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I would like to understand the second paragraph on the second page (marked page 320) of the article http://www.numdam.org/article/MSMF_1974__39-40__319_0.pdf on rigid analytic geometry by Michel Raynaud.

Let's take $K$ a complete nonarchimedean field and $R$ its valuation ring, and $a \in R$ a non-zero element of the unique maximal ideal of $R$ . Let $D:=\{z \in K \colon \vert z \vert \leq 1\}$ the closed unit disk. We can decompose $D$ into the smaller disk $D':=\{z \in K \colon \vert z \vert \leq \vert a \vert \}$ and the annulus $C:=\{z \in K \colon \vert a \vert \leq \vert z \vert \leq 1 \}$.

Then, the mentioned paragraph says that a power series converging on $D$ is the same as a pair consisting of a power series converging on $D'$ and a Laurent series converging on $C$ such that these two series agree on $C \cap D'$.

Now, I am aware that this follows from the Tate acyclicity theorem, but the article says it is immediate, so I would like to see how it is immediate. I am also aware how the corresponding fact is true in complex analysis, but that doesn't seem to help because in the rigid nonarchimedean setting we don't have a notion of differentiability.

So we know that the Laurent series is a power series on the circle $C \cap D'$. Is there an identity-theorem-type of argument to conclude that the Laurent series is a power series on the whole annulus $C$? Or is it even simpler to conclude?

EDIT: For Lubin's proof below to go through, we need to

1) have our functions take values in the algebraic closure $\overline{K}$ of $K$ (i.e. $D=\{z \in \overline {K} \colon \vert z \vert \leq 1\}$), OR

2) impose the extra condition that $K$ has an infinite residue field.

I too prefer the first option, because the second is rather exclusive since it excludes local fields.

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  • $\begingroup$ In what sense do we not have a notion of differentiability here? $\endgroup$ – Lubin Jan 11 at 3:38
  • $\begingroup$ My bad, we certainly do! I meant to say that we don’t have a rewarding notion, because there is no link between differentiable and analytic functions here. $\endgroup$ – Layer Cake Jan 11 at 23:12
  • $\begingroup$ Ah, of course. So the good notion is analyticity, rather than differentiability; and all proofs should depend on the former, without mention of the latter, I suppose. Let me think about this. I might be able to hack up an ugly proof for you. $\endgroup$ – Lubin Jan 12 at 2:49
  • $\begingroup$ Thank you for taking the time to think about it, I appreciate it. $\endgroup$ – Layer Cake Jan 12 at 9:43
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To clarify our ideas, let’s see what the Laurent series look like that are convergent on $\{z:|z|=1\}$ — the skin, so to speak, of the closed unit disk. These are the series $\sum_{-\infty<n<\infty}c_nx^n$ for which $\lim_{|n|\to\infty}|c_n|=0$ . That is, we need $|c_n|\to0$ for positive $n$ and negative $n$.

A worthwhile example is $\sum_{n\ge0}p^n(x^{-n^2}+x^{n^2})$. Draw the Newton picture and you see what’s going on: the points you draw are all $(\pm n^2,n)$. This is a series convergent only on the skin.

It’s the same thing for series about which you nothing more than that they are convergent on the skin of $D'$, namely the series $\sum_{-\infty<n<\infty}\frac{c_n}{a^n}x^n$ for which $\lim_{|n|\to\infty}|c_n|=0$; or if you like, the series $\sum_{-\infty<n<\infty}\gamma_nx^n$ for which $\lim_{|n|\to\infty}|a^n\gamma_n|=0$.

But the series we’re concerned with are power series, that is, of form $\sum_0^\infty\gamma_nx^n$; and since the series is convergent on the outer skin of $C$, even at $z=1$, so the coefficients $\gamma_n$ must have $|\gamma_n|\to0$. This is just the condition that our series converges on $D$.

EDIT:
Let’s see whether I can give a satisfactory answer to your very valid objection. It depends on making a careful distinction between a series $G(x)$ and the $p$-adic function $g$ defined by $G$.

While we’re at it, I want to change the coordinatization, sending a point $z$ to $z/a$, so that now our old set $D'$ is the closed unit disk $D$, and the original $D$ becomes what I’ll call $D^+$, namely $\{z:|z|\le|1/a|\}$. Then the original $C$ becomes $\{z:1\le|z|\le|1/a|\}$. Our intersection-set is just the units $U$ of $K$, considered as an analytic space. All this just to make the typing easier for me.

Our four rings of power series now are $S^{[0,1]}\{\sum_0^\infty c_nx^n: c_n\to0\}$ for $D$, $S^{[0,1/|a|]}=\{\sum_0^\infty c_nx^n:c_n/a^n\to0\}$ for $D^+$, $S^{[1,1/|a|]}=\{\sum_{-\infty,\infty}c_nx^n:\lim_{n\to-\infty}c_n=0\text{ and }\lim_{n\to\infty}c_n/a^n=0\}$ for our new annulus $C$, and $S^{\{1\}}=\{\sum_{-\infty<n<\infty}c_nx^n:\lim_{|n|\to\infty}c_n=0\}$ for $U$.

My first task is to show that a nonzero series $G(x)\in S^{\{1\}}$, which you recall may be evaluated at any $z\in U$, to give a numerical value, must define a function which is not identically zero on $U$.

Well, without loss of generality, we may assume that all coefficients of $G$ are in $R$, and indeed some of them are in $R^\times$, the unit group of $R$. But only finitely many of them! Now, by multiplying by a monomial, we may assume that $G(x)$ reduces to the nonzero $\Gamma(x)\in(R/\mathfrak m)[x]$, for $\mathfrak m$ the maximal ideal of $R$, and even, if you like, that $\Gamma$ is monic. But even over an algebraically closed field containing $R/\mathfrak m$, $\Gamma$ has at most finitely many roots. Thus, we may find $\xi$ in either $R/\mathfrak m$ or an algebraic closure for which $\Gamma(\xi)\ne0$, and when we lift $\xi$ to $z_0$ in $R$ or a finite unramified extension, if necessary, we find that $G(z_0)\ne0$, so that the function $g$ defined on $U$ by the series $G$ is not identically zero.

That was the hard part, if any such there was. (I’m sure you can see the rest of the argument.) We now consider an analytic function $f$ on $D$, given by $F(x)\in S^{[0,1]}$ and analytic function $h$ on our annulus $C$, given by $H(x)\in S^{[1,1/|a|]}$, such that $f$ and $h$ agree on $U$. But we have set-theoretic inclusions of $S^{[0,1]}$ and $S^{[1,1/|a|]}$ into $S{\{1\}}$ and since $f$ and $h$ agree on the set $U$, their difference (an element of $S^{[1,1/|a|]}$) is identically zero on $U$, so that $G$ and $H$ are equal, coefficient by coefficient.

I believe that the earlier argument I tried to give now applies, to yield our result.

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  • $\begingroup$ I understand what you wrote, but I am not sure whether it fully answers my question, let me state it more precisely: Let $f\colon D' \to K$ be a an analytic function (i.e. given by a power series $F$ converging on $D'$) and $g\colon C \to K$ a function given by a Laurent series $G$ converging on $C$, and suppose that $f$ and $g$ agree on $C \cap D'$. Obviously, we can define a function $h \colon D \to K$ by setting $h$ to be $f$ on $D'$ and $g$ on $C$. We now want to show that $h$ is given by a power series $H$ converging on $D$. Continued in my next comment... $\endgroup$ – Layer Cake Jan 12 at 16:19
  • $\begingroup$ Now, you've shown that, if we know that the Laurent series $G$ is in fact a power series, we can use the fact that $G$ converges on the outer skin of $C$ to deduce that $G$ converges on $D$. So if we know that $G=F$ (as formal series), then we are done, we can set $H:=G$. But, we only know that $F(z)=G(z)$ for all $z \in C \cap D'$. Can we deduce $F=G$ from that? $\endgroup$ – Layer Cake Jan 12 at 16:21
  • $\begingroup$ Yes, I confess that issues of this kind tend to confuse me. I guess in this case, we need to know that restriction of analytic functions from $D'$ to the skin of $D'$ is an injection. Let me think further. $\endgroup$ – Lubin Jan 12 at 18:56
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    $\begingroup$ I think I have it. But it’s late, and I’m fighting a cold. May have it for you by tomorrow afternoon. $\endgroup$ – Lubin Jan 13 at 4:49
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    $\begingroup$ On the other hand, because of the applications I have in mind, I like to take my functions to be defined over a finite extension of $\Bbb Q_p$ but to allow the coordinates of the points to be in $\Bbb Q_p^{\text{ac}}$. (you don’t need to pass to the completion of the algebraic closure in this case). You really do need to have a clear idea of what the underlying set is, and its topology, while you’re thinking about these things. $\endgroup$ – Lubin Jan 14 at 16:30

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