# Ideals are the same thing as kernels of ring homomorphisms

In ring the­ory, the no­tion of “ideal” cor­re­sponds pre­cisely with the no­tion of “ker­nel of ring ho­mo­mor­phism”.

This re­sult is analo­gous to the fact from group the­ory that nor­mal sub­groups are the same thing as ker­nels of group ho­mo­mor­phisms (proof).

# Proof

## Ker­nels are ideals

Let $$f: R \to S$$ be a ring ho­mo­mor­phism be­tween rings $$R$$ and $$S$$. We claim that the ker­nel $$K$$ of $$f$$ is an ideal.

In­deed, it is clearly a sub­group of the ring $$R$$ when viewed as just an ad­di­tive group noteThat is, af­ter re­mov­ing the mul­ti­plica­tive struc­ture from the ring. be­cause $$f$$ is a group ho­mo­mor­phism be­tween the un­der­ly­ing ad­di­tive groups, and ker­nels of group ho­mo­mor­phisms are sub­groups (in­deed, nor­mal sub­groups). (Proof.)

We just need to show, then, that $$K$$ is closed un­der mul­ti­pli­ca­tion by el­e­ments of the ring $$R$$. But this is easy: if $$k \in K$$ and $$r \in R$$, then $$f(kr) = f(k)f(r) = 0 \times r = 0$$, so $$kr$$ is in $$K$$ if $$k$$ is.

## Ideals are kernels

re­fer to the quo­tient group, and there­fore in­tro­duce the quo­tient ring

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