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I'm sort of stuck on this problem, I think I've been able to leverage that R is an integral domain to show that the units of R[X] are all constant functions and so the homomorphism maps the units of R to themselves I guess? I'm not really sure how to proceed from here or whether I'm even moving in the right direction.

trgjtk
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  • @AnneBauval errr i'm not really sure but i don't think so? if it does and i'm missing something obvious feel free to expand – trgjtk Jan 14 '24 at 07:12
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    You already noticed the injective group homomorphism $R^\times \rightarrow R[X]^\times,r\mapsto r$, and the answer in the linked post proves it is surjective when $R$ is an integral domain. – Anne Bauval Jan 14 '24 at 07:14
  • @AnneBauval maybe I'm misunderstanding the question but I was under the impression that the homomorphism referenced was an arbitrary one. Would the trivial homomorphism not map from $R^\times$ to $R[X]^\times$ and not be bijective? – trgjtk Jan 14 '24 at 19:05
  • The "homomorphism referenced" is clearly the canonical one ($r\mapsto r$). An arbitrary one would generally not be bijective. The trivial one ($r\mapsto 1$) does map $R^\times$ into $R[X]^\times$ and is (except in a few uninteresting very special cases) neither injective nor surjective. – Anne Bauval Jan 14 '24 at 19:52
  • @AnneBauval ahhhhh thank you! this was the primary source of my confusion. Much appreciated – trgjtk Jan 14 '24 at 20:24

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