Kernel of ring homomorphism

https://arbital.com/p/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 $S$.

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