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

Parents: