# Identity of the kernel of a ring homomorphism

Say we have a ring homomorphism from a ring $$R$$ to $$S$$ with multiplicative identities $$1_R$$ and $$1_S$$ respectively, is it true that the identity of the kernel which is a subring of $$R$$ is also $$1_R$$ if and only if $$1_R$$ maps to the $$0$$ of ring $$S$$ (the additive identity of $$S$$). Or otherwise the kernel doesn’t have an identity. Is that correct? Thanks in advance

Take $$R=S=\mathbb Z_6$$ and the ring homomorphism $$f$$ given by $$x \mapsto 4x$$.

Then $$\ker f = \{0,3\}$$ is a ring with $$3$$ as an identity.

($$f$$ does not send $$1$$ to $$1$$ but it is multiplicative.)

• @Ihf How is that a ring homomorphism f(xy) isn’t equal to f(x)f(y) or am I mistaken? – Eden Hazard Apr 15 at 16:35
• @EdenHazard, $(4x)(4y) = 16xy \equiv 4 xy$. – lhf Apr 15 at 16:48
• Is see so f(xy)= 4xy which is equivalent to 16xy=f(x)f(y). Thanks for this – Eden Hazard Apr 15 at 16:52
• Alternatively, take $S=\Bbb Z_3$ – Hagen von Eitzen Apr 16 at 3:11

Let $$J$$ be the kernel of $$f\colon R\to S$$. If $$1_R\in J$$, then clearly $$f=0$$.

In theory, $$J$$ might have an identity $$1_J$$ which is not an identity for all of $$R$$.

• Any examples? All the examples I have is that it is either 1R or there isn’t any. – Eden Hazard Apr 15 at 16:30
• I don't think $1_J$ can be different from $1_R$, since both $1_J$ and $1_R$ are a multiplicative identity element, which is unique. – Strichcoder Apr 15 at 16:30
• @strichcoder that’s what I thought aswell – Eden Hazard Apr 15 at 16:33
• @strichcoder I get what they’re saying now, 1R doesn’t have to be in the kernel , if it is then they are equal , but if it doesn’t like in above example by Ihf you might have another identity element. Or in other cases there might not even be one. – Eden Hazard Apr 15 at 17:08
• Ok, sorry. And thanks. – Strichcoder Apr 15 at 17:21