Kernel of ring homomorphism

by Patrick Stevens Aug 3 2016 updated Aug 4 2016

The kernel of a ring homomorphism is the collection of things which that homomorphism sends to 0.

[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$~$.



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$~$.