Proposition 3

Let $A$ be a noetherian ring, $M$ a finitely generated $A$-module, and $N$ an $A$-module.

a) One has

where $p$ ranges over the set of prime ideals (resp. maximal ideals) of $A$.

b) The mapping $p \mapsto \mathrm{dp}_{A_p}(M_p)$ of $\mathrm{Spec}(A)$ into $\overline{\mathbf{Z}}$ is upper semi-continuous.

Let us prove a). Let $n$ be an integer $\geqslant 0$. Suppose that one has $\mathrm{dp}_A(M) < n$. For every prime ideal $p$ of $A$ and every $A_p$-module $Q$, the $A_p$-module $\mathrm{Ext}_A^n(M_p, Q)$ is isomorphic to $(\mathrm{Ext}_A^n(M, Q))_p$ (Prop. 2), hence is zero; one deduces the inequality

Let us prove b). Let p be a prime ideal of A and n = dp_{A_p}(M_p). Let us show that there exists a neighbourhood U of p in Spec(A) such that one has dp_{A_q}(M_q) \leq n for every q \in U. This is clear if n = +\infty ; if n = -\infty , this follows from the fact that the support of M is closed. Suppose now that n is finite and choose an exact sequence of A-modules

where the P_i are free of finite type (A, X, p. 53, Prop. 6). Put P = Ker d_{n-1} ; this is a module of finite presentation. The A_p-module P_p is projective (A, X, p. 134, Prop. 1), hence free (II, § 3, No. 2, Cor. 2 of Prop. 5). By II, § 5, No. 1, Cor. of Prop. 2, there exists an element f of A - p such that the A_f-module P_f is free; the A_q-module P_q is then free for every element q of the open set U of Spec(A) formed by the prime ideals not containing f . This proves b).