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Let $R$, $S$ be commutative rings, and let $\Phi:R\to S$ be a homomorphism of rings. Prove that if $a\in R$ is a unit, then $\Phi(a^{-1})=\Phi(a)^{-1}$. Deduce that $\Phi$ maps units of $R$ to units of $S$.

I could really use some help on this problem. I know how to prove $\Phi(-a)=-\Phi(a)$ by saying $\Phi(-a)+\Phi(a)=\Phi(-a+a)=\Phi(0)=0$, therefore $\Phi(-a)=-\Phi(a)$. I know how to do that, but I can't figure out how to do this one. Any help would be greatly appreciated. Thanks

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Do the same thing but with multiplication instead of addition.

(The analogue of $\Phi(-a)+\Phi(a)$ would be $\Phi(a^{-1})\times\Phi(a)$)

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    $\begingroup$ When you put an entire sentence inside parentheses, it needs a period, and the period goes on the inside. Thus the edit actually made the post less grammatical. I am amused by it however. :) $\endgroup$
    – anon
    Nov 5 '13 at 23:42

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