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Ler $R, S$ be non-zero rings with identity. Is it possible to have $R[x;\sigma] \cong S[[x;\sigma']]$ for some endomorphisms $\sigma, \sigma'$ of $R$ and $S$ respectively ?

Note: In another post link text it is proved that if $\sigma$ and $\sigma'$ are identity then the answer is negative.

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    It would be helpful to include the definition of skew polynomial and skew power series rings (which I for one have forgotten). – Qiaochu Yuan Dec 29 '12 at 21:57
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    Could you choose more informative question titles? I know that a year from now I'm going to see "Polynomial Rings II" in the Related sidebar, and have to click on it just to remind myself what it was about. – arsmath Dec 29 '12 at 22:13
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    Following up on @arsmath's comment, it is much better MO style to put complete questions in the title. For example, "Is it possible for skew polynomials over one ring to agree with skew power series over another ring?" is plenty short enough to fit in a title. – Theo Johnson-Freyd Jan 01 '13 at 04:01

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Definitions first:

Let $R$ be a ring with unity, and $\sigma$ an endomorphism of $R$. The (left) Ore extension $R[x;\sigma]$ is the free left $R$-module with basis $\{1,x,x^2, \dots\}$, made into a ring by the multiplication $xr = \sigma(r)x$ for every $r \in R$. Intuitively, it's a non-commutative version of a polynomial ring.

The skew power series ring $R[[x;\sigma]]$ is defined in the same way but now infinite sums of powers of $x$ are allowed (but still no negative powers).

The following is an approach towards the answer but I'm a bit unsure of the details towards the end.

So let's take our two rings $R[x;\sigma]$ and $S[[y;\sigma']]$ and suppose there exists an isomorphism $\varphi: R[x;\sigma] \rightarrow S[[y;\sigma']]$.

We have $xr = \sigma(r)x$ and $ys = \sigma'(s)y$ for all $r \in R$, $s \in S$, and the isomorphism has to respect these identities.

Set $a = \varphi^{-1}(y)$. Now for any $r \in R$, we can look at how $r$ and $a$ interact:

\[ar = \varphi^{-1}(ys) = \varphi^{-1}(\sigma'(s)y) = \varphi^{-1}(\sigma'(s))a = \varphi^{-1}(\sigma'(\varphi(r)))a\] where $\varphi^{-1}(r) = s \in S[y;\sigma']$.

Let $\psi = \varphi^{-1}\circ \sigma' \circ \varphi$, which is an endomorphism of $R$. So $R[a;\psi]$ is also a skew polynomial ring, and this should? be isomorphic to $R[x;\sigma]$ (Maybe we need to assume $y$ is not a zero-divisor?)

Now we have $R[a;\psi] \cong R[x;\sigma] \cong S[[y;\sigma']]$; composing gives us an isomorphism $R[a;\psi] \rightarrow S[[y;\sigma']]$ which sends $a$ to $y$, which seems unlikely.

(Sorry, quite a bit of hand-waving towards the end there! You can also show that $1+ax$ must be invertible in $R[x;\sigma]$, similarly to the post linked to in the question, and then maybe use the existence of a commuting rule for $a$ to produce a contradiction?)

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