On index divisors and monogenity of certain sextic number fields defined by $x^6+ax^5+b$

On index divisors and monogenity of certain number fields defined by x12+axm+b

“…Thus there is two prime ideals of Z K lying above 3 with residue degrees 1 and 2 provided by φ 2 and φ 3 respectively. For i = 1, we have the following cases: (6,11), (15,2), (15,20), (24,11), (24,20)} (mod 27), then N + φ 1 (F) = S 1 has a single side joining (0, 2) and (3, 0). Thus 3Z (15,11), (15,38), (42,11), (42, 65), (69, 38), (69, 65)} (mod 81), then N + φ 1 (F) = S 1 has a single side joining (0, 3) and (3, 0) with (3,11), (3,20), (12,2), (12,20), (21,2), (21,20)} (mod 27), then N + φ 2 (F) = S 2 has a single side joining (2, 0) and (3, 0).…”

“…Since ∆(F) = 12 12 b 11 − 11 11 a 12 , ν 7 (∆) ≥ 1 if and only if (a, b) ∈ {(0, 0), (1,4), (2,4), (3,4), (4,4), (5,4), (6, 4)} (mod 7). Thanks to the index formula (1.1), 7 can divides the index i(K) only if (a, b) ∈ {(0, 0), (1,4), (2,4), (3,4), (4,4), (5,4), (6,4)} (mod 7).…”

“…(1) For (a, b) ∈ {(1, 4), (2,4), (3,4), (4,4), (5,4), (6,4)} (mod 7), one can easily check that F(x) ≡ φ i1 • φ i2 • φ 2 i3 (mod 7) with deg(φ i1 ) = 7, deg(φ i2 ) = 3, deg(φ i3 ) = 1, and φ i j is irreducible over F 7 for every i = 1, . .…”

“…Hence ν 3 (i(K)) = 0. (d) If (a, b) ∈ {(15, 65),(42, 38),(69,11),(6, 47),(33,20), (60, 74),(24, 56), (51, 29), (78, 2)} (mod 81), then ν 3 (∆(F)) ≥ 16. As in the proof of Theorem 2.3, let a 3 Since 3 does not divide 11a 3 ,then u ∈ Z 3 .…”

“…Hence ν 3 (i(K)) = 0. (iv) If d = 4, then N φ (F) = S 1 has a single side joining (0, 4) and (12,0) with 7 ) and a ≡ 0 (mod 3 6 ), then N φ (F) = S 1 has a single side joining (0, 6) and (12, 0) with…”