Joubert's theorem fails in characteristic 2

Abstract

Abstract. Let L/K be a separable field extension of degree 6. An 1867 theorem of P. Joubert asserts that if char(K) = 2 then L is generated over K by an element whose minimal polynomial is of the form t 6 + at 4 + bt 2 + ct + d for some a, b, c, d ∈ K. We show that this theorem fails in characteristic 2.

Generators of finite fields with powers of trace zero and cyclotomic function fields

“…We consider the following problem which is a question of Z. Reichstein, asked in connection with the results of [7]. Specifically, he asked when for a given m is there y ∈ F q n with F q (y) = F q n and Tr(y) = • • • = Tr(y m ) = 0, (where Tr is the F q n /F q trace).…”

“…Reichstein was particularly interested in the case n = 6, m = 3, p = 2. Here one can proceed directly as follows, as already indicated in [7]. To find y in F q 6 , not in a smaller field with Tr(y) = Tr(y 3 ) = 0 (where the trace is to F q ) it is enough to find x, z ∈ F q 6 with y = x q − x, y 3 = z q − z and F q (y) = F q 6 .…”