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A REVIEW OF COMMUTATIVE RING THEORY
MATHEMATICS UNDERGRADUATE SEMINAR: TORIC VARIETIES
ADRIANO FERNANDES
Contents
1. Basic Definitions and Examples
2. Ideals and Quotient Rings
3. Properties and Types of Ideals
4. C-algebras
References
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1. Basic Definitions and Examples
In this first section, I define a ring and give some relevant examples of rings we have
encountered before (and might have not thought of as abstract algebraic structures.) I
will not cover many of the intermediate structures arising between rings and fields (e.g.
integral domains, unique factorization domains, etc.) The interested reader is referred to
Dummit and Foote.
Definition 1.1 (Rings). The algebraic structure “ring” R is a set with two binary operations + and ·, respectively named addition and multiplication, satisfying
• (R, +) is an abelian group (i.e. a group with commutative addition),
• is associative (i.e. 8a, b, c 2 R, (a · b) · c = a · (b · c)) ,
• and the distributive law holds (i.e. 8a, b, c 2 R, (a + b) · c = a · c + b · c, a · (b + c) =
a · b + a · c.)
Moreover, the ring is commutative if multiplication is commutative. The ring has an
identity, conventionally denoted 1, if there exists an element 1 2 R s.t. 8a 2 R, 1 · a =
a · 1 = a.
From now on, all rings considered will be commutative rings (after all, this is a review
of commutative ring theory...) Since we will be talking substantially about the complex
field C, let us recall the definition of such structure.
Definition 1.2 (Fields). A commutative ring in which every nonzero element has a multiplicative inverse is called a field, i.e. if 8a 2 R there exists b 2 R s.t. a · b = 1.
Date: 02/10/2016.
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ADRIANO FERNANDES
It is trivial to see that the integers form a ring, that integers in modular arithmetic form
a ring, and that functions spaces (with point-wise addition and multiplication) are rings.
The rational, real and complex numbers are fields.
The concept of subsets of a set gives rise to a natural extension to a subring of a ring.
Definition 1.3 (Subring). A subring S of a ring R is any subset of R in which the
operations of addition and multiplication defined in R make S a ring when restricted to S.
Hence, a subring is a ring in itself sitting inside another ring. Transitivity is trivial by
the definition.
Example 1.4. Z
2Z
4Z
···
2n Z
· · · , for n 2 Z, n > 2. This example also
shows we can have an infinite chain of subrings inside a ring.
The rings we will be mostly interested in for this seminar will be Polynomial Rings.
Definition 1.5 (Polynomial Rings). Fix a commutative ring R with identity. Let x be
an indeterminate. P
The polynomial ring R[x] (read R adjoined x) is the set of all formal
sums of the form ni=0 ai xi for some n 0 and each ai 2 R. The formal sum is called a
polynomial in x with coefficients in the ring R.
If an 6= 0, then the polynomial is said to have degree n. The polynomial is monic
if the coefficient of the largest power is the multiplicative unit. Addition
Pn and multii
plication
are
defined
component-wise
in
polynomial
rings.
Let
p(x)
=
i=0 ai x and
Pn
Pn
i
i
q(x) = i=0 bi x be two polynomials. Then, p(x)·q(x) := i=0 (ai +bi )x and p(x)+q(x) :=
Pn Pk
k
k=0 [ i=0 (ai bk i )]x .
The ring over which the polynomials are formed make a substantial di↵erence, as the
following example shows.
Example 1.6. Consider p(x) = x+1. If p(x) 2 Z[x], 2p(x) = 2x+2 and p(x)2 = x2 +2x+1.
If p(x) 2 Z/2Z[x], 2p(x) = 0, and p(x)2 = x2 + 1.
Before proving some essential facts about polynomial rings, we need some more definitions.
Definition 1.7 (Integral Domain). An integral domain is any commutative ring R with
identity in which there are no zero divisors, i.e. 8a, b 2 R, a, b 6= 0, ab 6= 0.
Definition 1.8 (Units of a Ring). An element a 2 R is called a unit of the ring if there
exists an element b 2 R such that ab = 1. The units of a ring form a ring in itself, called
the Ring of Units, denoted R⇥
Now we are ready to state an prove some key facts about polynomial rings.
Proposition 1.9. Let R be an integral domain, and let p(x), q(x) 6= 0, p(x), q(x) 2 R[x].
Then,
(1) deg(pq) = deg(p) + deg(q)
(2) The units of R and R[x] are the same
(3) R[x] is an integral domain.
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Proof. If p(x) and q(x) have leading terms an xn and bm xm respectively, the leading term
of pq(x) is an bm xn+m . Since R has no zero divisors, an bm 6= 0. Hence the (3) and (1) hold.
For (2), suppose pq(x) = 1 ) deg(p) + deg(q) = 0 ) deg(p) = deg(q) = 0, hence both p, q
are elements of R and thus are units in R.
⇤
One can establish operation-preserving mappings between rings. We now focus on such
mappings.
Definition 1.10 (Ring Homomorphism). Let R and S be rings.
(1) A ring homomorphism is a map : R 7! S satisfying (a + b) = (a) + (b) and
(ab) = (a) (b), 8a, b 2 R
(2) The kernel of a ring homomorphism is the set of elements in R mapping to the
addition identity of S, i.e. ker = {r 2 R| (r) = 0S }
(3) A ring isomorphism is a bijective ring homomorphism.
Example 1.11.
Let n : C[x] 7! Cbedef inedas (p(x)) = p(0). This maps a polynomial to its constant
term. Since addition and multiplication of the constant terms are clearly preserved under
this mapping, this is a ring homomorphism. The kernel is the set of all polynomials with
constant term 0 in the complex field.
Proposition 1.12. Let
: R 7! S be a ring homomorphism. Then,
(1) The image of is a subring of S;
(2) ker is a subring of R and is closed under multiplication by elements of R.
Proof. For (1), if s1 , s2 2 im ) s1 = (r1 ) , s2 = (r2 ) for some r1 , r2 2 R. Since is a
homomorphism, s1 s2 = (r1 r2 ) and s1 s2 = (r1 r2 ), so s1 s2 , s1 s2 2 S. For (2),
if a , b 2 ker ) (a) = (b) = 0 ) (ab) = (a b) = 0, so the kernel is closed under
subtraction and multiplication. Now, for any r 2 R, (ra) = (r) (a) = (r)0 = 0 =
0 (r) = (a) (r) = (ar), so indeed the kernel is closed under multiplication by elements
of R.
⇤
Kernels are examples of structures appearing more generally in rings, so called ideals.
We now focus on these structures.
2. Ideals and Quotient Rings
Definition 2.1 (Ideals). Let R be a ring, r 2 R and I ⇢ R. I is an ideal of R if I is a
subring of R that is closed under multiplication by the elements of R, i.e. if a 2 I, then
ra, ar 2 I; one writes rI ⇢ I8r 2 R.
It turns out that there is a vital connection between ideals and ring homomorphisms.
Ideals are kernels for some ring homomorphism of the ring in which they sit in. The cosets
of any element of R, say r +I, r 2 R, I an ideal in R form a ring in itself if we appropriately
define addition and multiplication of cosets for the set of cosets. Ideals are in fact what we
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ADRIANO FERNANDES
require for those operations as we here present to make the set of cosets into a ring. We
define addition and multiplication for cosets of elements of R by an ideal I as follows:
(r + I) + (s + I) = r + s + I
(r + I)(s + I) = rs + I
It is trivial to show that the distributive law holds whence operations are defined as above.
Let r, s, t 2 R and I an ideal in R. Then
(r + I)[(s + I) + (t + I)] = (r + I)[s + t + I]
= r(s + t) + I = (rs + rt) + I
= (rs + I) + (rt + I) = (r + I)(s + I) + (r + I)(t + I)
So it seems that we can indeed form a ring by considering the cosets of elements of a ring
by the ideals of a ring. The discussion above proved the following
Proposition 2.2. Let R be a ring and I an ideal. If addition and multiplication are defined
as above, the set of cosets of elements of R by the ideal I form a ring. Conversely, any
subgroup I of R in which the above operations are well-defined form an ideal of R.
I skip the proof of the converse direction for succinctness.
Definition 2.3 (Quotient Rings). Let R be a ring and I be an ideal of it. The ring R/I
with operations defined above form the quotient ring of R, whose elements of the form
r + I for r 2 R.
The following theorem summarizes the present discussion:
Theorem 2.4 (The First Ring Isomorphism Theorem). Let : R 7! S be a homorphism
of rings. Then, ker is an ideal of R, im is a subring of S and R/ker is isomorphic as
a ring to (R). Alternatively, if I is any ideal of R, then ⇡ : R 7! R/I is a surjective ring
homomorphism with kernel I, called the natural projection of R onto R/I. Hence, every
ideal is the kernel of some ring homomorphism and vice-versa.
Ideals have set operations rigorously defined:
Definition 2.5 (Sum and Product of Ideals). Let I, J be two ideals of a ring R. Define
(1) the sum of I and J as I + J = {i + j|i 2 I, j 2 J};
(2) the product of I and J as IJ, the set of all finite sum of elements of the form ij with
i 2 I, j 2 J;
(3) the n power of I, denoted I n , the set of all finite sums of elements of the form i1 i2 · · · in
with ik 2 I , 8k = 1, ..., n
We finish this first section on ideals with the idea of a generator for an ideal.
Proposition 2.6 (Ideals and Generators). Let R be a ring and let I be an ideal of R. A
set J ⇢ R generates an ideal by setting
\
(J) = {I|J ⇢ I, I ⇢ R}
Namely, the intersection of arbitrarily many ideals is an ideal in itself.
A REVIEW OF COMMUTATIVE RING THEORYMATHEMATICS UNDERGRADUATE SEMINAR: TORIC VARIETIES
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Proof. We need only T
to show the last claim. So let Ik be arbitrarily many ideals of a ring
R, and consider I = Ik . Let i 2 I ) i 2 Ik 8k. Hence ri 2 Ik , 8k, thus ri 2 I.
⇤
To better understand the concept of a set generator for an ideal, notice that (J) is the
smallest ideal containing
Pn J as a set, and its elements are finite R-linear combinations of
elements of J, i.e.
i=1 ri ji for ri 2 R, ji 2 J 8i. The following example would be one of
interest going forward in this seminar. Naming-wise, an ideal generated by a single element
is called a Principal Ideal, whereas an ideal generated by a finite set is called a Finitely
Generated Ideal.
Example 2.7. Consider the two-indeterminate polynomial ring over the complex field
C[x, y]. Consider further the polynomials without constant terms. Clearly, any polynomial
in C[x, y] can be made into a “zero” constant polynomial by multiplying the entire polynomial by one of the indeterminates. Hence, we can form an ideal I of C[x, y] given by the
set of all zero constant polynomials. A representative element of such ideal takes the form
xP (x, y) + yQ(x, y), for P, Q 2 C[x, y]. Thus, (x, y) = I.
3. Properties and Types of Ideals
Definition 3.1 (Maximal Ideals). An ideal M in a ring R is called a Maximal Ideal if
M 6= R and no ideal I exists such that M ⇢ I ⇢ R. Equivalently, if M ⇢ I ⇢ R ) M = I.
Definition 3.2 (Prime Ideals). An ideal P in a ring R is called a Prime Ideal if
(1) P 6= R;
(2) if ab 2 R for a, b 2 R, then a 2 P or b 2 P
The following propositions (which we state without proof) make it easy to identify
maximal and prime ideals.
Proposition 3.3. Assume a commutative ring R and an ideal I of R.
(1) I is maximal in R if the quotient ring R/I is a field;
(2) I is prime in R if the quotient ring R/I is an integral domain.
Example 3.4. The quotient ring Z/2Z is a field, so the ideal generated by (2, x) is maximal
in Z[x]. The ideal (x) is not a maximal ideal in Z[x], since Z is not a field. However, (x)
is a prime ideal, since Z is an integral domain.
p
Definition 3.5 (Radical Ideals). An ideal I of R is called a Radical Ideal if I = I =
{r 2 R|rn 2 I for some n > 0}.
The first ring isomorphism theorem preserves the types of ideals. That is, if I is a prime,
maximal or radical ideal in the domain ring, its image in the codomain ring is also a prime,
maximal, or radical ideal, respectively. We can now solve some exercises from Smith et al.
Exercise 2.1.1.
Proof. (1) We want to show first that every maximal ideal is prime. From the proposition,
let R be our ring and M be a maximal ideal of R, so M/R is a field, hence an integral
domain, therefore M is also prime.
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ADRIANO FERNANDES
(2) Now, we show that every prime ideal is radical. So consider M as a prime ideal. I first
show that if r 2 R and n 2 N, n > 0, then rn 2 M ) r 2 M . Proceed by induction.
If n = 1 (base case), there is nothing to show. So let n 1 and rn 2 M ) r 2 M .
n+1 = r n r 2 M ) r 2 M orr n 2 M , in which case r 2 M regardless. Now, let
Then,
pr
p
r 2 M , so rn 2 M for some n > 0, n 2 N. By the above, r 2 M , so I ⇢ I. The
converse inclusion is trivial.
p
(3) The final claim is that a radical ideal
is an ideal. So let I be an ideal. That I ⇢ I is
p
trivial, for a = a1 2 I. Let a, b 2 I ) an 2 I and bm 2 I for some n, m > 0, n, m 2
N. A binomial expansion shows that
(a + b)
m+n
=
m+n
X✓
k=0
◆
m + n k m+n
a b
k
k
If k
n, then ak 2 I. If k < n, then bm+n k 2 I (since I is an ideal). So every
n
n n
term in the binomial expansion
p is in I. Moreover, 8r 2 R, (ra) = r a 2 I (and in
n
particular ( a) 2 R. Hence, I ⇢ I.
⇤
Exercise 2.1.2.
Proof.(1) First we show M is a maximal ideal i↵ R/M is a field. To show ((), suppose
R/M is a field, and J is an ideal of R that properly contains M , i.e. j + M is a nonzero
element of R/M . Hence, there exists an element k+M such that (k+M )(j +M ) = M +1.
Since j 2 J, kj 2 J, so 1 kj 2 M ⇢ J. Then 1 = (1 kj) + kj 2 J, so J=R, for an
ideal generated by unity is the ring itself. For (), suppose M is maximal, and consider
j 2 R M . Let J = {rj + k|r 2 R, k 2 M }, which is an ideal properly containing M .
For M is maximal, R = M , hence 1 + M = jr + M = (j + M )(r + M ), so units can
indeed be found for arbitrary elements.
(2) Now we show that if P is a prime ideal of R, then R/P is an integral domain. For
sufficiency, consider x + P, y + P such that (x + P )(y + P ) = xy + P = P (since P is the
zero in R/P ). Since P is prime, either x 2 P or y 2 P . WLOG, suppose x 2 P . Then
x + P = P , as we wanted to show. For necessity, let R/P be an integral domain, and
let a, b 2 R such that ab 2 P . Then, P = ab + P = (a + P )(b + P ). Since R/P is an
integral domain, a + P = P or b + P = P . WLOG, let a + P = P , hence a 2 P , so P is
prime.
⇤
Exercise 2.1.3.
Proof. First, we want to show that if I ⇢ S is an ideal, any ring map : R 7! S induces
an injective homomorphism of rings R 1 (I) 7! S/I. A representative element of R 1 (I)
is r + 1 (I), so by the structure preserving map, (r + 1 (I)) = (r) + I. So if (r +
1 (I)) = (s +
1 (I)) ) r = s, showing the map is injective. From previous discussion,
we have argued that the pre-image of kernels are also kernels under ring homomorphisms,
which completes the proof
⇤
Exercise 2.1.4.
A REVIEW OF COMMUTATIVE RING THEORYMATHEMATICS UNDERGRADUATE SEMINAR: TORIC VARIETIES
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Proof. For sufficiency, suppose the zero ideal is radical, i.e. {0} = (0) = {r 2 R|rn 2 (0)
for some n > 0}. Hence, 8r 2 R, rn = 0 i↵ r = 0, proving that R is reduced. For necessity,
suppose the ring is reduced, so rn = 0 i↵ r = 0. Thus, the zero ideal is also its radical,
trivially.
⇤
Exercise 2.1.5.
Proof. For necessity, suppose R/I is reduced, so (r + I)n = rn + I = I i↵ r + I = I, i.e.
rn 2 I i↵ r 2 I, so the ideal I is clearly radical. For sufficiency, let I be a radical ideal, and
let i 2 I = sqrtI, so that in = 0 for some n > 0 in R. Thus, in + I = I in R/I if i + I 2 I.
Since I is the zero of R/I, R/I is reduced. Conversely, let i 2 I, and since the ideal is
radical, in 2 I, so in R/I, i + I = I = in + I.
⇤
4. C-algebras
Definition 4.1 (C-algebra). A ring R is a C-algebra if C ⇢ R as a subring.
Every C-algebra is a vector space over R, where vector addition is defined as the addition
in R, and scalar multiplication for a given z 2 C and a vector r 2 R is given by multiplication in R. We extend ideas from previous section to the specific case of C-algebras.
Definition 4.2 (Generators
for a C-algebra). The C-SUBalgebra generated by a subset J
T
of the C-algebra S is {A|J ⇢ A, A ⇢ S, S a C-algebra }.
This subalgebra is the smallest C-subalgebra containing the subset J. Its elements are
all elements of S that can be expressed as polynomials with elements in S and coefficients
in C.
Definition 4.3 (Finitely generated subalgebras). An algebra R is finitely generated if
there exists a finite subset J ⇢ R generating R.
Example 4.4. In the polynomial ring C[x, y], we see a C-algebra, for C ⇢C[x, y] by setting
x = y = 0. It is finitely generated since (x, y) generate C[x, y].
Definition 4.5 (C-algebra Homomorphism). If R, S are two C-algebras, a map : R 7! S
is a C-algebra Homomorphism if it is a ring map that is linear over C, i.e. (zr) = z (r),
8z 2 C, r 2 R.
Example 4.6. Consider the C-algebra
C[x, y]
7! C[z]
x2 + y 3
. The images of the generators must be preserved under the projection homomorphism, so
a candidate for is (x) = z 3 and (y) = z 2 , for these satisfy the original relationships
between ideal generators in the kernel.
References
David S. Dummit, Richard M. Foote, Abstract Algebra, Third Edition, Wiley.
[0] [1] Smith et al., Algebraic Geometry, Springer, 2000