# On the Newton Polygon for $p-$adic Power series

I'm studyng a Book about $$p-$$adic numbers, and I have troubles with a "degenerate" case of a Newton polygon. Let $$f(X)=\sum a_{i}X^{i}\in\mathbb{Q}_{p}[\![X]\!]$$, we define the Newton poligon of $$f$$ as the convex hull of the set of points $$(i,v_{p}(a_{i}))$$ (where $$v_{p}$$ is the $$p-$$adic valuation) and if $$a_{i}=0$$ we introduce the point at infinity on the positive Y-axis $$Y_{+\infty}=(a_{i},v_{p}(a_{i}))$$. My trouble comes from this series: $$\sum_{n=0}^{\infty}p^{-n^{2}}X^{n}$$

I have a first infinite line, the positive part of the Y-axis, and since I have a strictly decreasing sequence on the $$v_{p}(p^{n-^{2}})'s$$ we cannot have a convex hull, so I only have the positive Y-axis? But the convex hull must cointain all of these points, so, my intuition suggest that the Newton poligon must be all the Y-axis

What is the newton polygon in this "degenerate" case? In the book there is not a discussion about this kind of examples, only treats the case of convergent power series, so we have a finite or infinite number of sides in our Newton polygon ...

Thanks a lot; I hope to have explained myself

• I'd say at first it is for polynomials and it is extended to convergent power series on $p^m \mathbb{Z}_p$ by uniform convergence – reuns Feb 7 at 23:16
• I was going to write an answer to this interesting question, but what I wrote seemed lacking in real information. I think that you and I may define the N-polygon slightly differently. I like to think of a vertical half-line going upward from every point $(n, v(a_n))$ for which $a_n\ne0$, and then take the closed convex hull of the union of these half-lines. In this definition, the N-pol. is all the right-hand half-plane. In a way, just the opposite of the picture that you drew. No matter, though: its shape tells us that the series is not convergent except for $z=0$. – Lubin Feb 8 at 5:38
• @Lubin Yes, I was thinking that, but from another (equivalent I think) definition; I consider the partial sums and their Newton polygons, so the Newton polygon of the series will be the "limit" of the Newton polygons of the partial sums and from this definition I have the same as you. In the book (also other books) doesn't care this kind of behaviour, only deals with convergent series, which at the end of the day those are the truly important for the results – Camacho Camachito Feb 9 at 0:30
• I don’t know whether you’re familiar with the Newton copolygon, which in the right viewpoint is a dual convex body to the polygon. The situation, as here, when the polygon is (for all intents and purposes) all of the ambient space, makes the copolygon empty (maybe reduced to a single point — I’m blanking on that). Since the domain of convergence of the original series can be read off from the copolygon, as its projection to the horizontal coordinate axis, you get a degenerate domain of convergence. – Lubin Feb 9 at 2:52