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 \(S\).


  • 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.


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