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. 
 A: 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).
A: I found the paper Isomorphic polynomial rings 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 $R[x]\cong S[x]$.
Added
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.
A: 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


A: 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 $R \cong S$.
