# 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? Apr 15, 2019 at 16:35
• @EdenHazard, $(4x)(4y) = 16xy \equiv 4 xy$.
– lhf
Apr 15, 2019 at 16:48
• Is see so f(xy)= 4xy which is equivalent to 16xy=f(x)f(y). Thanks for this Apr 15, 2019 at 16:52
• Alternatively, take $S=\Bbb Z_3$ Apr 16, 2019 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. Apr 15, 2019 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. Apr 15, 2019 at 16:30
• @strichcoder that’s what I thought aswell Apr 15, 2019 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. Apr 15, 2019 at 17:08
• Ok, sorry. And thanks. Apr 15, 2019 at 17:21