Proposition 6

Let $A$ be a noetherian ring, $N$ an $A$-module, $n$ an integer $\geqslant 0$. The following conditions are equivalent:

(i) one has $\operatorname{di}_A(N) < n$ ; (ii) for every prime ideal $p$ of $A$, one has $\operatorname{Ext}_A^n(A/p, N) = 0$ ; (iii) for every prime ideal $p$ of $A$, one has $\operatorname{Ext}_{A_p}^n(\kappa(p), N_p) = 0$.

If moreover the $A$-module $N$ is of finite type, these conditions are equivalent to:

(iv) for every maximal ideal $m$ of $A$, one has $\operatorname{Ext}_A^i(A/m, N) = 0$ for $n \leqslant i \leqslant n + \operatorname{ht}(m)$.

Observe that condition (iii) is equivalent to

(iii') for every prime ideal $p$ of $A$, one has $\operatorname{Ext}_A^n(A/p, N) \otimes_A \kappa(p) = 0$.

In fact, since $\operatorname{Ext}_A^n(A/p, N)$ is annihilated by $p$, the $A$-module $\operatorname{Ext}_A^n(A/p, N) \otimes_A \kappa(p)$ is isomorphic to $\operatorname{Ext}_A^n(A/p, N) \otimes_A A_p$, hence to $\operatorname{Ext}_{A_p}^n(\kappa(p), N_p)$ (No. 2, Proposition 2).

The implications (i) $\Rightarrow$ (ii) and (i) $\Rightarrow$ (iv) result from Proposition 1 of No. 1, and the implication (ii) $\Rightarrow$ (iii') is clear.

(iii) $\Rightarrow$ (i): for every $A$-module $M$, let us put $T(M) = \operatorname{Ext}_A^n(M, N)$. Suppose that (i) is not satisfied. Then there exists (No. 1, Proposition 1) an ideal $a$ of $A$ such that $T(A/a) \neq 0$. Let $p$ be an ideal of $A$ maximal among the ideals having this property; let us prove that $p$ is prime. By IV, § 1, No. 4, Theorem 1 and Theorem 2, there exists a composition sequence $(M_i)_{0 \leqslant i \leqslant m}$ of $A/p$ such that each quotient $M_i/M_{i+1}$ is isomorphic to a module $A/p_i$, where $p_i$ ($0 \leqslant i \leqslant m-1$) is a prime ideal containing $p$. If $T(A/p_i)$ were zero for every $i$, one would deduce by induction on $i$ from the exact sequences

that $T(M_0/M_m) = T(A/p)$ is zero, which is not so. There therefore exists an index $i$ such that $T(A/p_i) \neq 0$. In view of the maximal character of $p$, one has $p = p_i$, so that $p$ is prime.

Let $x$ be an element of $A - p$; from the exact sequence

one deduces an exact sequence

where $u = \operatorname{Ext}_A^n(x_A, 1_N) = x_{T(A/p)}$ (A, X, p. 89, Proposition 6). Because of the maximal character of $p$, one has $T(A/(p + xA)) = 0$, so that the homomorphism $u$ is injective. The nonzero $A/p$-module $T(A/p)$ is therefore torsion-free. This implies that $T(A/p) \otimes_A \kappa(p)$ is nonzero (A, II, p. 117, Corollary 1), which contradicts (iii').

Suppose that the $A$-module $N$ is of finite type, and let us prove that (iv) implies (iii). Let $p$ be a prime ideal of $A$ and $m$ a maximal ideal of $A$ containing $p$. There exists a saturated chain of prime ideals of $A$ with end terms $\mathfrak{p}$ and $\mathfrak{m}$. The length $r$ of this chain is less than $\mathrm{ht}(\mathfrak{m})$; under assumption (iv), one therefore has

It then follows from Lemma 3 of § 1, No. 7 that one has $\mathrm{Ext}_{A_{\mathfrak{p}}}^{n}(\kappa(\mathfrak{p}), N_{\mathfrak{p}}) = 0$, which proves (iii).