I'm stuck with an understanding of what should be the Newton polygon for a Laurent series. I'm reading ''An introduction to G-function" by Dwork and he dedicates only three pages to Newton polygons for Laurent series; he says what is a Laurent series, how we may ''decompose" it as the sum of two power series, and only deals the case when we have an annulus of convergence. The purpose of that section is to give a sketch of the proof of the generalized Weierstrass preparation theorem (A version of that theorem but for Laurent series). I'm filling the gaps, completing things and adding information. I have everything, even the theorem which in some sense the proof is similar to the case of power series; but what I dont understand is how the Newton polygon should be. Here is my attempt:

Recall that a Laurent series, is a series of the form $$f(X)=\sum_{n=-\infty}^{\infty}a_{n}X^{n}, \quad a_{n}\in\mathbb{Q}_{p}$$

($v$ is the additive valuation on $\mathbb{Q}_{p}$)

We can write $f(X)=f^{+}(X)+f^{-}(1/X)$ where $$f^{+}(X)=\sum_{n=0}^{\infty}a_{n}X^{n}\quad\text{and}\quad f^{-}(X)=\sum_{n=1}^{\infty}a_{-n}X^{n}\text{.}$$

We define the Newton polygon of $f$ as in the case of power series. This is his ''definition''. So I tried to understand what it means:

So, the Newton polygon of $f$ is the convex hull of the set of point $(j,v(a_{j}))$. In power series if $a_{j}=0$ we think the point $(j,v(a_{j}))$ as the point at infinity on the upper half plane but now the problem is consider the points when $n<0$. Lets take this examples:

  1. Consider $\sum_{n\in\mathbb{Z}}pX^{n}$, we have $(j,v(a_{j}))=(j,0)$ for every $n\in\mathbb{Z}$, so, I infer that the Newton polygon must be all the x-axis.

Recall that in power series the set of points are on or above the Newton polygon, but how this condition translates into Laurent series? this question comes from the next example:

  1. Consider $f=\sum_{n=-\infty}^{\infty}p^{n}X^{2^{n}}$. Here its more complicated. I tried to obtain the Newton polygon of $f$ from the Newton polygons of $f^{+}$ and $f^{-}$. The Newton polygon of $f^{+}$ is the positive $x-$axis. But what happens with the Newton polygon of $f^{-}$? The points $(j,v(a_{j}))$ with $j<0$ are on the negative $x-$axis and if $n$ is not a power of $2$ the point $(j,v(a_{j}))$ seems to be the point at minus infinity. I think that the Newton polygon of $f^{-}$ must be the negative part of the $x-$axis so that the Newton polygon of $f$ must be all the $x-$axis.

I'm starting with known power series. These examples comes from my examples on power series, and the Newton polygon of these have a finite number of sides. Also, in power series I start with an infinite vertical line, then one constructs the Newton polygon, but in Laurent series I don't have any vertical line (we only deal with convergent Laurent series). But what happens if we start from a power series having an infinite number of sides and trying to expand over $n\in\mathbb{Z}$? For example, the Newton polygon of the logarithm is well known, and looking at what ''should'' be the Newton polygon of $f^{-}$ it makes me to think the following:

I can define the Newton polygon of a Laurent series $f$ by the following: Let $C(f^{+})$ the convex hull of the set of points $(j,v(a_{j}))$ where $j\geq 0$ and let $C(f^{-})$ the convex hull of the set of points $(j,v(a_{j}))$ with $j<0$. Then the Newton polygon of $f$ is the convex hull of $C(f^{+})\cup C(f^{-})$. If $a_{j}=0$ for $j\geq 0$ we put $(j,v(a_{j}))=+\infty$ and if $a_{j}=0$ for $j<0$ we put $(j,v(a_{j}))=-\infty$. If $n<0$ the set of points $(j,v(a_{j}))$ are on or below $C(f^{-})$. Explicitly for $C(f^{-})$ we start from the positive $y-axis$ and we turn it clockwise and we follow the rules of the process with $f^{-}$ but in this direction, so I think that $C(f^{-}(X))=C(f^{+}(-X))$.

Is this ''definition'' (conjecture) valid? I was looking for books having this material, but I only found "A course in $p$-Adic Analysis" by Alain Robert. This is the only reference I have in English. In French I have ''Les nombres p-adiques'' by Yvette Amice. Here defines the Newton polygon of Laurent series as the graph of certain function and then proves that is equivalent to: The Newton polygon of $f$ is the boundary of the convex upper envelope of the set of points $(j,v(a_{j}))$. Then there are some examples of Newton polygons, but the examples are polynomials or power series, she never considers Laurent series.

I really appreciate any reference, discussion, if my ''definition'' is valid, anything. As I said, I'm filling the gaps and other stuffs, but also I want to have examples of what I'm saying.

Thanks a lot, also forgive me if I have misprints, I'm from Mexico and I'm not an expert on English. Greetings


Your examples (1) and (2) aren’t so interesting. Why don’t you consider something like $f(x)=\sum_{n\in\Bbb Z}p^{n^2}x^n\,$? In a case like this, you don’t need to go through any contortions at all: just identify all the points $(n,n^2)$, and draw the upper convex hull. You see that the slopes are all the numbers $(n+1)^2-n^2=2n+1$, thus all odd numbers. So you should expect that there is a single $\Bbb Q_p$-rational root of this $f$ of every odd valuation.

Coming back to your example (2), I think there’s something wrong with it as you’ve written it. For $n=-1$, for instance, your monomial looks to me like $p^{-1}X^{1/2}$, and the other negative values of $n$ are even worse. I’m sure you didn’t mean what you wrote. Maybe you can make a change in this example, maybe to conform with the geometric (Newton) picture you had in mind?


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