let $R$ and $S$ be unitary rings and $\phi:R\rightarrow S$ a ring homomorphism. is the following correct:

$\phi(1_R)=\phi(1_R\cdot1_R)=\phi(1_R)\cdot\phi(1_R)$ so $\phi(1_R)(1_S-\phi(1_R))=0_S$ and so $\phi(1_R)$ could be anything in $S$ when $S$ is a general ring, i mean we can not conclude what values $\phi(1_R)$ could take in $S$ . But when $S$ is an integral domain then we can say that we have $\phi(1_R)=0_S$ or $1_S-\phi(1_R)=0_S$ i.e, $\phi(1_R)=1_S$. Moreover when $\phi$ is a monomorphism then since only $0_R$ maps to $0_S$ then necessarily $\phi(1_R)=1_S$.

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    $\begingroup$ Your second step is wrong. Note that you're substracting things that are in different rings. $\endgroup$ – Weltschmerz Nov 2 '11 at 13:03
  • $\begingroup$ i made the correction, thanks ! $\endgroup$ – palio Nov 2 '11 at 13:04
  • $\begingroup$ This is correct. For example, $\phi : \mathbb Z\to \mathbb Z\oplus \mathbb Z$, $x\mapsto (x,0)$ is additive and multiplicative, but doesn't map $1$ to $(1,1)$. $\endgroup$ – user18119 Nov 2 '11 at 13:07
  • $\begingroup$ this is because $\mathbb Z\oplus \mathbb Z$ is not an integral domain since $(1,0).(0,1)=(0,0)$. but note that this is a monomorphism from the integral domain $\mathbb Z$ to the unitary ring $\mathbb Z\oplus \mathbb Z$ $\endgroup$ – palio Nov 2 '11 at 13:10
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    $\begingroup$ Your title is weird: usually "unitary ring homomorphis" means that the image of $1$ is $1$! $\endgroup$ – Mariano Suárez-Álvarez Nov 2 '11 at 13:31

As you correctly proved, it's true when $S$ is an integral domain.

Another fact is that if the homomorphism is surjective, then $\phi(1_R)$ is the identity in $S$, regardless of what $S$ is like. To prove it just check what it does to any other element of $S$.

Also true: if $R$ and $S$ are non-trivial rings, $S$ has an identity $1_S$, $\phi$ is injective and $1_S$ is in $\phi(R)$ then $R$ has an identity and $\phi(1_R)=1_S$.

  • $\begingroup$ the example that @Qil gave is surjective but $\phi(1)$ is not $(1,1)$ $\endgroup$ – palio Nov 2 '11 at 13:15
  • $\begingroup$ @palio: I don't think it's surjective. $\endgroup$ – Weltschmerz Nov 2 '11 at 13:20
  • $\begingroup$ i'm sorry it is clearly not surjective $\endgroup$ – palio Nov 2 '11 at 13:30
  • $\begingroup$ i could'nt see why being surjective $\phi$ must map $1_R$ to $1_S$? $\endgroup$ – palio Nov 2 '11 at 18:29
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    $\begingroup$ @palio: Pick any $y\in S$. Then $\phi(1_R)y=\phi(1_R)\phi(x)$ for some $x$ in $R$ because $\phi$ is surjective. Now, $\phi(1_R)\phi(x)=\phi(1_R x)=\phi(x)=y$. And you know that the identity in a ring is unique. Hence $\phi(1_R)=1_S$. $\endgroup$ – Weltschmerz Nov 2 '11 at 19:55

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