On monogenity of certain number fields defined by trinomials

On common index divisors and monogenity of certain number fields defined by $x^5+ax+b$

“…Assume also that l ≥ 2 and R 1 (F)(y) = ±(y − t) 2 is the residual polynomial of F(x) associated to this side for some integer t. Then we can construct an element s ∈ Z such that s ≡ u (mod p) and F(x) is x − s-regular with respect to p. Such an element s is called a regular element of F(x) with respect to φ and p. How to construct such a regular element s? By theorem of the polygon, 2 is the residual polynomial of F 2 associated to this side. Let s 0 = t, s 1 = s 0 + p k t, and 2 for some integer t 1 ∈ Z, then we can repeat the same process.…”

“…Therefore Jones's and Khanduja's results cover partially the study of monogenity of number fields defined by trinomials. In [2], Ben Yakkou and El Fadil gave sufficient conditions on coefficients of a trinomial which guarantee the non-monogenity of the number field defined by such a trinomial. Also in [13], Gaál studied the multi-monogenity of number fields defined by some sextic irreducible trinomials.…”

On index divisors and monogenity of certain number fields defined by trinomials $x^6+ax+b$

“…Therefore Jones's and Khanduja's results cover partially the study of monogenity of number fields defined by trinomials. In [1], Ben Yakkou and El Fadil gave sufficient conditions on coefficients of a trinomial which guarantee the non-monogenity of the number field defined by such a trinomial. Also in [13], Gaál studied the multi-monogenity of number fields defined by some sextic irreducible trinomials.…”

“…x], by Remark 3, 2 is a common index divisor of K. (b) If a ≡ 0 (mod 8) and b ≡ 3 (mod 8), then N φ 2 (F) = S 21 has a single side joining (0, 2),(1,1), and (2, 0) with R 1 (F)(y) = xy 2 + (x − 1)y + 1 its attached residual polynomial of F(x). Since R 1 (y) = xy 2 +(x−1)y+1 = x(y−1)(y−x 2 ), φ 2 provides two distinct prime ideals of Z K lying above 2 with residue degree 2 each.…”

“…Thus there are exactly three prime ideals of Z K lying above 3. For a ∈ {15, 24} (mod 27) and b ≡ −1 + a (mod 81), we have N + φ 2 (F) = S 21 + S 22 has two sides joining (0, v),(1,2), and(3, 0) with v ≥ 4 such that R 1 (F)(y) = y 2 − y ± 1 is the residual polynomial of F(x) associated to S 22and S 21 is of degree 1. Thus there are exactly three prime ideals of Z K lying above 3.…”