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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.

E. KOW
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    One can see this directly if you assume that $I$ is finitely generated which, let’s be honest, essentially every real world example is. The point is that you want to commute the inverse limit and tensor product in the computaiton $(\varprojlim R/I^n[\zeta])\otimes_R (R/I)$. This can then be done via the Mittag-Leffler condition. This is almost certainly explained in Fujiwara—Kato or Bosch’s book somewhere, but Matthew Emerton also explains it here: https://math.stackexchange.com/questions/181004/inverse-limit-of-modules-and-tensor-product. Let me know if that is enough for you. – Alex Youcis Jan 02 '21 at 18:38
  • Excellent, thanks! – E. KOW Jan 02 '21 at 23:53
  • Should I post this as an answer? – Alex Youcis Jan 06 '21 at 00:56
  • yes maybe post it as an answer so I'll be able to mark it as answered. btw in the non f.g. case - is the statement itself still true? – E. KOW Jan 06 '21 at 01:06
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    To be honest, I’m not sure. I think it depends on how you define $R\langle \zeta\rangle$—do you define it as restricted power series, or the completion of $R[t]$? Do those even coincide when $I$ is not f.g. (or the topology is not defined by a f.g. ideal is the more operative question)—in the latter case, is the latter even complete? – Alex Youcis Jan 06 '21 at 01:08
  • Oh I see, I didn't think about it. Thank you! – E. KOW Jan 10 '21 at 18:54

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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.

Alex Youcis
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