Given a ring_homomorphism 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 ${r \in R ∣ f (r) = 0_{S}}$ where $0_{S}$ is the zero element of $S$ .

Examples

Given the "identity" (or "do nothing") ring homomorphism $i d : Z \to Z$ , which sends $n$ to $n$ , the kernel is just ${0}$ .
Given the ring homomorphism $Z \to Z$ taking $n \mapsto n (mod 2)$ (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 $f (r) = f (r \times 1) = f (r) \times f (1) = f (r) \times 0 = 0$ .

Kernel of ring homomorphism