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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.

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    $\begingroup$ Sub-question: Does $R[x]\cong R[[x]]$ imply $R[x][y]\cong R[x][[y]]$? $\endgroup$ Jan 5, 2016 at 18:33
  • $\begingroup$ We know that if $R$ has finite Krull dimension $n=dim (R)$, then $n<dim (R [x])$. So in this case the claim is not true. Also, for semi local ring $R$, since $R [x]$ has infinitely many maximal ideals, is not true. $\endgroup$ Jan 7, 2016 at 15:58
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    $\begingroup$ Wait, $R[[y]][x]$ is only a subring of $R[x][[y]]$, right? For example, $1+xy+x^2 y^2 +\dotsc$ is contained in $R[x][[y]]$, but not in $R[[y]][x]$. (Therefore, I also don't understand the example mentioned in the other question.) $\endgroup$ Jan 7, 2016 at 16:32

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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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    $\begingroup$ How is the example of $R=S[x_1, x_2, \ldots][[y_1, y_2, \ldots]]$ incorrect then? $\endgroup$
    – Wojowu
    Jan 7, 2016 at 17:30
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    $\begingroup$ See my comment above: We cannot exchange $[-]$ with $[[-]]$. $\endgroup$ Jan 7, 2016 at 18:28

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