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