I know the opposite is not true in general as was discussed here: Does $R[x] \cong S[x]$ imply $R \cong S$?
Rings are assumed to be commutative with identity.
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2$R\cong S$ means that after renaming we have $R=S$. So it is obvious. – Dietrich Burde Mar 15 '19 at 13:18
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Yes. When you realise that $R \cong S$ means that $R$ and $S$ are the same ring just with the elements relabeled, the result is obvious.
If you want to exhibit an example of a ring isomorphism $R[X] \rightarrow S[X]$, then start with a ring isomorphism $f: R \rightarrow S$ and define $\overline{f} : R[X] \rightarrow S[X]$ by $$\overline{f} \left( \sum_{i=0}^{k} a_i X^i \right) = \sum_{i=0}^{k} f(a_i) X^i.$$
Joseph Martin
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I think it does, could you please define your "equivalence symbol"? If it means "Isomorph" then your claim is absolutely true:
Let $f:R\to S$ be a ring isomorphism. We define $g:R[x]\to S[x]$ as follows:
- $g(a)=f(a)$ for every $a \in R$
- $g(x)=x$.
You can easily check that $g$ is an isomorphism.
