Let $R$ and $S$ be non-zero rings with identity. Is it possible to have $R[x] \cong S[[x]]$ ?



Polynomial Rings 
Here's a proof that no such *commutative* rings $R$, $S$ exist. (See the edit for an extension to noncommutative rings.) 


Suppose we have an isomorphism $\phi: R[x] \to S[[x]]$; let $a = \phi^{-1}(x)$. First we claim that for all $b \in R[x]$, the element $1 + ab$ is invertible in $R[x]$. Indeed, the element $\phi(1 + ab) = 1 + (\phi(b))x$ has an inverse given by a formal geometric series, so $\phi^{-1}$ applied to this element must also be invertible. 


In particular, $1 + ax$ must be invertible in $R[x]$. But it is well-known (in the commutative case) that any invertible polynomial $a\_0 + a\_1 x + \ldots + a\_n x^n$ has $a\_0$ invertible and the other $a\_i$ nilpotent. It follows quickly that $a$ must be nilpotent. But then $\phi(a) = x$ is nilpotent in $S[[x]]$. We have reached an absurdity. 


**Edit:** Aided by Martin's excellent suggestion in his comment, we may easily extend the argument to noncommutative rings. Indeed, since $x$ is central in $S[[x]]$, we have that $a$ is central in $R[x]$. In particular, $a$ commutes with scalars; writing $a$ as a polynomial, it follows that each coefficient of $a$ is central in $R$. This is true also of the polynomial $1 + ax$, and now the proof that all but the unit coefficient of $1 + ax$ is nilpotent goes through as in the commutative case (see for example the nice argument given [here](http://ysharifi.wordpress.com/2010/09/02/units-in-polynomial-rings/)). Thus $a$ is nilpotent. 


