[summary: The kernel of a [-ring_homomorphism] is the collection of elements which the homomorphism sends to $~$0$~$.]

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 [ideal_ring_theory ideals]. (Proof.) In a way, "ideal" is to "ring" as "Subgroup" is to "group", and certainly [subring_ring_theory 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$~$.