Pierre Holmes

Answered

2022-07-18

Newton polygons

This question is primarily to clear up some confusion I have about Newton polygons.

Consider the polynomial ${x}^{4}+5{x}^{2}+25\in {\mathbb{Q}}_{5}[x]$. 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 ${x}^{4}+5{x}^{2}+25$ 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 $f(x)={a}_{n}{x}^{n}+...+{a}_{0}$ $\in {\mathbb{Q}}_{p}[x]$ (with ${a}_{n}{a}_{0}\ne 0$) 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 ${x}^{4}+5{x}^{2}+25$ also qualify as a pure polynomial?

This question is primarily to clear up some confusion I have about Newton polygons.

Consider the polynomial ${x}^{4}+5{x}^{2}+25\in {\mathbb{Q}}_{5}[x]$. 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 ${x}^{4}+5{x}^{2}+25$ 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 $f(x)={a}_{n}{x}^{n}+...+{a}_{0}$ $\in {\mathbb{Q}}_{p}[x]$ (with ${a}_{n}{a}_{0}\ne 0$) 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 ${x}^{4}+5{x}^{2}+25$ also qualify as a pure polynomial?

Answer & Explanation

akademiks1989rz

Expert

2022-07-19Added 16 answers

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.

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.

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