Pierre Holmes

2022-07-18

Newton polygons
This question is primarily to clear up some confusion I have about Newton polygons.
Consider the polynomial ${x}^{4}+5{x}^{2}+25\in {\mathbb{Q}}_{5}\left[x\right]$. I have to decide if this polynomial is irreducible over ${\mathbb{Q}}_{5}$
So, I compute its Newton polygon. On doing this I find that the vertices of the polygon are (0,2), (2,1) and (4,0). The segments joining (0,2) and (2,1), and (2,1) and (4,0) both have slope $-\frac{1}{2}$, and both segments have length 2 when we take their projections onto the horizontal axis.
Am I correct in concluding that the polynomial ${x}^{4}+5{x}^{2}+25$ factors into two quadratic polynomials over ${\mathbb{Q}}_{5}$, and so is not irreducible?
I am deducing this on the basis of the following definition of a pure polynomial given in Gouvea's P-adic Numbers, An Introduction (and the fact that irreducible polynomials are pure):
A polynomial is pure if its Newton polygon has one slope.
What I interpret this definition to mean is that a polynomial $f\left(x\right)={a}_{n}{x}^{n}+...+{a}_{0}$ $\in {\mathbb{Q}}_{p}\left[x\right]$ (with ${a}_{n}{a}_{0}\ne 0$) is pure, iff the only vertices on its Newton polygon are $\left(0,{v}_{p}\left({a}_{0}\right)\right)$ and $\left(n,{v}_{p}\left({a}_{n}\right)\right)$. Am I right about this, or does the polynomial ${x}^{4}+5{x}^{2}+25$ also qualify as a pure polynomial?

Expert

Explanation:
There is no vertex at (2,1). In my opinion, the right way to think of a Newton Polygon of a polynomial is as a closed convex body in ${\mathbb{R}}^{2}$ with vertical sides on both right and left. A point P is only a vertex if there's a line through it touching the polygon at only one point. So this polynomial definitely is pure, and N-polygon theory does not help you at all. Easiest, I suppose, will be to write down what the roots are and see that any one of them generates an extension field of ${\mathbb{Q}}_{5}$ of degree 4: voilà, your polynomial is irreducible.

Do you have a similar question?