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I am trying to show that given $\phi: R\rightarrow R'$ is a ring homomorphism $R, R'$ are rings, there exist a homomorphism between the group of units that is $\phi': R^\times\rightarrow R'^\times$ is a group homomorphism. I was able to show that the the ring homomorphism maps units to units but I am not sure how to check the homomorphism condition explicitly.

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  • $\begingroup$ I am sorry.. what is your question? $\endgroup$ – user87543 Nov 17 '14 at 14:39
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    $\begingroup$ The homomorphism condition for $\phi'$ is that $\phi'(xy) = \phi'(x)\phi'(y)$ for all units $x,y \in R^\times$, correct? You seem to be using the map $\phi$ restricted to $R^\times$ for your $\phi'$, so what do we know about the multiplicative properties of $\phi$? $\endgroup$ – Joshua Mundinger Nov 17 '14 at 14:41
  • $\begingroup$ @Alqatrkapa: We know that $\phi(xy)=\phi(x)\phi(y)$ so this condition will also hold for our restricted map? $\endgroup$ – Rutherford Mark Nov 17 '14 at 14:45
  • $\begingroup$ @RutherfordMark Precisely. $\endgroup$ – Joshua Mundinger Nov 17 '14 at 14:46
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(I assume you're using $\phi'=\phi\rvert_{R^\times}$, that is, the restriction of $\phi$ to $R^\times$.)

Then, the statement follows trivially from the fact that $\phi$ is a ring homomorphism: From $$\forall x,y\in R.\quad \phi(x\cdot y)=\phi(x)\cdot\phi(y) $$ and $R^\times\subseteq R$, it is obvious that $$\forall x,y\in R^\times.\quad \phi'(x\cdot y)=\phi'(x)\cdot\phi'(y) \text.$$

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