On prolongations of valuations via Newton polygons and liftings of polynomials

Abstract

a b s t r a c tLet v be a real valuation of a field K with valuation ring R v . Let K (θ ) be a finite separable extension of K with θ integral over R v and F (x) be the minimal polynomial of θ over K . Using Newton polygons and residually transcendental prolongations of v to a simple transcendental extension K (x) of K together with liftings with respect to such prolongations, we describe a method to determine all prolongations of v to K (θ ) along with their residual degrees and ramification indices over v. The problem is classical but our approach uses new ideas. The paper gives an analogue of Ore's Theorem when the base field is an arbitrary rank-1 valued field and extends the main result of [S.D. Cohen, A. Movahhedi, A. Salinier, Factorization over local fields and the irreducibility of generalized difference polynomials, Mathematika 47 (2000) .

On the discriminant of pure number fields

“…With the above notation, we prove the following lemma which extends Lemma 2.2 to general n. As shown in equation (10),…”

Discriminant and Integral basis of sextic fields defined by x^6+ax+b

“…In view of (25) and the hypothesis v 2 (a) = 4, b 16 ≡ 1 (mod 4), we see that V 2 (ψ 2 + 1) = 2 3 , i.e., V 2 (θ 6 + 16) = 14 3 . Keeping in mind the strong triangle law, the last equality immediately gives V 2 (θ 6 − 16) = 14 3 and hence…”

“…Using (25) and the hypothesis v 2 (a) = 4, b 16 ≡ 3 (mod 4), we see that V 2 ( θ 6 16 − 1) = 2 3 , i.e., V 2 (θ 6 − 16) = 14 3 . The last equality by virtue of the strong triangle law immediately gives (33).…”

“…Next we write the φ 2 -Newton polygon of f (x) where φ 2 (x) = x − δ using the φ 2 -expansion of f given by (14). Consider first the subcase when v 2 (f (δ)) = 2u + 2.…”

“…≡ bD 2 − 1 (mod 8) and bD 2 − 1 ≡ 2 (mod 4) by hypothesis. Keeping in mind the values of v 2 (f (δ)), v 2 (f (δ)) and using (14), it can be easily seen that the φ 2 -Newton polygon of f (x) has a single edge of positive slope, say S joining the point (4, 0) with (6, 2u + 1) and the polynomial associated to f (x) with respect to (φ 2 , S ) is linear. So f (x) is 2-regular with respect to φ 1 , φ 2 .…”