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Example of ring "quasi-homomorphism" not conserving multiplicative identity?

I'm using a definition of homomorphism that requires f(1)=1 and I'm trying understand the reasons and consequence of this point.

I mean: if we're connecting two rings with identity, does this not force f(1)=1?

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3 Answers

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Take any idempotent $e\notin\{0,1\}$ in any ring $R$.

The inclusion of $eRe\rightarrow R$ does not preserve identity.


To add some details, it is easy to check that $eRe$ is a subring of $R$ with identity $e$. The inclusion $x\mapsto x$ would not send $e$ to 1 in $R$, and so it does not preserve identity. There is no need for $e$ to be central.

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You should ask for $e$ to be central, or else this need not be a map of rings. –  David Speyer Jan 17 '13 at 14:16
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@DavidSpeyer No, that's superfluous. It is an inclusion map of a subring into a ring... I don't see why centrality of $e$ matters to map $ere\mapsto ere$. –  rschwieb Jan 17 '13 at 14:57
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I'm guessing @David was thinking of the projection $R \to eRe : x \mapsto exe$. –  Hurkyl Jan 17 '13 at 15:16
    
My bad, you're right. –  David Speyer Jan 17 '13 at 23:04
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You can have multiplicative and additive maps between rings that do not preserve identity. For instance $i: \mathbb Z \rightarrow \mathbb Z \times \mathbb Z$ given by $i(n)=(0,n)$. One might call such a map a rng homomorphism.

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consider $f:\mathbb Z \to \mathbb Z$ given by $f(a)=2\cdot a$.

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This is not a multiplicative map, because $4=f(1)f(1)\neq f(1)$. –  JSchlather Jan 17 '13 at 8:50
    
I apologize (going to sleep now). –  Ittay Weiss Jan 17 '13 at 8:53
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