Solvable Extensions, p-cyclic over Characteristic p

p-cyclic over Characteristic p

Let F be an extension of a field K having characteristic p. By the previous lemma, if F = K(u), where u^p-u-a = 0, then either F = K, or F is an extension of dimension p, F includes all the conjugates of u, F is normal and separable, F is galois, and F is a cyclic extension of order p.

Now let's prove the converse. Assume F is a cyclic extension of order p, and note that the trace of 1 = 0. If c generates the galois group G, then we can set 1 = v-c(v) for some v in F (thanks to an earlier theorem).

Set u = -v, and note that c(u) = u+1. Since c(u) ≠ u, u does not lie in K.

There are no proper subgroups of G, and no intermediate fields of F, so K(u) = F.

Finally, c fixes u^p-u, so u^p-u = a for some a in K. Therefore u is a root of x^p-x-a, which is irreducible over K.

When K has characteristic p, F/K is cyclic of order p iff F = K(u) for some u that is a root of the irreducible polynomial x^p-x-a.