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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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    Your second step is wrong. Note that you're substracting things that are in different rings.2011-11-02
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    i made the correction, thanks !2011-11-02
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    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 $(1,1)$.2011-11-02
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    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$2011-11-02
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    I mean "we can not conclude what values $\phi(1_R)$ could take in S" is correct.2011-11-02
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    @palio: you don't need $\phi$ to be injective, it suffices that it be non-zero.2011-11-02
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    Your title is weird: usually "unitary ring homomorphis" means that the image of $1$ is $1$!2011-11-02
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    @Weltschmerz : yes absolutely since if $\phi(1_R)=0_S$ then $\forall r\in R$, we have $\phi(r)=\phi(1_R.r)=\phi(1_R).\phi(r)=0_S.\phi(r)=0_S$.2011-11-02
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    @Mariano Suárez-Alvarez : i just mean the rings are unitary so we can speak of $1_R$ and $1_S$ but my problem is to know $\phi(1_R)$ under different conditions. Moreover i'm not able to change the title so if you have access please do it .. thanks2011-11-02
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    Why can't you change the title? Right below the body of your question, there is a link which reads «edit» which allows you to edit the question *and* the title.2011-11-02
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    @Mariano Suárez-Alvarez : i hope it is more clear now.2011-11-02

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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 $\phi(1_R)=1_S$.

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    the example that @Qil gave is surjective but $\phi(1)$ is not $(1,1)$2011-11-02
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    @palio: I don't think it's surjective.2011-11-02
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    i'm sorry it is clearly not surjective2011-11-02
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    i could'nt see why being surjective $\phi$ must map $1_R$ to $1_S$?2011-11-02
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    @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 $\phi(1_R)=1_S$.2011-11-02
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    yes absolutely!!! since for any $y\in S$ we have that $\phi(1_R).y=y$ then $\phi(1_R)$ is an identity in $S$ and so necessarily $\phi(1_R)=1_S$2011-11-03