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This question already has an answer here:

Let $R$ and $S$ be commutative rings. Let $x, y$ be indeterminates, and assume that one has an isomorphism $R[x] \rightarrow S[y]$ (not necessarily mapping $x$ to $y$ of course). Does this imply $R \cong S$? If not, what is a counterexample?

This may seem like a homework problem, but is not.

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marked as duplicate by Martin Brandenburg, user38268, user26857, Qiaochu Yuan, Dennis Gulko May 10 '13 at 9:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This was settled in the negative by M. Hochster in 1972. His paper can be found here.

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  • $\begingroup$ Wonderful! Thanks! $\endgroup$ – Rankeya May 10 '13 at 7:14
  • $\begingroup$ settled in the sense: There are counterexamples. $\endgroup$ – Martin Brandenburg May 10 '13 at 9:00

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