Bourbaki

Exercise 29, § 1

Commutative Algebra, V, 362

Let k be a field of characteristic ≠2\neq 2, A the polynomial ring k[T]k[T] in one indeterminate and a,ba, b two elements of k such that a≠0,b≠0a \neq 0, b \neq 0 and a≠ba \neq b. Let K=k(T)K = k(T) be the field of fractions of A and L, M the extensions of K adjoining to K respectively a root of X2−T(T−a)X^2 - T(T - a) and a root of X2−T(T−b)X^2 - T(T - b). Let B, C be the integral closures of A in L and M respectively, which are finitely generated free A-modules. Show that L⊗KML \otimes_K M is a field, a separable finite extension of K, but that B⊗ACB \otimes_A C, which is canonically identified with a subring of L⊗KML \otimes_K M, is not the integral closure of A in L⊗KML \otimes_K M.

Solution

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Transcribed from the printed volume. Not translated.