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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.


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.

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up vote 46 down vote accepted

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).

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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

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.

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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 alt text alt text

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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 $R \cong S$.

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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 2 days ago
@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 2 days ago

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