# Kernel of ring homomorphism

Given a ring homomorphism $$f: R \to S$$ between rings $$R$$ and $$S$$, we say the kernel of $$f$$ is the collection of elements of $$R$$ which $$f$$ sends to the zero element of $$S$$.

Formally, it is $$\{ r \in R \mid f(r) = 0_S \}$$\$ where $$0_S$$ is the zero element of $$S$$.

# Examples

• Given the “identity” (or “do nothing”) ring homomorphism $$\mathrm{id}: \mathbb{Z} \to \mathbb{Z}$$, which sends $$n$$ to $$n$$, the kernel is just $$\{ 0 \}$$.

• Given the ring homomorphism $$\mathbb{Z} \to \mathbb{Z}$$ taking $$n \mapsto n \pmod{2}$$ (using the usual shorthand for modular arithmetic), the kernel is the set of even numbers.

# Properties

Kernels of ring homomorphisms are very important because they are precisely ideals. (Proof.) In a way, “ideal” is to “ring” as “subgroup” is to “group”, and certainly subrings are much less interesting than ideals; a lot of ring theory is about the study of ideals.

The kernel of a ring homomorphism always contains $$0$$, because a ring homomorphism always sends $$0$$ to $$0$$. This is because it may be viewed as a group homomorphism acting on the underlying additive group of the ring in question, and the image of the identity is the identity in a group.

If the kernel of a ring homomorphism contains $$1$$, then the ring homomorphism sends everything to $$0$$. Indeed, if $$f(1) = 0$$, then $$f(r) = f(r \times 1) = f(r) \times f(1) = f(r) \times 0 = 0$$.

Parents: