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Suppose $R$ and $S$ are integral domains. And let $\phi:R[x]\longrightarrow S[x]$ be an isomorphism between their polynomial rings. Is it possible that $\phi|_R$, the restriction of the isomorphism to $R$, is not an ismorphism between $R$ and $S$?

ABC
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  • Also, the answer in [this question](https://math.stackexchange.com/questions/387345) refers to a paper with more discussion. – Tesla Daybreak Feb 19 '21 at 09:35
  • Or since we could have $R\cong S$ just not with this $\phi\vert_R$, let $R=k[x_1,x_2,\dots]=S$ and $\phi$ changes $x,x_1$. – user10354138 Feb 19 '21 at 09:39
  • @Testa Daybreak Thanks, that was useful. It was a more profound question than I first thought. I was just asking beacause I was trying to prove that if $\phi$ is an isomorphism between $R[x]$ and $S[x]$ whose restriction to $R$ is an ismorphism between $R$ and $S$, then a polynomial $p(x)$ is ireducible iff $\phi(p(x))$ is irreducible. In the proof I did, I did not need the fact about the restriction and I wondered why that hypothesis was there. Maybe there was something wrong with my proof. – ABC Feb 19 '21 at 09:44

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