Bourbaki

Exercise 1, § 5

Tag 06J8, permanent. Cite it as [Bourbaki, Tag 06J8].

Algebra, V, p. 150

a) Let E be an extension of a field K, x∈Ex \in E a transcendental element over K and let F=K(xn)F = K(x^n) for an integer n>1n > 1. Show that if the intersection F∩K[(x−1)(xn−1)]F \cap K[(x-1)(x^n-1)] is distinct from K, then K is of characteristic p>0p > 0 and n=pen = p^e, so that K(x)K(x) is a p-radical extension of F. (Writing y=x−1y = x-1 and using Ex. 12, b) of V, p. 149, show that (y+1)n−1=ayn(y+1)^n - 1 = a y^n for an a∈Ka \in K.) b) Show that if K⊂F′⊂K(x)K \subset F' \subset K(x) and if K(x)K(x) is not a p-radical extension of F′F' and if K≠F′≠K(x)K \neq F' \neq K(x), then there exists z∈K[x]z \in K[x] such that z∉Kz \notin K and F′∩K[z]=KF' \cap K[z] = K. (Argue by contradiction, using a) and Ex. 12, b) of V, p. 149.)

c) Under the same hypotheses as in b), suppose that F′=K(y)F' = K(y), where y∈K[x]y \in K[x]; show that there exists u∈K[x]u \in K[x] such that u∉Ku \notin K and F′∩K(u)=F′∩K[u]=KF' \cap K(u) = F' \cap K[u] = K. (Observe that if this were false, we should have F′∩K[z]≠KF' \cap K[z] \neq K for all z∈K[x]z \in K[x] such that z∉Kz \notin K, using Ex. 12, c) of V, p. 149.)

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