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For an arbitrary ID (integral domain) $R$ with subring $S$, assume that $S$ has an unity. Then does $R$ have a unity too? If not, please provide a counter-example.

man_in_green_shirt
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jawlang
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1 Answers1

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If $S$ is nontrivial (that is, has a nonzero identity $e$) then yes.

A nonzero idempotent in a domain must act as the identity for the domain.

To see this, just examine what $e(ex-x)$ and $(xe-x)e$ must be, where $x$ is an arbitrary element of the domain.

This came up under slightly different circumstances here.

Community
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rschwieb
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  • If $e$ is the identity, then won't $ex=x$ and therefore $ex-x=0$? – man_in_green_shirt Aug 06 '15 at 12:55
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    @man_in_green_shirt: As a general rule, it's unproductive to assume that which you are attempting to prove. – WillO Aug 06 '15 at 12:57
  • Ok, but I don't see the significance of $e(ex-x)$. Ok, if you multiply $e$ by $e(ex-x)$, then you get $e(ex-x)$ because $e$ is idempotent. So if any element can be written as $ex-x$, then $e$ is an identity. However, how can that be possible when $ex-x$ is always $0$? – man_in_green_shirt Aug 06 '15 at 13:07
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    @man_in_green_shirt: What's being claimed here is not that any element can be written as $ex-x$, but that every element can be written as $x$. – WillO Aug 06 '15 at 13:29
  • @man_in_green_shirt The conclusion is that $ex-x=xe-x=0$, and therefore $e$ is the identity. It is not the hypothesis... – rschwieb Aug 06 '15 at 14:01
  • Ah ok got it - using the distributive property, $e(ex-x)=e^2x-ex=ex-ex=0$. As it's an integral domain and $e\neq 0$, we must have $ex-x=0$ and hence $ex=x$, so $e$ is the identity. Thanks – man_in_green_shirt Aug 06 '15 at 20:43