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

Pierre Holmes

Pierre Holmes



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 Q 5 [ x ]. I have to decide if this polynomial is irreducible over 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 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 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 Q p [ x ] (with a n a 0 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




2022-07-19Added 16 answers

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 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 Q 5 of degree 4: voilà, your polynomial is irreducible.

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