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.
Best Answer
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.