Exercise 1, § 7

a) Let $K$ be a field of characteristic $p > 0$. An algebraic extension $F$ of $K$ of finite degree is said to be exceptional if it is not separable over $K$ and if it contains no element x \notin K which is p-radical over K, in other words if F^p \cap K = K^p (V, p. 151, Ex. 3). Show that if F is exceptional and if E = F_s, z \in F, z \notin E and z^p \in E, then for each t \in K we have z^p \notin E^p(t) (observe that otherwise we should have E^p(t) = E^p(t^p) and so t \in F^p).

b) Conversely, let E be a separable algebraic extension of K of finite degree, and let x \in E be such that for each t \in K we have x \notin E^p(t). Let F = E(x^{1/p}), so that E = F_s. Show that F is an exceptional extension of K. (If there existed y \in F such that y \notin E and y^p = z \in K, show that we would have z \in E^p(x); this would imply z \in E^p, hence y \in E, which is a contradiction.)

A separable extension E of K of finite degree is called strongly separable if the union of the fields E^p(t), where t ranges over K, is distinct from E; it thus comes to the same to say that E = F_s, where F is a inseparable exceptional extension of K. A strongly separable extension of K cannot be a perfect field.

c) Show that if E is a strongly separable extension of K then it is impossible that [E:E^p] = p. (Argue by contradiction, showing that [E:E^p] = p would imply E^p(t) = E for all t \in K such that t \notin K^p; observe on the other hand that if K \subset E^p, E cannot be strongly separable over K unless it is perfect.)

d) Show that if [K:K^p] = p, then there is no strongly separable extension of K, and so neither is there an exceptional extension of K.