# Does $R\cong S$ imply $R[X] \cong S[X]$?

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.

• $R\cong S$ means that after renaming we have $R=S$. So it is obvious. – Dietrich Burde Mar 15 at 13:18

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.

• Yes it means isomorph. – AromaTheLoop Mar 15 at 13:08

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 $$\overline{f} \left( \sum_{i=0}^{k} a_i X^i \right) = \sum_{i=0}^{k} f(a_i) X^i.$$