Chapter 4

Rings

The ring axioms and the arithmetic they force; the units as a group and why a unit is never a zero divisor; finite rings where every non-zero-divisor is a unit; the subring criterion and the centre; characteristic, forced to 0 or prime in a domain; and the binomial theorem with the freshman's dream giving the Frobenius map. Unit groups computed for Z/nZ, matrix rings, polynomial rings, the Gaussian integers and the quaternions.

Introduction

A group, as built in Groups, has one operation. That is enough to describe symmetry, but not enough to describe arithmetic: the integers carry an addition and a multiplication, and every interesting fact about them — that $12 = 2^{2} \cdot 3$ , that $g cd (a, b)$ is an integer combination of $a$ and $b$ , that a polynomial of degree $n$ has at most $n$ roots — is a fact about how the two operations interact. A ring is the axiomatisation of that interaction, and the interaction is carried by exactly one axiom: distributivity.

It is worth being clear about how little is assumed. Addition is required to be an abelian group, so subtraction always works. Multiplication is required only to be associative. It need not be commutative, it need not have an identity, and it certainly need not have inverses. Everything else in this chapter is a consequence of that short list, and the surprise is how much follows. From the axioms alone we shall prove that multiplying by $0$ gives $0$ — a fact so familiar it looks like an axiom but is not — that the sign rules $(- a) b = - (ab)$ and $(- a) (- b) = ab$ are theorems rather than conventions, and that the invertible elements of any ring form a group, which is the bridge back to Groups and the single most reusable result here.

The chapter runs in five movements. The first extracts the arithmetic that the axioms force. The second builds a catalogue of examples wide enough that no false generalisation survives it: $Z$ , the modular rings $Z / n Z$ , matrix rings, polynomial rings, the Gaussian integers, rings of functions, and Hamilton's quaternions. The third isolates the two extremes of multiplicative behaviour — units, which you can divide by, and zero divisors, which annihilate something nonzero — and proves they never coincide. The fourth is the substructure theory: the subring criterion, intersections, and centres. The fifth is the characteristic, which measures how many times you must add $1$ to itself to get back to $0$ , and which culminates in the freshman's dream $(a + b)^{p} = a^{p} + b^{p}$ — an identity that is false in every ring you met before this chapter and true in a great many of the ones you will meet after it.

Two conventions, fixed here and used throughout. First, ring will mean ring with identity: we assume an element $1$ with $1 a = a 1 = a$ . Structures satisfying the other axioms but lacking a $1$ (the even integers, for instance) are common enough to need a name, and the standard joke-name is a rng — a ring with the "i" of identity removed. Second, we write $ab$ for $a \cdot b$ , and $a - b$ for $a + (- b)$ .

4.1The axioms and what they force

Definition 4.1 (Ring). A ring is a triple $(R, +, \cdot)$ consisting of a set $R$ with two binary operations satisfying:

$(R, +)$ is an abelian group. Addition is associative and commutative, there is an element $0 \in R$ with $a + 0 = a$ for all $a$ , and each $a \in R$ has an additive inverse $- a$ with $a + (- a) = 0$ .
Multiplication is associative. $(ab) c = a (b c)$ for all $a, b, c \in R$ .
Multiplication has an identity. There is $1 \in R$ with $1 a = a 1 = a$ for all $a$ .
The distributive laws hold. For all $a, b, c \in R$ ,

a (b + c) = ab + a c and (b + c) a = ba + c a .

The two distributive laws are stated separately because multiplication is not assumed commutative, and neither one implies the other. Axioms 1 and 2 say nothing about how the operations relate; axiom 4 is the entire relationship. Delete it and you are left with an abelian group and a monoid sharing a set by accident, and not one theorem below survives.

Definition 4.2 (Commutative ring, division ring, field). A ring $R$ is commutative if $ab = ba$ for all $a, b \in R$ . A ring in which $1 \neq = 0$ and every nonzero element has a multiplicative inverse is a division ring; a commutative division ring is a field.

Intuition. A ring is a number system in which you may always add, subtract and multiply, but not necessarily divide. Everything that can go wrong with division is the subject of this chapter and the next two: an element may fail to be invertible ( $2$ in $Z$ ), or may be worse than non-invertible and actually destroy information ( $2$ in $Z /6 Z$ , since $2 \cdot 3 = 0$ ). Fields are the rings where nothing goes wrong, and they are studied in Integral Domains and Fields.

Before any examples, we ask what the axioms alone give. The answer is more than one expects, and the proofs are short exercises in using distributivity as the only link between the two operations.

Theorem 4.3 (Arithmetic forced by the axioms). Let $R$ be a ring and $a, b, c \in R$ . Then

$a 0 = 0 a = 0$ ;
$a (- b) = (- a) b = - (ab)$ ;
$(- a) (- b) = ab$ ;
$a (b - c) = ab - a c$ and $(b - c) a = ba - c a$ ;
$(- 1) a = - a$ .

Proof. (1). Since $0 + 0 = 0$ , distributivity gives

a 0 = a (0 + 0) = a 0 + a 0.

Now $a 0$ has an additive inverse, because $(R, +)$ is a group. Adding $- (a 0)$ to both sides yields $0 = a 0$ . The identical argument on the other side, using $(0 + 0) a = 0 a + 0 a$ , gives $0 a = 0$ . Note what was used: distributivity to split the product, and the existence of additive inverses to cancel. Nothing else.

(2). We must show $a (- b)$ is the additive inverse of $ab$ , i.e. that the two sum to $0$ . Indeed

ab + a (- b) = a (b + (- b)) = a 0 = 0

by distributivity and part (1). Since inverses in an abelian group are unique, $a (- b) = - (ab)$ . Symmetrically $(- a) b + ab = (- a + a) b = 0 b = 0$ , so $(- a) b = - (ab)$ .

(3). Apply (2) twice:

(- a) (- b) = - (a (- b)) = - (- (ab)) = ab,

the last step because $- (- x) = x$ in any group.

(4). $a (b - c) = a (b + (- c)) = ab + a (- c) = ab - a c$ by distributivity and (2); likewise on the right.

(5). Take $b = 1$ in (2): $(- 1) a = - (1 a) = - a$ .∎

Part (3) deserves a second look, because "minus times minus is plus" is usually taught as a rule to memorise. It is not a rule; it is forced. Any system in which you can add, subtract and multiply, and in which multiplication distributes over addition, must satisfy $(- a) (- b) = ab$ . There is no consistent alternative arithmetic in which two negatives multiply to a negative — such a system would violate distributivity, and hence could not be a ring at all.

Example 4.4 (Deriving $(- 1) (- 1) = 1$ from nothing). Using only the ring axioms and no properties of $Z$ , show $(- 1) (- 1) = 1$ in every ring.

Solution.

By Theorem Arithmetic forced by the axioms part (5), $(- 1) x = - x$ for every $x$ . Put $x = - 1$ :

(- 1) (- 1) = - (- 1) .

In the abelian group $(R, +)$ , the inverse of $- 1$ is $1$ , since $(- 1) + 1 = 0$ and inverses are unique. Hence $- (- 1) = 1$ .
Combining, $(- 1) (- 1) = 1$ .
The argument used distributivity (inside part 5), the uniqueness of additive inverses, and nothing about numbers. So it holds in $Z$ , in $M_{2} (R)$ , in $Z /6 Z$ and in every other ring alike.

□

Identities, once they exist, are unique — a fact inherited from monoid theory but worth recording because it licenses the notation $1$ and $0$ as if they named single objects, which is exactly what we have been doing.

Proposition 4.5 (Uniqueness of the identities and of inverses). In any ring $R$ :

the additive identity $0$ is unique, and the multiplicative identity $1$ is unique;
each $a \in R$ has exactly one additive inverse;
if $a$ has a multiplicative inverse, it has exactly one, written $a^{- 1}$ .

Proof. (1). Suppose $1$ and $1^{'}$ both satisfy the identity axiom. Then $1 = 1 \cdot 1^{'}$ because $1^{'}$ is an identity, and $1 \cdot 1^{'} = 1^{'}$ because $1$ is an identity. Hence $1 = 1^{'}$ . The same two-line argument with $+$ in place of $\cdot$ gives $0 = 0^{'}$ .

(2). If $b$ and $b^{'}$ are both additive inverses of $a$ , then

b = b + 0 = b + (a + b^{'}) = (b + a) + b^{'} = 0 + b^{'} = b^{'},

using associativity of addition.

(3). Suppose $ba = ab = 1$ and $b^{'} a = a b^{'} = 1$ . Then

b = b 1 = b (a b^{'}) = (ba) b^{'} = 1 b^{'} = b^{'},

using associativity of multiplication. Note that the argument needs $b$ to be a left inverse and $b^{'}$ a right inverse; a two-sided inverse is therefore unique even in a noncommutative ring.∎

There is one degenerate ring, and it is degenerate for a reason the axioms make precise.

Proposition 4.6 (The trivial ring, and why $1 \neq = 0$ is a real hypothesis). Let $R$ be a ring. Then $1 = 0$ in $R$ if and only if $R = {0}$ .

Proof. If $R = {0}$ then $0$ is the only candidate for either identity, and indeed $0 \cdot 0 = 0$ makes $0$ a multiplicative identity, so $1 = 0$ .

Conversely suppose $1 = 0$ and let $a \in R$ be arbitrary. Then

a = a 1 = a 0 = 0,

the last equality by Theorem Arithmetic forced by the axioms. So every element is $0$ and $R = {0}$ .∎

The one-element ring ${0}$ is called the zero ring or trivial ring. It is a perfectly legitimate ring; it is simply useless, because it has no room for a distinction between "true" and "false" statements about its elements. That is why definitions of integral domain, division ring and field all carry the clause $1 \neq = 0$ : without it the trivial ring would qualify as a field, and the theorem "a field has no zero divisors" would be vacuously misleading rather than informative.

Pitfall. The hypothesis $1 \neq = 0$ is not cosmetic. When you construct a quotient ring $R / I$ in Ideals and Quotient Rings and find that it is trivial, the correct conclusion is that $I = R$ — and Proposition The trivial ring, and why $1 \neq = 0$ is a real hypothesis is the reason the two statements are equivalent.

Example 4.7 (Verifying the axioms for $Z$ ). Check that $(Z, +, \cdot)$ is a commutative ring, naming which axiom each familiar property supplies.

Solution.

Additive group. $Z$ is closed under $+$ , addition is associative and commutative, $0$ is neutral, and $- a \in Z$ for every $a$ . So $(Z, +)$ is an abelian group — in fact the infinite cyclic group generated by $1$ , as in Groups.
Multiplicative associativity. $(ab) c = a (b c)$ holds for integers.
Identity. $1 \cdot a = a \cdot 1 = a$ .
Distributivity. $a (b + c) = ab + a c$ ; commutativity of multiplication then gives the right-hand law for free.
Hence $Z$ is a commutative ring with $1$ . It is not a field: $2$ has no inverse in $Z$ , since $2 x = 1$ has no integer solution.
Sanity check on the theorem above. $(- 3) (- 4) = 12 = 3 \cdot 4$ , as Theorem Arithmetic forced by the axioms part (3) predicts.

□

Example 4.8 (A rng: the even integers). Show that $2 Z = {\dots, - 4, - 2, 0, 2, 4, \dots}$ satisfies every ring axiom except the existence of a multiplicative identity.

Solution.

Additive group. $2 a - 2 b = 2 (a - b) \in 2 Z$ , so $2 Z$ is a subgroup of $(Z, +)$ , abelian because $Z$ is.
Closure under multiplication. $(2 a) (2 b) = 2 (2 ab) \in 2 Z$ .
Associativity, commutativity and distributivity are inherited: they are identities holding for all integers, hence in particular for the even ones.
No identity. Suppose $e \in 2 Z$ satisfied $e x = x$ for all $x \in 2 Z$ . Taking $x = 2$ gives $2 e = 2$ , so $e = 1$ , which is not even. Contradiction.
So $2 Z$ is a rng and not a ring under our convention. It reappears in Ideals and Quotient Rings as the ideal $(2)$ , which is the context in which it is genuinely useful.

□

Exercises for §4.1 →

4.2A catalogue of rings

A definition is only as informative as the examples that separate it from its neighbours. The list below is chosen so that every plausible false generalisation about rings dies against at least one entry: some are commutative and some are not, some have zero divisors and some do not, some are finite, some are infinite, and one of them is a division ring that is not a field.

Example 4.9 (The modular rings $Z / n Z$ ). Describe the ring structure on residue classes modulo $n$ , and locate its zero divisors for $n = 6$ .

Solution.

The set. For $n \geq 1$ , $Z / n Z = {\overline{0}, \overline{1}, \dots, \overline{n - 1}}$ , where $\overline{a}$ is the class of integers congruent to $a$ modulo $n$ .
The operations. $\overline{a} + \overline{b} = \overline{a + b}$ and $\overline{a} \overline{b} = \overline{ab}$ . These are well defined: if $a \equiv a^{'}$ and $b \equiv b^{'}$ then $n ∣ (a - a^{'})$ and $n ∣ (b - b^{'})$ , so $n$ divides $(a + b) - (a^{'} + b^{'})$ and also $ab - a^{'} b^{'} = a (b - b^{'}) + b^{'} (a - a^{'})$ .
The axioms. Every axiom is inherited from $Z$ by pushing it through the bars: associativity, commutativity and distributivity are identities in $Z$ , hence identities among classes. So $Z / n Z$ is a commutative ring with $1 = \overline{1}$ , of order $n$ .
Zero divisors at $n = 6$ . $\overline{2} \cdot \overline{3} = \overline{6} = \overline{0}$ , yet neither factor is $\overline{0}$ . Likewise $\overline{3} \cdot \overline{4} = \overline{12} = \overline{0}$ . So $\overline{2}, \overline{3}, \overline{4}$ are zero divisors and only $\overline{1}, \overline{5}$ are not.
Contrast. In $Z /5 Z$ no such collapse occurs, because $5$ is prime: $5 ∣ ab$ forces $5 ∣ a$ or $5 ∣ b$ . Reducing modulo a composite manufactures zero divisors; reducing modulo a prime does not.

□

Proposition 4.10 (Matrix rings). Let $R$ be a ring and $n \geq 1$ . The set $M_{n} (R)$ of $n \times n$ matrices with entries in $R$ , under entrywise addition and the usual matrix product, is a ring with identity the matrix $I_{n}$ . For $n \geq 2$ it is noncommutative and has zero divisors, even when $R$ is a field.

Proof. $(M_{n} (R), +)$ is an abelian group because addition is entrywise and $(R, +)$ is one; the zero matrix is the identity and $(- A)_{ij} = - a_{ij}$ . Associativity of the product is the computation

((A B) C)_{i l} = j \sum (k \sum a_{ik} b_{kj}) c_{j l} = j, k \sum a_{ik} b_{kj} c_{j l} = k \sum a_{ik} (j \sum b_{kj} c_{j l}) = (A (BC))_{i l},

where rearranging the double sum is legitimate because addition in $R$ is associative and commutative and each $\sum$ is finite. Distributivity follows from distributivity in $R$ entry by entry, and $I_{n} A = A I_{n} = A$ is immediate.

For the last claim it suffices to work in the top-left $2 \times 2$ corner, padding with zeros. Put

E_{12} = (0010), E_{22} = (0001) .

Then $E_{12} E_{22} = E_{12}$ while $E_{22} E_{12} = 0$ , so the ring is noncommutative, and the second product exhibits two nonzero matrices with product zero, so it has zero divisors. Moreover $E_{12}^{2} = 0$ , so $E_{12}$ is a nonzero nilpotent.∎

Matrix rings are the standard antidote to the belief that multiplication "should" commute. They are also the reason ring theory bothers to state left and right versions of everything: $A B = 0$ tells you nothing about $B A$ , as the proof just showed.

Proposition 4.11 (Polynomial rings). Let $R$ be a commutative ring. The set $R [x]$ of formal expressions $a_{0} + a_{1} x + \dots + a_{n} x^{n}$ with $a_{i} \in R$ , added coefficientwise and multiplied by

(i \sum a_{i} x^{i}) (j \sum b_{j} x^{j}) = k \sum (i + j = k \sum a_{i} b_{j}) x^{k},

is a commutative ring with $1$ . If $R$ is an integral domain then so is $R [x]$ , and $de g (f g) = de g f + de g g$ for nonzero $f, g$ .

Proof. The additive group is a direct sum of copies of $(R, +)$ , hence abelian. Associativity of the product reduces to the identity $\sum_{i + j + k = m} a_{i} b_{j} c_{k}$ being computable in either grouping, which holds because $R$ is associative and commutative; commutativity of the product follows from $\sum_{i + j = k} a_{i} b_{j} = \sum_{i + j = k} b_{j} a_{i}$ . Distributivity is coefficientwise. The constant polynomial $1$ is the identity.

For the degree claim, let $f$ have degree $m$ with leading coefficient $a_{m} \neq = 0$ and $g$ degree $n$ with leading coefficient $b_{n} \neq = 0$ . The coefficient of $x^{m + n}$ in $f g$ is $a_{m} b_{n}$ , and every higher coefficient vanishes because it is a sum of products $a_{i} b_{j}$ with $i > m$ or $j > n$ . If $R$ is a domain then $a_{m} b_{n} \neq = 0$ , so $de g (f g) = m + n$ and in particular $f g \neq = 0$ ; hence $R [x]$ has no zero divisors, and it is commutative with $1 \neq = 0$ , so it is a domain.∎

Pitfall. The degree formula needs $R$ to be a domain. In $(Z /4 Z) [x]$ we have $(2 x + 1) (2 x + 1) = 4 x^{2} + 4 x + 1 = 1$ , a product of two degree-one polynomials of degree zero — and, incidentally, a nonconstant unit, which Proposition Units of a polynomial ring below rules out precisely when $R$ is a domain.

Example 4.12 (The Gaussian integers). Show that $Z [i] = {a + bi : a, b \in Z} \subseteq C$ is a commutative ring, and that the norm $N (a + bi) = a^{2} + b^{2}$ satisfies $N (z w) = N (z) N (w)$ .

Solution.

Closure. $(a + bi) \pm (c + d i) = (a \pm c) + (b \pm d) i$ and

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i,

and all four of $a \pm c$ , $b \pm d$ , $a c - b d$ , $a d + b c$ are integers. So $Z [i]$ is closed under the operations of $C$ . 2. Axioms. Associativity, commutativity and distributivity hold in $C$ , hence in any closed subset; $0$ and $1$ lie in $Z [i]$ and additive inverses stay inside. So $Z [i]$ is a commutative ring with $1$ , and being a subset of $C$ it has no zero divisors: it is an integral domain. 3. Multiplicativity of the norm. $N (z) = z \overset{z}{ˉ} = ∣ z ∣^{2}$ , and conjugation is multiplicative, so

N (z w) = z w \overline{z w} = z \overset{z}{ˉ} w \overset{w}{ˉ} = N (z) N (w) .

Directly: $(a c - b d)^{2} + (a d + b c)^{2} = (a^{2} + b^{2}) (c^{2} + d^{2})$ , an identity worth recognising — it is the two-squares identity of Diophantus. 4. Check. $N (2 + i) = 5$ , $N (1 + i) = 2$ , and $(2 + i) (1 + i) = 1 + 3 i$ with $N = 1 + 9 = 10 = 5 \cdot 2$ . ✓□

Figure 4.13. The Gaussian integers form a square lattice in the complex plane. The four points on the unit circle are exactly the elements of norm

1

, and these are exactly the units.

The norm converts a question about invertibility in $Z [i]$ into a question about integers, which is the whole reason it is introduced; we cash that in when we compute $Z [i]^{\times}$ below. The same device drives the theory of quadratic fields and reappears in Polynomial Rings and Unique Factorization.

Example 4.14 (Functions under pointwise operations). Let $X$ be any nonempty set and $R$ a ring. Show that $R^{X}$ , the set of all functions $f : X \to R$ with

(f + g) (x) = f (x) + g (x), (f g) (x) = f (x) g (x),

is a ring, and determine when it has zero divisors.

Solution.

Axioms. Every axiom is checked pointwise and therefore reduces to the corresponding axiom in $R$ . For instance $(f (g + h)) (x) = f (x) (g (x) + h (x)) = f (x) g (x) + f (x) h (x) = (f g + f h) (x)$ , and since two functions agreeing at every point are equal, $f (g + h) = f g + f h$ .
Identities. The zero of $R^{X}$ is the constant function $0$ ; the one is the constant function $1$ . Note $R^{X}$ is commutative exactly when $R$ is.
Zero divisors. Suppose $X$ has at least two points $p \neq = q$ and $R \neq = {0}$ . Let $f$ be $1$ at $p$ and $0$ elsewhere, and $g$ be $1$ at $q$ and $0$ elsewhere. Both are nonzero, and $(f g) (x) = 0$ for every $x$ , so $f g = 0$ .
Conclusion. $R^{X}$ has zero divisors whenever $∣ X ∣ \geq 2$ , no matter how well behaved $R$ is. Taking $R = R$ and $X = R$ : the continuous functions $max (x, 0)$ and $max (- x, 0)$ are nonzero with product identically zero. Rings of functions are never domains unless $X$ is a single point.

□

Example 4.15 (Hamilton's quaternions: a division ring that is not a field). The quaternions are $H = {a + bi + c j + d k : a, b, c, d \in R}$ with multiplication determined by

i^{2} = j^{2} = k^{2} = ijk = - 1.

Show that $H$ is a noncommutative division ring.

Solution.

The multiplication table. From $ijk = - 1$ and $k^{2} = - 1$ one gets $ij = k$ ; then $ji = - k$ , and cyclically $jk = i$ , $kj = - i$ , $ki = j$ , $ik = - j$ . In particular $ij \neq = ji$ , so $H$ is noncommutative.
Ring axioms. Addition is componentwise on $R^{4}$ , hence an abelian group; multiplication is defined by extending the table bilinearly, which builds distributivity in by construction, and associativity is checked on the basis ${1, i, j, k}$ .
Conjugate and norm. Write $\overset{q}{ˉ} = a - bi - c j - d k$ . Then

q \overset{q}{ˉ} = \overset{q}{ˉ} q = a^{2} + b^{2} + c^{2} + d^{2} =: N (q) \in R_{\geq 0},

since all the cross terms cancel in pairs. 4. Inverses. If $q \neq = 0$ then $N (q) > 0$ , and

q \cdot \frac{q ˉ}{N ( q )} = \frac{q ˉ}{N ( q )} \cdot q = 1.

So every nonzero quaternion is invertible and $H$ is a division ring, but not a field because multiplication does not commute. 5. Check. $q = 1 + i$ has $N (q) = 2$ and $q^{- 1} = (1 - i) /2$ ; indeed $(1 + i) (1 - i) /2 = (1 - i^{2}) /2 = 2/2 = 1$ . ✓□

Remark. $H$ shows that "division ring" is strictly weaker than "field". Wedderburn's theorem says the gap closes in the finite case: every finite division ring is commutative, hence a field. So $H$ has to be infinite, and indeed it is.

Proposition 4.16 (Products of rings). For rings $R$ and $S$ , the set $R \times S$ with componentwise operations is a ring with zero $(0, 0)$ and identity $(1, 1)$ . If $R$ and $S$ are both nontrivial, $R \times S$ has zero divisors; in particular it is never an integral domain.

Proof. Each axiom is a statement about a pair of components and holds because it holds in each factor: $(R \times S, +)$ is the direct product of two abelian groups, multiplication is associative componentwise, $(1, 1)$ is a two-sided identity, and both distributive laws are checked coordinate by coordinate.

For the second claim put $u = (1, 0)$ and $v = (0, 1)$ . Since $1 \neq = 0$ in $R$ and in $S$ , both $u$ and $v$ are nonzero, yet

uv = (1 \cdot 0, 0 \cdot 1) = (0, 0)

by Theorem Arithmetic forced by the axioms. So $u$ and $v$ are zero divisors.∎

Note also that $u^{2} = u$ : the element $(1, 0)$ is a nontrivial idempotent. Idempotents other than $0$ and $1$ are exactly the fingerprints of a product decomposition, which is what the Chinese Remainder Theorem exploits when it writes $Z /6 Z ≅ Z /2 Z \times Z /3 Z$ .

The catalogue sorts itself into a chain of ever stronger conditions, and it is worth seeing the whole hierarchy at once before the next section starts proving things about it.

Figure 4.17. Each class is obtained from the one outside it by an extra demand: commutativity, then no zero divisors, then inverses for every nonzero element. The quaternions sit in the outermost band, being noncommutative, yet have inverses — so the picture is a chain, not a total ordering of desirability.

Exercises for §4.2 →

4.3Units, zero divisors, and the group of units

Multiplication in a ring can behave in two opposite ways. At best an element is invertible, so multiplying by it is a reversible operation and no information is lost. At worst it annihilates something nonzero, so multiplying by it destroys information irretrievably. These are the two notions that organise everything that follows, and the first theorem of the section is that they are mutually exclusive.

Definition 4.18 (Unit, zero divisor, nilpotent). Let $R$ be a ring.

$u \in R$ is a unit if there exists $v \in R$ with $uv = vu = 1$ . The set of units is written $R^{\times}$ .
$a \in R$ is a left zero divisor if $a \neq = 0$ and $ab = 0$ for some $b \neq = 0$ ; right zero divisor is defined symmetrically, and in a commutative ring the two coincide and we simply say zero divisor.
$a \in R$ is nilpotent if $a^{m} = 0$ for some integer $m \geq 1$ , and idempotent if $a^{2} = a$ .

Definition 4.19 (Integral domain). An integral domain is a commutative ring with $1 \neq = 0$ and no zero divisors: $ab = 0$ implies $a = 0$ or $b = 0$ .

Theorem 4.20 (The units of a ring form a group). Let $R$ be a ring. Then $R^{\times}$ is a group under the multiplication of $R$ , with identity $1$ . If $R$ is commutative, $R^{\times}$ is abelian.

Proof. We verify the group axioms of Groups one at a time.

Closure. Let $u, w \in R^{\times}$ with inverses $u^{- 1}, w^{- 1}$ . Then

(u w) (w^{- 1} u^{- 1}) = u (w w^{- 1}) u^{- 1} = u 1 u^{- 1} = u u^{- 1} = 1,

and symmetrically $(w^{- 1} u^{- 1}) (u w) = 1$ . So $u w$ is a unit with $(u w)^{- 1} = w^{- 1} u^{- 1}$ — note the reversal of order, which matters in a noncommutative ring.

Associativity. Inherited from the associativity of multiplication in $R$ .

Identity. $1 \cdot 1 = 1$ , so $1 \in R^{\times}$ , and $1 u = u 1 = u$ for every $u$ .

Inverses. If $u \in R^{\times}$ with inverse $v$ , then the equations $uv = vu = 1$ read symmetrically in $u$ and $v$ , so $v$ is itself a unit with $v^{- 1} = u$ . Hence $u^{- 1} \in R^{\times}$ .

That is all four axioms, so $R^{\times}$ is a group; and by Proposition Uniqueness of the identities and of inverses the inverse of each unit is unique, so $u \mapsto u^{- 1}$ is a well-defined map. If $R$ is commutative then so is the restricted multiplication, so $R^{\times}$ is abelian.∎

This is the load-bearing result of the chapter. It says that inside every ring, however badly behaved, there sits a group — and every theorem of Groups and Subgroups, Cosets, and Quotient Groups immediately applies to it. Lagrange's theorem applied to $(Z / n Z)^{\times}$ is Euler's theorem $a^{φ (n)} \equiv 1 (mod n)$ ; the same theorem applied to $M_{n} (F)^{\times} = G L_{n} (F)$ is the starting point of the representation theory of finite groups. The construction is also the reason the definition of a unit insists on a two-sided inverse: one-sided inverses need not compose into a group.

Proposition 4.21 (A unit is never a zero divisor). Let $R$ be a ring and $u \in R^{\times}$ . Then $u$ is neither a left nor a right zero divisor. Consequently $R^{\times}$ and the set of zero divisors are disjoint, and in a field the zero divisors are absent entirely.

Proof. Suppose $u b = 0$ for some $b \in R$ . Multiply on the left by $u^{- 1}$ :

b = 1 b = (u^{- 1} u) b = u^{- 1} (u b) = u^{- 1} 0 = 0,

using associativity and Theorem Arithmetic forced by the axioms for the last step. So $u b = 0$ forces $b = 0$ , and $u$ is not a left zero divisor. Multiplying on the right by $u^{- 1}$ handles $b u = 0$ the same way.

For the consequence, note that in a field every nonzero element is a unit, so no nonzero element can be a zero divisor; a field is therefore an integral domain.∎

Proposition 4.22 (Cancellation in a domain). Let $R$ be an integral domain and $a \neq = 0$ . If $ab = a c$ then $b = c$ .

Proof. From $ab = a c$ we get $ab - a c = 0$ , and by Theorem Arithmetic forced by the axioms part (4) this is $a (b - c) = 0$ . Since $R$ is a domain and $a \neq = 0$ , the other factor must vanish: $b - c = 0$ , i.e. $b = c$ .∎

Cancellation is exactly what fails in $Z /6 Z$ : there $2 \cdot 1 = 2 \cdot 4$ but $1 \neq = 4$ . Notice that cancellation does not require $a$ to be invertible — only that it be a non-zero-divisor. That distinction disappears in the finite case, which is the next theorem and is genuinely surprising.

Theorem 4.23 (In a finite ring, non-zero-divisors are units). Let $R$ be a finite ring and let $a \in R$ be neither $0$ nor a left zero divisor. Then $a \in R^{\times}$ . Consequently every finite integral domain is a field.

Proof. Consider the map $λ_{a} : R \to R$ given by $λ_{a} (x) = a x$ .

$λ_{a}$ is injective. If $a x = a y$ then $a (x - y) = 0$ , and since $a$ is not a left zero divisor this forces $x - y = 0$ .

$λ_{a}$ is surjective. An injective map from a finite set to itself is surjective, by the pigeonhole principle.

So there is $b \in R$ with $ab = 1$ . It remains to show $ba = 1$ as well. Compute

a (ba - 1) = (ab) a - a = 1 a - a = a - a = 0,

using distributivity and part (1) of Theorem Arithmetic forced by the axioms. Since $a$ is not a left zero divisor, $ba - 1 = 0$ , i.e. $ba = 1$ . Hence $a$ is a unit.

For the consequence: a finite integral domain is commutative with $1 \neq = 0$ and has no zero divisors, so every nonzero element satisfies the hypothesis and is therefore a unit. That is precisely the definition of a field.∎

Intuition. Finiteness turns "does not destroy information" into "can be undone". Multiplication by a non-zero-divisor is an injective self-map of $R$ ; on a finite set injective and bijective are the same thing, and a bijection has an inverse. On an infinite set they are not: multiplication by $2$ on $Z$ is injective and very far from surjective, which is why $2$ is a non-zero-divisor that is not a unit.

Proposition 4.24 (The units of $Z / n Z$ ). Let $n \geq 2$ . Then

(Z / n Z)^{\times} = {\overline{a} : g cd (a, n) = 1}, (Z / n Z)^{\times} = φ (n),

and every class that is neither $\overline{0}$ nor a unit is a zero divisor. In particular $Z / n Z$ is a field if and only if $n$ is prime.

Proof. Coprime implies unit. If $g cd (a, n) = 1$ then by Bézout there are integers $x, y$ with $a x + n y = 1$ . Reducing modulo $n$ gives $\overline{a} \overline{x} = \overline{1}$ , so $\overline{a}$ is a unit.

Unit implies coprime. If $\overline{a} \overline{x} = \overline{1}$ then $a x = 1 + kn$ for some integer $k$ , so $a x - kn = 1$ and any common divisor of $a$ and $n$ divides $1$ . Hence $g cd (a, n) = 1$ .

The count is the definition of Euler's totient: $φ (n)$ is the number of $a \in {1, \dots, n}$ coprime to $n$ .

Non-units are zero divisors. Let $\overline{a} \neq = \overline{0}$ with $d = g cd (a, n) > 1$ . Put $b = n / d$ , so $1 \leq b < n$ and $\overline{b} \neq = \overline{0}$ . Then

ab = \frac{a}{d} \cdot n \equiv 0 (mod n),

since $a / d$ is an integer. So $\overline{a}$ is a zero divisor.

The field criterion. If $n = p$ is prime then every $a$ with $1 \leq a \leq p - 1$ is coprime to $p$ , so all $p - 1$ nonzero classes are units and $Z / p Z$ is a field. If $n$ is composite, $n = d e$ with $1 < d, e < n$ , then $\overline{d} \overline{e} = \overline{0}$ exhibits a zero divisor, and by Proposition A unit is never a zero divisor the ring is not a field.∎

Figure 4.25. The twelve classes of

Z /12 Z

on a dial. Four are units — the classes coprime to

12

, so

φ (12) = 4

— seven are zero divisors, and

\overline{0}

is neither. Every nonzero class falls into exactly one of the two camps.

Example 4.26 (The unit group of $Z /12 Z$ , and its zero divisors). List $(Z /12 Z)^{\times}$ , identify it as an abstract group, and count the zero divisors.

Solution.

Units. The classes coprime to $12 = 2^{2} \cdot 3$ are those divisible by neither $2$ nor $3$ : $\overline{1}, \overline{5}, \overline{7}, \overline{11}$ . So $φ (12) = 4$ .
Orders. $5^{2} = 25 \equiv 1$ , $7^{2} = 49 \equiv 1$ , $1 1^{2} = 121 \equiv 1 (mod 12)$ . Every non-identity element has order $2$ .
Identification. A group of order $4$ with no element of order $4$ is the Klein four-group $C_{2} \times C_{2}$ , not $C_{4}$ . So $(Z /12 Z)^{\times} ≅ C_{2} \times C_{2}$ .
Zero divisors. The remaining nonzero classes are $\overline{2}, \overline{3}, \overline{4}, \overline{6}, \overline{8}, \overline{9}, \overline{10}$ — seven of them — and by Proposition The units of $Z / n Z$ each is a zero divisor. In general the count is $n - φ (n) - 1$ , here $12 - 4 - 1 = 7$ . ✓
Check closure. $5 \cdot 7 = 35 \equiv 11 (mod 12)$ , and $\overline{11}$ is in the list, as Theorem The units of a ring form a group requires.

□

Example 4.27 (Units of $Z /8 Z$ versus $Z /10 Z$ ). Both groups have order $4$ . Show they are not isomorphic.

Solution.

$(Z /8 Z)^{\times} = {1, 3, 5, 7}$ and $3^{2} = 9 \equiv 1$ , $5^{2} = 25 \equiv 1$ , $7^{2} = 49 \equiv 1 (mod 8)$ . Every element squares to $1$ , so the group is $C_{2} \times C_{2}$ .
$(Z /10 Z)^{\times} = {1, 3, 7, 9}$ and $3^{2} = 9$ , $3^{3} = 27 \equiv 7$ , $3^{4} = 81 \equiv 1 (mod 10)$ . So $\overline{3}$ has order $4$ and the group is cyclic, $C_{4}$ .
A cyclic group of order $4$ has an element of order $4$ ; the Klein group has none. Order of an element is an isomorphism invariant, so the two groups are not isomorphic even though $φ (8) = φ (10) = 4$ .
The moral: $φ (n)$ determines the size of the unit group, never its structure.

□

Proposition 4.28 (Units of a matrix ring). Let $F$ be a field and $n \geq 1$ . Then $A \in M_{n} (F)$ is a unit if and only if $det A \neq = 0$ , and $M_{n} (F)^{\times} = G L_{n} (F)$ . More generally, over a commutative ring $R$ , $A \in M_{n} (R)^{\times}$ if and only if $det A \in R^{\times}$ .

Proof. The adjugate matrix satisfies $A adj (A) = adj (A) A = (det A) I_{n}$ , an identity of polynomials in the entries and therefore valid over any commutative ring.

If $det A \in R^{\times}$ , set $B = (det A)^{- 1} adj (A)$ ; then $A B = B A = I_{n}$ , so $A$ is a unit. Conversely if $A B = I_{n}$ then multiplicativity of the determinant gives $det A \cdot det B = det I_{n} = 1$ , so $det A$ is a unit in $R$ .

Over a field, $R^{\times} = F ∖ {0}$ , so the criterion reads $det A \neq = 0$ .∎

Example 4.29 (Counting the units of $M_{2} (Z /3 Z)$ ). How many invertible $2 \times 2$ matrices are there over the field of three elements?

Solution.

By Proposition Units of a matrix ring, the units are the invertible matrices, i.e. those whose two rows are linearly independent over $F_{3}$ .
First row. Any nonzero vector: $3^{2} - 1 = 8$ choices.
Second row. Any vector not in the line spanned by the first row. That line has $3$ elements, so $9 - 3 = 6$ choices.
Hence $∣ G L_{2} (F_{3}) ∣ = 8 \cdot 6 = 48$ .
Check against the general formula. $∣ G L_{n} (F_{q}) ∣ = \prod_{k = 0}^{n - 1} (q^{n} - q^{k})$ ; for $n = 2, q = 3$ this is $(9 - 1) (9 - 3) = 48$ . ✓ Out of $3^{4} = 81$ matrices, $48$ are units and the other $33$ are zero divisors or zero.

□

Proposition 4.30 (Units of a polynomial ring). Let $R$ be an integral domain. Then $R [x]^{\times} = R^{\times}$ : the only units of $R [x]$ are the constant polynomials that are units in $R$ .

Proof. Every unit of $R$ is clearly a unit of $R [x]$ . Conversely suppose $f g = 1$ in $R [x]$ with $f, g \neq = 0$ . By Proposition Polynomial rings, $de g f + de g g = de g 1 = 0$ , and degrees are non-negative, so $de g f = de g g = 0$ . Thus $f = a$ and $g = b$ are constants with $ab = 1$ , i.e. $a \in R^{\times}$ .∎

In particular $Z [x]^{\times} = {\pm 1}$ and $Q [x]^{\times} = Q ∖ {0}$ . The hypothesis cannot be dropped: in $(Z /4 Z) [x]$ the polynomial $2 x + 1$ is its own inverse, because $4 = 0$ kills the leading term.

Example 4.31 (The units of the Gaussian integers). Determine $Z [i]^{\times}$ .

Solution.

Suppose $z w = 1$ with $z, w \in Z [i]$ . Apply the norm, which is multiplicative by the example The Gaussian integers:

N (z) N (w) = N (1) = 1.

Both $N (z)$ and $N (w)$ are non-negative integers, so $N (z) = 1$ .
Writing $z = a + bi$ , the equation $a^{2} + b^{2} = 1$ over the integers has exactly four solutions: $(\pm 1, 0)$ and $(0, \pm 1)$ .
Conversely each of $1, - 1, i, - i$ is invertible: $i \cdot (- i) = - i^{2} = 1$ and $(- 1) (- 1) = 1$ . Hence

Z [i]^{\times} = {1, - 1, i, - i},

a cyclic group of order $4$ generated by $i$ , since $i^{1} = i$ , $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ . 5. These are exactly the lattice points on the unit circle in the figure The Gaussian integers form a square lattice in the complex plane.□

Proposition 4.32 (Nilpotents in a commutative ring). Let $R$ be commutative and $a \in R$ nilpotent with $a^{m} = 0$ . Then $1 + a \in R^{\times}$ , and $a$ itself is a zero divisor whenever $a \neq = 0$ . In an integral domain the only nilpotent is $0$ .

Proof. $1 + a$ is a unit. The finite geometric sum telescopes:

(1 + a) (1 - a + a^{2} - \dots + (- 1)^{m - 1} a^{m - 1}) = 1 + (- 1)^{m - 1} a^{m} = 1,

since $a^{m} = 0$ . The second factor is therefore a two-sided inverse ( $R$ being commutative).

$a$ is a zero divisor. Take $m \geq 1$ minimal with $a^{m} = 0$ and suppose $a \neq = 0$ , so $m \geq 2$ . Then $a^{m - 1} \neq = 0$ by minimality, and $a \cdot a^{m - 1} = a^{m} = 0$ .

Domains. A domain has no zero divisors, so the previous paragraph forces $a = 0$ .∎

Example 4.33 (Nilpotents in $Z /72 Z$ ). Find every nilpotent element of $Z /72 Z$ and count them.

Solution.

$\overline{a}$ is nilpotent iff $72 ∣ a^{m}$ for some $m$ . Since $72 = 2^{3} \cdot 3^{2}$ , this happens iff every prime dividing $72$ also divides $a$ — that is, iff $6 ∣ a$ .
Why. If $6 ∣ a$ then $a^{3}$ is divisible by $6^{3} = 216$ , which is a multiple of $72$ , so $\overline{a}^{3} = \overline{0}$ . If some prime $p \in {2, 3}$ fails to divide $a$ , then $p$ never divides $a^{m}$ , so $72 ∤ a^{m}$ for any $m$ .
The list. $\overline{0}, \overline{6}, \overline{12}, \dots, \overline{66}$ : the multiples of $6$ , of which there are $72/6 = 12$ .
General rule. The nilpotents of $Z / n Z$ are the multiples of $rad (n) = \prod_{p ∣ n} p$ , and there are $n / rad (n)$ of them. Here $rad (72) = 6$ and $72/6 = 12$ . ✓
Check the proposition. $\overline{6}$ is nilpotent, and $\overline{7} = \overline{1} + \overline{6}$ should be a unit: $7 \cdot 31 = 217 = 3 \cdot 72 + 1 \equiv 1 (mod 72)$ . ✓

□

Exercises for §4.3 →

4.4Subrings

Every algebraic structure comes with a notion of substructure, and in each case the point is the same: checking that a subset is again a structure ought not to require re-verifying the identities, because identities are inherited. What genuinely has to be checked is closure.

Definition 4.34 (Subring). A subset $S$ of a ring $R$ is a subring if $S$ is itself a ring under the operations of $R$ , with the same $1$ . Explicitly: $S$ is an additive subgroup of $R$ , closed under multiplication, and $1_{R} \in S$ .

Theorem 4.35 (Subring criterion). A subset $S \subseteq R$ is a subring if and only if

$1 \in S$ ;
$a - b \in S$ for all $a, b \in S$ ;
$ab \in S$ for all $a, b \in S$ .

Proof. ( $\Rightarrow$ ) A subring is an additive subgroup containing $1$ and closed under multiplication, which is exactly (1)–(3).

( $\Leftarrow$ ) Assume (1)–(3). By (1), $S \neq = \emptyset$ . Taking $a = b = 1$ in (2) gives $0 = 1 - 1 \in S$ ; taking $a = 0$ gives $- b = 0 - b \in S$ for every $b \in S$ ; and then $a + b = a - (- b) \in S$ . So $S$ is closed under addition, contains $0$ , and contains additive inverses: it is a subgroup of $(R, +)$ , abelian because $R$ is.

Multiplication restricted to $S$ lands in $S$ by (3), and is associative and distributive over addition because those are identities holding for all elements of $R$ , hence in particular for elements of $S$ . The element $1 \in S$ acts as a multiplicative identity for the same reason. Therefore $S$ satisfies every ring axiom.∎

The criterion replaces eight axioms by three conditions, and the mechanism deserves naming: identities are inherited, existence statements are not. Associativity, commutativity and distributivity are universally quantified equations, so they pass to any subset for free. The existence of $0$ , of negatives and of $1$ are existence statements, and those must be checked — which is precisely what conditions (1) and (2) do.

Pitfall. Condition (1) is a genuine convention and not every textbook adopts it. Under our convention $2 Z$ is not a subring of $Z$ , because $1 \in / 2 Z$ ; under the no-identity convention it is. Worse, a subset can contain an identity of its own that is not $1_{R}$ : in $Z \times Z$ the subset $S = Z \times {0}$ is closed under subtraction and multiplication and has identity $(1, 0) \neq = (1, 1)$ . It is a ring, but not a subring of $Z \times Z$ .

Example 4.36 (The Gaussian integers as a subring of $C$ ). Use the criterion to confirm that $Z [i]$ is a subring of $C$ , and decide whether $Z [2] = {a + b 2 : a, b \in Z}$ is a subring of $R$ .

Solution.

$1 \in Z [i]$ : take $a = 1, b = 0$ .
Differences: $(a + bi) - (c + d i) = (a - c) + (b - d) i$ , with integer components.
Products: $(a + bi) (c + d i) = (a c - b d) + (a d + b c) i$ , with integer components.
All three conditions hold, so $Z [i]$ is a subring of $C$ by Theorem Subring criterion.
$Z [2]$ : $1 = 1 + 02$ is in it; differences are componentwise; and

(a + b 2) (c + d 2) = (a c + 2 b d) + (a d + b c) 2,

with integer components because $(2)^{2} = 2 \in Z$ . So it is a subring of $R$ . 6. Contrast: ${a + b 32 : a, b \in Z}$ is not a subring, since $(32)^{2} = 34$ is not of that form — closure under multiplication fails, and one has to enlarge to ${a + b 32 + c 34}$ .□

Proposition 4.37 (Intersections of subrings). Let ${S_{α}}_{α \in A}$ be any nonempty family of subrings of $R$ . Then $⋂_{α} S_{α}$ is a subring of $R$ . The union of two subrings, by contrast, is a subring only when one contains the other.

Proof. Write $S = ⋂_{α} S_{α}$ and check the three conditions of Theorem Subring criterion.

$1 \in S$ . Each $S_{α}$ is a subring, so $1 \in S_{α}$ for every $α$ , hence $1 \in S$ .

Differences. Let $a, b \in S$ . For each $α$ we have $a, b \in S_{α}$ , so $a - b \in S_{α}$ . As this holds for every $α$ , $a - b \in S$ .

Products. Identically: $ab \in S_{α}$ for every $α$ , so $ab \in S$ .

For the second claim, suppose $S \cup T$ is a subring with $S \neq \subseteq T$ and $T \neq \subseteq S$ . Choose $s \in S ∖ T$ and $t \in T ∖ S$ . Then $s + t \in S \cup T$ , so either $s + t \in S$ — whence $t = (s + t) - s \in S$ , a contradiction — or $s + t \in T$ , whence $s = (s + t) - t \in T$ , also a contradiction.∎

The intersection property is what makes "the subring generated by a set" well defined: given $X \subseteq R$ , the intersection of all subrings containing $X$ is the smallest subring containing $X$ , written $Z [X]$ when $R$ is commutative. Concretely it consists of all integer-coefficient polynomial expressions in the elements of $X$ , and $Z [i]$ and $Z [2]$ above are exactly this construction for $X = {i}$ and $X = {2}$ .

Proposition 4.38 (The centre of a ring). Let $R$ be a ring. The centre

Z (R) = {z \in R : zr = rz for all r \in R}

is a commutative subring of $R$ .

Proof. $1 \in Z (R)$ because $1 r = r = r 1$ for every $r$ .

Closed under differences. If $z, w \in Z (R)$ and $r \in R$ then, using Theorem Arithmetic forced by the axioms part (4),

(z - w) r = zr - w r = rz - r w = r (z - w) .

Closed under products. $(z w) r = z (w r) = z (r w) = (zr) w = (rz) w = r (z w)$ , moving $r$ leftwards past $w$ and then past $z$ .

So $Z (R)$ is a subring by Theorem Subring criterion, and it is commutative because any two of its elements commute with everything, in particular with each other.∎

Example 4.39 (The centre of $M_{n} (F)$ ). Show that the centre of the matrix ring over a field consists exactly of the scalar matrices $λ I_{n}$ .

Solution.

Scalar matrices are central: $(λ I_{n}) A = λ A = A (λ I_{n})$ .
Conversely let $Z$ be central and let $E_{ij}$ be the matrix with $1$ in position $(i, j)$ and $0$ elsewhere. Direct computation gives

E_{ij} Z = (row j of Z, placed in row i), Z E_{ij} = (column i of Z, placed in column j) .

Equating the two for all $i, j$ : comparing entry $(i, j)$ gives $z_{jj} = z_{ii}$ , so all diagonal entries agree; comparing any off-diagonal entry gives $z_{kj} = 0$ for $k \neq = j$ .
Hence $Z = λ I_{n}$ with $λ = z_{11}$ , and $Z (M_{n} (F)) ≅ F$ .
Interpretation. A noncommutative ring can have a large commutative subring — but the centre measures how close to commutative the whole ring is, and for $M_{n} (F)$ with $n \geq 2$ the answer is "as far as possible short of having a trivial centre".

□

Example 4.40 (Every subring of $Z$ ). Determine all subrings of $Z$ , and all additive subgroups.

Solution.

Additive subgroups. $(Z, +)$ is cyclic, so by the classification of subgroups of a cyclic group in Subgroups, Cosets, and Quotient Groups, every subgroup is $n Z$ for a unique $n \geq 0$ .
Which are closed under multiplication? All of them: $(na) (nb) = n (nab) \in n Z$ .
Which contain $1$ ? Only $n = 1$ , since $1 \in n Z$ forces $n ∣ 1$ .
So $Z$ has exactly one subring, namely $Z$ itself. Every other $n Z$ with $n \geq 2$ is a rng, and these are exactly the ideals of $Z$ — the objects of Ideals and Quotient Rings.
This is the cleanest illustration of the difference the $1 \in S$ convention makes: one subring under our convention, infinitely many under the other.

□

Exercises for §4.4 →

4.5The characteristic

Adding $1$ to itself repeatedly either runs forever through distinct elements or eventually returns to $0$ . Which of the two happens is the single most important coarse invariant of a ring, and it is the one place where the additive and multiplicative structures visibly constrain each other.

Notation. For $m \in Z$ and $a \in R$ write $m \cdot a$ for the $m$ -fold sum $a + \dots + a$ when $m > 0$ , for $0$ when $m = 0$ , and for $- ((- m) \cdot a)$ when $m < 0$ . This is not a product in $R$ — $m$ need not be an element of $R$ — but it obeys $m \cdot (ab) = (m \cdot a) b = a (m \cdot b)$ and $(m \cdot a) (n \cdot b) = (mn) \cdot (ab)$ , both by repeated distributivity.

Definition 4.41 (Characteristic). The characteristic of a ring $R$ , written $char R$ , is the least integer $n \geq 1$ such that $n \cdot 1 = 0$ , if such an $n$ exists, and $0$ otherwise.

Proposition 4.42 (Characteristic as an additive order). Let $R$ be a ring with $char R = n$ . Then $n$ is the order of $1$ in the group $(R, +)$ , and $n \cdot a = 0$ for every $a \in R$ . If $R$ is finite then $char R$ divides $∣ R ∣$ and is in particular nonzero.

Proof. The first statement is the definition of order restated: $n$ is the least positive integer with $n \cdot 1 = 0$ .

For the second, let $a \in R$ . Then by repeated distributivity

n \cdot a = n \cdot (1 a) = (n \cdot 1) a = 0 \cdot a = 0,

using Theorem Arithmetic forced by the axioms at the last step. So the additive order of every element divides $n$ .

If $R$ is finite, then $1$ has finite order in the finite group $(R, +)$ , so $char R \neq = 0$ , and by Lagrange's theorem that order divides $∣ R ∣$ .∎

Example 4.43 (Characteristics of the standard rings). Compute the characteristic of $Z$ , $Q$ , $Z / n Z$ , $M_{2} (Z /5 Z)$ , $(Z /3 Z) [x]$ , and $Z /4 Z \times Z /6 Z$ .

Solution.

$Z$ and $Q$ : no positive multiple of $1$ is $0$ , so the characteristic is $0$ .
$Z / n Z$ : the least $m \geq 1$ with $m \overline{1} = \overline{m} = \overline{0}$ is $m = n$ . So the characteristic is $n$ .
$M_{2} (Z /5 Z)$ : the identity is $I_{2}$ , and $m I_{2} = 0$ iff $m \equiv 0 (mod 5)$ . Characteristic $5$ — note the ring has $5^{4} = 625$ elements, and $5 ∣ 625$ as Proposition Characteristic as an additive order requires.
$(Z /3 Z) [x]$ : the identity is the constant $\overline{1}$ , of additive order $3$ . Characteristic $3$ , in an infinite ring — so finiteness is sufficient but not necessary for positive characteristic.
$Z /4 Z \times Z /6 Z$ : here $m (1, 1) = (0, 0)$ iff $4 ∣ m$ and $6 ∣ m$ , so the least such $m$ is $lcm (4, 6) = 12$ . Characteristic $12$ , while the ring has $24$ elements.
The last two show that the characteristic is not the number of elements; it is the additive order of $1$ , nothing more.

□

Theorem 4.44 (The characteristic of an integral domain). Let $R$ be an integral domain. Then $char R$ is either $0$ or a prime number.

Proof. Suppose $n = char R > 0$ and $n$ is not prime. Since $1 \neq = 0$ in a domain we have $n \neq = 1$ , so $n$ is composite: write $n = ab$ with $1 < a, b < n$ .

Using the notational identity $(a \cdot 1) (b \cdot 1) = (ab) \cdot 1$ , we get

(a \cdot 1) (b \cdot 1) = (ab) \cdot 1 = n \cdot 1 = 0.

Because $R$ is a domain, one of the factors vanishes: $a \cdot 1 = 0$ or $b \cdot 1 = 0$ . But $1 \leq a < n$ and $1 \leq b < n$ , and $n$ was the least positive integer with $n \cdot 1 = 0$ . That is a contradiction.

Hence $n$ is prime.∎

Corollary 4.45 (Prime subrings). Every ring $R$ contains a smallest subring, namely ${m \cdot 1 : m \in Z}$ . It is isomorphic to $Z$ if $char R = 0$ and to $Z / n Z$ if $char R = n > 0$ . In particular a field has prime subfield $Q$ or $Z / p Z$ .

Proof. Let $P = {m \cdot 1 : m \in Z}$ . It contains $1 = 1 \cdot 1$ ; it is closed under differences since $m \cdot 1 - k \cdot 1 = (m - k) \cdot 1$ ; and it is closed under products since $(m \cdot 1) (k \cdot 1) = (mk) \cdot 1$ . So $P$ is a subring by Theorem Subring criterion, and any subring contains $1$ and is closed under differences, hence contains every $m \cdot 1$ : $P$ is the smallest one.

The map $Z \to R$ , $m \mapsto m \cdot 1$ , respects both operations by the identities just used, and its image is $P$ . Its kernel is the set of $m$ with $m \cdot 1 = 0$ , which is a subgroup of $Z$ , hence $n Z$ where $n = char R$ . The first isomorphism theorem of Group Homomorphisms and Isomorphisms, applied to the additive groups and checked to respect multiplication, gives $P ≅ Z / n Z$ , read as $Z$ when $n = 0$ .

If $R$ is a field it is a domain, so by Theorem The characteristic of an integral domain $n$ is $0$ or prime; in the first case $P ≅ Z$ and the smallest subfield containing it is $Q$ , in the second $P ≅ Z / p Z$ is already a field.∎

Exercises for §4.5 →

4.6The binomial theorem and the freshman's dream

Expanding $(a + b)^{n}$ requires moving $a$ past $b$ , so the familiar formula is a theorem about commutative rings — or, more precisely, about pairs of commuting elements. Once it is available, characteristic $p$ produces an identity that looks like a student's error and is in fact one of the most useful facts in algebra.

Theorem 4.46 (Binomial theorem in a commutative ring). Let $R$ be a ring and let $a, b \in R$ commute, $ab = ba$ . Then for every integer $n \geq 1$

(a + b)^{n} = k = 0 \sum n (k n) \cdot a^{k} b^{n - k},

where $(k n) \cdot x$ denotes the $(k n)$ -fold sum of $x$ with itself, as in the notation above.

Proof. Induction on $n$ . For $n = 1$ the right-hand side is $(0 1) \cdot b + (1 1) \cdot a = a + b$ . ✓

Assume the formula for $n$ . Then, using distributivity on both sides,

(a + b)^{n + 1} = (a + b) k = 0 \sum n (k n) \cdot a^{k} b^{n - k} = k = 0 \sum n (k n) \cdot a^{k + 1} b^{n - k} + k = 0 \sum n (k n) \cdot a^{k} b^{n - k + 1} .

Here the step $b a^{k} = a^{k} b$ used the hypothesis $ab = ba$ repeatedly — this is the only place commutativity enters, and it is indispensable. Re-indexing the first sum with $j = k + 1$ and combining the coefficient of $a^{j} b^{n + 1 - j}$ gives

(j - 1 n) + (j n) = (j n + 1),

Pascal's rule, valid as an identity of integers and therefore valid after applying it to the integer multiples. The extreme terms $j = 0$ and $j = n + 1$ have coefficient $1$ on both sides. Hence the formula holds for $n + 1$ , and by induction for all $n \geq 1$ .∎

Pitfall. In $M_{2} (R)$ the formula fails outright. With $A = E_{12}$ and $B = E_{21}$ one has $(A + B)^{2} = A B + B A = I_{2}$ , whereas $A^{2} + 2 A B + B^{2} = 0 + 2 E_{11} + 0 \neq = I_{2}$ . The correct general expansion of $(A + B)^{2}$ is $A^{2} + A B + B A + B^{2}$ , and no amount of care recovers the binomial coefficients without commutativity.

Lemma 4.47 (Prime divisibility of binomial coefficients). Let $p$ be prime and $0 < k < p$ . Then $p ∣ (k p)$ .

Proof. Start from the identity

k! (p - k)! (k p) = p! .

The right-hand side is divisible by $p$ , since $p$ is one of the factors of $p!$ . So $p$ divides the product $k! (p - k)! (k p)$ .

Now $p$ is prime, so by Euclid's lemma it must divide one of the factors. It does not divide $k!$ : that is a product of integers $1, 2, \dots, k$ , all strictly between $0$ and $p$ , and a prime dividing a product must divide one of them. For the same reason $p ∤ (p - k)!$ , because $0 < p - k < p$ .

Therefore $p ∣ (k p)$ .∎

The hypothesis $0 < k < p$ is exactly right: $(0 p) = (p p) = 1$ are the two coefficients that survive, and they are the two terms that appear in the theorem below. Primality is also essential — $(2 4) = 6$ is not divisible by $4$ — which is why the next result is a statement about characteristic $p$ and not about characteristic $n$ .

Theorem 4.48 (The freshman's dream). Let $R$ be a commutative ring with $char R = p$ prime. Then for all $a, b \in R$ and all $m \geq 1$ ,

(a + b)^{p} = a^{p} + b^{p}, (a + b)^{p^{m}} = a^{p^{m}} + b^{p^{m}} .

Proof. By Theorem Binomial theorem in a commutative ring,

(a + b)^{p} = a^{p} + k = 1 \sum p - 1 (k p) \cdot a^{k} b^{p - k} + b^{p} .

Fix $k$ with $0 < k < p$ . By Lemma Prime divisibility of binomial coefficients we may write $(k p) = p c_{k}$ for some integer $c_{k}$ . Then, for any $x \in R$ ,

(k p) \cdot x = (p c_{k}) \cdot x = c_{k} \cdot (p \cdot x) = c_{k} \cdot 0 = 0,

because $p \cdot x = 0$ for every $x$ by Proposition Characteristic as an additive order. So every middle term of the sum vanishes, leaving $(a + b)^{p} = a^{p} + b^{p}$ .

For the second identity, induct on $m$ . The case $m = 1$ is done. If $(a + b)^{p^{m}} = a^{p^{m}} + b^{p^{m}}$ , then applying the case $m = 1$ to the elements $a^{p^{m}}$ and $b^{p^{m}}$ gives

(a + b)^{p^{m + 1}} = ((a + b)^{p^{m}})^{p} = (a^{p^{m}} + b^{p^{m}})^{p} = a^{p^{m + 1}} + b^{p^{m + 1}} .

∎

Intuition. The identity is startling because it is the one algebraic error every student is trained out of. What makes it true here is not that squaring has become linear, but that all the cross terms carry a coefficient divisible by $p$ , and in characteristic $p$ such a coefficient annihilates whatever it multiplies. The middle terms of the expansion are still there; they are all zero.

Corollary 4.49 (The Frobenius endomorphism, and Fermat's little theorem). Let $R$ be a commutative ring of prime characteristic $p$ . The map $F : R \to R$ , $F (a) = a^{p}$ , satisfies

F (a + b) = F (a) + F (b), F (ab) = F (a) F (b), F (1) = 1.

If $R = Z / p Z$ then $F$ is the identity map, which is to say $a^{p} \equiv a (mod p)$ for every integer $a$ .

Proof. Additivity is Theorem The freshman's dream. Multiplicativity is $(ab)^{p} = a^{p} b^{p}$ , valid because $R$ is commutative, and $1^{p} = 1$ . So $F$ preserves both operations and the identity — it is a ring homomorphism, in the sense given in Ideals and Quotient Rings.

Now take $R = Z / p Z$ . Its unit group $(Z / p Z)^{\times}$ has order $p - 1$ by Proposition The units of $Z / n Z$ , so by Lagrange's theorem applied to that group, $\overline{a}^{p - 1} = \overline{1}$ for every $\overline{a} \neq = \overline{0}$ . Multiplying by $\overline{a}$ gives $\overline{a}^{p} = \overline{a}$ , and the identity also holds trivially for $\overline{a} = \overline{0}$ . Hence $F = id$ , which is Fermat's little theorem.∎

Fermat's little theorem has just been deduced from two structural facts — that the units of a ring form a group, and that Lagrange's theorem applies to any finite group. That is the payoff of Theorem The units of a ring form a group: an arithmetic statement proved without any arithmetic.

Example 4.50 (The freshman's dream in $(Z /3 Z) [x]$ ). Expand $(x + 1)^{3}$ and $(x + 2)^{9}$ in $(Z /3 Z) [x]$ .

Solution.

The ring $(Z /3 Z) [x]$ is commutative with characteristic $3$ , by the example Characteristics of the standard rings, so Theorem The freshman's dream applies with $p = 3$ .
Therefore $(x + 1)^{3} = x^{3} + 1^{3} = x^{3} + 1$ .
Verification the long way. $(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1$ , and $3 \equiv 0 (mod 3)$ kills the two middle terms, leaving $x^{3} + 1$ . ✓
For the second, $9 = 3^{2}$ , so the iterated form gives $(x + 2)^{9} = x^{9} + 2^{9}$ .
Reduce the constant: $2^{9} = 512 = 170 \cdot 3 + 2$ , so $2^{9} \equiv 2 (mod 3)$ . Alternatively, Fermat gives $2^{3} \equiv 2$ , hence $2^{9} = (2^{3})^{3} \equiv 2^{3} \equiv 2$ . So $(x + 2)^{9} = x^{9} + 2$ .

□

Example 4.51 (A binomial coefficient modulo a prime). Compute $(3 10) mod 5$ and $(3 7) mod 7$ , and say which one Lemma Prime divisibility of binomial coefficients predicts.

Solution.

$(3 7) = 35 = 5 \cdot 7$ , so $(3 7) \equiv 0 (mod 7)$ . This is exactly the lemma at work: $p = 7$ and $0 < 3 < 7$ .
$(3 10) = 120$ , and $120 = 24 \cdot 5$ , so $(3 10) \equiv 0 (mod 5)$ as well — but not because of the lemma, whose hypothesis requires the upper index to be the prime itself.
That the lemma does not extend blindly is shown by $(5 10) = 252 = 50 \cdot 5 + 2 \equiv 2 (mod 5)$ : here the upper index is a multiple of $5$ and the coefficient is not divisible by $5$ .
The correct general statement is Lucas' theorem: $(k m) mod p$ is the product of $(k _{i} m _{i})$ over the base- $p$ digits. For $m = 10 = (20)_{5}$ and $k = 5 = (10)_{5}$ this gives $(1 2) (0 0) = 2$ . ✓

□

Exercises for §4.6 →

Common Mistakes

Assuming multiplication commutes. It does not in $M_{n} (R)$ or $H$ , and the binomial theorem, the formula $(ab)^{- 1} = b^{- 1} a^{- 1}$ and the phrase "the inverse" all have to be stated with the order kept straight.
Treating $a \cdot 0 = 0$ or $(- a) (- b) = ab$ as axioms. They are theorems, proved from distributivity and the additive group structure alone, and knowing that is what lets you use them in an unfamiliar ring without apology.
Confusing a non-unit with a zero divisor. In $Z$ the element $2$ is not invertible, yet it is not a zero divisor either. Only in a finite ring do the two notions become complementary, by Theorem In a finite ring, non-zero-divisors are units.
Cancelling in a ring with zero divisors. From $2 x = 2 y$ in $Z /6 Z$ you may not conclude $x = y$ : $2 \cdot 1 = 2 \cdot 4$ . Cancellation is a property of domains, not of rings.
Forgetting the $1 \in S$ clause in the subring test. $2 Z$ is closed under subtraction and multiplication but is not a subring under our convention; $Z \times {0} \subseteq Z \times Z$ even has an identity of its own, just not the right one.
Confusing the characteristic with the number of elements. $Z /4 Z \times Z /6 Z$ has $24$ elements and characteristic $12$ ; $(Z /3 Z) [x]$ is infinite and has characteristic $3$ .
Applying the freshman's dream outside characteristic $p$ . $(a + b)^{2} = a^{2} + b^{2}$ is false in $Z$ , false in $Z /4 Z$ , and false in any commutative ring where $2$ is not $0$ . It also fails for $(a + b)^{4}$ in characteristic $2$ only in the sense that the exponent must be a power of $p$ — $4 = 2^{2}$ is fine, $6$ is not.
Assuming every ring has a $1$ , or that $1 \neq = 0$ . Both are conventions with content: the trivial ring satisfies $1 = 0$ , and by Proposition The trivial ring, and why $1 \neq = 0$ is a real hypothesis that single equation collapses the whole ring to one point.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter