This question is primarily to clear up some confusion I have about Newton polygons.
Consider the polynomial $x^4 + 5x^2 +25 \in \mathbb{Q}_{5}[x]$. 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 $x^4 +5x^2 +25$ 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 $f(x) = a_nx^n + … + a_0$ $\in \mathbb{Q}_{p}[x]$ (with $a_na_0 \neq 0$) is pure, iff the only vertices on its Newton polygon are $(0,v_p(a_0))$ and $(n, v_p(a_n))$. Am I right about this, or does the polynomial $x^4 + 5x^2+25$ also qualify as a pure polynomial?
Best Answer
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.