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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

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2 Answers 2

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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.)

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  • $\begingroup$ @Ihf How is that a ring homomorphism f(xy) isn’t equal to f(x)f(y) or am I mistaken? $\endgroup$ Apr 15, 2019 at 16:35
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    $\begingroup$ @EdenHazard, $(4x)(4y) = 16xy \equiv 4 xy$. $\endgroup$
    – lhf
    Apr 15, 2019 at 16:48
  • $\begingroup$ Is see so f(xy)= 4xy which is equivalent to 16xy=f(x)f(y). Thanks for this $\endgroup$ Apr 15, 2019 at 16:52
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    $\begingroup$ Alternatively, take $S=\Bbb Z_3$ $\endgroup$ Apr 16, 2019 at 3:11
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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 $R$.

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  • $\begingroup$ Any examples? All the examples I have is that it is either 1R or there isn’t any. $\endgroup$ Apr 15, 2019 at 16:30
  • $\begingroup$ 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. $\endgroup$ Apr 15, 2019 at 16:30
  • $\begingroup$ @strichcoder that’s what I thought aswell $\endgroup$ Apr 15, 2019 at 16:33
  • $\begingroup$ @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. $\endgroup$ Apr 15, 2019 at 17:08
  • $\begingroup$ Ok, sorry. And thanks. $\endgroup$ Apr 15, 2019 at 17:21

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