# Nonarchimedean convergent power series

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.

• In what sense do we not have a notion of differentiability here? – Lubin Jan 11 at 3:38
• 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. – Layer Cake Jan 11 at 23:12
• 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. – Lubin Jan 12 at 2:49
• Thank you for taking the time to think about it, I appreciate it. – Layer Cake Jan 12 at 9:43

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 $$\sum_{-\infty for which $$\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 $$\sum_{-\infty for which $$\lim_{|n|\to\infty}|c_n|=0$$; or if you like, the series $$\sum_{-\infty 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 $$S^{\{1\}}=\{\sum_{-\infty for $$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.

• 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... – Layer Cake Jan 12 at 16:19
• 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? – Layer Cake Jan 12 at 16:21
• 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. – Lubin Jan 12 at 18:56
• I think I have it. But it’s late, and I’m fighting a cold. May have it for you by tomorrow afternoon. – Lubin Jan 13 at 4:49
• 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. – Lubin Jan 14 at 16:30