On the factors of a polynomial

Abstract

In this article, we show that if ffalse(xfalse)=anxn+an−1xn−1+⋯+a0,a0≠0 is a polynomial with rational coefficients and if there exists a prime p whose highest power ri dividing ai (where ri=∞ if ai=0) satisfies rn=0, nri⩾false(n−ifalse)r0>0 for 0⩽i⩽n−1, then f(x) has at most gcd(r0,n) irreducible factors over the field boldQ of rational numbers and each irreducible factor has degree at least n/trueprefixgcdfalse(r0,nfalse) over boldQ. This result extends the famous Eisenstein–Dumas irreducibility criterion. In fact, we prove our result in a more general setup for polynomials with coefficients in arbitrary valued fields.

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

“…We find the discriminant and an integral basis of K = Q(θ) where θ is a root of the polynomial f (x) = x 6 + 4x + 4. In view of a simple result 2 proved in [10], each irreducible factor of f (x) over Q has degree at least 3 from which it can be easily deduced that f (x) is irreducible over Q. Here D = discr(f (x)) = −2…”

“…by k 0 . Applying Proposition 3.E with φ(x) replaced by x − β and j by 1, we see that the element (10), it now follows that the element…”

“…Since v 2 (D) ≥ 9, it follows from (9) that s 0 ≥ 3. Using this fact together with ( 8) and (10), it can be easily seen that the φ 2 -Newton polygon of f (x) has a single edge of positive slope, say S 2 joining the points (4, 0) and (6, s 0 ) with slope s 0 2 . In view of ( 9) and the hypothesis v 2 (D) odd, we conclude that s 0 is odd.…”

“…Therefore 1 16 [8lθ + 4z 1 θ 2 + θ 5 ] ∈ S (2) . This is not possible because by virtue of (35), we have V 2 (θ 3 − 4) = V 2 (θ 3 + 4) = 5 2 and hence V 2 (8lθ + 4z 1 θ 2 + θ 5 ) equals 10 3 or 11 3 or 23 6 according as the pair (z 1 , l) belongs to {0, 2} × {0, 1} or to {1, −1} × {1} or to {1, −1} × {0}. This contradiction proves the claim in the present subcase.…”

Erratum

“…Between September 2019 and February 2020, twenty‐one research articles 1–21 were published in the Bulletin of the London Mathematical Society with incorrect copyright statements. The copyright in these articles is stated as belonging to the London Mathematical Society; however, that is incorrect and the copyright remains with the authors of those articles.…”