Abstract AlgebraRings and Ideals

Rings

A ring is an abelian group under addition with an associative multiplication that distributes over it; ℤ, ℤₙ, matrices and polynomials are rings, and the units are the elements with multiplicative inverses.

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Integers, matrices, polynomials and residues mod n can all be added and multiplied. A ring is the structure with both operations, linked by the distributive law. Everything you learned about groups applies to the addition.

Definition. A ring is a set $R$ with two operations, $+$ and $\cdot$, such that

  • $(R, +)$ is an abelian group (identity $0$, inverses $-a$);
  • multiplication is associative: $(ab)c = a(bc)$;
  • multiplication distributes over addition: $a(b + c) = ab + ac$ and $(a + b)c = ac + bc$.

A ring is commutative if $ab = ba$ always, and has an identity (or unity) if some $1 \ne 0$ has $1a = a1 = a$. A unit is an element with a multiplicative inverse; the units form a group $U(R)$ under multiplication. Examples: $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{Z}_{n}$ (units $U(n)$), polynomials $\mathbb{R}[x]$, the $2 \times 2$ real matrices (not commutative; units are the invertible matrices), and $2\mathbb{Z}$ (no identity).

Why it works

Rules that follow from the axioms. In any ring, $0a = 0$ and $(-a)b = -(ab)$.
Proof. $0a = (0 + 0)a = 0a + 0a$ by distributivity; adding $-(0a)$ to both sides gives $0 = 0a$. Next, $ab + (-a)b = (a + (-a))b = 0b = 0$, so $(-a)b$ is the additive inverse of $ab$. $\square$ So "a negative times a positive is negative" is not a convention: it is forced by distributivity. Applying the second rule twice gives $(-a)(-b) = ab$.
The multiplication table of the ring $\mathbb{Z}_{6}$. The units $1$ and $5$ have a $1$ in their rows; $2 \cdot 3 = 0$ although neither factor is $0$.

Worked example: Units of a product ring

How many units does $\mathbb{Z}_{4} \times \mathbb{Z}_{9}$ have (operations coordinate by coordinate)?
  1. $(a, b)$ has an inverse exactly when $a$ is a unit of $\mathbb{Z}_{4}$ and $b$ is a unit of $\mathbb{Z}_{9}$.
  2. $|U(4)| = 2$ and $|U(9)| = 6$.

Answer: $2 \cdot 6 = 12$.

Worked example: Matrices over $\mathbb{Z}_{2}$ do not commute

In the ring of $2 \times 2$ matrices over $\mathbb{Z}_{2}$, compare $AB$ and $BA$ for $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.
  1. $AB = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ (mod 2).
  2. $BA = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.

Answer: $AB \ne BA$: a finite noncommutative ring.

The characteristic of a ring with identity is the least $n \ge 1$ with $n \cdot 1 = 1 + \cdots + 1 = 0$, or $0$ if there is none: $\mathbb{Z}_{n}$ has characteristic $n$, $\mathbb{Z}$ and $\mathbb{Q}$ have characteristic $0$, and $\mathbb{Z}_{4} \times \mathbb{Z}_{6}$ has characteristic $\operatorname{lcm}(4, 6) = 12$.

Common mistake

Assuming a ring's multiplication commutes, or that a ring has a $1$. Matrix rings fail the first ($AB \ne BA$ above), and $2\mathbb{Z}$ fails the second: no even number $u$ has $u \cdot 2 = 2$. Many statements about "rings" in this course say "commutative ring with identity" for exactly this reason.

Key idea

A ring is an abelian group under $+$ with an associative, distributive multiplication. Units are the invertible elements. From the axioms, $0a = 0$ and $(-a)b = -(ab)$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 16, Rings (section 16.1, Rings)
  • MIT OpenCourseWare 18.702 Algebra II (Spring 2011), lecture notes on rings
  • John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 5.1, Commutative Rings
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