Chapter 11

Medium

Rings — Definition and Examples

00 · Symbol Glossary

$(R,+,\times)$A ring

A set $R$ with two operations: addition and multiplication, interacting through the distributive law. The central object of the second half of this course, the way $(G,\ast)$ was for the first half.

$0$Additive identity

The identity element for $+$ : $a+0=a$ for all $a\in R$ . Guaranteed by $(R,+)$ being an abelian group.

$1$Multiplicative identity (unity)

An element with $1\cdot a=a\cdot1=a$ for all $a\in R$ , when one exists. A ring possessing such an element is called a ring with unity; this course's rings are unital by convention unless stated otherwise.

$-a$Additive inverse

The inverse of $a$ under $+$ , guaranteed since $(R,+)$ is a group. Always exists — unlike multiplicative inverses, which are special.

$R^\times$Group of units

$R^\times = \{a\in R : a \text{ has a multiplicative inverse}\}$ . Always a group under multiplication (Theorem 11.3) — the ring-theoretic analog of "the invertible elements."

$M_n(\mathbb{R})$n by n real matrices

All $n\times n$ real matrices (not just invertible ones — contrast with $GL_n(\mathbb{R})$ from Chapter 02), under matrix addition and multiplication. The standard example of a noncommutative ring.

01 · The Definition

Groups modeled a single operation. Rings model two operations at once — addition and multiplication — linked by the requirement that multiplication distributes over addition, exactly as ordinary arithmetic behaves.

Definition — Ring

A ring is a set $R$ with two binary operations $+$ and $\times$ satisfying:

R1. $(R,+)$ is an abelian group (identity $0$ , every element has an additive inverse, $+$ is commutative).

R2. $\times$ is associative: $(ab)c=a(bc)$ .

R3. Distributive laws: $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$ , for all $a,b,c\in R$ .

If additionally $ab=ba$ for all $a,b$ , $R$ is commutative. If there exists $1\in R$ with $1\neq0$ and $1\cdot a=a\cdot1=a$ for all $a$ , $R$ has unity (or is unital). This course's rings are commutative with unity unless stated otherwise.

Multiplication need not have inverses, or even be commutative

Notice how much weaker R2–R3 are compared to the group axioms: multiplication need not have an identity, need not have inverses, and need not commute. A ring guarantees a full group's worth of structure for $+$ , but only associativity (linked to $+$ via distributivity) for $\times$ . Most of this chapter is about which of these extra multiplicative properties a given ring happens to have.

Example — The integers, the motivating case

$(\mathbb{Z},+,\times)$ : $(\mathbb{Z},+)$ is an abelian group (Chapter 02), $\times$ is associative and commutative, distributivity is ordinary arithmetic, and $1\in\mathbb{Z}$ is the multiplicative identity. A commutative ring with unity — the model every other ring in this chapter is compared against.

02 · First Consequences of the Axioms

A few facts hold in every ring, before any specific example is chosen — direct analogs of Chapter 02's first consequences of the group axioms.

Theorem 11.1 — Multiplication by Zero

For every $a\in R$ : $0\cdot a = a\cdot0 = 0$ .

Proof

$0\cdot a = (0+0)\cdot a = 0\cdot a + 0\cdot a$ , using $0+0=0$ (identity law in $(R,+)$ ) and distributivity. Let $x=0\cdot a$ ; the equation reads $x=x+x$ . Add $-x$ to both sides: $x+(-x) = (x+x)+(-x)$ , i.e. $0 = x+(x+(-x)) = x+0 = x$ (associativity in $(R,+)$ , then inverse and identity laws). So $0\cdot a=0$ . The argument for $a\cdot0=0$ is symmetric, using $a(0+0)=a0+a0$ .

Theorem 11.2 — Signs Under Multiplication

For all $a,b\in R$ : (a) $(-a)b = a(-b) = -(ab)$ ; (b) $(-a)(-b) = ab$ .

Proof

(a) $ab + (-a)b = (a+(-a))b = 0\cdot b = 0$ by distributivity and Theorem 11.1. So $(-a)b$ satisfies the defining property of the additive inverse of $ab$ ; by uniqueness of inverses in the group $(R,+)$ (Theorem 1.2), $(-a)b=-(ab)$ . The argument for $a(-b)=-(ab)$ is symmetric.

(b) Apply part (a) twice: $(-a)(-b) = -(a(-b))$ (treating $-a$ as the " $-a$ " in part (a) applied to $a(-b)$ ... more directly: $(-a)(-b) = -(a \cdot(-b))$ by part (a) with $b$ replaced by $-b$ , and $a\cdot(-b) = -(ab)$ by part (a) directly. So $(-a)(-b) = -(-(ab)) = ab$ , using that the additive inverse of an additive inverse returns the original element (a direct consequence of uniqueness of inverses in $(R,+)$ ).

Example — Sanity check in Z

Theorem 11.2 predicts $(-3)(-5)=3\times5=15$ and $(-3)(5) = -(3\times5)=-15$ — exactly ordinary arithmetic. The theorem's content is that these familiar sign rules are forced by the ring axioms alone (distributivity plus additive inverses), not a special feature of the integers.

03 · A Gallery of Rings

Example — $\mathbb{Z}_n$ as a ring

$(\mathbb{Z}_n, +, \times)$ : additive structure from Chapter 01's $\mathbb{Z}_n$ , multiplication $[a][b]:=[ab]$ well-defined by Theorem 1.5. Commutative, unity $[1]$ . A finite commutative ring with unity, for every $n\geq2$ .

Example — $M_n(\mathbb{R})$, a noncommutative ring

All $n\times n$ real matrices under matrix addition and multiplication. $(M_n(\mathbb{R}),+)$ is abelian (componentwise addition). Matrix multiplication is associative and distributes over addition (standard linear algebra facts). Unity: the identity matrix $I$ . Noncommutative for $n\geq2$ (Chapter 02's example), and — unlike $GL_n(\mathbb{R})$ — this ring includes all matrices, not just invertible ones, so most elements have no multiplicative inverse.

Common mistake — 2Z has no unity

$(2\mathbb{Z}, +, \times)$ (even integers): closed under both operations (even+even=even, even $\times$ even=even), abelian additive group (Chapter 01, Exercise 1.1's style argument), associative and distributive (inherited from $\mathbb{Z}$ ). But is there $u\in2\mathbb{Z}$ with $ua=a$ for every even $a$ ? Only $u=1$ could work, and $1\notin2\mathbb{Z}$ . $2\mathbb{Z}$ is a ring without unity — a genuine example showing R1–R3 alone do not force a multiplicative identity to exist.

04 · Units and Zero Divisors

Definition — Unit

An element $a\in R$ (ring with unity) is a unit if there exists $b\in R$ with $ab=ba=1$ . Write $R^\times$ for the set of all units.

Theorem 11.3 — R× Is a Group

$(R^\times, \times)$ is a group.

Proof

Closure: if $a,c\in R^\times$ with inverses $a^{-1},c^{-1}$ , then $(ac)(c^{-1}a^{-1}) = a(cc^{-1})a^{-1} = a\cdot1\cdot a^{-1}=aa^{-1}=1$ , and similarly on the other side, so $ac\in R^\times$ (this is exactly Theorem 2.3's argument, transplanted from groups). Associativity: inherited from $R$ 's multiplication (R2). Identity: $1\in R^\times$ since $1\cdot1=1$ . Inverses: by definition of $R^\times$ , every element already has an inverse inside $R^\times$ (the inverse $a^{-1}$ is itself a unit, with inverse $a$ ). All group axioms hold.

Example — Units of Z

$\mathbb{Z}^\times = \{1,-1\}$ : these are the only integers with an integer multiplicative inverse ( $1\cdot1=1$ , $(-1)(-1)=1$ ; no other integer $n$ has $1/n\in\mathbb{Z}$ unless $n=\pm1$ ).

Theorem 11.4 — Units of $\mathbb{Z}_n$

$[a]\in\mathbb{Z}_n^\times$ if and only if $\gcd(a,n)=1$ .

Proof

( $\Leftarrow$ ) If $\gcd(a,n)=1$ , Bézout's identity gives integers $x,y$ with $ax+ny=1$ . Reducing mod $n$ : $[a][x] + [n][y] = [1]$ , and $[n]=[0]$ , so $[a][x]=[1]$ : $[x]$ is a multiplicative inverse for $[a]$ .

( $\Rightarrow$ ) If $[a][x]=[1]$ for some $x$ , then $ax\equiv1\pmod n$ , so $ax-1=ny$ for some integer $y$ , i.e. $ax-ny=1$ . Any common divisor $d$ of $a$ and $n$ divides the left side $ax-ny$ , hence divides $1$ , forcing $d=1$ : $\gcd(a,n)=1$ .

Example — Units of $\mathbb{Z}_{12}$

$\gcd(a,12)=1$ for $a\in\{1,5,7,11\}$ (the integers in $\{1,\ldots,11\}$ sharing no common factor with $12=2^2\times3$ ). So $\mathbb{Z}_{12}^\times = \{1,5,7,11\}$ , a group of order 4 under multiplication mod 12 — matching Exercise 1.5's direct computation for $\mathbb{Z}_5$ as a special case where $n$ is prime (every nonzero remainder is automatically coprime to a prime).

Definition — Zero Divisor

A nonzero $a\in R$ is a zero divisor if $ab=0$ for some nonzero $b\in R$ .

Example — Zero divisors in $\mathbb{Z}_4$ and $M_2(\mathbb{R})$

In $\mathbb{Z}_4$ : $[2][2]=[4]=[0]$ , and $[2]\neq[0]$ — so $[2]$ is a zero divisor (Chapter 01's FailBlock on $\mathbb{Z}_4$ , now named). In $M_2(\mathbb{R})$ : $\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$ , two nonzero matrices multiplying to the zero matrix — zero divisors exist even in this infinite, familiar ring.

Definition — Integral Domain

A commutative ring with unity ( $1\neq0$ ) having no zero divisors is called an integral domain.

Theorem 11.5 — Cancellation in Domains

In an integral domain $R$ , if $ab=ac$ and $a\neq0$ , then $b=c$ .

Proof

$ab=ac \implies ab-ac=0 \implies a(b-c)=0$ (distributivity and Theorem 11.2). Since $R$ has no zero divisors and $a\neq0$ , the factor $(b-c)$ must be $0$ (otherwise $a$ would be a zero divisor, witnessed by the nonzero $b-c$ ). So $b-c=0$ , i.e. $b=c$ .

Common mistake — Cancellation fails outside domains

In $\mathbb{Z}_4$ : $[2][1]=[2]$ and $[2][3]=[6]=[2]$ , so $[2][1]=[2][3]$ — but $[1]\neq[3]$ . Cancellation fails because $[2]$ is a zero divisor, so $\mathbb{Z}_4$ is not an integral domain, and Theorem 11.5 simply doesn't apply.

05 · Subrings

Definition — Subring

A subset $S\subseteq R$ is a subring if $S$ is itself a ring under $R$ 's operations (restricted to $S$ ).

Theorem 11.6 — Subring Test

A nonempty subset $S\subseteq R$ is a subring if and only if: (S1) $a,b\in S \implies a-b\in S$ ; (S2) $a,b\in S \implies ab\in S$ .

Proof

( $\Rightarrow$ ) A subring satisfies its own R1 (in particular, closed under $+$ and additive inverses, so closed under subtraction) and R2/closure of $\times$ .

( $\Leftarrow$ ) S1 is exactly the subgroup test (Theorem 3.1) applied to $(S,+)$ inside $(R,+)$ — recall subtraction closure is equivalent to closure under $+$ and inverses simultaneously (if $a,b\in S$ then $0=a-a\in S$ , then $-b=0-b\in S$ , then $a+b=a-(-b)\in S$ ). So $(S,+)$ is an abelian subgroup of $(R,+)$ . S2 gives closure of $\times$ . Associativity of $\times$ and both distributive laws are inherited from $R$ , since they hold for all elements of $R$ , hence for the subset $S$ . All ring axioms hold for $S$ .

Example — $\mathbb{Z} \le \mathbb{Q} \le \mathbb{R} \le \mathbb{C}$, a chain of subrings

Each is closed under subtraction and multiplication within the next: integers are closed under both inside $\mathbb{Q}$ , and so on up the chain. Each number system in this familiar chain is a subring of the next.

Common mistake — Verifying only closure under multiplication is not enough

Is $\mathbb{N}=\{0,1,2,\ldots\}$ a subring of $\mathbb{Z}$ ? Closed under multiplication: yes. Closed under subtraction: no — $2-5=-3\notin\mathbb{N}$ . Fails S1, so $\mathbb{N}$ is not a subring, even though it's closed under $\times$ and contains $0,1$ . Both parts of Theorem 11.6 are required.

06 · Exercises

EXERCISE 11.1

Apply Theorem 11.2 directly to a familiar ring, treating $-3$ and $-5$ as $-(3)$ and $-(5)$ .

By Theorem 11.2(b), $(-a)(-b)=ab$ for any ring. With $a=3,b=5$ in $\mathbb{Z}$ : $(-3)(-5) = 3\times5 = 15$ — forced by the ring axioms (distributivity + additive inverses), not a separate arithmetic fact.

Use Theorem 11.2 to justify why $(-3)\times(-5) = 15$ in $\mathbb{Z}$ , citing the specific theorem rather than "common knowledge."

EXERCISE 11.2

Apply Theorem 11.4: which elements of $\{1,\ldots,7\}$ share no common factor with 8?

$8=2^3$ . Elements coprime to 8 among $\{1,\ldots,7\}$ : exclude even numbers ( $2,4,6$ ), leaving $\{1,3,5,7\}$ . By Theorem 11.4, $\mathbb{Z}_8^\times = \{1,3,5,7\}$ .

Find $\mathbb{Z}_8^\times$ using Theorem 11.4.

EXERCISE 11.3

Check the definition of integral domain directly: does $\mathbb{Z}_6$ have any zero divisors?

$[2][3] = [6] = [0]$ in $\mathbb{Z}_6$ , with both $[2]$ and $[3]$ nonzero. So $[2]$ (and $[3]$ ) are zero divisors. $\mathbb{Z}_6$ is not an integral domain. (This matches Theorem 11.4-style reasoning: $6$ is not prime, so some nonzero remainders share a common factor with $6$ — namely $2$ and $3$ — and it is exactly those shared-factor pairs that multiply to $[0]$ .)

Is $\mathbb{Z}_6$ an integral domain? Exhibit a zero divisor if not.

EXERCISE 11.4

Apply Theorem 11.6: check closure under subtraction and multiplication for matrices with even integer entries.

Let $S = \{A \in M_2(\mathbb{Z}) : \text{every entry of } A \text{ is even}\}$ . Subtraction: entrywise, even minus even is even, so $A-B\in S$ . Multiplication: each entry of $AB$ is a sum of products of entries from $A$ and $B$ ; since every entry of $A$ (and $B$ ) is even, each such product is a multiple of $4$ (hence even), and a sum of even numbers is even, so $AB\in S$ . Both conditions of Theorem 11.6 hold: $S$ is a subring of $M_2(\mathbb{Z})$ — notably without unity, since $I\notin S$ (its diagonal entries are $1$ , odd).

Show that the set of $2\times2$ integer matrices with all-even entries is a subring of $M_2(\mathbb{Z})$ .

07 · Chapter Summary

Concept	Statement
Ring	$(R,+,\times)$ : $(R,+)$ abelian group, $\times$ associative, distributive over $+$
$0\cdot a = a\cdot0=0$	Theorem 11.1
Sign rules	$(-a)b=a(-b)=-(ab)$ ; $(-a)(-b)=ab$ (Thm 11.2)
Unit	$a$ with a multiplicative inverse; $R^\times$ is always a group (Thm 11.3)
Units of $\mathbb{Z}_n$	$[a]\in\mathbb{Z}_n^\times \iff \gcd(a,n)=1$ (Thm 11.4)
Zero divisor	Nonzero $a$ with $ab=0$ for some nonzero $b$
Integral domain	Commutative, unital, no zero divisors
Cancellation in domains	$ab=ac,\,a\neq0 \implies b=c$ (Thm 11.5)
Subring test	Closed under subtraction and multiplication (Thm 11.6)

Next: Chapter 12 — Ideals & Quotient Rings identifies the ring-theoretic analog of a normal subgroup — a subset absorbing multiplication from the whole ring — and uses it to build quotient rings the same way Chapter 05 built quotient groups.