“…For α ∈ N r , we have g α (X) = (X − x)f α (X) ∈ D, giving that there exists some K ∈ [k, L] such that g α = f K with L = K because f L (X) = X − x. It follows that K ⊂ L is minimal by [31,Lemma 5.7]. The monic minimal polynomial of x over k is f (x) := X 6 − 2 = (X −x)(X +x)(X 2 +xX +x 2 )(X 2 −xX +x 2 ), which is its decomposition into irreducible polynomials over L. Set f 1 (X) := X + x, f 2 (X) := X 2 + xX + x 2 , f 3 (X) := X 2 − xX + x 2 and g α (X) := (X − x)f α (X), for α = 1, 2, 3.…”