# Are rings of power series over a local field complete?

Let $$K$$ be a finite extension of $$\mathbb{Q}_p$$, and let $$D$$ be some disk $$D = \{ x\in \overline{K} \mid |x| < c < 1\}$$. Is the set of power series in $$K[[T]]$$ which converge on $$D$$ and are bounded on $$D$$ complete with respect to the supremum norm $$\|f\|_\infty=\sup_{x\in \overline{K}, |x|?

• I used your answer to interpret your question, which is not good, though the question is interesting Commented Nov 28, 2021 at 5:12

Your disc $$D$$ depends on $$c$$ and you never said anything about what kind of number $$c$$ is, other than it is in $$(0,1)$$. In particular, whether or not $$c$$ is in $$|K^\times|$$ could be an important distinction.

If $$c$$ is in $$|K^\times|$$, say $$c=|\alpha|_p$$, then by rescaling (replacing $$f(T)$$ with $$f(\alpha T)$$) we can put ourselves in the case $$c=1$$, so we can then suppose $$D$$ is the open unit disc in $$\overline{K} = \overline{\mathbf Q_p}$$. Or if we start with a series $$f(T)$$ converging on the open unit disc in $$\overline{K}$$, then $$f(T/\alpha)$$ converges on all $$x$$ with $$|x|_p < c$$. So when $$c$$ is in $$|K^\times|$$, there is no real difference between $$c \not= 1$$ and $$c=1$$. In any case, I have a surprise for you: the sup-norm $$||f||_{\infty}$$ need not be finite. So you need to be explicit about looking at series with a finite sup-norm on $$D$$. (Edit: the original version of the OP's question did not include finiteness of $$||f||_\infty$$ as a hypothesis, but the OP now fixed that, which makes this answer no longer directly relevant except to point out why the finiteness of $$||f||_\infty$$ has to be an assumption, just like in the definition of $$L^p$$-spaces.) I think your mistake is possibly due to overlooking a contrast between discs in finite extensions of $$\mathbf Q_p$$ and discs in the algebraic closure: your disc $$D$$ is not compact, so continuous functions on $$D$$ need not be bounded.

Specifically, the $$p$$-adic logarithm series $$\log(1+T) = T - T^2/2 + \cdots$$ in $$\mathbf Q[[T]]$$ converges on the open unit disc $$\Delta$$ in $$\overline{\mathbf Q_p}$$ since it converges on the open unit disc in $$L$$ for each finite extension $$L$$ of $$\mathbf Q_p$$ and its image on that disc is all of $$\overline{\mathbf Q_p}$$. Let’s prove that. Pick $$z$$ in $$\overline{\mathbf Q_p}$$. It lies in a finite extension $$L$$ of $$\mathbf Q_p$$. For big enough $$n$$, $$|p^nz|_p < (1/p)^{1/(p-1)}$$, so $$p^nz = \log(1+y)$$ for some $$y$$ in $$L$$ where $$|y|_p < (1/p)^{1/(p-1)} < 1$$ since the $$p$$-adic logarithm is a bijection (even an isometry) between the open discs in $$L$$ of radius $$(1/p)^{1/(p-1)}$$ around $$1$$ and $$0$$. Write $$1+y$$ as a $$p^n$$-th power in $$\overline{\mathbf Q_p}$$, say $$1+y = w^{p^n}$$. Then $$|w|_p \leq 1$$, and looking at both sides of the equation $$1+y = w^{p^n}$$ in the residue field of a finite extension of $$\mathbf Q_p$$ containing $$y$$ and $$w$$ shows $$|w-1|_p < 1$$. Then
$$p^nz = \log(1+y) = \log(w^{p^n}) = p^n\log(w),$$ so $$z = \log(w) = \log(1 + (w-1))$$ where $$w-1 \in \Delta$$. Thus the surjectivity of $$\log(1+T)$$ from $$\Delta$$ to $$\overline{\mathbf Q_p}$$ is proved.

Instead of reducing to the case $$c=1$$, when $$c \in (0,1)$$ is in $$|K^\times|$$, say $$c=|\alpha|_p$$, let $$f(T) = \log(1+T/\alpha)$$, which has coefficients in $$K$$ and converges at all $$x$$ in $$\overline{K}$$ with $$|x|_p < c$$ since $$|x/\alpha|_p < 1$$. Then $$f(T)$$ on your disc $$D$$ has image $$\overline{K}$$ and thus its sup-norm on $$D$$ is infinite.

Similar results could be obtained for many other power series besides the logarithm by using Newton polygon arguments, but for the logarithm we can “see” its surjectivity onto $$\overline{K}$$ in the more direct way above.

• Yes obviously. You should have pinged me, only the last sentence of my answer is unclear/wrong Commented Nov 28, 2021 at 13:57
• @reuns I think for the OP, the possible unboundedness of the sup-norm of a power series on $D$ may not have been obvious. I had not seen the end of your initial answer before I wrote mine. I will edit the part where I say the “whole premise” of the question is flawed.
– KCd
Commented Nov 28, 2021 at 14:08
• I just edited thanks to you! I missed that the OP had in mind the ring of convergent power series, not that with bounded sup norm, and indeed I didn't clarify in my reasonning that they were different. You are right that this is the main answer to the question. Commented Nov 28, 2021 at 14:11
• Hello @KCd I apologize for not including the boundedness condition in the premise. Thank you for pointing out this is necessary in order for the space to be a metric space. Commented Nov 28, 2021 at 17:02

I think the main theorem is that on the ring $$R_K$$ of power series $$\in K[[T]]$$ that converge on the closed unit disk $$x\in \overline{K},|x|\le 1$$ the sup norm of the coefficients and the sup norm of the values taken on the closed unit disk give the same norm.

This is because $$R_K = \{f= \sum_{n\ge0} a_n T^n\in K[[T]], v(a_n)\to \infty\}$$. If $$f\ne 0$$, with $$v(f)=\inf_n v(a_n)$$, then $$p^{-v(f)} f\in O_{K(p^{v(f)})}/(\pi_{K(p^{v(f)})})[T]$$ is a non-zero polynomial, it will be non-zero for some root of unity $$\zeta\bmod (\pi_{K(p^{v(f)})})$$ so that $$v(p^{-v(f)} f(\zeta))=0$$ ie. $$\inf_{|x|\le 1} v(f(x))\le v(f(\zeta))=v(f)\le \inf_{|x|\le 1} v(f(x))$$

In particular $$R_K$$ is complete for both norm.

Then your ring is just $$\{f\in K[[T]],\forall d\in \Bbb{Q}\cap (0,c), f(p^d T) \in R_{K(p^d)}\}$$ which will be complete for the topology induced by the collection of $$\sup_{|x|\le d} |f(x)|$$ norms, but as kCd said $$\sup_{|x| doesn't need being finite so it is not a norm on your ring. The subring with $$\sup_{|x| finite is complete.

• So it is complete if we restrict the space to being bounded power series? Commented Nov 28, 2021 at 15:51
• Yes ${}{}{}{}{}$ Commented Nov 28, 2021 at 15:58

Ok here is my shot at an answer, but I am not certain it is correct.

Suppose $$|f_n(x) - f_m(x)| < \epsilon$$ for all $$|x|. I show all terms of this difference must have absolute value less than $$\epsilon$$ implying an upper bound on the coefficients. Suppose $$f_n(x)-f_m(x)$$ has terms of absolute value greater than $$\epsilon$$ when evaluated at some $$z$$. Let $$S = \{c_{i_1}z^{i_1},\ldots, c_{i_k}z^{i_k}\}$$ be the list of all terms which have lowest valuation $$u$$, implying the valuation of $$\sum_{j=1}^k c_{i_j}z^{i_j}$$ is greater than $$u$$. There is some finite extension $$K'$$ of $$K$$ in which all of the elements of $$S$$ live. Now take $$L$$ to be some unramified extension of $$K'$$ such that the size of the residue field of $$L$$ is greater than $$i_k$$. Now suppose $$\lambda$$ is some uniformizer of $$L$$, so that there exists a power $$\lambda^{e_1}$$ such that the valuation of $$\lambda^{e_1}$$ is $$u$$. Then $$c_{i_j}z^{i_j}\lambda^{-e_1}$$ are all units, and the polynomial $$g(T) = \sum_{j=1}^kc_{i_j}z^{i_j}\lambda^{-e_1}T^{i_j}$$ must vanish for all $$t$$ in $$L/\lambda$$ else we can find $$x = zt \in L$$ with $$|f_n(x)-f_m(x)| > \epsilon$$ and $$|x| < c$$. However $$g$$ cannot possibly vanish on all of $$L/\lambda$$ because $$|L/\lambda| > i_k$$ and $$\deg g = i_k$$. Thus all terms of $$f_n(z)-f_m(z)$$ have absolute value at most $$\epsilon$$, and this proves the result.