# 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 \(f(r) = f(r \times 1) = f(r) \times f(1) = f(r) \times 0 = 0\).

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