A ring carries two operations — addition and multiplication — laced together by distributivity. A group homomorphism is a map that respects a single operation. So the natural question is: what should a structure-preserving map between rings respect? The answer is the obvious one — both operations at once. A map that honours addition and multiplication is a ring homomorphism, and it is the tool that lets us compare one ring with another, transplant theorems, and — most spectacularly — recognise when two rings that look different are secretly the same.

You already know the founding example, even if you have never named it. When you say "the remainder of 17 on division by 6 is 5," you are applying the map \mathbb{Z} \to \mathbb{Z}_6 that sends each integer to its clock position. Reduction mod n respects sums (17 + 4 \equiv 5 + 4) and products (17 \times 4 \equiv 5 \times 4) — it is a ring homomorphism, and its whole job is to collapse \mathbb{Z} down onto a small clock without breaking the arithmetic.

This page pins down the definition, extracts the two objects every homomorphism hands you for free — its kernel and its image — and builds up to the crown jewel: the First Isomorphism Theorem, which says every homomorphism is really a quotient in disguise. By the end you will be able to test whether a map is a homomorphism, compute its kernel, and read off an isomorphism from it.

The definition

Compare this with a group homomorphism, which preserves only one operation. A ring homomorphism is, in particular, a homomorphism of the additive groups (R,+) \to (S,+) — so all the group facts come along free: \varphi(0_R) = 0_S and \varphi(-a) = -\varphi(a). But it must also respect multiplication, and that second condition is the whole story — it is what separates a ring homomorphism from a mere additive map.

Picture it: reduction mod 6

The canonical homomorphism is \pi : \mathbb{Z} \to \mathbb{Z}_6, a \mapsto a \bmod 6. Below, the top row is a stretch of integers 0, 1, \dots, 11; the bottom row is the six residues of \mathbb{Z}_6. Each arrow sends an integer to its remainder, and every residue class gets its own colour. Watch how 0 and 6 — the multiples of 6 in view — both land on 0: that fibre over 0 is the kernel, the full set 6\mathbb{Z}.

The map is surjective — every residue is hit — and it is spectacularly not injective: infinitely many integers pile onto each of the six targets. The homomorphism laws say those collapsing arrows never break the arithmetic: if you add two integers upstairs and then reduce, you get the same answer as reducing first and adding downstairs. Reduction is a wormhole from the infinite integers to a six-element clock that preserves every sum and product.

Kernel and image

Every homomorphism \varphi : R \to S ships with two subsets that measure exactly how it distorts R.

Here is why the kernel is an ideal, not merely a subring. Take a \in \ker\varphi and any r \in R. Then \varphi(ra) = \varphi(r)\varphi(a) = \varphi(r)\cdot 0 = 0, so ra is back in the kernel — the kernel "swallows" multiplication by anything in R. That absorbing property is precisely the definition of an ideal, and it is the ring-theoretic upgrade of the fact that a group homomorphism's kernel is a normal subgroup.

For reduction \pi : \mathbb{Z} \to \mathbb{Z}_6, the kernel is \ker\pi = 6\mathbb{Z} = \{\dots, -6, 0, 6, 12, \dots\} and the image is all of \mathbb{Z}_6. The kernel is fat, so \pi is far from injective — exactly what the picture showed.

Worked examples: is it a homomorphism?

Example 1 — reduction really works. Check \pi : \mathbb{Z} \to \mathbb{Z}_5, a \mapsto a \bmod 5. Modular arithmetic gives (a + b) \bmod 5 = (a \bmod 5) + (b \bmod 5) and the same for products, and \pi(1) = 1. All three conditions hold — it is a homomorphism, surjective, with kernel 5\mathbb{Z}.

Example 2 — an additive map that fails. Consider f : \mathbb{Z} \to \mathbb{Z}, f(a) = 2a. It preserves addition: f(a + b) = 2(a+b) = 2a + 2b = f(a) + f(b). But multiplication breaks: f(ab) = 2ab while f(a)f(b) = (2a)(2b) = 4ab. These disagree (take a = b = 1: 2 \ne 4), so f is not a ring homomorphism, even though it is a perfectly good group homomorphism of (\mathbb{Z}, +).

Example 3 — evaluation. Fix c \in \mathbb{R} and define \operatorname{ev}_c : \mathbb{R}[x] \to \mathbb{R} by p \mapsto p(c) — "plug in c." Since (p + q)(c) = p(c) + q(c) and (pq)(c) = p(c)\,q(c), and the constant polynomial 1 evaluates to 1, evaluation is a ring homomorphism. Its kernel is the set of polynomials with p(c) = 0 — precisely the multiples of (x - c), the ideal (x - c). And it is surjective, since every real r is the value of the constant polynomial r.

The First Isomorphism Theorem

Now the payoff. A homomorphism \varphi : R \to S crushes the kernel to a point and lays the rest of R down as the image. The claim is that the image is completely determined by how the kernel folds R — that is, by the quotient ring R/\ker\varphi.

Read the theorem as a factory: hand it a surjective homomorphism and it hands you back an isomorphism. Apply it to reduction \pi : \mathbb{Z} \to \mathbb{Z}_n, which is surjective with kernel n\mathbb{Z}, and out drops the identity you have been using all along: \mathbb{Z} / n\mathbb{Z} \;\cong\; \mathbb{Z}_n. The abstract quotient ring \mathbb{Z}/n\mathbb{Z} — cosets of n\mathbb{Z} — is the same ring as the concrete clock \mathbb{Z}_n. The two names for "arithmetic mod n" finally shake hands.

Apply it instead to evaluation \operatorname{ev}_c : \mathbb{R}[x] \to \mathbb{R}, surjective with kernel (x - c), and you learn \mathbb{R}[x] / (x - c) \;\cong\; \mathbb{R}. Quotienting the polynomials by "x - c = 0" — i.e. forcing x = c — collapses the whole polynomial ring down to just the real numbers. The isomorphism theorem turns "plug in a number" into a statement about quotient rings.