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11: we no longer suppose that A is a normal ring, nor that L/K is separable. ] According to th. 2, A is integral over a subalgebra C isomorphic to WI,... ,-Ll, and B is obviously the integral closure of C in L . It thus suffices to do the proof when A = k[Xl, . . , Xn] . Moreover, being free to extend L , we can suppose that the extension L/K is quasi-Galois; if we write M for the largest purely inseparable extension of K contained in L , the extension L/M is separable. Let D be the integral closure of A in M ; if we know that D is finitely generated as a module over A , proposition 11, applied to L/M , shows that B is finitely generated as a module over D , whence over A.

6. Remark. For a proof of Cohen’s theorem, see [C&h], expo& 17: and also [Bour], Chap. IX, 53. Let us consider first the case where M = A[X] 8~ N i N being an A-module (we write A4 = N[X] ): since A[X] is A-flat, we have proj dim,Ixl N[X] < proj dim, N , cf. prop. 19. Moreover, it is clew that proj dim, N = proj dim, M Now if M is an arbitrary A[X] -module, it is also an A-module, and we will let MIX] denote the A[X] -module defined by the A-module M We have an exact sequence (cf. j rni) = c xi+’ @‘a m< - c X” @a xm<.

6, or [Bow], Chap. X, p. ‘21, Rem. 1. for all i Since B is regular, these modules are 0 for i > dim B This implies that the same is true for To$(k, k) , because of the following lemma: Lemma 2. 4. A criterion for normality 83 ==a M=O. Indeed, if A4 were # 0, it would have a quotient which is isomorphic to k = A/m(A) and we would have B @a k = 0, contradicting the fact that m(A) is contained in m(B), Hence Torf (k. k) = 0 for large i , which shows that A is regular. Theorem 13. Assume B is finitely generated as an A -module.