Newton polygonsThis question is primarily to clear up some confusion I have about Newton polygons.Consider...
This question is primarily to clear up some confusion I have about Newton polygons.
Consider the polynomial . I have to decide if this polynomial is irreducible over
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 , and both segments have length 2 when we take their projections onto the horizontal axis.
Am I correct in concluding that the polynomial factors into two quadratic polynomials over , 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 (with ) is pure, iff the only vertices on its Newton polygon are and . Am I right about this, or does the polynomial also qualify as a pure polynomial?