
N : L, 27
P -primary, 132
P (n) , 136
R/I, 11
RY , 23
R[B], 17
R[X], 16
R[[X]], 20
R[ u1 ], 64
RP , 75
T (M ), 139
U −1 M , 65
U −1 R, 63
U −1 f , 66
V (B), 123
XB , 123
Ann, 27
Ass, 128
Cok, 25
HomR , 30
Im, 25
Ker, 11, 25
Spec, 123
Supp, 126
det, 89
C, 10
Fp , 19
N, 10
Q, 10
R, 10
Z, 10
ι, 63
`R , 102
⊕, 28
⊗, 35
⊗
QR , 34
, 28
rank
√ R , 91
I, 14
aM , 22
b|a, 15
c(f ), 71
char, 12
eα , 29
ei , 87
evx , 33
f 0 , 18
f (P ), 77
f ∨ , 33
fP , 77
k(P ), 75
o(f ), 20
v :, 142
a
φ, 123
0-sequence, 41
abelian group, 9
addition, 9
alternating:, 90
annihilator, 27
artinian module, 105
artinian modules, 105
artinian ring, 107
artinian rings, 107
ass of modules, 128
associated map, 123
associated prime ideal, 128
associative, 9, 21
basis, 29
147
148
bilinear, 21
binomial formula, 10
canonical homomorphism, 65
canonical ring homomorphism, 63
Cayley-Hamilton’s theorem, 91
change of ring, 37
change of rings, 37
characteristic, 12
Chinese remainder theorem, 14
Chinese remainders, 14
coefficient, 16, 20
cofactor matrix, 89
cokernel, 25
colon ideal, 11, 27
comaximal ideals, 14
commutative, 9
composition series, 102
constants, 16
content of polynomial, 71
contracted ideal, 11
contravariant, 31
cosets, 9
Cramer’s rule, 90
decomposable, 30
decomposition of ideals, 136
decomposition of modules, 134
Dedekind domain, 143
Dedekind domains, 143
degree, 16
derivative, 18
determinant, 89
direct product, 28
direct sum, 28, 29
discrete valuation ring, 141
discrete valuation ring of v, 142
discrete valuation rings, 141
distributive, 9
divisible module, 58
division, 15
domain, 10
dual homomorphism, 33
dual module, 33
embedded prime, 130
essential extension, 59
evaluation, 33
evaluation map, 17
exact functor, 50, 51
exact sequence, 41
exact sequences, 41
exactness of fractions, 67
exactness of hom, 50
exactness of tensor, 53
extended ideal, 11
factor group, 9
INDEX
factor module, 24
factor ring, 11
faithfully flat, 83
faithfully flat ring homomorphism, 84
faithfully flat ring homomorphisms, 83
field, 10
field extension, 19
fields, 19
finite field extension, 19
finite ideal, 11
finite length, 102
finite module, 87
finite modules, 87
finite presented module, 94
finite presented modules, 94
finite ring extension, 99
finite ring homomorphism, 99
finite ring homomorphisms, 99
finite type ring, 17
finite type rings, 117
finitely generated ring, 17
five lemma, 48
flat module, 60
flat modules, 60
flat ring homomorphism, 81
flat ring homomorphisms, 81
fraction field, 64
free module, 29
free modules, 89
Frobenius homomorphism, 12
functor, 31
Gauss’ lemma, 71
generated, 23
going-down, 85
going-up, 100
Gorenstein ring, 112
greatest common divisor, 16
Hilbert’s basis theorem, 117
homomorphism, 9, 21
homomorphism module, 30
homomorphism modules, 30
homomorphism modules of fractions, 70
ideal, 11
ideal generated by, 11
ideal product, 11
ideals, 11
idempotent, 15
identity, 9, 21
identity isomorphism, 9, 21
image, 25
indecomposable, 30
induced module, 39
injections, 28
injective envelope, 59
injective module, 56
INDEX
149
injective modules, 56
irreducible component, 125
irreducible element, 15
irreducible principal ideal, 15
irreducible space, 125
irreducible subset, 125
isomorphism, 9, 21
noncommutative ring, 9
nontrivial idempotent, 15
nonzero divisor, 10, 22
normed:, 90
notherian modules, 113
Jacobson radical, 77
polynomial, 16
polynomial ring, 16
polynomials, 16
power series, 20
power series ring, 20
power series rings, 118
primary decomposition, 134
primary ideal, 132
primary modules, 132
primary submodule, 132
prime divisor, 15
prime element, 15
prime fields, 19
prime filtrations of modules, 121
prime ideal, 13
prime ideals, 13, 73
principal ideal, 11
principal ideal domain, 15
principal ideal domains, 139
principal open subsets, 123
product ring, 10
projection, 9, 24
projections, 28
projective module, 55
projective modules, 55
proper ideal, 11
kernel, 11, 25
kernel and cokernel, 25
Krull’s intersection theorem, 117, 120
Krull’s theorem, 73
leading coefficient, 16
least common multiple, 16
left exact contravariant functor, 50
left exact functor, 50
length, 102
linear map, 22
local artinian ring, 110
local parameter, 141
local ring, 75
local ring homomorphism, 75
localization, 109
localization of modules, 77
localization of noetherian rings, 120
localization of rings, 75
localized homomorphism, 77
localized module, 77
localized ring, 75
locally free module, 79
maximal ideal, 13
minimal prime, 126
minimal prime ideal, 74
minor, 89
module, 21
module of fractions, 65
modules and homomorphisms, 21
modules of fractions, 65
monic polynomial, 16
monomial, 16
multilinear:, 89
multiplication, 9
multiplication of principal ideals, 15
multiplicative subset, 63
multiplicity, 18
Nakayama’s lemma, 93
natural homomorphism, 31
natural isomorphism, 31
negative, 9
nilpotent, 14
nilradical, 14
noetherian module, 113
noetherian ring, 115
noetherian rings, 115
order, 20
radical, 14
rank, 91
reduced, 14
reduced primary decomposition, 134
reflexive module, 33
residue field, 75
residue homomorphism, 77
restriction of scalars, 22
retraction, 43
right exact contravariant functor , 50
right exact functor, 50
ring, 9
ring extension, 9
ring generated, 17
ring of fractions, 63
rings, 9
rings of fractions, 63
root, 18
roots, 18
scalar multiplication, 21, 22
section, 43
semi-local ring, 128
150
short exact sequence, 43
simple module, 101
simple modules, 101
simple root, 18
snake homomorphism, 46
snake lemma, 47
spectrum, 123
split exact sequence, 44
standard basis, 29
subfield, 19
subgroup, 9
submodule, 21
submodule generated, 23
submodules and factor modules, 23
subring, 9
sum and product, 28
support, 126
support of modules, 126
symbolic power, 136
tensor modules of fractions, 69
tensor product, 34
tensor product modules, 34
tensor product ring, 39
the length, 102
The local-global principle, 79
the polynomial ring is factorial, 71
the snake lemma, 45
torsion element, 139
torsion free module, 139
torsion module, 139
torsion submodule, 139
total ring of fractions, 64
uniformizing parameter, 141
unique factorization, 15
unique factorization domain, 15
unit, 10
valuation, 142
vector space, 22
windmill lemma, 49
Zariski topology, 123
zero, 9
zero divisor, 10, 22
zero ideal, 11
zero module, 21
zero submodule, 21
INDEX