Rings: Definition and Examples

Ring theory is the common language underlying an enormous range of mathematics and its applications. In algebraic geometry, polynomial rings $k[x_1,\dots,x_n]$ and their quotients encode geometric objects (varieties), with ring-theoretic operations translating directly into geometric ones. In algebraic number theory, rings of integers in number fields generalize $\mathbb{Z}$, and their ideal structure controls questions of unique factorization. In cryptography, the ring $\mathbb{Z}/n\mathbb{Z}$ from Example 1 is the literal arithmetic engine of RSA encryption, where the failure of general cancellation and the structure of units modulo a composite $n$ are exploited directly. In linear algebra and physics, matrix rings such as $M_n(\mathbb{R})$ or $M_n(\mathbb{C})$ model linear operators, quantum observables, and symmetry transformations, with noncommutativity ($AB\ne BA$) reflecting physically meaningful phenomena such as the uncertainty principle. In computer science, finite rings like $\mathbb{Z}/2^{32}\mathbb{Z}$ literally model fixed-width integer arithmetic on computer hardware, wraparound overflow included. In functional analysis and topology, rings of continuous or measurable functions $\mathrm{Fun}(X,\mathbb{R})$ encode topological or measure-theoretic information about the space $X$ in purely algebraic terms.

Ask the tutor: How does the definition of a ring change once we additionally require every nonzero element to have a multiplicative inverse, and why does that jump take us all the way to fields?

Think of a ring as a machine with two dials, addition and multiplication, that satisfy the arithmetic reflexes ingrained from working with ordinary numbers: you can add and subtract freely (that is the abelian group axiom), you can multiply associatively, and multiplication distributes over addition so that "expanding brackets" always works. What the axioms deliberately omit is just as important as what they include. There is no axiom forcing $ab = ba$ — matrices under multiplication famously fail this — and there is no axiom forcing the existence of a multiplicative inverse, or even of $1$ itself. A useful mental picture is a layered diagram: abelian groups form the base layer (just addition), rings add a compatible multiplication on top, and further conditions (commutativity, existence of $1$, existence of inverses) carve out smaller and smaller sub-classes inside the ring world. The diagram below shows this nesting using concrete inhabitants of each region.

Reading the diagram from outside in: dropping the identity requirement gives the widest class (e.g. $2\mathbb{Z}$); requiring an identity but not commutativity gives matrix rings; requiring both commutativity and an identity gives $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$; requiring every nonzero element to be invertible narrows all the way down to fields such as $\mathbb{Q}$. Each ring example in this article can be located in exactly one of these nested regions.

Example 1: The ring $\mathbb{Z}/12\mathbb{Z}$, its zero-divisors, and the failure of cancellation. Let $R = \mathbb{Z}/12\mathbb{Z} = \{\bar 0, \bar 1, \dots, \bar{11}\}$, the set of congruence classes modulo $12$, with

These operations are well-defined: if $a \equiv a' \pmod{12}$ and $b \equiv b' \pmod{12}$, write $a' = a + 12k$, $b' = b + 12\ell$; then $a'+b' = (a+b) + 12(k+\ell) \equiv a+b$, and

so both classes $\overline{a'+b'}$ and $\overline{a'b'}$ agree with $\overline{a+b}$ and $\overline{ab}$ respectively. The group $(\mathbb{Z}/12\mathbb{Z}, +)$ is abelian (inherited from $\mathbb{Z}$), and associativity, commutativity of $\cdot$, and both distributive laws all follow immediately by lifting to representatives in $\mathbb{Z}$ and using that these laws hold in $\mathbb{Z}$. The identity is $\bar 1$. So $\mathbb{Z}/12\mathbb{Z}$ is a commutative ring with identity, matching the general construction $\mathbb{Z}/n\mathbb{Z}$. Now compute a nontrivial product: take $\bar a = \bar 4$ and $\bar b = \bar 9$ (chosen asymmetrically, neither dividing $12$ alone in an obvious way, neither equal to $\pm1 \bmod 12$). Then

since $36 = 3\cdot 12$. So $\bar 4$ and $\bar 9$ are both nonzero, yet their product is zero: they are zero-divisors. This shows the ring axioms alone (Theorem above) do not forbid $ab = 0$ with $a,b \ne 0$ — that is an extra property (studied under "integral domains", a topic of Commutative Rings and Units). A consequence: cancellation can fail. Note $\bar 4 \cdot \bar 3 = \overline{12} = \bar 0 = \bar 4 \cdot \bar 9 - \bar 4\cdot\bar 9$... more directly, compare $\bar 4 \cdot \bar 3$ and $\bar 4 \cdot \bar 9$ minus each other:

so $\bar4\cdot \bar 3 = \bar 4 \cdot \bar 9$ even though $\bar 3 \ne \bar 9$ in $\mathbb{Z}/12\mathbb{Z}$. One cannot cancel the common factor $\bar 4$. This is a genuinely ring-theoretic phenomenon absent from $\mathbb{Z}$ itself (where $ab=0 \Rightarrow a=0$ or $b=0$), illustrating that passing from $\mathbb{Z}$ to a quotient ring can introduce new pathologies even though both are commutative rings with identity.

Example 2: A concrete noncommutative computation in $M_2(\mathbb{R})$. Let $R = M_2(\mathbb{R})$, the set of $2\times 2$ real matrices, with entrywise addition and the usual matrix product. Addition makes $(R,+)$ an abelian group (componentwise, inherited from $(\mathbb{R},+)$), with zero element the zero matrix $O$ and additive inverse $-A$ given by negating every entry. Matrix multiplication is associative (a standard linear algebra fact, provable directly from the summation definition of the product using associativity and distributivity of real number arithmetic), and both distributive laws $A(B+C) = AB+AC$, $(B+C)A = BA + CA$ hold by direct entrywise computation. The identity element is $I = \begin{pmatrix}1&0\\0&1\end{pmatrix}$. Hence $M_2(\mathbb{R})$ is a ring with identity. To see it is not commutative, take asymmetric matrices

Compute:

Since $AB = \begin{pmatrix}5&2\\1&1\end{pmatrix} \ne \begin{pmatrix}3&6\\1&3\end{pmatrix} = BA$, we have $AB \ne BA$. This single computation certifies that $M_2(\mathbb{R})$ is a genuinely noncommutative ring, and it also illustrates Theorem part (b) concretely: with $-A = \begin{pmatrix}-1&-2\\0&-1\end{pmatrix}$,

confirming $(-A)B = -(AB)$ as guaranteed abstractly, here checked entry by entry: e.g. the $(1,1)$ entry is $(-1)(3)+(-2)(1) = -5$, matching $-(5)$.