Let A be a noetherian Macaulay ring and M a finitely generated A-module of finite projective dimension.
a) Let p∈Spec(A); let us denote by C(p) the set of irreducible components of Supp(M) containing p. We have
dimAp(Mp)−profAp(Mp)=dpAp(Mp)−X∈C(p)infcodim(X,Spec(A)).
b) The mapping p↦dimAp(Mp)−profAp(Mp) of Spec(A) into Z is upper semi-continuous.
c) The set of prime ideals p of A such that the Ap-module Mp is Macaulay is open and dense in Spec(A). Its intersection with Supp(M) is dense in Supp(M).
a) We may suppose that p∈Supp(M). Put φ(p)=dim(Ap)−dimAp(Mp). By cor. 2 above, we have
dimAp(Mp)−profAp(Mp)=dpAp(Mp)−φ(p).
Since A is a Macaulay ring, we have
dim(Ap)=codim(V(p),Spec(A))=codim(V(p),X)+codim(X,Spec(A))
dimAp(Mp)=codim(V(p),Supp(M))=X∈C(p)supcodim(V(p),X)
and consequently
φ(p)=X∈C(p)infcodim(X,Spec(A)).
b) Let p∈Spec(A), and let F be the union of the irreducible components of Supp(M) which do not contain p. For every element q of Spec(A)−F, we have C(q)⊂C(p), whence φ(q)≥φ(p) by the formula above. Consequently the function φ is lower semi-continuous; assertion b) then follows from prop. 3 of No. 2.
c) Let U be the set of elements p of Spec(A) such that Mp is Cohen-Macaulay. The condition p∈U is equivalent to dim(Mp)−prof(Mp)≤0, so that U is open by b). Since U contains Spec(A)−Supp(M), it is enough to prove that U∩Supp(M) is dense in Supp(M). For every minimal prime ideal p of Supp(M), the Ap-module Mp is of finite length (IV, § 2, No. 5, Corollary 2 of Proposition 7 and § 1, No. 3, Corollary 1 of Proposition 7), hence Cohen-Macaulay; consequently U meets all the irreducible components of Supp(M). One concludes by means of Proposition 1 of II, § 4, No. 1.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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