Restricted power series and tensor product Let $R$ be a complete adic ring (not necessarily Noetherian) with an ideal of definition $I$. Let $R\left<\zeta\right> := \underset{\longleftarrow}{\lim}_n R/I^n \left[\zeta\right]$ be the ring of restricted power series in $R$. Is it true that $R\left<\zeta\right>\otimes_R R/I \simeq R/I\left[\zeta\right]$?
Edit: I think I see why it is true by the explicit description of $R\left<\zeta\right>$, but for some reason I can't prove it using only formal properties of tensor, limits etc.
 A: Just to get this off the unanswered list.
Assume that $I$ is finitely generated one can show using the Mittag—Leffler condition (e.g. see this post by Matt Emerton) that the natural map
$$(\varprojlim R[z]/I^n)\otimes_R (R/I^m)\to \varprojlim(R[z]/I^n\otimes_R R/I^m)$$
is an isomorphism. But, evidently
$$R[z]/I^n\otimes_R R/I^m$$
forms the constant projective system $R[z]/I^m=(R/I^m)[z]$ and so we see the desired conclusion.
What happens if $I$ is not finitely generated? If $R$ is of finite ideal type, which means that $R$ possesses an ideal of definition which is finitely generated, then one can in practice replace $I$ by this different ideal of definition and thus one can assume WLOG that $I$ is finitely generated. If $R$ is not even of finite ideal type the one gets into a sticky question of what precisely $R\langle \zeta\rangle$ means. Namely, there are two candidates for this adic ring
$$\left\{\sum_n a_n \zeta^n\in R[[\zeta]]: \lim a_n=0\text{ in }R\right\}$$
and
$$\varprojlim R[\zeta]/I^n=\text{the }I\text{-adic completion of }R[\zeta]$$
If $R$ is of finite ideal type, then these are isomorphic. If $R$ is not of finite ideal type, I’m not even sure if these are isomorphic or if, even, the former is complete (e.g. see [1, Example 7.2.10])—but I have not checked in detail.
Note that, in practice, in the context of rigid geometry/formal geometry that appears naturally in nature are all of finite ideal type, so I wouldn’t worry too much.
