On integral basis of pure number fields

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

“…and V 2 (8θ 4 + 16θ) = 5. Since V 2 (16θ 2 ) = 14 3 , it now follows from the strong triangle law that V 2 (η 2 1 ) = 13 3 which proves (29).…”

“…In 2020, Jakhar et al gave an explicit construction of an integral basis of all those n-th degree pure fields Q( n √ a) which are such that for each prime p dividing n, either p a or p does not divide v p (a), where v p (a) stands for the highest power of p dividing a; clearly this condition is satisfied when either a, n are coprime or a is squarefree (cf. [12], [13]). A different approach using p-integral basis defined below has been followed by A. Alaca, S. Alaca and K. S. Williams in [1], [2], [3] to construct integral bases of all cubic fields and all those quartic as well as quintic fields which are generated over Q by a root of an irreducible trinomial of the type x n + ax + b belonging to Z[x] with n = 4, 5.…”

“…The above equation shows that S (2) contains an element of the type 1 8 [4x +4y θ±2θ 2 +2θ 4 +θ 5 ] for some x , y ∈ {0, 1}. Keeping in mind that V 2 (θ 3 ± 2) = 3 2 in view of (20), it can be seen that V 2 (4x + 4y θ + 2θ 4 ± 2θ 2 + θ 5 ) equals 2 or 13 6 according as x = 1 or x = 0. This contradiction proves assertion (ii) in the present subcase.…”

“…Thus f (x) is not 2-regular with respect to φ 1 , x − β. So we now consider another rational number δ with v 2 (δ) = 0 for which f (x) turns out to be 2-regular with respect to (13). Define an integer u ≥ 1 by u = s 0 2 .…”

“…2 and hence V 2 (4x + 4y θ ± 2θ 2 + θ 5 ) equals 2 or 13 6 according as x = 1 or x = 0 . This contradiction proves assertion (ii) in the present subcase.…”