(i) one has di_A(M) ≤ n (in other words, M possesses an injective resolution of length ≤ n) ; (ii) for every A-module N and every integer i > n, one has Ext_A^i(N, M) = 0 ; (iii) for every ideal a of A, one has Ext_A^{n+1}(A/a, M) = 0 ; (iv) for every exact sequence of A-modules

where the $I^i$ are injective, the A-module Q is injective.

(i) ⇒ (ii) : this follows from A, X, p. 100, Th. 1. (ii) ⇒ (iii) : this is clear. (iii) ⇒ (iv) : in the situation of (iv), for every A-module N one has an isomorphism of Ext_A^1(N, Q) onto Ext_A^{n+1}(N, M) (A, X, p. 128, Cor. 4) ; under assumption (iii), the A-module Ext_A^1(A/a, Q) is zero for every ideal a of A, and Q is injective (A, X, p. 93, Prop. 11). (iv) ⇒ (i) : consider the exact sequence (A, X, p. 52)

if condition (iv) is satisfied, the A-module $K^{n-1}(M)$ is injective, whence (i).

Recall (A, X, p. 138, Def. 2) that the homological dimension of the ring $A$, denoted by $\mathrm{dh}(A)$, is the least upper bound in $\overline{\mathbf{Z}}$ of the set of integers $n$ for which there exist two $A$-modules $M$ and $N$ such that $\mathrm{Ext}^n_A(M, N)$ is nonzero. It is therefore also the least upper bound of the set of projective dimensions (or injective dimensions) of all the $A$-modules; when $A$ is noetherian, it is enough to restrict oneself to finitely generated $A$-modules (A, X, p. 139, Cor.).

If the ring $A$ is noetherian and the $A$-module $M$ finitely generated, one has a canonical isomorphism of $S^{-1}A$-modules

Since the $A$-module $S^{-1}A$ is flat, this follows from A, X, p. 110, Prop. 9 and p. 111, Prop. 10.

a) The support of $\mathrm{Tor}_i^A(M, N)$ is contained in $\mathrm{Supp}(M) \cap \mathrm{Supp}(N)$, and the same is true of the support of $\mathrm{Ext}_A^i(M, N)$ if $A$ is noetherian and $M$ finitely generated.

b) Suppose that $A$ is noetherian, and that the modules $M$ and $N$ are finitely generated; if the $A$-module $M \otimes_A N$ is of finite length, the same is true of $\mathrm{Tor}_i^A(M, N)$ and of $\mathrm{Ext}_A^i(M, N)$.

If $p$ is a prime ideal of $A$ not belonging to $\mathrm{Supp}(M) \cap \mathrm{Supp}(N)$, one of the modules $M_p$ or $N_p$ is zero, which implies a) in view of Prop. 2.

In order that a finitely generated module over a noetherian ring be of finite length, it is necessary and sufficient that its support consist of maximal ideals (IV, § 2, No. 5, Prop. 7). Under assumption b), the $A$-modules $\mathrm{Tor}_i^A(M, N)$ and $\mathrm{Ext}_A^i(M, N)$ are finitely generated (A, X, p. 108, Cor.) ; the assertion b) therefore follows from a).

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).

In fact, the subspace X of Supp(M) formed by the maximal ideals is quasi-compact (II, § 4, No. 2, Prop. 8 and 9 and No. 3, Cor. 7 of Prop. 11) ; the mapping m \mapsto dp_{A_m}(M_m) of X into \overline{\mathbf{R}} is upper semi-continuous (Prop. 3), hence bounded (TG, IV, p. 30, Cor. of Th. 3).

(i) one has $\operatorname{dp}_A(M) < n$ ; (ii) one has $\operatorname{Tor}_n^A(M, \kappa_A) = 0$ ; (iii) one has $\operatorname{Ext}_A^n(M, \kappa_A) = 0$ ; (iv) every minimal projective resolution of M has length $< n$.

Assertions (i) $\Rightarrow$ (ii) and (i) $\Rightarrow$ (iii) are immediate (A, X, p. 100, Th. 1). Let L be a minimal projective resolution of M ; if (ii) or (iii) is satisfied, one has $L_n = 0$ by (1). Since every minimal projective resolution of M is isomorphic to L (A, X, p. 54, Prop. 8), one deduces (iv). Implication (iv) $\Rightarrow$ (i) is trivial.

(i) one has $\operatorname{dh}(A) < n$ ; (ii) one has $\operatorname{Ext}_A^i(M, N) = 0$ and $\operatorname{Tor}_i^A(M, N) = 0$ for every pair $(M, N)$ of A-modules and every integer $i \geq n$ ; (iii) one has $\operatorname{Tor}_n^A(\kappa_A, \kappa_A) = 0$ ; (iv) one has $\operatorname{Ext}_A^n(\kappa_A, \kappa_A) = 0$ ; (v) one has $\operatorname{dp}_A(\kappa_A) < n$.

It is clear that (i) implies (ii) and that (ii) implies (iii) and (iv). By Prop. 4 applied to the A-module $\kappa_A$, each of conditions (iii) and (iv) implies (v). Let us prove that (v) implies (i) : if $\operatorname{dp}_A(\kappa_A) < n$, one has $\operatorname{Tor}_n^A(M, \kappa_A) = 0$ for every A-module M ; hence every finitely generated A-module is of projective dimension $< n$ (Prop. 4), which implies $\operatorname{dh}(A) < n$ (A, X, p. 138, Prop. 4).

(i) we have $\mathrm{dp}_A(M) < n$ ; (ii) for every maximal ideal $m$ of $A$, we have $\mathrm{Ext}_A^n(M, A/m) = 0$ (resp. we have $\mathrm{Tor}_n^A(M, A/m) = 0$) ; (iii) for every maximal ideal $m$ of $A$, we have $\mathrm{Ext}_{A_m}^n(M_m, A/m) = 0$ (resp. we have $\mathrm{Tor}_n^{A_m}(M_m, A/m) = 0$).

(i) $\Rightarrow$ (ii) : this is clear. (ii) $\Rightarrow$ (iii) : this follows from Prop. 2 of No. 2. (iii) $\Rightarrow$ (i) : by Prop. 4, condition (iii) implies the inequality $\mathrm{dp}_{A_m}(M_m) < n$ for every maximal ideal $m$ of $A$, and one concludes by Prop. 3.

(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).

deduced from the exact sequences (4), (5) and (6) are isomorphisms.

It is enough to treat the case $r = 1$. Put $x = x_1$. The $A$-modules $\mathrm{Ext}^n_{A/xA}(M/xM, \kappa_A)$ and $\mathrm{Ext}^n_A(M, \kappa_A)$ are isomorphic for every integer $n$ (prop. 7); the equality $\mathrm{dp}_{A/xA}(M/xM) = \mathrm{dp}_A(M)$ follows from prop. 4 of No. 3. Analogously, the $A$-modules $\mathrm{Ext}^{n-1}_{A/xA}(\kappa_A, M/xM)$ and $\mathrm{Ext}^n_A(\kappa_A, M)$ are isomorphic for every integer $n$, and the equality $\mathrm{di}_{A/xA}(M/xM) = \mathrm{di}_A(M) - 1$ follows from prop. 6 of No. 3.

Let us reason by induction on $\mathrm{dp}_A(M)$.

a) If $\mathrm{dp}_A(M)$ is zero, $M$ is a nonzero finitely generated free $A$-module; its depth is equal to $\mathrm{prof}(A)$ (§ 1, No. 1, remark 4).

b) Suppose $\mathrm{dp}_A(M) = 1$ and choose a minimal projective resolution

of $M$ (No. 3, prop. 4, (iv)). The $A$-modules $L_0$ and $L_1$ are nonzero finitely generated free modules, hence of depth $\mathrm{prof}(A)$ (§ 1, No. 1, remark 4). The mapping $1_{\kappa_A} \otimes d_1 : \kappa_A \otimes_A L_1 \to \kappa_A \otimes_A L_0$ is zero, which implies that $d_1$ belongs to $m_A \mathrm{Hom}_A(L_1, L_0)$. By remark 5 of § 1, No. 1, one has $\mathrm{prof}_A(M) = \mathrm{prof}(A) - 1$.

c) Suppose $\mathrm{dp}_A(M) > 1$. Choose an exact sequence

where $L$ is a finitely generated free $A$-module. One then has $\mathrm{prof}_A(L) = \mathrm{prof}(A)$ (§ 1, No. 1, remark 4), $\mathrm{dp}_A(N) = \mathrm{dp}_A(M) - 1$ (A, X, p. 135, cor. 2 c)), whence $\mathrm{prof}_A(N) = \mathrm{prof}(A) - \mathrm{dp}_A(N)$ (induction hypothesis), and in particular $\mathrm{prof}_A(N) < \mathrm{prof}_A(L)$. By prop. 1 of § 1, No. 1, one then has $\mathrm{prof}_A(M) = \mathrm{prof}_A(N) - 1$, which completes the proof.

a) One has $dp_A(M) \leq \operatorname{prof}(A)$. For there to be equality, it is necessary and sufficient that the maximal ideal $m_A$ be associated to $M$.

b) One has $\operatorname{prof}_A(M) \leq \operatorname{prof}(A)$. For there to be equality, it is necessary and sufficient that $M$ be free.

a) In fact, “ $\operatorname{prof}_A(M) = 0$ ” is equivalent to “ $m_A \in \operatorname{Ass}(A)$ ” ($§ 1$, No. 1, remark 2).

b) In fact, “ $dp_A(M) = 0$ ” is equivalent to “ $M$ is free ”.

COROLLARY 2.: Retain the hypotheses of theorem 1 and suppose moreover that $A$ is a Macaulay ring. Then $dp_A(M)$ is the sum of the two positive integers $\dim(A) - \dim_A(M)$ and $\dim_A(M) - \operatorname{prof}(M)$.

In particular, $dp_A(M)$ is then greater than $\dim(A) - \dim_A(M)$, and there is equality if and only if $M$ is Cohen-Macaulay.

In fact, let $p \in F$. We have $\operatorname{Ext}_A^i(M_p, N_p) \neq 0$ (resp. $\operatorname{Tor}_i^A(M_p, N_p) \neq 0$) by prop. 2 of No. 2, hence $i \leq dp_{A_p}(M_p) \leq dp_A(M) < +\infty$ (No. 2, prop. 3). Theorem 1 implies $\operatorname{prof}(A_p) \geq i$. Consequently ($§ 1$, No. 5, prop. 8)

With the terminology of $§ 1$, No. 5, remark 4, the conclusion of cor. 3 means that the modules $\operatorname{Ext}_A^i(M, N)$ and $\operatorname{Tor}_i^A(M, N)$ are of grade $\geq i$. It implies that the codimension of their support in $\operatorname{Spec}(A)$ is $\geq i$ ($§ 1$, No. 7, prop. 12).

a) Let $p \in \operatorname{Spec}(A)$; let us denote by $\mathscr{C}(p)$ the set of irreducible components of $\operatorname{Supp}(M)$ containing $p$. We have

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

c) The set of prime ideals $p$ of $A$ such that the $A_p$-module $M_p$ is Macaulay is open and dense in $\operatorname{Spec}(A)$. Its intersection with $\operatorname{Supp}(M)$ is dense in $\operatorname{Supp}(M)$.

a) We may suppose that $p \in \operatorname{Supp}(M)$. Put $\varphi(p) = \dim(A_p) - \dim_{A_p}(M_p)$. By cor. 2 above, we have

Since $A$ is a Macaulay ring, we have

and consequently

b) Let $p \in \operatorname{Spec}(A)$, and let $F$ be the union of the irreducible components of $\operatorname{Supp}(M)$ which do not contain $p$. For every element $q$ of $\operatorname{Spec}(A) - F$, we have $\mathscr{C}(q) \subset \mathscr{C}(p)$, whence $\varphi(q) \geq \varphi(p)$ by the formula above. Consequently the function $\varphi$ is lower semi-continuous; assertion b) then follows from prop. 3 of No. 2.

c) Let $U$ be the set of elements $p$ of $\operatorname{Spec}(A)$ such that $M_p$ is Cohen-Macaulay. The condition $p \in U$ is equivalent to $\dim(M_p) - \operatorname{prof}(M_p) \leq 0$, so that $U$ is open by b). Since $U$ contains $\operatorname{Spec}(A) - \operatorname{Supp}(M)$, it is enough to prove that $U \cap \operatorname{Supp}(M)$ is dense in $\operatorname{Supp}(M)$. For every minimal prime ideal $p$ of $\operatorname{Supp}(M)$, the $A_p$-module $M_p$ 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 $\operatorname{Supp}(M)$. One concludes by means of Proposition 1 of II, § 4, No. 1.