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At the beginning of Proposition 3.4.4 in Caruso's An introduction to $p$-adic period rings, it is simply stated that because $A_{\mu_0}$ is the $p$-adic completion of $A_{\inf}[\frac t p]$, it follows that $A_{\mu_0} \subset A_{\inf} + \frac t p A_{\mu_0}$. I do not seem to be able to prove this, although it "feels" right and forms the foundation of the later arguments. I suspect that it is a general but somewhat obscure result.

So let $A \subset B$ be a ring extension and $x \in B$ such that $B = \widehat{A[x]}$ and $\widehat A = A$. Does it follow that $B = A + xB$?

This notably amounts to showing that $A + xB$ is closed. However, the usual methods for this fail me in the above example: $A \cap xB$ is non-trivial; neither summand is an algebra over any field; and while $A$ and $xB$ are both closed, neither of them is compact.

mesz
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  • The statement looks slightly underspecified. I think you need to explain which completion of $A$ yields $A$ and which completion of $A[x]$ yields $B$… – Aphelli Dec 08 '22 at 18:50
  • It is the $p$-adic completion in both cases. – mesz Dec 08 '22 at 18:51

1 Answers1

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So, let’s rephrase: $p$ is a prime number, $A$ is a $p$-adically complete ring, $A \subset B$ is an inclusion, $x \in B$ such that if $A’=A[x]$, $A’ \subset B$ is a $p$-adic completion.

That assumption means that elements of $B$ can be seen as sequences $a_n \in A’$ such that $a_{n+1}-a_n \in p^nA’$.

So let $b \in B$ correspond to some sequence $a_n$. Then write $a_{n+1}-a_n=p^n(c_n+xd_n)$ with $c_n \in A$, $d_n \in A’$. Then define sequences $u_n \in A, v_n \in A’$ so that $u_{n+1}-u_n=p^nc_n$, $u_1+xv_1=a_1$, $v_{n+1}-v_n=p^nd_n$. Then, as $A$ is $p$-adically complete, $u_n$ defines some $\alpha \in A$, and by definition $v_n$ is some $\beta \in B$, and $\alpha+x\beta=b$.

Aphelli
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