Download 3. Localization.

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MODULES 3.1
27. februar 2002
3. Localization.
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(3.1) Construction. Let S be a multiplicatively closed subset of the ring A. For
every A-module M we define a relation !!∼ on the cartesian product M × S by
(x, s) ∼ (y, t) if there is an element r ∈ S such that r(tx − sy) = 0 in M . It is clear
that the relation ∼ is reflexive, that is x ∼ x, and symmetric, that is x ∼ y implies
that y ∼ x. It is transitive because if (x, s) ∼ (x0 , s0 ) and (x0 , s0 ) ∼ (x00 , s00 ) there are
elements t, t0 in S such that t(s0 x − sx0 ) = 0 and t0 (s00 x0 − s0 x00 ) = 0. Then we have
that tt0 s0 (s00 x − sx00 ) = tt0 s0 s00 x − tt0 s0 sx00 = t0 s00 tsx0 − tst0 s00 x0 = 0, and consequently
that (x, s) ∼ (x00 , s00 ).
Let !!S −1 M = M × S/∼ be the residue classes of M × S modulo the equivalence
relation ∼. The class of the element (x, s) we denote by x/s. There is a canonical
map!!
iSM : M → S −1 M
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defined by iSM (x) = x/1.
On the set S −1 M there is a unique addition such that S −1 M becomes a group and
such that the canonical map iSM is a group homomorphism. The sum of two elements
x/s and y/t in S −1 M is defined by x/s + y/t = (tx + sy)/st. We have that the
addition is independent of the choice of representative (x, s) for the class x/s because
if x/s = x0 /s0 there is an element r ∈ S such that r(s0 x − sx0 ) = 0. Consequently we
have that r(s0 t(tx + sy) − st(tx0 + s0 y)) = t2 rs0 x + rs0 tsy − rst2 x0 − rsts0 y = 0, and
thus (tx + sy)/st = (tx0 + s0 y)/s0 t. Symmetrically the addition is independent of the
choice of representative (y, t) of the class y/t. It is easily checked that S −1 M with
this addition becomes an abelian group with 0 = 0/1.
We define the product of an element f /s ∈ S −1 A with an element x/t ∈ S −1 M
by (f /s)(x/t) = (f x)/(st). A simple calculation shows that the multiplication is
independent of the choice of representatives (f, s) and (x, t) of the classes f /s, respectively x/t. In particular we obtain a multiplication on S −1 A and it is easily seen
that this multiplication toghether with the group structure on S −1 A makes S −1 A
into a ring. Moreover, with this ring structure the above operation of S −1 A on S −1 M
makes S −1 M into a (S −1 A)-module.
The canonical map!!
iSA : A → S −1 A
that maps an element f in A to f /1 is a ring homomorphism.
(3.2) Definition. We call S −1 M the localization of M by the multiplicative set S.
(3.3) Proposition. Let A be a ring and S a multiplicatively closed subset. Moreover, let M be an A-module and N an S −1 A-module. For every homomorphism
modules3
u : M → N[iSM ]
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of A-modules there is a unique S −1 A-module homomorphism
v : S −1 M → N[iSM ]
such that u = viSM .
The canonical map iSA : A → S −1 A has the universal property:
For every homomorphism of rings ϕ : A → B where ϕ(s) is invertible in B for
every element s ∈ S, there is a unique ring homomorphism χ : S −1 A → B such that
ϕ = χiSA .
Proof. If v exists we have, for all x ∈ M and s ∈ S, that v(x/s) = v((1/s)iSM (x)) =
(1/s)v(iSM (x)) = (1/s)u(x). Hence v is uniquely determined if it exists.
To show that v exists we let v(x/s) = (1/s)u(x) for all x ∈ M and s ∈ S. This
definition is independent of the choise of representative (x, s) for the class x/s because
if x/s = y/t with y ∈ M and t ∈ S there is an r ∈ S such that u(r(tx − sy)) =
ru(tx − sy) = 0. Hence we have that r(t(u(x) − su(y)) = 0 in N and consequently
that u(x)/s = u(y)/t in the S −1 A-module N . It is clear that v is an S −1 A-module
homomorphism and that u = viSM .
Finally when ϕ : A → B is a ring homomorphism such that ϕ(s) is invertible in
B for all s ∈ S we have that B is an S −1 A-module by the multiplication (f /s)g =
ϕ(f )ϕ(s)−1 g for all f /s ∈ S −1 A and g ∈ B. It is easily checked that the definition is
independent of the representative (f, s) of the element f /s and that B becomes an
S −1 A-module. Hence it follows from the first part of the Proposition that we have a
map χ : S −1 A → B of S −1 A-modules, and it is clear that χ is a ring homomorphism.
(3.4) Remark. The universal property characterizes iSA : A → S −1 A up to an
isomorphism of rings. In fact let ψ : A → T be a homomorphism of rings with the
same universal property as iSA . That is, for each homomorphism of rings ϕ : A → B
with ϕ(s) invertible in B for all s ∈ S there is a unique homomorphism τ : T → B
such that ϕ = τ ψ. Then the universal properties give unique ring homomorphisms
ω : S −1 A → T and τ : T → S −1 A such that ωiSA = ψ and τ ψ = iSA . Hence we have
that iSA = τ ωiSA and ψ = ωτ ψ. By unicity, we obtain that τ and ω are inverse maps.
(3.5) Example. Let S = Z \ {0}. Then S −1 Z are the rational numbers Q.
(3.6) Example. Let A be a ring and S a multiplicatively closed subset of A containing 0. Then S −1 A = 0.
(3.7) Example. Let A = Z/6Z and let S = {1, 2, 22 , . . . }. Then we have that
S −1 A = Z/3Z. The map iSA : Z/6Z → Z/3Z coincides with the canonical residue
map of Z/6Z modulo the ideal 3Z/6Z. In particular we have that iSA is not injective.
(3.8) Remark. Let A be a ring and let S be a multiplicatively closed subset that
consist of non-zero divisors different from 0. Then the map iSA : A → S −1 A is
injective. We often identify A with its image by iSA . When A is an integral domain
and S = A \ {0} we have that S −1 A is a field.
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(3.9) Definition. The total quotient ring, or total fraction ring of a ring A is the
localization S −1 A of A in the multiplicative set S consisting of all non-zero divisors
different from 0. When A is an integral domain we call the field S −1 A the quotient
field, or field of fractions of A.
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(3.10) Notation. Let f be an element of A. The set S = {1, f, f 2 , . . . , } is a
multiplicatively closed subset of A. We write !!S −1 M = Mf . Let p be a prime ideal
of A. Then the set T = A \ p is a multiplicatively closed subset of A. We write !!
T −1 M = Mp . The Ap -module Mp is called the localization of M at p. Moreover we
write iSM = ifM and iTM = ipM .
(3.11) Proposition. Let A be a ring and S a multiplicatively closed subset. For
every prime ideal p in A that does not intersect S we have that !!pS −1 A = {f /s ∈
S −1 A : f ∈ p} is a prime ideal in S −1 A. The correspondence that maps p to pS −1 A
is a bijection between the prime ideals in A that do not intersect S and the prime
ideals of S −1 A. The inverse correspondence associates to a prime ideal q in S −1 A
the ideal (iSA )−1 (q) in A.
Proof. Let q be a prime ideal in S −1 A. It is clear that (iSA )−1 (q) is a prime ideal in
A that does not intersect S.
Let p be a prime ideal in A that does not intersect S. If (f /s)(g/t) ∈ pS −1 A there
is an r ∈ S such that rf g ∈ p. Since r ∈
/ p we have that f or g are in p, and thus
that f /s or g/t is in pS −1 A. Moreover we have that (iSA )−1 (pS −1 A) = p since, if
iSA (f ) = g/t with g ∈
/ p, then there is an r ∈
/ p such that r(tf − g) = 0 in A. We
obtain that rtf ∈
/ p, and thus that f ∈
/ p.
It remains to prove that if p = (iSA )−1 (q) then pS −1 A = q. However it is clear
that pS −1 A ⊆ q. Conversely if f /s ∈ q we must have that f ∈ p.
(3.12) Corollary. Let p be a prime ideal in the ring A. Then the localization Ap
of A at p is a local ring with maximal ideal pAp .
Proof. In this case S = A \ p so p is maximal among the ideals in A that do not
intersect S.
(3.13) Remark. Let b be an ideal in S −1 A and let a = (iSA )−1 (b). Then, b =
aS −1 A = {f /s ∈ S −1 A : f ∈ a, s ∈ S}. It is clear that aS −1 A ⊆ b. Conversely, when
f /s ∈ b we have that f /1 ∈ b and consequently f ∈ a. Hence f /s = (f /1)(1/s) ∈
aS −1 A.
(3.14) Proposition. There is a canonical isomorphism of S −1 A-modules
M ⊗A S −1 A → S −1 M
(3.14.1)
that is uniquely determined by mapping x ⊗A (f /s) to (f x)/s for all f ∈ A, s ∈ S
and x ∈ M .
→
Proof. It follows from the explicit description of the map (3.14.1) that it is a map of
S −1 A-modules if it exists.
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To prove the existence we consider the map M × S −1 A → S −1 M that maps
(x, f /s) to (f x)/s. It is clear that this map is A-bilinear. Consequently we obtain
an A-linear map M ⊗A S −1 A → S −1 M that maps x ⊗A f /s to (f x)/s. It is clear
that this map is an (S −1 A)-homomorphism.
In order to show that the map is an isomorphism we construct an inverse S −1 M →
M ⊗A S −1 A by mapping x/s to x⊗A 1/s. The latter map is independent of the choice
of representative (x, s) of the class of x/s. In fact if x/s = y/t there is an r ∈ S
such that r(tx − sy) = 0 in A. We obtain that x ⊗A (1/s) = x ⊗A ((rt)/(rst)) =
rtx ⊗A (1/(rst)) = rsy ⊗A (1/(rst)) = y ⊗A ((rs)/(rst)) = y ⊗A (1/t).
It is clear that the two maps are inverses of each other.
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(3.15) Homomorphisms. Let S be a multiplicatively closed subset of A, and
let u : M → N be a homomorphism of A-modules. There is a canonical map of
S −1 A-modules:!!
S −1 u : S −1 M → S −1 N
that maps x/s to u(x)/s for all s ∈ S and x ∈ M . The map is independent of the
choice of representative (x, s) of the class x/s because if x/s = y/t there is an r ∈ S
such that r(tx − sy) = 0, and thus u(x)/s = u(y)/t. It follows from the explicit form
of S −1 u that it is an S −1 A-module homomorphism.
(3.16) Remark. When v : N → P is a homomorphism of A-modules we have that
S −1 (vu) = S −1 vS −1 u, and S −1 idM = idS −1 M . In other words, the correspondence
that maps an A-module M to the (S −1 A)-module S −1 M is a covariant functor from
A-modules to (S −1 A)-modules.
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(3.17) Notation. Let f be an element of A and let S = {1, f, f 2 , . . . , }. Moreover
let p be a prime ideal of A, and let T = A \ p. For every homomorphism u : M → N
we write !!uf = S −1 u and !!up = T −1 u. Moreover we write the canonical maps
!!iSA = ifA and !!iTA = ipA .
(3.18) Proposition. Let A be a ring and S a multiplicatively closed subset. Moreover let
u
v
0 → M0 −
→M −
→ M 00 → 0
be an exact sequence of A-modules. Then the sequence
S −1 u
S −1 v
0 → S −1 M 0 −−−→ S −1 M −−−→ S −1 M 00 → 0
is an exact sequence of S −1 A-modules.
Proof. We first show that S −1 u is injective. If x0 ∈ M 0 and s ∈ S, and S −1 u(x0 /s) =
u(x0 )/s = 0 there is a t ∈ S such that u(tx0 ) = tu(x0 ) = 0. Since u is injective we
have that tx0 = 0 and consequently x0 /s = 0.
It is clear that we have an inclusion Im(S −1 u) ⊆ Ker(S −1 v). We will show that
the opposite inclusion holds. Let x/s ∈ Ker(S −1 v), that is v(x)/s = 0. Then there
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is a t ∈ S such that v(tx) = tv(x) = 0. Since Im(u) = Ker(v) there is an x0 ∈ M 0
such that u(x0 ) = tx. Consequently S −1 u(x0 /(st)) = u(x0 )/(st) = (tx)/(st) = x/s,
and thus x/s ∈ Im(S −1 u).
Finally it is obvious that S −1 v is surjective.
→
(3.19) Remark. We paraphrase Proposition (3.17) by saying that the functor S −1
is exact.
(3.20) Proposition. Let A be a ring and let S be a multiplicatively closed subset.
Moreover let {Mα }α∈I be a collection of A-modules. Then there is a canonical
isomorphism of A-modules
∼
⊕α∈I S −1 Mα −→
S −1 (⊕α∈I Mα )
→
→
→
→
(3.20.1)
such that the composite of the map (3.20.1) with the canonical map vβ : S −1 Mβ →
⊕α∈I S −1 Mα to factor β is the localization S −1 uβ : S −1 Mβ → S −1 (⊕α∈I Mα ) of the
canonical map uβ : Mβ → ⊕α∈I Mα to factor β.
Proof. The canonical map uβ : Mβ → ⊕α∈I Mα gives a map S −1 uβ : S −1 Mβ →
S −1 (⊕α∈I Mα ) and by the universal property of direct products we obtain the map
⊕α∈I (S −1 Mα ) → S −1 (⊕α∈I Mα ) of (3.20.1).
To show that the map is an isomorphism we construct the inverse. The canonical
maps Mα → S −1 Mα for α ∈ I define a homomorphism ⊕α∈I Mα → ⊕α∈I S −1 Mα .
Consequently it follows from Proposition (3.3) that we have a canonical homomorphism S −1 (⊕α∈I Mα ) → ⊕α∈I S −1 Mα and it is clear that this map is the inverse of
the map (3.20.1).
(3.21) Proposition. Let A be a ring and let M be an A-module. The following
conditions are equivalent:
(1) M = {0}.
(2) Mp = {0} for all prime ideals p of A.
(3) Mm = {0} for all maximal ideals m of A.
Proof. (1) ⇒ (2) and (2) ⇒ (3) are clear.
(3) ⇒ (1) Let x ∈ M and let ax = {f ∈ A : f x = 0}. It is clear that ax is an ideal
in A. We shall show that ax = A, and hence in particular that 1x = x = 0. Assume
to the contrary that ax 6⊆ A. Then there is a maximal ideal m of A that contains ax .
Since Mm = 0 we can find an element s ∈ A \ m such that sx = 0. Then s ∈ ax which
is impossible since ax ⊆ m. This contradicts the assumption that ax 6⊆ A. Hence we
have proved that ax = A for all x ∈ M and consequently that M = 0.
(3.22) Proposition. Let A be a ring and u : M → N an A-linear homomorphism.
The following conditions are equivalent:
(1) u is injective, respectively surjective.
(2) up is injective, respectively surjective, for all prime ideal p of A.
(3) um is injective, respectively surjective, for all maximal ideals m of A.
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→
→
→
→
Proof. We prove the equivalence of the conditions for injective maps:
(1) ⇒ (2) It follows from Proposition (?) that condition (2) follows from conditon
(1).
(2) ⇒ (3) This implication is clear.
(3) ⇒ (1) Let L = Ker(u). When um is injective it follows from Proposition (?)
that Lm = 0 for all maximal primes m of A. Hence it follows from Proposition (?)
that L = 0 and thus that u is injective.
Similar arguments show the equivalence of the assertions for surjective maps.
(3.23) Corollary. Let f 6= 0 be an element of A. We have:
(1) If f is not a zero divisor in A then f /1 is not a zero divisor in the localization
Ap of A in p for all prime ideals p of A.
(2) If f /1 is not a zero divisor in Am for all maximal ideals m of A then f is not
a zero divisor in A.
Proof. We have that f is not a zero divisor in A if and only if the multiplication
map fA : A → A is injective, and f /1 is not a zero divisor in Ap if and only if the
multiplication map (f /1)Ap : Ap → Ap is not injective. Hence the Corollary follows
from the Proposition.
(3.24) Exercises.
1. Let K be a field and let K[u, v] be the polynomial ring in the variables u, v with
coefficients in K. Moreover let A = K[u, v]/(uv).
(1) Show that the ideal p = (u)/(uv) is a prime ideal in A.
(2) Describe the localization Ap .
2. Let M and N be A-modules and let S be a multiplicatively closed subset of A.
Show that the S −1 A-modules S −1 (M ⊗A N ) and S −1 M ⊗S −1 A S −1 N are canonically
isomorphic.
3. Let f be a nilpotent element in A, and M an A-module. Determine Mf .
4. For every f ∈ A and every prime ideal p of A we let f (p) be the image of f by the
ip
ϕAp /mp
A
composite map A −→
Ap −−−−−→ Ap /mp . Show that f (p) = 0 for all prime ideals p
if and only if f is contained in the radical r(A) of A.
5. Let ϕ : A → B be a homomorphism of rings, and let S be a multiplicatively
closed subset in A.
(1) Show that T = ϕ(S) is a multiplicatively closed subset of B.
(2) Show that there is a canonical isomorphism between the S −1 A-modules T −1 B
and S −1 B = B ⊗A S −1 A.
6. Let A be a ring and p a prime ideal.
(1) Show that if the local ring Ap has no nilpotent elements different from zero
for all prime ideals p in A then A has no nilpotent elements different from
zero.
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(2) Is it true that if A has no nilpotent elements different from zero then Ap has
no nilpotent elements different from zero for all prime ideals p of A?
7. Let M be a finitely generated A module, and let S be a multiplicatively closed
subset of M . Show that S −1 M = 0 if and only if there is an element s ∈ S such that
sM = 0.
8. Let !!P be the set of all prime number in Z
Q
(1) Show that the map Z → p∈P Z/pZ that sends an integer n to (n, n, . . . ) is
injective.
(2) Show that for all injective maps u : G → H of groups the map u ⊗Z idQ :
G ⊗Z Q → H
Q⊗idZ Q is injective.
(3) Show that ( p∈P Z/pZ) ⊗Z Q is not zero
Q
Q
(4) Show that ( p∈P Z/pZ) ⊗Z Q is not isomorphic to p∈P (Z/pZ ⊗Z Q).
9. Let A be a ring and S a multiplicatively closed subset. Moreover let M be an
A-module. Describe the kernel of the A-module homomorphism M → M ⊗A S −1 A
that maps x ∈ M to x ⊗A 1.
10. Let A 6= 0 be a ring and u : Am → An an A-linear map. Moreover let p be a
minimal prime ideal in A.
(1) Let f1 , f2 , . . . , fm be elements in p. Show that the ideal b in Ap generated by
the elements f1 /1, f2 /1, . . . , fn /1 is nilpotent, that is, we have bm = (0) for
some positive integer m.
(2) Let p be the integer such that bp 6= (0) and bp+1 = (0) in Ap . Show that for
all elements f ∈ bp we have that f s 6= 0 for all s ∈ A \ p, and that fi f t = 0
for some t ∈ A \ p for i = 1, 2, . . . , m.
(3) Show that if u is injective then the map
up : (A/p)m → (A/p)n
is injective, where the A/p-module homomorphism up is defined by
up ((uA/p (f1 ), uA/p (f2 ), . . . , uA/p (fn ))
= (uA/p (u1 (x)), uA/p (u2 (x)), . . . , uA/p (un (x)))
for all x = (f1 , f2 , . . . , fm ) in Am and where u(x) = (u1 (x), u2 (x), . . . , un (x))
in An .
(4) Show that when u is injective then m ≤ n.