Newton polygonsThis 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?

Question

Newton polygonsThis 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&#039;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?

akademiks1989rz · Accepted Answer

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                     R              2   with vertical sides on both right and left. A point P is only a vertex if there&#039;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.

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.

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