---
title: "Rings: Definitions and Examples"
module: Ring Theory
moduleNumber: 6
lessonNumber: 1
order: 601
summary: >
  A ring carries two operations: an abelian group under addition and an
  associative multiplication linked by the distributive laws. The named special
  cases — commutative rings, integral domains, division rings, and fields —
  differ only in how their multiplication behaves. Standard examples include
  quadratic integer rings, polynomial rings, matrix rings, and group rings.
topics: [Ring Theory]
sources:
  - book: Dummit & Foote
    ref: "Ch. 7 — Introduction to Rings; §7.1 Basic Definitions and Examples"
  - book: Dummit & Foote
    ref: "§7.2 Polynomial Rings, Matrix Rings, and Group Rings"
  - book: Judson
    ref: "Ch. 16 — Rings; §16.1 Rings, §16.2 Integral Domains and Fields"
draft: false
---

Group theory keeps one operation. Arithmetic keeps two: the integers can be added
and multiplied, and the two operations are tied together by the rule
$a(b + c) = ab + ac$. A **ring** is the axiomatization of exactly that structure —
an abelian group under addition, a second associative operation called
multiplication, and the distributive laws that connect them.[^df-def]

## The axioms

> **Definition (Ring).** A ring $R$ is a set with two binary operations $+$ and
> $\times$ such that
>
> - $(R, +)$ is an abelian group, with additive identity $0$ and additive
>   inverse $-a$ for each $a$;
> - $\times$ is associative: $(ab)c = a(bc)$ for all $a, b, c \in R$;
> - the distributive laws hold: $(a + b)c = ac + bc$ and $a(b + c) = ab + ac$.
>
> $R$ is **commutative** if $ab = ba$ for all $a, b$. $R$ **has an identity** (a
> $1$) if there is an element $1 \in R$ with $1 \cdot a = a \cdot 1 = a$ for all
> $a$.

We write $ab$ for $a \times b$. Two conventions deserve a word. First, addition is
required to be commutative, which looks like an extra demand but in a ring with $1$
is forced: expanding $(1 + 1)(a + b)$ by the two distributive
laws in the two possible orders gives $a + b + a + b$ one way and $a + a + b + b$
the other, and equating them yields $b + a = a + b$. Second, multiplication is
_not_ assumed to have inverses, or an identity, or commutativity. Those are the
degrees of freedom that separate one class of ring from another.

The distributive laws already dictate how $0$ and negatives interact with
multiplication.

> **Proposition (Arithmetic of $0$ and negatives).** In any ring $R$, for all
> $a, b \in R$:
>
> - $0a = a0 = 0$;
> - $(-a)b = a(-b) = -(ab)$;
> - $(-a)(-b) = ab$;
> - if $R$ has a $1$, that identity is unique and $-a = (-1)a$.

Each follows from cancellation in the additive group. For the first,
$0a = (0 + 0)a = 0a + 0a$, so $0a = 0$; the rest are one line apiece.[^df-def] The
familiar sign rules of elementary arithmetic are thus consequences of the axioms,
not separate assumptions.

## Units and zero divisors

Two features of the multiplicative structure organize the whole theory: whether an
element can be inverted, and whether it can multiply a nonzero element to zero.

> **Definition (Unit and zero divisor).** Let $R$ be a ring.
>
> - A nonzero $a \in R$ is a **zero divisor** if there is a nonzero $b \in R$
>   with $ab = 0$ or $ba = 0$.
> - Assume $R$ has $1 \neq 0$. A **unit** is an element $u$ with a two-sided
>   inverse $v$, meaning $uv = vu = 1$. The units form a group under
>   multiplication, denoted $R^\times$, the **group of units** of $R$.

These two conditions are mutually exclusive. If $a$ is a unit with $va = 1$ and
$ab = 0$, then $b = 1b = (va)b = v(ab) = 0$, so $a$ is not a zero divisor. A unit
can be inverted; a zero divisor obstructs cancellation. In the integers $\mathbb{Z}$ the only units are $\pm 1$ and there
are no zero divisors at all. In $\mathbb{Z}/6\mathbb{Z}$ both phenomena appear, and
the multiplication table shows where.

$$
% caption: The multiplication table of Z/6Z. Shaded cells are products of two
% nonzero elements that equal zero, so 2, 3, 4 are zero divisors; 1 and 5 are the
% units.
\begin{tikzpicture}[scale=0.82, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% shaded nontrivial-zero cells (nonzero times nonzero = 0)
\foreach \x/\y in {4/4, 3/3, 5/3, 4/2}{
  \fill[acc!16] (\x-0.5,\y-0.5) rectangle (\x+0.5,\y+0.5);
  \draw[acc] (\x-0.5,\y-0.5) rectangle (\x+0.5,\y+0.5);
}
% header labels
\foreach \c/\lab in {1/0,2/1,3/2,4/3,5/4,6/5}{
  \node[font=\bfseries] at (\c,7) {\lab};
  \node[font=\bfseries] at (0,{7-\c}) {\lab};
}
\node[font=\footnotesize] at (0,7) {$ab$};
% outer border and grid
\draw[black] (0.5,0.5) rectangle (6.5,6.5);
\foreach \i in {1,...,5}{
  \draw[black] (0.5,\i+0.5) -- (6.5,\i+0.5);
  \draw[black] (\i+0.5,0.5) -- (\i+0.5,6.5);
}
% entries: x=column (b+1), y=row (6-a), v=product
\foreach \x/\y/\v in {%
  1/6/0, 2/6/0, 3/6/0, 4/6/0, 5/6/0, 6/6/0,
  1/5/0, 2/5/1, 3/5/2, 4/5/3, 5/5/4, 6/5/5,
  1/4/0, 2/4/2, 3/4/4, 4/4/0, 5/4/2, 6/4/4,
  1/3/0, 2/3/3, 3/3/0, 4/3/3, 5/3/0, 6/3/3,
  1/2/0, 2/2/4, 3/2/2, 4/2/0, 5/2/4, 6/2/2,
  1/1/0, 2/1/5, 3/1/4, 4/1/3, 5/1/2, 6/1/1}{
  \node at (\x,\y) {\v};
}
\node[text=acc, anchor=west, font=\footnotesize] at (7.0,3.5) {zero divisors: $2$, $3$, $4$};
\node[anchor=west, font=\footnotesize] at (7.0,2.7) {units: $1$, $5$};
\end{tikzpicture}
$$

Every nonzero element of $\mathbb{Z}/6\mathbb{Z}$ is exactly one of the two. That
is not an accident of $6$: in $\mathbb{Z}/n\mathbb{Z}$ the class $\bar a$ is a unit
when $\gcd(a, n) = 1$ and a zero divisor otherwise, so the classes split cleanly
into units and zero divisors together with $0$.[^df-def] The residues coprime to
$n$ are the units, which recovers the group $(\mathbb{Z}/n\mathbb{Z})^\times$ met in
[modular arithmetic](/abstract-algebra/foundations/integers-and-modular-arithmetic).

$$
% caption: Every nonzero element of a finite commutative ring with identity is
% either a unit or a zero divisor; here the split for Z/6Z.
\begin{tikzpicture}[font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) rectangle (10,2.4);
\node[anchor=south west, font=\footnotesize\itshape] at (0.05,2.45) {the ring $Z_6$};
% zero
\draw[black, fill=black!6] (0.3,0.4) rectangle (2.0,2.0);
\node at (1.15,1.55) {$0$};
\node[font=\footnotesize, black] at (1.15,0.75) {zero};
% units
\draw[acc, fill=acc!10] (2.4,0.4) rectangle (5.4,2.0);
\node at (3.4,1.4) {$1$}; \node at (4.4,1.4) {$5$};
\node[font=\footnotesize, text=acc] at (3.9,0.72) {units};
% zero divisors
\draw[black, fill=acc!5] (5.8,0.4) rectangle (9.7,2.0);
\node at (6.7,1.4) {$2$}; \node at (7.75,1.4) {$3$}; \node at (8.8,1.4) {$4$};
\node[font=\footnotesize, black] at (7.75,0.72) {zero divisors};
\end{tikzpicture}
$$

> **Worked example.** In $\mathbb{Z}/12\mathbb{Z}$, sort the nonzero classes into
> units and zero divisors.
>
> A class $\bar a$ is a unit exactly when $\gcd(a, 12) = 1$. The residues coprime
> to $12$ are $1, 5, 7, 11$, so
> $$
> (\mathbb{Z}/12\mathbb{Z})^\times = \{1, 5, 7, 11\}, \qquad \lvert(\mathbb{Z}/12\mathbb{Z})^\times\rvert = \varphi(12) = 4.
> $$
> Each is its own inverse: $5^2 = 25 = 1$, $7^2 = 49 = 1$, and $11^2 = 121 = 1$
> modulo $12$. The remaining nonzero classes $2, 3, 4, 6, 8, 9, 10$ each share a
> factor with $12$ and kill a suitable partner:
> $$
> 3 \cdot 4 = 12 = 0, \qquad 2 \cdot 6 = 12 = 0, \qquad 8 \cdot 9 = 72 = 0, \qquad 6 \cdot 10 = 60 = 0.
> $$
> The tally — $4$ units, $7$ zero divisors, and $0$ — accounts for all $12$
> classes.

## The hierarchy of rings

The divisions are cumulative: each named class adds one requirement to the previous.

> **Definition (Domain, division ring, field).** Let $R$ have $1 \neq 0$.
>
> - An **integral domain** is a commutative ring with no zero divisors.
> - A **division ring** (or skew field) is a ring in which every nonzero element
>   is a unit.
> - A **field** is a commutative division ring.

A field is thus a commutative ring with $1 \neq 0$ in which $R^\times = R
\setminus \{0\}$: every nonzero element inverts. Fields have no zero divisors — a
unit never is one — so every field is an integral domain, and the classes nest.

$$
% caption: The four named classes nest by strength: each inner region adds a
% requirement to the one containing it.
\begin{tikzpicture}[font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) rectangle (11.2,6.2);
\node[anchor=north west, black] at (0.15,6.05) {rings};
\draw[black, fill=black!4] (0.55,0.5) rectangle (10.7,5.3);
\node[anchor=north west, black] at (0.7,5.15) {rings with $1$};
\draw[black, fill=acc!5] (1.15,1.0) rectangle (10.1,4.4);
\node[anchor=north west, black] at (1.3,4.25) {commutative rings with $1$};
\draw[acc, fill=acc!9] (2.0,1.5) rectangle (9.3,3.5);
\node[anchor=north west, text=acc] at (2.15,3.35) {integral domains (no zero divisors)};
\draw[acc, thick, fill=acc!16] (3.2,1.9) rectangle (8.1,2.9);
\node[text=acc, font=\footnotesize\bfseries] at (5.65,2.4) {f\/ields};
% example tags placed in the correct band
\node[black] at (8.7,3.9) {$Z_6$};
\node[black] at (8.7,2.4) {$Z$};
\node[black] at (7.3,2.4) {$Q$};
\node[black] at (5.5,0.24) {$2Z$};
\end{tikzpicture}
$$

Two small results tie the classes together in the finite case.

> **Proposition (Cancellation in a domain).** If $a$ is not a zero divisor and
> $ab = ac$, then $a = 0$ or $b = c$. In particular, cancellation holds for every
> nonzero element of an integral domain.

The proof is immediate: $ab = ac$ gives $a(b - c) = 0$, and absence of zero
divisors forces $b - c = 0$.[^df-def] What a zero divisor destroys is
cancellation.

> **Theorem (Finite domains are fields).** Every finite integral domain is a
> field.

Fix a nonzero $a$ in a finite domain $R$. Cancellation makes the map $x \mapsto ax$
injective, and an injective self-map of a finite set is surjective, so some $b$ has
$ab = 1$. Then $a$ is a unit; since $a$ was arbitrary, $R$ is a field.[^ju-domain]
Wedderburn's theorem sharpens this: a _finite division ring_ is automatically
commutative, hence a field, though that lies deeper.

| Class | $1$? | commutative? | every nonzero element | zero divisors |
| --- | --- | --- | --- | --- |
| Ring | maybe | maybe | — | possible |
| Commutative ring with $1$ | yes | yes | — | possible |
| Integral domain | yes | yes | cancellable | none |
| Division ring | yes | maybe | a unit | none |
| Field | yes | yes | a unit | none |

## Subrings and the characteristic

A **subring** of $R$ is a subgroup of $(R, +)$ closed under multiplication. To
check that $S \subseteq R$ is a subring it suffices that $S$ is nonempty and closed
under subtraction and multiplication.[^df-def] Thus $\mathbb{Z} \subset \mathbb{Q}
\subset \mathbb{R} \subset \mathbb{C}$ is a chain of subrings, and $2\mathbb{Z}$ is
a subring of $\mathbb{Z}$ (a ring without identity, since $2\mathbb{Z}$ has no $1$).

One numerical invariant records how the identity behaves under repeated addition.

> **Definition (Characteristic).** The **characteristic** of a ring $R$ with $1$
> is the smallest positive integer $n$ with $\underbrace{1 + \cdots + 1}_{n} = 0$;
> if no such $n$ exists the characteristic is $0$.

So $\mathbb{Z}/n\mathbb{Z}$ has characteristic $n$ and $\mathbb{Z}$ has
characteristic $0$. The map $k \mapsto k \cdot 1$ is a ring homomorphism
$\mathbb{Z} \to R$ whose kernel is $n\mathbb{Z}$ where $n$ is the characteristic,
which is why "characteristic $0$" (kernel $0$) is the right name.[^df-quot] An
integral domain has characteristic either $0$ or a prime $p$: if the characteristic
factored as $n = ab$ with $1 < a, b < n$, then $(a \cdot 1)(b \cdot 1) = n \cdot 1
= 0$ would exhibit zero divisors. In characteristic $p$ the map $a \mapsto a^p$
satisfies $(a + b)^p = a^p + b^p$, the Frobenius identity that drives much of finite
field theory.

## Quadratic fields and their integers

For a squarefree integer $D$ that is not a perfect square, the set
$$
\mathbb{Q}(\sqrt{D}) = \{\, a + b\sqrt{D} \mid a, b \in \mathbb{Q} \,\}
$$
is a subring of $\mathbb{C}$, and in fact a field: closure under multiplication is
the identity $(a + b\sqrt{D})(c + d\sqrt{D}) = (ac + bdD) + (ad + bc)\sqrt{D}$, and
the inverse of a nonzero element comes from rationalizing,
$$
(a + b\sqrt{D})^{-1} = \frac{a - b\sqrt{D}}{a^2 - Db^2},
$$
whose denominator is nonzero because $D$ is not a square. These **quadratic fields**
recur throughout field theory.[^df-def] Inside each sits a ring of integers
$$
\mathcal{O} = \mathbb{Z}[\omega], \qquad
\omega = \begin{cases} \sqrt{D}, & D \equiv 2, 3 \pmod 4, \\[2pt]
\dfrac{1 + \sqrt{D}}{2}, & D \equiv 1 \pmod 4, \end{cases}
$$
the analogue of $\mathbb{Z} \subset \mathbb{Q}$. When $D = -1$ this is the ring of
**Gaussian integers** $\mathbb{Z}[i]$. The **field norm** $N(a + b\sqrt{D}) = a^2 -
Db^2$ is multiplicative, $N(\alpha\beta) = N(\alpha)N(\beta)$, and it detects units
exactly: $\alpha \in \mathcal{O}$ is a unit if and only if $N(\alpha) = \pm 1$.
Solving $N(\alpha) = \pm 1$ is Pell's equation, so the arithmetic of units in
$\mathcal{O}$ is the arithmetic of an old Diophantine problem. This norm returns as
the Euclidean function that makes $\mathbb{Z}[i]$ a
[Euclidean domain](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds).

> **Worked example.** Find the units of the Gaussian integers $\mathbb{Z}[i]$.
>
> Here $D = -1$, so the norm is $N(a + bi) = a^2 + b^2$, a nonnegative integer. A
> unit satisfies $N(\alpha) = \pm 1$, and since the norm cannot be negative this
> forces $N(\alpha) = 1$:
> $$
> a^2 + b^2 = 1, \qquad a, b \in \mathbb{Z}.
> $$
> The only integer solutions are $(\pm 1, 0)$ and $(0, \pm 1)$, so
> $$
> \mathbb{Z}[i]^\times = \{1, -1, i, -i\},
> $$
> a cyclic group of order $4$ generated by $i$.

## Three constructions that manufacture rings

Fix a ring $R$. Three standard constructions produce larger rings from it. The
pattern is the same in all three: take formal objects
with coefficients in $R$, add componentwise, and multiply by the rule the objects'
names suggest.

$$
% caption: Three ways to build a larger ring on top of a base ring R: adjoin an
% indeterminate, arrange entries in a square array, or attach coefficients to
% group elements.
\begin{tikzpicture}[font=\footnotesize,
  box/.style={draw=black, minimum width=32mm, minimum height=20mm, align=center}]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw=acc, thick, fill=acc!8, minimum width=20mm, minimum height=11mm] (R) at (0,0) {base ring $R$};
\node[box] (poly) at (4.6,2.4) {\textbf{polynomial ring}\\[2pt] $R[x]$\\ terms $a_i x^i$};
\node[box] (mat)  at (4.6,0)   {\textbf{matrix ring}\\[2pt] $M_n(R)$\\ arrays $(a_{ij})$};
\node[box] (grp)  at (4.6,-2.4){\textbf{group ring}\\[2pt] $RG$\\ formal sums $a_g\, g$};
\draw[->, acc, thick] (R) -- (poly);
\draw[->, acc, thick] (R) -- (mat);
\draw[->, acc, thick] (R) -- (grp);
\node[black, anchor=west, font=\scriptsize] at (7.1,2.4) {adjoin an indeterminate};
\node[black, anchor=west, font=\scriptsize] at (7.1,0) {$n$ by $n$ entries};
\node[black, anchor=west, font=\scriptsize] at (7.1,-2.4) {scalars on a group};
\end{tikzpicture}
$$

### Polynomial rings

A **polynomial** in $R[x]$ is a formal sum $a_n x^n + \cdots + a_1 x + a_0$ with
each $a_i \in R$. Addition is componentwise; multiplication expands and collects,
$$
\Big(\sum_i a_i x^i\Big)\Big(\sum_j b_j x^j\Big) = \sum_k \Big(\sum_{i + j = k} a_i b_j\Big) x^k.
$$
The base ring sits inside as the constant polynomials. Behavior over a domain is
clean.

> **Proposition (Polynomials over a domain).** If $R$ is an integral domain and
> $p, q \in R[x]$ are nonzero, then
>
> - $\deg(pq) = \deg p + \deg q$;
> - $R[x]$ and $R$ have the same units;
> - $R[x]$ is an integral domain.

The degree formula holds because the leading coefficients multiply to $a_n b_m \neq
0$ in the domain $R$, which simultaneously rules out zero divisors in $R[x]$ and
forces any unit to have degree $0$.[^df-poly] The coefficient ring controls
everything: $x^2 + 1$ is not a square in $\mathbb{Z}[x]$, yet in
$\mathbb{Z}/2\mathbb{Z}[x]$ it equals $(x + 1)^2$. Polynomial rings over a field are
the subject of a
[later module](/abstract-algebra/factorization-and-polynomials/polynomial-rings-over-fields).

> **Worked example.** The degree formula and the unit description both need the
> domain hypothesis. Over $R = \mathbb{Z}/4\mathbb{Z}$, which is not a domain,
> $$
> (2x + 1)^2 = 4x^2 + 4x + 1 = 1,
> $$
> since $4 = 0$ in $R$. So $2x + 1$ is a unit of $R[x]$ of degree $1$, and it is
> its own inverse. The zero divisor $2 \in R$ makes a degree-$1$ polynomial
> invertible, which cannot happen over a domain, where the units of $R[x]$ are
> exactly the units of $R$.

### Matrix rings

$M_n(R)$ is the set of $n \times n$ matrices over $R$ with the usual matrix
addition and multiplication. For $n \geq 2$ and any nonzero $R$ it is
_noncommutative_ and _has zero divisors_: if $A$ has a single nonzero entry in
position $(1,1)$ and $B$ a single nonzero entry in position $(1,2)$, then $AB \neq
0$ while $BA = 0$. When $R$ has $1$, the identity matrix is the $1$ of $M_n(R)$ and
its units are the invertible matrices, the group $GL_n(R)$ met among the
[matrix groups](/abstract-algebra/groups-and-symmetry/matrix-and-quaternion-groups).
The scalar matrices form a subring isomorphic to $R$.

### Group rings

For a commutative ring $R$ with $1 \neq 0$ and a finite group $G = \{g_1, \ldots,
g_n\}$, the **group ring** $RG$ consists of all formal sums $\sum_{i} a_i g_i$ with
$a_i \in R$. Addition is componentwise; multiplication uses the group operation on
the basis and distributes, so the coefficient of $g_k$ in a product is $\sum_{g_i
g_j = g_k} a_i b_j$. The ring $RG$ is commutative exactly when $G$ is. Both $R$ (as
$R \cdot 1_G$) and $G$ (as $1 \cdot g$) sit inside $RG$, and $G$ lands inside the
unit group. Whenever $|G| > 1$ there are zero divisors: an element $g$ of order $m >
1$ gives
$$
(1 - g)(1 + g + \cdots + g^{m-1}) = 1 - g^m = 0.
$$
Group rings over $\mathbb{C}$ underlie representation theory. Note $RQ_8$ is _not_
the Hamilton quaternions: in the group ring the element $g_2$ of order $2$ is not
the additive inverse of $1$, whereas in the quaternions $-1$ is.[^df-poly] The
[dihedral](/abstract-algebra/groups-and-symmetry/dihedral-and-symmetric-groups) and
[quaternion](/abstract-algebra/groups-and-symmetry/matrix-and-quaternion-groups)
groups supply the first nontrivial group rings.

[^df-def]: **Dummit & Foote**, _Abstract Algebra_, §7.1 — Basic Definitions and Examples: the ring axioms, the forced commutativity of addition, units and zero divisors, integral domains and the cancellation law, subrings, and the quadratic field / quadratic integer ring examples with the field norm.
[^df-poly]: **Dummit & Foote**, _Abstract Algebra_, §7.2 — Examples: Polynomial, Matrix, and Group Rings: construction of $R[x]$, $M_n(R)$, and $RG$, degrees and units of $R[x]$ over a domain, noncommutativity and zero divisors in $M_n(R)$, and zero divisors in $RG$.
[^df-quot]: **Dummit & Foote**, _Abstract Algebra_, §7.3, Exercise 26 — the characteristic of a ring as the kernel of $\mathbb{Z} \to R$, and the Frobenius identity $(a+b)^p = a^p + b^p$ in characteristic $p$.
[^ju-domain]: **Judson**, _Abstract Algebra: Theory and Applications_, §16.2 — Integral Domains and Fields: the cancellation law and the theorem that every finite integral domain is a field.