Ring Theory

1. Rings and homomorphisms

After groups, rings are one of the most fundamental objects in algebra. They generalize

objects like the integers, n×n matrices, and function spaces more completely because these

objects carry multiple operations. In the case of integers and matrices, these two operations

are addition and multiplication. With function spaces, we have addition and composition.

Our goal initially will be to understand the basic structure of rings, their ideals (the analog

of normal subgroups) and modules (spaces that rings act on).

In addition, ring theory is an essential tool in algebraic geometry, which translates the study

of geometric objects into the language of algebra. In this course we will see the beginnings

of this correspondence.

Whilst studying rings, it makes sense to also study fields. Loosely, a field is a ring with

a well-defined inverse operation to multiplication (i.e., division). Think: rational, real, or

complex numbers. Field theory is inherently related to group theory via Galois Theory,

which was born out of the problem of solving polynomials by radicals.

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1.1. Basic definitions and examples. We begin by recalling some definitions from group

theory in order to formally define a ring.

Definition 1. Let G be a set and · a binary operation under which G is closed. If the binary

operation is associative, we say (G, ·) is a semigroup, that is,

(a · b) · c = a · (b · c) for alla, b, c ∈ G.

If in addition there exists an identity element e ∈ G (e · a = a = a · e for all a ∈ G) then

(G, ·) is said to be a monoid. If in addition there exists an inverse element for every element

in G (a−1 · a = e = a · a−1 for all a ∈ G.) then (G, ·) is said to be a group. The group (G, ·)is said to be abelian if · is commutative, a · b = b · a for all a, b ∈ G.

Definition 2. A ring is a set R together with two binary operations (commonly denoted +

and ·) such that

• (R,+) is an abelian group.

• (R, ·) is a semigroup.

• The left and right distributive laws hold:

a · (b+ c) = a · b+ a · c and (a+ b) · c = a · c+ b · c.

One reason I prefer Hungerford over a book like Artin is that this is the right definition of a

ring. Then again, I’m a noncommutative ring theorist. Note that there is no assumption in

the above definition there is no assumption that the multiplication operation is commutative.

Making this assumption rules out important examples like matrices and function spaces.

Definition 3. A ring (R,+, ·) is said to be commutative if a · b = b · a for all a, b ∈ R. If

(R, ·) is a monoid, then R is said to be a ring with identity or a ring with unit.

We will now consider many examples of rings.

Example 1. (1) The integers Z with the operations of addition and multiplication is

a commutative ring with unit. The even integers is also a commutative ring but

without unit. The integers mod n is also a ring.

(2) The rational numbers Q with the operations of addition and multiplication is a com-

mutative ring with unit. Note that every nonzero element in Q has a multiplicative

inverse, which is not the case with the integers.

(3) The set of polynomials in the variable x with coefficients in R is a commutative ring

with unit (the polynomial ring over R). It is denoted R[x].2

(4) Given an abelian group A and End(A) its set of endomorphisms. Then End(A) is a

ring (possibly noncommutative) with the operations

(f + g)(a) = f(a) + g(a)

(f ◦ g)(a) = f(g(a))

for all a ∈ A, f, g ∈ End(A).

(5) The set of n × n matrices over R, denoted Mn(R) with the operations of addition

and matrix multiplication is a (noncommutative if n > 1) ring with identity. In this

ring, some elements are invertible, those with nonzero determinant, but others are

not. The subset of invertible matrices is denoted GLn(R), and this forms a group

under matrix multiplication (called the general linear group) but it is not a ring.

Remark. Now’s as good a time as any to acknowledge the fact that there is what some

would consider an unnecessary hypothesis in our definition of a ring. What we have tech-

nically defined is an associative ring, meaning that the multiplication is associative. There

is a significant branch of mathematics dealing with nonassociative rings. In particular, Lie

Theory deals with such rings, called Lie algebras which have an additional operation of scalar

multiplication over some field, as well as Lie groups. The multiplication is actually a bilinear

bracketing operation, [a, b]. A standard example is the set of n×n matrices with real entries

and trace 0 where [a, b] = ab − ba. We won’t spend time on such examples so the above

definition of a ring is sufficient.

For R to be a ring, (R, ·) need only be a monoid. It is a ring with unit when (R, ·) is a

semigroup. What about when (R, ·) is a group? This leads to the notion of a field.

Definition 4. Let a be a nonzero element in a ring R. The element a is said to be a left zero

divisor (resp. right zero divisor) if there exists a nonzero b ∈ R such that ab = 0 (resp. ba = 0).

A zero divisor is an element which is both a left and right zero divisor. A commutative ring

with identity 1R 6= 0 and no zero divisors is called an integral domain.

In an integral domain the zero property holds. That is, ab = 0 implies a = 0 or b = 0.

Example 2. The matrix

(0 1

1 0

)is a left and right zero divisor in Mn(R).

Example 3. The integers is an example of an integral domain, as well as Zp for p prime.

The integers mod n for n composite is not an integral domain.3

Proposition. A ring R satsifies the left cancellation law (ab = ac and a 6= 0 implies b = c)

if and only if it has left zero divisors.

Proof. Let a, b, c ∈ R with a 6= 0. Then ab = ac⇔ ab− ac = 0⇔ a(b− c) = 0. Hence, if R

has no left zero divisors, then b− c = 0 so b = c. Conversely, left cancellation implies �

Definition 5. The element a is said to be left invertible (resp. right invertible) if there exists

c ∈ R such that ca = 1R (resp. ac = 1R). The element c is called a left inverse (resp. right

inverse) of a. An element is invertible (or a unit) if it has a left and right inverse.

Definition 6. A ring D with identity 1D 6= 0 in which every nonzero element is a unit is

called a division ring. A field is a commutative division ring.

Exercise. Prove that the set of units of a ring R forms a group.

Example 4. The rings Q, R, C, and Zp with p prime are all fields.

Exercise. Verify that Zp, p prime, is an integral domain and every element is invertible

(under multiplication).

Proposition. Every field F is an integral domain.

Proof. Let a ∈ F is nonzero and suppose there exists b ∈ F such that ab = 0. Then a is

invertible so

b = 1F b = (a−1a)b = a−1(ab) = a−10 = 0.

�

1.2. New rings from old ones. We construct several new examples of rings from those

that we already understand. This gives us a way of realizing one of the most important

objects in commutative algebra: the polynomial ring.

Definition 7. Let R be a ring and S a nonempty subset of R. We say S is a subring of R

if it is a ring under the operations inherited from R.

A subring of a ring R may have significantly different structure than that of R. For example,

2Z is a subring of Z but without identity. We saw previously that Q is a field but Z is a

subring of Q and not a field.

Example 5. The center of a ring R, denoted C(R) = {x ∈ R | ax = xa for all a ∈ R}, is a

subring of R.4

Definition 8. Let R and S be rings. The direct product of R and S, denoted R × S, is a

ring whose underlying set is the cartesian product of R and S and operations given by

(r1, s1) + (r2, s2) = (r1 + r2, s1 + s2)

(r1, s1) · (r2, s2) = (r1 · r2, s1 · s2)

where the operations in each coordinate are those from the individual rings.

This should be familiar from the direct product of rings. Once we have discussed homomor-

phisms and the category of rings we can generalize this to arbitrary products of rings.

Exercise. R× S is a commutative ring [resp. ring with identity] if and only if R and S are

commutative [resp. ring with identity].

If R × S is an integral domain, then R and S are integral domains. The converse of this

statement does not hold. In fact, the direct product of two rings is never an integral domain.1

Definition 9. Let R be a ring. Denote by Mn(R) the set of n× n matrices with entries in

R. ThenMn(R) is a ring, possibly noncommutative, with the operations of matrix addition

and matrix multiplication.

Previously we discussed the ring R[x] of polynomials in the variable x with coefficients in

x. The following definition generalizes this idea to any ring R. A quick comment before we

start: this is not the formal definition of a polynomial ring. We will come back to that later.

Definition 10. Let R be a ring with identity. The polynomial ring R[x] is defined by

the underlying set {p0 + p1x + p2x2 + · · · + pdx

d | pi ∈ R} of elements called polynomials

with coefficients in R. The elements xi are formal symbols called indeterminates and satisfy

xi · xj = xi+k. By convention, x0 = 1R, x1 = x, 1Rx = x and 0Rx = 0R. If p ∈ R[x], the

largest d such that pd 6= 0 is known as the degree of p, and denoted deg(p).

The operations are addition and multiplication, defined below. Let a, b ∈ R[x] and write

a = a0 + a1x+ a2x2 + · · ·+ anx

d

b = b0 + b1x+ b2x2 + · · ·+ bmx

d′ .

Let ` = max{d, d′} and k = d+ d′.

Addition is given by

a+ b = (a0 + b0) + (a1 + b1)x+ · · ·+ (a` + b`)x`,

1Thanks to Brad H. for reminding me of this.

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where ai = 0 if i > d and aj = 0 if j > d′.

Multiplication is given by

ab = c0 + c1x+ · · ·+ ckxk

where cn =∑n

i=0 an−ibi.

A final important example of a ring which ties us back into group theory is a group ring.

Loosely, one thinks of this as a way of tying together group and ring structures.

Example 6. Let G be a group (written multiplicatively) and R a ring. We denote the group

ring of R and G by R(G) and describe it as follows.

The underlying set of R(G) is the set of formal sumsn∑i=1

rigi for ri ∈ R, gi ∈ G. Addition is

given byn∑i=1

rigi +n∑i=1

sigi = (ri + si)gi.

Note that, in the above, some of the ri or si may be zero so there is no loss in taking the

sums to be of the same length. Multiplication is given by(n∑i=1

rigi

)(m∑j=1

sjhj

)=

n∑i=1

m∑j=1

(risj)(gihj).

You are encouraged to read through the example on page 117 of Hungerford on the real

quaternions, that is, the group ring R(Q) where Q is the quaternions.

1.3. Ring homomorphisms. As with groups, we should think about homomorphisms as

structure-preserving maps. Hence, a ring homomorphism should preserve the additive and

multiplicative structure of a ring.

Definition 11. Let R and S be rings. A function f : R → S is a homomorphism of rings

provided that

f(a+ b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a, b ∈ R.

Exercise. The class of all rings together with all ring homomorphisms forms a category.

Definition 12. Let f : R→ S be a ring homomorphism. Then f is a

• monomorphism if f is injective;

• epimorphism if f is surjective;

• isomorphism if f is bijective.6

An isomorphism with S = R is an automorphism of R.

Definition 13. Let f : R→ S be a ring homomorphism.

The kernel of f is the set ker f = {f ∈ R | f(r) = 0}.

The image of f is the set im f = {s ∈ S | f(r) = s for some r ∈ R}.

Exercise. Let f : R → S be a homomorphism of rings and a, b ∈ ker f . Prove that

a+ b ∈ ker f and ra ∈ ker f .

The previous exercise proves that ker f is an ideal of the ring R. One can think of this as

the analog of normal subgroups in group theory. We will study ideals in more detail in the

next section.

Example 7. (1) The map

Z→ Zm

k 7→ k

is an epimorphism of rings.

(2) The map

Z3 → Z6

k 7→ 4k

is a (well-defined) epimorphism of rings.

(3) Let f : R→ S be a homomorphism of rings. The map

f : R[x]→ S[x]

n∑i=0

rixi 7→

n∑i=0

f(ri)xi

is homomorphism. Moreover, f is a monomorpism (resp. epimorphism, isomorphism)

if and only if f is.

Definition 14. Let R be a ring. The characteristic of R is defined as

charR = min{n ∈ Z+ | na = 0 for all a ∈ R}

if such an integer exists. Otherwise, R is said to have characteristic zero (charR = 0).

Proposition. Let R be a ring with identity and charR = m. Then charR = min{n ∈ Z+ |n1R = 0}.

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Proof. Let n be the smallest positive integer such that n1R = 0. Since charR = m then

m ≤ n. Then for all a ∈ R,

na = n(1ra) = (n1r)a = 0a = 0.

Hence, n ≤ m so n = m. �

Theorem 8 (Theorem III.1.10). Every ring R may be embedded in a ring S with identity.

The ring may be chosen to be either of characteristic zero or of the same characteristic as R.

Proof. Let S be the additive abelian group R⊕ Z. Define multiplication in S by

(r1, k1) · (r2, k2) = (r1r2 + k2r1 + k1r2, k1k2) for ri ∈ R, ki ∈ Z.

It is clear that (0, 1R) = 1S. Moreover, for any n ∈ Z, we have n(r1, k1) = (nr1, nk1) and

since Z has characteristic zero, nk1 6= 0 for k1 6= 0. Hence, char(S) = 0. Finally, the map

R→ S

r 7→ (r, 0)

is clearly an embedding.

If char(R) = n > 0 then use the same argument but take S = R ⊕ Zn. Then char(S) = n

(check!). �

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2. Ideals

One of our primary goals in group theory was to study the structure of groups. We could

do this by decomposing the group in different ways. The most basic method was to write

G = NH where N and H are disjoint normal subgroups. Hence, the study of normal

subgroups is essential to understanding the structure of the group itself. Our goal will be

similar with the study of ideals in ring theory, and we will see many similarities with normal

subgroups.

Definition 15. A nonempty subset I of a ring R is a left ideal (resp. right ideal) if for all

a, b ∈ I and r ∈ R,

• a, b ∈ I ⇒ a− b ∈ I;

• a ∈ I, r ∈ R⇒ ra ∈ I (resp. ar ∈ I).

I is said to be an ideal (for emphasis, a two-sided ideal) if it is both a left and right ideal.

The definition implies a left or right ideal I of a ring R is a subring of R, but not all subrings

are ideals (e.g., the center of a ring).

If R is commutative, then every left ideal or right ideal is an ideal. That is, there is no

distinction.

Example 9. • Every ring R has two ideals: R itself and the trivial ideal 0.

• For each n ∈ Z, nZ = 〈n〉 is an ideal in Z.

• If f : R→ S is a homomorphism of rings, then ker f is an ideal of R. (Note: im f is

a subring of S but not an ideal). This is important in the study of quotient rings.

• Let D be a division ring. Define Ik ⊂ Mn(D) to be the set of all matrices that have

nonzero entries only in column k. Then Ik is a left ideal of Mn(D) but not a right

ideal. Similarly, the set Jk of all matrices that have nonzero entries only in row k is

a right ideal of Mn(D) but not a left ideal.

• If I is a left ideal of a ring R, then I[x] = {∑akx

k | ak ∈ I} (polynomials with

coefficients in I) is a left ideal of R[x].

• Let R be a ring and a ∈ R. Then Ra = {ra | r ∈ R} is a left ideal of R. Similarly,

aR = {ar | r ∈ R} is a right ideal of R. If a is the center of R, then aR = Ra.

Definition 16. A [left, right, two-sided] ideal I of R is said to be proper if I 6= 0 and I 6= R.

If I is a [left, right, two-sided] ideal of R and 1R ∈ I, then I = R. Consequently, if D is a

division ring then D has no proper [left, right, two-sided] ideals.9

Proposition. If I and J are [left, right, two-sided] ideals of a ring R, then

(1) I ∩ J is a [left, right, two-sided] ideal

(2) I + J is a [left, right, two-sided] ideal.

(3) IJ is a [left, right, two-sided] ideal (IJ = {∑aibi | ai ∈ I, bi ∈ J}).

Proof. (1) Suppose I and J are left ideals and let a, b ∈ I ∩ J . Then a, b ∈ I and a, b ∈ J .

Hence, a−b ∈ I and a−b ∈ J because I and J are left ideals. Hence, a−b ∈ I∩J . Similarly,

let r ∈ R. Then ra ∈ I and ra ∈ J because I and J are left ideals. Hence, ra ∈ I ∩ J . So

I ∩ J is a left ideal.

Nearly identical arguments prove the cases when I and J are right ideals or two-sided ideals.

(2) Similarly, suppose I and J are left ideals and let a, b ∈ I+J . Then a = i1+j1, b = i2+j2

with i1, i2 ∈ I and j1, j2 ∈ J . By associativity of addition,

a+ b = (i1 + j1) + (i2 + j2) = (i1 + i2) + (j1 + j2) ∈ I + J.

Let r ∈ R. Since I and J are left ideals then ri1 ∈ I and rj1 ∈ J . Thus, by the distributive

law,

ra = r(i1 + j1) = ri1 + rj1 ∈ I + J.

Hence, I + J is a left ideal and a similar argument proves the claim when I and J are right

or two-sided ideals.

(3) Exercise �

Exercise. Extend (1) previous example to arbitrary families of [left, right, two-sided] ideals.

Extend (2) to finite families.

Exercise. Show that the union of two [left, right, two-sided] ideals need not be an ideal.

Definition 17. Let R be a ring and X a subset of R. Let {Ik | k ∈ K} be the set of [left,

right, two-sided] ideals containing X. The [left, right, two-sided] ideal⋂k∈K

Ik is the [left,

right, two-sided] ideal generated by X and the elements of X are called the generators of the

ideal. When |X| <∞, we say the ideal is finitely generated.

In the case of two-sided ideals, we denote the ideal generated by the set X by (X). Your

textbook is a little ambiguous on this.

Definition 18. A (two-sided) ideal in a ring R generated by a single element a ∈ R is called

a principal ideal and is denoted (a). A ring in which every ideal is principal is called a principal

ideal ring or a principal ideal domain (PID) when R is an integral domain.10

Lemma. Let I be a nonzero ideal in Z such that a, b ∈ I are nonzero and relatively prime.

Then I = Z.

Proof. By the division algorithm, there exists integers r, s such that 1 = ra+ sb ∈ I. Hence,

1 ∈ I and so I = Z. �

Proposition. The ring Z is a principal ideal domain.

Proof. Let I be a proper ideal in Z. Let x = min{a ∈ I | a > 0}. Let y ∈ I, y > 0. Then

gcd(x, y) > 1 so gcd(x, y) = x. Similarly, if z < 0 then −z > 0 and so gcd(x,−z) = x. Thus,

I = (x). �

Say I is an ideal in a ring R (finitely) generated by elements {x1, . . . , xn}. Then elements of

I are finite linear combinations of elements of the form nxi (n ∈ Z), rixi, xisi, and rixisi. If

R has identity then this simplifies to just those terms of the last form.

2.1. Quotient Rings. As we noted previously, the kernel of a ring homomorphism is an

ideal. In group theory we used this fact (with normal subgroups in place of ideals) to develop

quotient groups which led to the Isomorphism Theorems. It is good to keep the analogy in

mind in what follows.

Let I be an ideal (that is, a two-sided ideal) of a ring R. Since I is a normal subgroup of

(R,+), it makes sense to define the coset a + I for a ∈ R. By results from group theory,

(R/I,+) is a group with addition defined by

(a+ I) + (b+ I) = (a+ b) + I for any a, b ∈ R.

Theorem 10 (Theorem III.2.7). Let R be a ring and I an ideal of R. Then the additive

quotient group R/I is a ring with multiplication given by

(a+ I)(b+ I) = (ab) + I for all a, b ∈ R.

If R is commutative or has an identity, then the same is true of R/I.

Proof. The group structure on (R/I,+) comes for free from the fact that I is a normal

subgroup of (R,+).

As is the case whenever we are dealing with cosets, the challenge is in proving well-definedness

(of multiplication). Suppose a + I = a′ + I and b + I = b′ + I, that is, a and a′ are two11

representatives of the same equivalence class (similarly for b and b′). But then a′ = a+ i and

b′ = b+ j for some i, j ∈ I. Then

(a′b′) = (a+ i)(b+ j) = ab+ aj + ib+ ij.

Since I is a (two-sided) ideal, then aj, ib, ij ∈ I so a′b′ ∈ ab+ I. Hence, a′b′ + I = ab+ I.

We check that the left distributive law holds. The right distributive law is similar.

(a+ I) [(b+ I) + (c+ I)] = (a+ I) [(b+ c) + I]

= a(b+ c) + I

= (ab+ ac) + I

= (ab+ I) + (ac+ I)

= (a+ I)(b+ I) + (a+ I)(c+ I).

The remaining comment on commutivity and identity are left as an (easy) exercise. �

The next result is the first step on the path to the First Isomorphism Theorem for Rings.

Theorem 11. Let f : R → S be an homomorphism of rings. Then ker f is an ideal of

R. Moreover, if I is an ideal of a ring R, then I = ker f for some homomorphism f . In

particular, I is the kernel of the canonical projection

π : R→ R/I

r 7→ r + I.

Proof. We have already shown that kerπ is an ideal. Because ker π is a normal subgroup of

(R,+), it follows that π is an epimorphism of groups. We need only check that it is also a

ring homomorphism. Let a, b ∈ R, then

π(a)π(b) = (a+ I)(b+ I) = ab+ I = π(ab).

Hence, π is a ring epimorphism. �

In the same sense that the canonical projection of a group onto its quotient group was

universal, so is the canonical projection for rings.

Theorem 12. If f : R→ S is a ring homomorphism and I an ideal of R which is contained

in ker f , then there is a unique ring homomorphism f : R/I → S such that f(a+ I) = f(a)

for all a ∈ R. Moreover, im f = im f , ker f = ker f/I, and f is an isomoprhism if and only

if f is an epimorphism and I = ker f .12

Theorem 13. Isomorphism Theorem for Rings Let f : R→ S be a homomorphism of rings

and I, J ideals of R.

• (First Isomorphism Theorem) f induces an isomorphism of rings R/ ker f ∼= im f .

• (Second Isomorphism Theorem) There is an isomorphism of rings I/(I ∩ J) ∼= (I +

J)/J ;

• (Third Isomorphism Theorem) If I ⊂ J , then J/IO is an ideal of R/I and there is

an isomorphism of rings (R/I)/(J/I) ∼= R/J .

2.2. Prime and maximal ideals. We have already seen that in Z, all of the ideals can be

expressed as nZ for some integer n. But certain ideals are special. In particular, if I = pZfor some prime p, then every element in I can be written pk for some integer k. This isn’t

true in general. For example, 12 ∈ 6Z but 12 = 4 · 3. Moreover, 6Z can be decomposed as

the product of ideals 4Z and 3Z. Thus, ideals generated by a prime in Z are the irreducible

ideals in the sense of ideal multiplication. To understand the structure of ideals in a ring, it

is generally not necessary to find all of the ideals, just the irreducible ones.

I’m not going to follow the text very closely in this section. One major reference for this

material is the book Noncommutative Noetherian Rings by Goodearl and Warfield.

Definition 19. Let R be a ring. An ideal P of R is said to be prime if P 6= R and whenever

AB ⊂ P for ideals A and B of R, A ⊂ P or B ⊂ P .

A ring in which the trivial ideal is prime is said to be a prime ring.

When R is commutative, the definition of a prime ideal is equivalent to an element-wise

condition that is reminiscent of the integers. For R commutative, an ideal P is prime if and

only if for a, b ∈ R with ab ∈ P , a ∈ P or b ∈ P . Hence, it is clear that a commutative ring

is an integral domain if and only if it is a prime ring.

Theorem 14. Let R be a ring and P a proper ideal. The following are equivalent.

(a) P is a prime ideal.

(b) If I and J are any ideals of R properly containing P , then IJ 6⊂ P .

(c) R/P is a prime ring.

(d) If x, y ∈ R with xRy ⊂ P , then either x ∈ P or y ∈ P .

Proof. (a) ⇒ (b) Suppose I and J are ideals such that P ( I and P ( J . If IJ ⊂ P then

I ⊂ P or J ⊂ P , a contradiction.13

(b) ⇒ (c) Choose ideals I, J in R/P . By definition of R/P , there exists ideals I ′, J ′ of R

with P ⊂ I ′ and P ⊂ J ′, such that I ′/P = I and J ′/P = J . If IJ = 0 in R/P , then

I ′J ′ ⊂ P . Thus, I ′ = P or J ′ = P , so I = 0 or J = 0.

(c) ⇒ (a) Choose ideals I, J of R such that IJ ⊂ P . It follows that I + P/P and J + P/P

are ideals of R/P such that their product is zero. But then I + P/P = 0 or J + P/P = 0,

whence I ⊂ P or J ⊂ P .

(a) ⇔ (b) Exercise. �

A consequence of (c) is that in a nontrivial commutative ring R with identity, an ideal P is

prime if and only if R/P is an integral domain.

The question now is where can we find these prime ideals. It’s fairly obvious in a ring like

Z because we already know a lot about prime numbers. In general for a ring this can be a

difficult problem. One place to look is maximal ideals.

Definition 20. A [left,right,two-sided] ideal M in a ring R is said to be maximal if M 6= R

and for every [left,right,two-sided] ideal N such that M ⊂ N ⊂ R, either N = M or N = R.

Theorem 15. Every maximal (two-sided) ideal M of a ring R is a prime ideal.

Proof. Let M be a maximal ideal of R and let I and J be any two ideals of R properly

containing M . Then I +M = J +M = R and

R = (I +M)(J +M) ⊂ IJ +M.

Hence, IJ 6⊂M . Thus, by the previous theorem, M is prime. �

The next theorem is a sort of existence theorem for maximal ideals. Recall that Zorn’s

Lemma states that if A is a partially ordered set such that every chain in A has an upper

bound in A, then A contains a maximal element. This is equivalent to the Axiom of Choice

and the Well-Ordering Principle.

The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who

can tell about Zorn’s lemma? - Jerry Bona

Theorem 16 (Theorem III.2.18). In a nonzero ring R with identity, maximal [left,right,two-

sided] ideals always exist. In fact, every [left,right,two-sided] ideal in R (except R itself) is

contained in a maximal [left,right,two-sided] ideal.14

Proof. Let A be a left ideal of R, A 6= R. Let S be the set of all left ideals B in R such that

A ⊂ B 6= R. Since A ∈ S, then S is nonempty. Moreover, S is a partially ordered set via

set inclusions. We claim that every chain of left ideals in S has an upper bound and we may

then apply Zorn’s Lemma.

Let C = {Ci | i ∈ I} be a chain in S and set C =⋃Ci. We claim C is the upper bound of

C and C ∈ S.

If a, b ∈ C, then for some i, j ∈ I, a ∈ Ci and b ∈ Cj. Since C is a chain, then assume

WLOG that Cj ⊂ Ci. Hence, a, b ∈ Ci and because Ci is a left ideal, a− b ∈ Ci and ra ∈ Cifor any r ∈ R. Thus, a − b, ra ∈ C so C is a left ideal. It follows that A ⊂ Ci for every i

and so A ⊂ C.

Each Ci ∈ S so Ci 6= R for all i ∈ I. Hence, 1R /∈ C. Thus, C 6= R so C ∈ S. Therefore,

the hypothesis of Zorn’s Lemma is satisfied and so S contains a maximal element which is a

maximal left ideal of R containing A. �

We already know that when P is a prime ideal of a commutative ring R, then R/P is an

integral domain. There is a similar correspondence for maximal ideals.

Theorem 17 (Theorem III.2.20). Let M be an ideal in a nontrivial ring R with identity.

(i) If M is maximal and R is commutative, then the quotient ring R/M is a field.

(ii) If the quotient ring R/M is a division ring, then M is maximal.

Proof. (i) Let M be maximal and R commutative so that R/M is an integral domain. We

need only show that every nonzero element in R/M has a multiplicative inverse in R/M .

Choose a ∈ R such that a + M 6= M . That is, a /∈ M so M ⊂ M + (a) ⇒ M + (a) = R.

Thus, 1R = m+ ra for some m ∈M , r ∈ R. Therefore, 1r − ra = m ∈M . Thus,

1R +M = ra+M = (r +M)(a+M).

That is, r +M is the multiplicative inverse of a+M in R/M .

(ii) Suppose R/M is a division ring. Let N be an ideal such that M ( N . We claim N = R.

Choose a ∈ N −M . Since R/M is a divsion ring, a + M has a multiplicative inverse, say

b + M . Thus, ab + M = 1R + M and so ab − 1R ∈ M ⊂ N . Since ab ∈ N (it is an ideal),

then 1R ∈ N so N = R. Hence, M is maximal. �

15

3. Polynomial Rings

We begin by (formally) defining a polynomial ring. This has the advantage of removing some

of the ambiguity of our previous definition. Specifically in speaking of the ‘formal symbol’

x.

Definition 21. Let R be a ring and let R[x] denote the set of all sequences of elements of

R, (a0, a1, a2, . . .), such that ai = 0 for all but a finite number of indices i. The operations

are defined as

(a0, a1, . . .) + (b0, b1, . . .) = (a0 + b0, a1 + b1, . . .) and

(a0, a1, . . .) · (b0, b1, . . .) = (c0, c1, . . .) where cn =n∑i=0

an−ibi

The ring R[x] is known as the polynomial ring over R or the ring of polynomials of R.

One should check that the above definition is valid, that is, our operations of addition and

multiplication define a ring.

The two definitions of polynomial rings we are equivalent (define isomorphic rings). This is

a (slightly less trivial exercise). When R has identity, then xn is the sequence of zeros with

1R in the nth spot.

What about polynomials in several variables? (For that matter, what about polynomials in

infinitely many variables? That’s a whole other story.) One needs a change in perspective,

that is, to treat sequences as functions f : N → R. Then f(k) gives the kth entry in the

sequence or the coefficient of xk as a polynomial.

Definition 22. Let R be a ring and denote by R[x1, . . . , xn] the set of functions f : Nn → R

such that f(u) 6= 0 for only finitely many u ∈ Nn.

Then R[x1, . . . , xn] is a ring with addition and multiplication defined by

(f + g)(u) = f(u) + g(u) and (fg)(u) =∑

v+w=u

f(v)g(w)

where f, g ∈ R[x1, . . . , xn] and u, v, w ∈ Nn.

The ring R[x1, . . . , xn] is called the ring of polynomials in n indeterminates over R.

As before, one should check that R[x1, . . . , xn] is well-defined. Moreover, this coincides

with the expected (but unstated) definition/realization of polynomials over R in the actual

variables xi.16

Exercise. Prove that (R[x])[y] ∼= R[x, y].

Definition 23. Let φ : R → S be a homomorphism of rings and s1, . . . , sn ∈ S. For

f ∈ R[x1, . . . , xn] we write

f =m∑i=0

aixki11 · · ·xkinn ai ∈ R; kij ∈ N.

(If R is a ring without identity then we simply omit those xi with exponent zero.) We define

the evaluation map as

Φ : R[x1, . . . , xn]→ S

f 7→m∑i=0

φ(ai)ski11 · · · skinn .

Proposition. If R and S are commutative, then the evaluation map is a homomorphism of

rings.

Proof. We do this for n = 1. The general case is left as an exercise.

Let φ : R→ S be a homomorphism of rings and choose s ∈ S.

Let f, g ∈ R[x] and write

f =n∑i=0

aixi and g =

n∑i=0

bixi.

(Some of the ai and bi may be zero.) Then

Φ(f + g) = Φ

(n∑i=0

(ai + bi)xi

)=

n∑i=0

φ(ai + bi)si

=n∑i=0

φ(ai)si +

n∑i=0

φ(bi)si = Φ(f) + Φ(g).

Similarly,

Φ(fg) = Φ

(n∑i=0

(ci)xi

)with ci =

i∑k=0

akbi−k

=n∑i=0

φ(ci)si =

n∑i=0

φ

(i∑

k=0

akbi−k

)si

=n∑i=0

i∑k=0

φ(ak)φ(bi−k)si = Φ(f)Φ(g).

�17

These definitions aren’t completely arbitrary. In fact, polynomial rings satisfy a certain

universal mapping property which we now state.

Theorem 18 (Theorem 5.5). Let R and S be commutative rings with identity and φ : R→ S

a homomorphism of rings such that φ(1R) = 1S. If s1, s2, . . . , sn ∈ S, then the evaluation

map is the unique homomorphism of rings ψ : R[x1, . . . , xn] → S such that ψ |R= φ and

ψ(xi) = si for i = 1, 2, . . . , n. This property completely determines the polynomial ring

R[x1, . . . , xn] up to isomorphism.

Our next goal is to prove that F [x] is a PID for a field F (and another property called UFD

which we have yet to define.) For this, we will need a notion of degree. We have discussed

degree of polynomials in one variable. Once we pass to multiple variables this is a bit more

complex.

Definition 24. Let R be a ring. The degree of a monomial axk11 · · ·xknn ∈ R[x1, . . . , xn]

(a 6= 0) is the nonnegative integer k1 + · · ·+ kn.

Let f ∈ R[x1, . . . , xn] be nonzero. Write f =∑aix

ki1 · · ·xkin . The total degree of f is the

maximum degree of a monomial summand such that ai 6= 0. (To be clear, a summand is

generally defined as a monomial appearing with nonzero coefficient.) The zero polynomial

is defined to have degree −∞.

The polynomial f is said to be homogeneous if each summand is of degree k. The degree of

f in xk is is the degree of f considered as a polynomial in R[x1, . . . , xk−1, xk+1, . . . , xn].

Theorem 19. Let R be a ring and f, g ∈ R[x1, . . . , xn].

• deg(f + g) ≤ max{deg f, deg g}• deg(fg) ≤ deg f + deg g with equality when R has no zero divisors.

• If n = 1 and the leading coefficient of f or g is not a zero divisor in R, then deg(fg) =

deg f + deg g.

Theorem 20 (Division Algorithm). Let R be a ring with identity and f, g ∈ R[x] nonzero

polynomials such that the leading coefficient of g is a unit in R. Then there exist unique

polynomials q, r ∈ R[x] such that

f = qg + r and deg r < deg g.

Proof. (Existence) If deg g > deg f , let q = 0 and r = f . If deg g ≤ deg f , then write

f =n∑i=0

aixi and g =

m∑i=0

bixi

18

with an, bm 6= 0 and m ≤ n, and bm a unit in R. We will use induction on the degree of

f . If n = 0, then m = 0, f = a0, g = b0 and b0 is a unit. Set q = a0b−10 and r = 0; then

deg r < deg g and

qg + r = (a0b−10 )b0 = a0 = f.

Assume the hypothesis holds for polynomial of degree less than n. Then (anb−1m xn−m)g is

a polynomial of degree n with leading coefficient an. Hence, deg(f − (anb−1m xn−m)g) < n.

Thus, there exists polynomials q′ and r such that

f − (anb−1m xn−m)g = q′g + r and deg r < deg g.

Now set q = (anb−1m xn−m)g + q′ and the result follows.

(Uniqueness) Suppose f = q1g + r1 and f = q2g + r2 with the deg r1, deg r2 < deg g. Then

(q1 − q2)g = r2 − r1. Because the leading term of g is a unit and not a zero divisor,

deg(q1 − q2) + deg(g) = deg((q1 − q2)g) = deg(r2 − r1).

Since deg(r2 − r1) ≤ max{deg r2, deg r1} < deg g, this equality is true if and only if

deg(q1, q2) = −∞ = deg(r2 − r1).

Thus, q1 − q2 = 0 and r1 − r2 = 0. �

Theorem 21 (Remainder Theorem). Let R be a ring with identity and f(x) ∈ R[x]. For

any c ∈ R, there exists a unique q(x) ∈ R[x] such that f(x) = q(x)(x− c) + f(c).

Proof. If f = 0 then let q = 0. Now suppose f 6= 0. Then the Division Algorithm implies

that there exist unique polynomials q(x), r(x) ∈ R[x] such that f(x) = q(x)(x − c) + r(x)

and deg(r(x)) < deg(x− c) = 1. Thus, r(x) = r is a constant polynomial. It is left to show

that f(c) = r.

If q(x) =∑n−1

j=0 bjxj, then

f(x) = q(x)(x− c) + r = −b0c+n−1∑k=0

(−bkc+ bk − 1)xk + bn−1xn + r.

It follows that

f(c) = −b0c+n−1∑k=0

(−bkc+ bk − 1)ck + bn−1cn + r

= −n−1∑k=0

bkck+1 +

n∑k=1

bk−1ck + r = 0 + r = r.

�19

Corollary 22. Let F be a field. The polynomial ring F [x] is a principal ideal domain.

Proof. Because F is a field, it is an integral domain and so is F [x]. Let I be a nonzero ideal

of F . If I contains a unit, then I = F .

Suppose I 6= 0 and I 6= F . Let S = {deg(f) | f ∈ J, f 6= 0} ⊂ N. By the Well-Ordering

Principle, S contains a least element, call it d. Because I is proper, d > 0.

Let g be a polynomial in F [x] of degree d. We claim I = (g). Clearly (g) ⊂ I, we must show

the reverse inclusion.

Let f ∈ J . By the division algorithm there exist unique polynomials q, r ∈ F [x] such that

f = qg + r, deg r < deg g = d. Since f, qg ∈ J , then r = f − qg ∈ J . By the minimality of

d, deg(r) = 0. Since J does not contain a unit, then r = 0 so f ∈ (g). �

It is worth pointing out here, as Hungerford does, that F [x1, . . . , xn] is not a PID. We can’t

iterate the above proof because F [x1] is not a field. In particular, the ideal (x1, . . . , xn) is

not principal. It does, however, satisfy a weaker condition.

Definition 25. Let R be a commutative ring. A nonzero element a ∈ R is said to divide an

element b ∈ R (a | b) if there exists x ∈ R such that ax = b. Elements a, b ∈ R are said to

be associates if a | b and b | a.

A nonzero, nonunit element c ∈ R is said to be irreducible provided c = ab implies a or b is

a unit.

A nonzero, nonunit element p ∈ R is said to be prime provided p = ab implies p | a or p | b.

Definition 26. An integral domain R is a unique factorization domain (UFD) provided that

• every nonzero nonunit a ∈ R can be written a = c1c2 · · · cn with ci irreducible.

• If there are two such factorizations a = c1c2 · · · cn and a = d1d2 · · · dm then n = m

and for some permutation σ ∈ Si, ci and dσ(i) are associates for every i.

The polynomial ring F [x1, . . . , xn] is a UFD. This takes a bit of work to show and we won’t

do that here. We are missing one big technical tool in the proof: the ability to pass from a

ring R to its ring of quotients. That is, the ring formed by inverting all nonzero elements

(think: passing from Z to Q). One can use this to show that for an UFD D, the polynomial

ring D[x] is a UFD. The result then follows by induction. Instead, we will next prove that

all PIDs are UFDs.

20

Theorem 23. Every PID is a UFD.

Proof. Let R be a PID. Suppose that factoring into irreducibles holds, that is, for a ∈ R,

a = c1c2 · · · cn = d1d2 · · · dm with each ci, di irreducible. Because c1 is irreducible and R is a

PID, c1 is prime (Exercise). Then c1 divides some di. Since c1 is a nonunit, it is an associate

of di. Now use induction.

We will first show that every element of R can be factored into irreducibles. Secondly we

will show that the factoring is unique up to associates.

Let S be the set of elements which cannot be factored into irreducibles. We claim S = ∅.Suppose otherwise and choose a ∈ S. But then (a) is a proper ideal and is contained in a

maximal ideal (c). The element c is irreducible. If it weren’t, then c = c1c2 implies (c) ⊂ (c1),

so (c1) = (c) or (c1) = R. If (c1) = R, then c1 is a unit. If (c1) = (c), then c1 = cb and so

c = c1c2 = cbc2 so bc2 = 1 (R is an integral domain). Hence, b is a unit.

Since (a) ⊂ (c), c divides a. Thus, for each a ∈ S there exists an irreducible divisor ca of a

(Axiom of Choice) and ca uniquely determines xa ∈ R such that caxa = a.

Since a is not irreducible, xa is not a unit. We claim xa ∈ S. If it weren’t, then xa would

have a factorization into irreducibles and hence so would a. Thus, xa ∈ S.

Since xa | a, then (a) ⊂ (xa). We claim this inclusion is proper. If it were not, then xa = ay

for some y ∈ R. Then a = xaca = ayca and yca = 1. But ca is irreducible and hence a

nonunit. Thus the claim holds.

Set a0 = a and a1 = xa. Repeating the above argument. gives a divisor a2 ∈ S of a1 and

(a1) ( (a2). Continuing in this way we arrive at an infinite chain of ideals

(a) ( (a1) ( (a2) ( · · · .

We claim this is a contradiction in R.

Let A =⋃i≥0(ai). We claim A is an ideal. Let u, v ∈ A, then u ∈ (ai) and v ∈ (aj) for

some i, j. WLOG, assume i ≥ j. Then (aj) ⊂ (ai) so u − v ∈ (ai) ⊂ A. If r ∈ R, then

rv ∈ (aj) ⊂ A. Hence, the claim holds.

Since R is a PID, then A = (w) for some w ∈ R. But then w ∈ (an) for some n and so

(w) ⊂ (an). Thus, for every j ≥ n,

(w) ⊂ (an) ⊂ (aj) ⊂ A = (w)

so (aj) = (an). This contradicts the fact that the chain is infinite, so S is empty. �21

Definition 27. Let N be the set of nonnegative integers and R a commutative ring. R is a

Euclidean ring if there exists a function φ : R\{0} → N such that

• if a, b ∈ R and ab 6= 0, then φ(a) ≤ φ(ab);

• if a, b ∈ R and b 6= 0, then there exists q, r ∈ R such that a = qb + r with r = 0, or

r 6= 0 and φ(r) < φ(b).

A Euclidean ring which is an integral domain is called a Euclidean domain.

Example 24. (1) The ring Z with φ(x) = |x| is a Euclidean domain.

(2) The polynomial ring F [x] with F and φ(f) = deg(f) is a Euclidean domain.

(3) The Gaussian integers

Z[i] = {a+ bi | a, b ∈ Z} ⊂ C

with φ(a+ bi) = a2 + b2 is a Euclidean domain.

Theorem 25. Every Euclidean ring R is principal ideal ring with identity. Consequently

every Euclidean domain is a UFD.

Proof. Let I be a nonzero ideal of R. Let S = {φ(x) | x 6= 0, x ∈ I} ⊂ N. Then S has a least

element n. Choose a ∈ I such that φ(a) = n. We claim I = (a). By the division algorithm,

if b ∈ I, then b = qa + r with r = 0 or r 6= 0 and φ(r) < φ(a). Then r = b − qa ∈ I so we

must have r = 0. Thus, I ⊂ Ra ⊂ (a) ⊂ I, so I = Ra = (a) and R is a principal ideal ring.

Since R is an ideal, R = Ra for some a ∈ R. Hence, a = ea = ae for some e ∈ R. If

b ∈ R = Ra, then b = xa for some x ∈ R. Then be = (xa)e = x(ae) = xa = b. Therefore, e

is a multiplicative identity for R. �

So what is an example of something that is not a UFD? Consider the ring R = Z[√−5]. This

is like the Gaussian integers above, it consists of elements of the form a+b√−5 with a, b ∈ Z.

The only units in this ring are 1 and −1. Why you ask? Suppose (a+b√−5)(c+d

√−5) = 1.

Then we have (ac − 5bd) + (ad + bc)√−5 = 1. This implies ac − 5bd = 1 and ad + bc = 0.

This has a solution only when a = c/(c2 + 5d2) and b = −d/(c2 + 5d2). But a and b must be

integers, implying one of c or d is zero. If d = 0, then a = 1c∈ Z, so a = c = ±1. If c = 0,

then b = − 15d∈ Z, a contradiction.

Note that

6 = 2 · 3 = (1 +√−5)(1−

√−5).

All of 2, 3, 1 +√−5, 1 −

√−5 are irreducible (check!) and are not associates. Hence, R is

not a UFD.22

Let’s end this chapter with something fun: factoring! What are the irreducible polynomials

in Z2[x]. (Remember this is a UFD because Z2 is a field.) We want to use a method similar

to sieving prime numbers. Let’s start listing polynomials in Z2[x].

1, x, x+ 1, x2, x2 + x, x2 + 1, x2 + x+ 1.

Clearly 1, x, and x + 1 are irreducible, but x2 and x2 + x are not, nor is x2 + 1. Then

x2 + x+ 1 is irreducible.

We can use this idea to determine whether polynomials are irreducible over Z[x]. Let p be

a prime. There is a homomorphism of rings (called reduction mod p)

Z[x]→ Zp[x]∑anx

n 7→∑

anxn.

We call the image of a polynomial f ∈ Z[x] under the map the residue of f mod p. If f factors

over Z[x], say f = gh, then f factors over Zp, f = gh.

Definition 28. A polynomial f ∈ Z[x] is called primitive if its coefficients have no common

integer factor except for the units ±1.

Proposition. Let f ∈ Z[x] and let p be a prime integer which does not divide the leading

coefficient of f . If the reside of f modulo p is irreducible, then f is irreducible in Q[x].

Example 26. The polynomial f = x2 + 4x− 10 is irreducible.

Reduce mod 3 to the polynomial x2 + x− 1. Using the sieving technique above, we see that

this polynomial is irreducible in Z3[x]. Hence, f is irreducible in Q[x].

That’s not a typo at the end. The statement actually tells us about reducibility over the

rationals, even though our polynomial comes from the integers. But this comes from the

fact that if a polynomial factors over the rationals, it factors over the integers.

This leads to a method for factoring over the integers.

Proposition (Eisenstein Criterion). Let f(x) =∑aix

i ∈ Z[x] with deg(f) = n and let p

be a prime integer. Suppose the ai satisfy the following.

• p - an;

• p | ai for 0 ≤ i ≤ n− 1;

• p2 - a0.

Then f is irreducible in Q[x]. If f is primitive, it is irreducible in Z[x].23

Example 27. The polynomial x4 + 50x3 + 30x2 + 20 is irreducible in Q[x] and in Z[x].

What have we not talked about? Well, lots of things. The biggest topic I had hoped to

cover was Modules. One can think of this as an analog of vector spaces but for rings instead

of fields. Understanding the module structure of a ring is almost as important as studying

the ring itself. Remember last semester when we spent one day discussing Representation

Theory? This is the analog in Ring Theory. Modules are the spaces on which a ring acts.

I’ll hopefully have a chance to come back to this later in the semester.

Another big topic that I hinted at earlier is Localization. This is the idea of transferring

from a ring to its ring of quotients. Hungerford does this in the commutative case and that

is the right place to start. But localization works in the noncommutative setting as well with

the right hypotheses. We probably won’t need this during the remainder of the semester,

but it is a topic we could discuss later if there is interest.

24