If $R[x]$ and $R[[x]]$ are isomorphic, then are they isomorphic to $R$ as well? [duplicate]

examples-counterexamples ring-theory commutative-algebra polynomials formal-power-series

There are examples of commutative rings $R \neq 0$ such that $R[x]$ is isomorphic to $R[[x]]$ (see this question; an example would be $R=S[x_1, x_2, \ldots][[y_1, y_2, \ldots]]$, with $S \neq 0$ any commutative ring). This is false, see Martin Brandenburg's answer.

The following question was asked as a comment on the thread linked above: if $R$ is such that $R[x]\cong R[[x]]$, must we have that $R\cong R[x] \cong R[[x]]$?

This is clearly true for the example above, which is (essentially) the only family of examples I could come up with.


Yes, because $R[x] \cong R[[x]]$ implies $R=0$ (see my answer to the previous question here).

(I make this community wiki because this answer is rather trivial now.)

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