Theorem 1

Let k be a field and A a local k-algebra essentially of finite type. Put n = \dim(A) + \deg.\mathrm{tr}k(\kappa_A).*

a) One has [\kappaA \otimes_A \Omega_k(A) : \kappa_A] \geq n.*

b) The following conditions are equivalent: (i) the k-algebra A is absolutely regular; (ii) the A-module \Omegak(A) is free of rank n;* (iii) one has [\kappaA \otimes_A \Omega_k(A) : \kappa_A] = n.*

Consider the assertion (iii') one has [\kappaA \otimes_A \Omega_k(A) : \kappa_A] \leq n;* it is clear that (ii) implies (iii) and that (iii) implies (iii'). To prove the theorem it is enough to prove the implications (i) \Rightarrow (ii) and (iii') \Rightarrow (i).

Choose a k-algebra of finite type B and a prime ideal q of B such that A is isomorphic to $B_q$ (No. 1, Example 2). One has \dim_q(B) = n; replacing B by a ring $B_f$ , with $f \in B - q$, one may impose that the spectrum of B is connected and of dimension n, which we shall henceforth suppose.

(i) \Rightarrow (ii): Suppose the k-algebra A absolutely regular. As in the proof of Prop. 8, let us denote by \mu : A \otimes_k A \to A and \nu : \kappa_A \otimes_k A \to \kappa_A the homomorphisms deduced from the multiplication of A; put I = \mathrm{Ker}(\mu) and n = \mathrm{Ker}(\nu). The ideal I is completely secant; the ideal n is maximal and the local ring (\kappa_A \otimes_k A)_n is regular (loc. cit.). By Theorem 1 of § 5, No. 2, the finitely generated A-module I/I^2, which is none other than \Omega_k(A), is projective, hence free. But by the remark of § 5, No. 6, the ideal n identifies with \kappa_A \otimes_A 1, so that n/n^2 is isomorphic to \kappa_A \otimes_A \Omega_k(A). One therefore has

Let us denote by B' the ring \kappa_A \otimes_k B, and by S the image in B' of B - q. The ring \kappa_A \otimes_k A identifies with $B' \otimes_B B_q$, that is to say with $S^{-1}B'$. There therefore exists a maximal ideal q' of B' not meeting S such that one has n = $S^{-1}q'$; the ring (\kappa_A \otimes_k A)_n is then isomorphic to $B'_{q'}$ , and one has

Since n contains \kappa_A \otimes_k qB_q , q' contains the ideal \kappa_A \otimes q of B' , hence its inverse image in B contains q ; it is equal to q since q' does not meet S . By VIII, § 2, No. 4, cor. to prop. 5, one has

which proves (ii).

(iii') \Rightarrow (i) : Suppose that one has [\kappa_A \otimes_A \Omega_k(A) : \kappa_A] \leq n , that is to say [\kappa(q) \otimes_B \Omega_k(B) : \kappa(q)] \leq n , and let us prove that the k-algebra B is absolutely regular. Let (x_1, \ldots , x_n) be a sequence of elements of B such that 1 \otimes dx_1, \ldots , 1 \otimes dx_n generate the \kappa(q)-vector space \kappa(q) \otimes_B \Omega_k(B) . Replacing B by B_f , for a suitable element f of B - q , one may suppose that dx_1, \ldots , dx_n generate the B-module \Omega_k(B) (Nakayama lemma and II, § 5, No. 1, prop. 2). Let \bar{k} be an algebraically closed extension of k . It is enough to prove that the \bar{k}-algebra B_{(\bar{k})} is regular, since this will imply that B is absolutely regular (cor. 4 to prop. 7 of No. 4). For every canonical factor C of B_{(k)} , the differentials d(1 \otimes x_i) generate the C-module \Omega_{\bar{k}}(C) (A, III, § 10, No. 12, prop. 20). The theorem therefore follows from the two following lemmas: