Does $R[x] \cong S[x]$ imply $R \cong S$?

This is a very simple question but I believe it's nontrivial.

I would like to know if the following is true:

If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ are isomorphic.

Thanks!

If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true.

Here is a counterexample.

Let $$R=\dfrac{\mathbb{C}[x,y,z]}{\big(xy - (1 - z^2)\big)}$$, $$S=\dfrac{\mathbb{C}[x,y,z]}{\big(x^2y - (1 - z^2)\big)}$$. Then, $$R$$ is not isomorphic to $$S$$ but, $$R[T]\cong S[T]$$.

In many variables, this is called the Zariski problem or cancellation of indeterminates and is largely open. Here is a discussion by Hochster (problem 3).

• May I ask what the isomorphism $f\!:R[T]\!\rightarrow\!S[T]$ is, and how do we know that $R\!\ncong\!S$? I'm hoping for an elementary answer...
– Leo
Sep 18 '11 at 14:37
• Additionally -- what is $T$? An arbitrary ring? A copy of Z ... ? Mar 27 '18 at 20:47
• @AlexMeiburg : it denotes an intedeterminate ; $R[T]$ is the polynomial ring of 1 variable over $R$. Said differently, it is the monoid algebra of $\Bbb N$ over $R$. May 11 '18 at 12:05

I found this paper by Brewer and Rutter that discusses related matters. They cite a forthcoming paper by Hochster which proves there are non-isomorphic commutative integral domains $R$ and $S$ with $R[x]\cong S[x]$.

Hochster's paper is M. Hochster, Nonuniqueness of coefficient rings in a polynomial ring, Proc. Amer. Math. Soc. 34 (1972), 81-82, and is freely available.

• The link in the post seems to be that, but the paper by Brewer and Rutte is most likely this one: Brewer, J. W.; Rutter, Edgar A., Isomorphic polynomial rings, Arch. Math. 23, 484-488 (1972). ZBL0254.16001, MR320068 Jan 12 at 20:39

There's been much work on this problem since the mentioned seminal work in the early seventies. Searching on the buzzword "stably equivalent" should help locate most of it. Below is a helpful introduction from Jon L Johnson: Cancellation and Prime Spectra  • If we assume that $R$ and $S$ are subrings of $\Bbb R$ (or of $\Bbb C$), do you know if there is an example where $R[x] \cong S[x]$ but $R \not \cong S$? Aug 18 '16 at 14:25

Along the lines of "particular cases where the claim is true," if $$R$$ and $$S$$ are fields, then they must be isomorphic since they are distinguishable as the units (excepting zero of course) of $$R[x]$$ and $$S[x]$$, respectively, and an isomorphism conserves units.

More generally if we only know that one of $$R$$ or $$S$$ is a field, the claim is still true.

Suppose $$R$$ is a field and $$R[X] \cong S[Y]$$, where $$X,Y$$ are distinct indeterminates for clarity, and $$f:R[X] \rightarrow S[Y]$$ is an isomorphism. Since the nonzero elements of $$f(R)$$ are units in $$S[Y]$$, they must be degree zero, i.e. elements of $$S$$. Also the inverse image of $$S$$ is a subring of Euclidean domain $$R[X]$$, so $$S$$ is an integral domain.

Knowing that $$f(R) \subseteq S$$, we need only show $$f$$ maps $$R$$ onto $$S$$. Suppose $$s \in S$$. Then there exists polynomial $$p(X) \in R[X]$$ s.t. $$f(p(X)) = s$$. Let $$p(X) = \sum_{i=0}^n \enspace r_i X^i$$ where coefficients $$r_i \in R$$.

Now $$f(X) = q(Y)$$ for some polynomial $$q(Y) \in S[Y]$$, and degree of $$q(Y)$$ must be positive for $$f(R[X]) = S[Y]$$. Apply $$f$$ to $$p(X)$$ and compare degrees:

$$s = \sum_{i=0}^n \enspace f(r_i) q(Y)^i$$

Since $$S$$ is an integral domain, the degrees of $$q(Y)^i$$ are positive for $$i \gt 0$$. Thus the only nonzero coefficient of $$p(X)$$ is $$r_0$$, which shows $$f(r_0) = s$$. Therefore $$f$$ maps $$R$$ onto $$S$$ and $$R \cong S$$.

• I'm not sure to understand the reasoning in your first paragraph : $(R[X])^{\times} = R^{\times} \cong S^{\times} = (S[X])^{\times}$ doesn't imply $R \cong S$ as fields. You can even see problem 10E in Halmos Problems for Mathematicians, Young and Old. Jul 24 '16 at 13:41
• @Watson: Here we have an isomorphism of rings $R[X]$ and $S[Y]$ and subrings $R,S$ resp. which both happen to be fields. A degree argument is meant to show that the image of $R$ is $S$ under the isomorphism, and therefore that these are isomorphic as rings. I'll look for the Halmos exercise, but I suspect the point there is one knows only a isomorphism of (multiplicative) groups. Jul 24 '16 at 13:52