# 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 $R=\frac{\mathbb{C}[x,y,z]}{(xy - (1 - z^2))}$, $S=\frac{\mathbb{C}[x,y,z]}{(x^2y - (1 - z^2))}$. 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... – Leon Sep 18 '11 at 14:37
• Additionally -- what is $T$? An arbitrary ring? A copy of Z ... ? – Alex Meiburg 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 $R$. – Watson 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.

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 $R \not \cong S$? – Watson 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. – Watson 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. – hardmath Jul 24 '16 at 13:52