Rings: Definitions and Examples

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.¹

The axioms

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.

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.¹ 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.

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.

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 $0$.¹ The residues coprime to $n$ are the units, which recovers the group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ met in modular arithmetic.

The hierarchy of rings

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

A field is thus a commutative ring with $1\ne 0$ 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.

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

The proof is immediate: $ab=ac$ gives $a(b-c)=0$, and absence of zero divisors forces $b-c=0$.¹ What a zero divisor destroys is cancellation.

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.² Wedderburn's theorem sharpens this: a finite division ring is automatically commutative, hence a field, though that lies deeper.

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.¹ 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.

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.³ 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

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,

whose denominator is nonzero because $D$ is not a square. These quadratic fields recur throughout field theory.¹ Inside each sits a ring of integers

the analogue of $\mathbb{Z}\subset \mathbb{Q}$. When $D=-1$ this is the ring of Gaussian integers $\mathbb{Z}[i]$. The field norm $N(a+b\sqrt{D})=a^2-D{b}^2$ 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.

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.

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,

The base ring sits inside as the constant polynomials. Behavior over a domain is clean.

The degree formula holds because the leading coefficients multiply to $a_n{b}_m\ne 0$ in the domain $R$, which simultaneously rules out zero divisors in $R[x]$ and forces any unit to have degree $0$.⁴ 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.

Matrix rings

$M_n(R)$ is the set of $n\times n$ matrices over $R$ with the usual matrix addition and multiplication. For $n\ge 2$ 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 $AB\ne 0$ 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 $G{L}_n(R)$ met among the matrix groups. The scalar matrices form a subring isomorphic to $R$.

Group rings

For a commutative ring $R$ with $1\ne 0$ and a finite group $G=\{g_1,\dots ,g_n\}$, 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

Group rings over $\mathbb{C}$ underlie representation theory. Note $R{Q}_8$ 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.⁴ The dihedral and quaternion groups supply the first nontrivial group rings.