Let $x = (x_1, \ldots, x_n)$ be a system of coordinates of $A$ (VIII, § 5, No. 1, def. 1). The sequence $x$ generates $m_A$ and is completely secant for $A$ (loc. cit., No. 2, th. 1). The Koszul complex $K_\bullet(x, A)$ is a free resolution of $\kappa_A$ (A, X, p. 159, remark 3), whose differential is zero modulo $m_A$. Hence, for every integer $i \geqslant 0$, one has (§ 3, No. 3, formula (1))

It then follows from cor. 1 of prop. 4 of No. 3, § 3 that $\mathrm{dh}(A) = n$.

By prop. 1, every finitely generated module over a regular noetherian local ring admits a projective resolution of finite length by finitely generated projective modules, hence free modules (II, § 3, No. 2, cor. 2 of prop. 5). It then follows from VII, § 4, No. 7, cor. 3 of prop. 16 that such a ring is factorial.

In fact, $M$ has finite projective dimension (prop. 1), and one has $\mathrm{prof}(A) = \dim(A)$ since $A$ is a Macaulay ring (§ 2, No. 5, example 7). One then applies th. 1 of § 3, No. 5.

In fact, a regular noetherian local ring is integrally closed (VIII, § 5, No. 2, cor. 1 of th. 1). Corollary 3 therefore follows from corollary 2 and § 1, No. 10, prop. 16.

In fact, one has $\dim(B) = \dim_A(B)$ (VIII, § 2, No. 3, th. 1); moreover, $B$ is a Macaulay ring if and only if it is a Macaulay $A$-module (§ 2, No. 6, prop. 8). It is therefore enough to apply corollaries 1 and 2.

We have seen that a regular noetherian local ring is of finite homological dimension (prop. 1).

Conversely, let $A$ be a noetherian local ring of finite homological dimension $n$; by § 3, No. 3, Corollary 2 of Proposition 4 and No. 5, Theorem 1, one has

If $n = 0$, the $A$-module $\kappa_A$ is free, hence $m_A = 0$ and $A$ is a field. Suppose $n > 0$ and argue by induction on $n$. Since $\mathrm{prof}(A) > 0$, the ideal $m_A$ is not associated to $A$ (§ 1, No. 1, Remark 2), hence is not contained in the union of $m_A^2$ and the ideals associated to $A$ (II, $§ 1$, No. 1, Proposition 2). Consequently (IV, $§ 1$, No. 1, Corollary 2 of Proposition 2), one can find an element $x$ of $m_A - m_A^2$ such that the homothety $x_A$ is injective. Let $B$ denote the noetherian local ring $A/xA$ and consider the sequence of $A$-modules where the mapping $i$ is deduced by passing to quotients from the mapping $a \mapsto ax$ of $A$ into $m_A$, and where $p$ is the canonical surjection; it is exact. Since the class of $x$ in the $\kappa_A$-vector space $m_A/m_A^2$ is non-zero, there exists an $A$-linear mapping $\phi : m_A \to \kappa_A$ with $\phi(x) = 1$; by passing to the quotient, one deduces from $\phi$ a retraction of $i$, so that the preceding exact sequence is split. This implies the relations (Corollary 2 of Proposition 7 of § 3, No. 4 and $A$, X, p. 135, Corollary 1). Corollary 2 of loc. cit. applied to the exact sequence of $B$-modules $0 \to m_B \to B \to \kappa_B \to 0$ implies $\mathrm{dp}_B(\kappa_B) < +\infty$. The ring $B$ is therefore of finite homological dimension (§ 3, No. 3, Corollary 2 of Proposition 4), and of depth $n - 1$ (§ 1, No. 4, Proposition 7 and No. 3, Corollary of Proposition 4). It follows from the induction hypothesis that $B$ is regular, hence that $A$ is regular (VIII, $§ 5$, No. 3, Corollary 1 of Proposition 2).

Consequently, if $A$ is a noetherian local ring, there is an equivalence between the following three properties: (i) $A$ is regular; (ii) the $A$-module $\kappa_A$ is of finite projective dimension; (iii) every finitely generated $A$-module is of projective dimension $< +\infty$.

Let us prove that (i) implies (ii). Let $M$ be a finitely generated $A$-module. Under hypothesis (i), one has $\mathrm{dp}_{A_m}(M_m) \leq \mathrm{dh}(A_m) < +\infty$ for every maximal ideal $m$ of $A$ (No. 1, prop. 1); therefore $M$ has projective dimension $< +\infty$ ($§ 3$, No. 2, cor. 2 of prop. 3), whence (ii).

(i) one has $\mathrm{dh}(A) < +\infty$ ; (ii) $A$ is regular and one has $\dim(A) < +\infty$.

If these conditions are satisfied, one has $\dim(A) = \mathrm{dh}(A)$.

If the ring $A$ is regular, one has for every maximal ideal $m$ of $A$ the equality $\dim(A_m) = \mathrm{dh}(A_m)$ (No. 1, prop. 1), and therefore

($§ 3$, No. 2, prop. 3 and VIII, $§ 1$, No. 3, prop. 8). On the other hand, if $\mathrm{dh}(A) < +\infty$, the ring $A$ is regular by prop. 4. The corollary follows.

There exist regular noetherian rings of infinite dimension (VIII, $§ 5$, exerc. 6).

In fact, a regular local ring is integrally closed (VIII, $§ 5$, No. 2, cor. 1 of th. 1), Gorenstein ($§ 3$, No. 9, example 4) and Macaulay ($§ 2$, No. 5, example 6).

a) In order that the ring $\hat{A}$ be regular, it is necessary and sufficient that, for every maximal ideal $m$ of $A$ containing $J$, the ring $A_m$ be regular.

b) If the ring $A$ is regular, the ring $\hat{A}$ is regular. If the ring $\hat{A}$ is regular and the ideal $J$ contained in the radical of $A$, the ring $A$ is regular.

Let $p$ be a prime ideal of $S_A(P)$ and $q$ its inverse image in $A$. The local ring $S_A(P)_p$ is a ring of fractions of the ring $S_A(P)_q$, which is isomorphic to $S_{A_q}(P_q)$ (A, III, p. 72, prop. 7); it is enough to prove that the latter is regular. This reduces us to the case where $A$ is local; but then $P$ is free of finite type. By prop. 1 of No. 1 and A, X, p. 143, cor. 1, one has $dh(S_A(P)) = dh(A) + rg_A(P) < +\infty$, and $S_A(P)$ is regular by prop. 4.

This follows from cor. 4 and cor. 3 b).

The point is to prove that, for every maximal ideal $m$ of $A$, the $A_m$-module $N_m$ is free (II, § 5, No. 2, th. 1). The $A$-module $B/\mathrm{Ann}_B(N)$ is a submodule of the finitely generated $A$-module $\mathrm{End}_A(N)$, and is therefore finitely generated. If one replaces $B$ by $B/\mathrm{Ann}_B(N)$, the hypotheses of the proposition are still satisfied ($§ 2$, No. 1, example 5); one may therefore suppose that $B$ is a finitely generated $A$-module and that $\mathrm{Supp}_B(N) = \mathrm{Spec}(B)$.

Let $m$ be a maximal ideal of $\mathrm{Supp}_A(N)$; put $n = \dim(A_m)$. By Cor. 2 of Prop. 3 of No. 1, it is enough to prove that $N_m$ is a Macaulay $A_m$-module of dimension $n$. Every maximal ideal of $B_m$ is of the form $nB_m$, where $n$ is a prime ideal of $B$ lying over $m$ (V, § 2, No. 1, Lemma 1 and Prop. 1). Let $p$ be a minimal prime ideal of $\mathrm{Supp}_B(N)$, contained in $n$. The closed subset $V(pB_n)$ of $\mathrm{Supp}_{B_n}(N_n)$ is then of codimension zero; since the $B_n$-module $N_n$ is Macaulay, the Cor. of Prop. 2, § 2, No. 2 yields the equality $\dim_{B_n}(N_n) = \dim(B_n/pB_n)$. But $p$ lies over a minimal prime ideal of $A$, contained in $m$, so that

If the ring B is a Macaulay ring, the A-module B is projective by Prop. 5.

Suppose conversely that the A-module B is projective. The A-module A is Macaulay (Cor. 2 of Prop. 4); the A-module B is a direct factor of a free A-module of finite type, hence is Macaulay (§ 2, No. 1, Example 2). One then applies Cor. 1 of Prop. 8 of § 2, No. 6.

b) Let A be a presentable ring and J an ideal of A. The separated completion $\widehat{A}$ of A for the J-adic topology is presentable.

c) Every complete local noetherian ring is presentable.

d) Let A be a presentable local ring. There exists a regular local ring R and a surjective local homomorphism of R onto A.

Let A be a presentable ring; choose a regular ring R and a surjective homomorphism $\rho : R \to A$.

a) Let S be a multiplicative subset of A; put T = $\rho^{-1}(S)$. The homomorphism $T^{-1}R \to S^{-1}A$ deduced from $\rho$ is surjective and the ring $T^{-1}R$ is regular (No. 3, Example 1), hence $S^{-1}A$ is presentable.

Let B be an A-algebra of finite type; there exist a finite set I and a surjective homomorphism $A[(T_i)_{i \in I}] \to B$, hence a surjective homomorphism $R[(T_i)_{i \in I}] \to B$. Since the ring $R[(T_i)_{i \in I}]$ is regular (No. 2, Cor. 5 of Prop. 4), the ring B is presentable.

b) Put $I = \rho^{-1}(J)$ and denote by $\hat{R}$ the separated completion of R for the I-adic topology; for each integer $n \geq 0$, the canonical mapping $I^n/I^{n+1} \to J^n/J^{n+1}$ is surjective. Consequently, the homomorphism $\hat{R} \to \hat{A}$ deduced from $\rho$ is surjective (III, § 2, No. 8, Cor. 2 of Th. 1); since $\hat{R}$ is regular (No. 2, Cor. 3 of Prop. 4), the ring $\hat{A}$ is presentable.

c) This follows from IX, § 2, No. 5, Th. 3 a) and IX, § 3, No. 3, Th. 2 a).

d) Let $m_A$ be the maximal ideal of A; then $p = \rho^{-1}(m_A)$ is a prime ideal of R, the local ring $R_p$ is regular and the homomorphism $R_p \to A$ deduced from $\rho$ is local and surjective.

Since fields and Dedekind rings are regular, hence presentable, Proposition 6 implies that most rings ordinarily encountered in algebraic geometry are presentable.

a) The ring $A$ is noetherian and catenary.

b) Let M be a finitely generated A-module. The mapping

of $\operatorname{Spec}(A)$ into $\mathbf{Z}$ is upper semicontinuous.

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

Choose a regular ring R and a surjective homomorphism $R \to A$.

a) The ring R is a Macaulay ring (No. 2, Cor. 2 of Prop. 4), hence A is catenary ($§ 2$, No. 2, Prop. 2 and VIII, § 1, No. 3, Rem. 2).

b) The R-module M is finitely generated and of projective dimension $< +\infty$ (No. 1, Prop. 1). Let us identify $\operatorname{Spec}(A)$ with a closed subset of $\operatorname{Spec}(R)$. Then the function $p \mapsto \dim_{A_p}(M_p) - \operatorname{prof}_{A_p}(M_p)$ on $\operatorname{Spec}(A)$ is the restriction of the function $q \mapsto \dim_{R_q}(M_q) - \operatorname{prof}_{R_q}(M_q)$ on $\operatorname{Spec}(R)$; it is then enough to apply Cor. 4 of Th. 1 of § 3, No. 5.

c) This is proved as in part c) of loc. cit.

a) For every finitely generated A-module M, one has $dp_A(M) = dp_B(B \otimes_A M)$.

b) If the ring B is regular, the ring A is regular.

where the $A$-module $L$ is free of finite type. One has $dp_A(N) = dp_A(M) - 1$ (A, X, p. 135, Cor. 2 of Prop. 1). Since $B$ is flat over $A$, the sequence

is exact and one has $dp_B(B \otimes_A N) = dp_B(B \otimes_A M) - 1$. The induction hypothesis applied to $N$ enables one to conclude. This proves a); assertion b) follows from a) and Prop. 4 of No. 2.

(i) $A$ is regular ; (ii) $B$ is a projective $A$-module ; (iii) $B$ is a flat $A$-module ; (iv) $B$ is a faithfully flat $A$-module.

(i) $\Rightarrow$ (ii) : this follows from the Cor. of Prop. 5 of No. 3. (ii) $\Rightarrow$ (iii) : this follows from I, § 3, No. 1, Prop. 1. (iii) $\Rightarrow$ (iv) : for every prime ideal $\mathfrak{p}$ of $A$, one has $\mathfrak{p}B \neq B$ (V, § 2, No. 1, Cor. 1 of Theorem 1). It is then enough to apply I, § 3, No. 1, Prop. 1. (iv) $\Rightarrow$ (i) : this follows from Prop. 8, b).

For every noetherian local ring $A$, let us denote by $\delta(A)$ the integer

Recall (VIII, § 5, No. 1) that $\delta(A)$ is always positive and that its vanishing characterises regular local rings.

Let $\rho : A \to B$ be a local homomorphism of noetherian local rings; from $\rho$ one deduces a $\kappa_A$-linear homomorphism $\mathfrak{m}_A / \mathfrak{m}_A^2 \to \mathfrak{m}_B / \mathfrak{m}_B^2$, whence a $\kappa_B$-linear homomorphism

Lemma 1.— One has

Let $C$ denote the local ring $\kappa_A \otimes_A B$. Consider the exact sequence of $B$-modules

by tensor product with $\kappa_B$, one obtains an exact sequence of $\kappa_{13}$-vector spaces

whence one deduces the equality

which implies the lemma.

(i) the ring $B$ is regular and the $\kappa_B$-linear mapping

is injective; (ii) the rings $B$ and $\kappa_A \otimes_A B$ are regular and the $A$-module $B$ is flat; (iii) the rings $A$ and $\kappa_A \otimes_A B$ are regular and the $A$-module $B$ is flat; (iv) the rings $A$ and $\kappa_A \otimes_A B$ are regular, and one has

One has $\dim(B) \leq \dim(A) + \dim(\kappa_A \otimes_A B)$ (VIII, § 3, No. 4, Corollary 1 of Proposition 7); the equivalence of (i) and (iv) therefore follows from Lemma 1. Under hypothesis (ii), the $A$-module $B$ is faithfully flat since one has $\mathfrak{m}_A B \subset \mathfrak{m}_B \neq B$ (I, § 3, No. 1, Proposition 1), which implies (iii) by Proposition 8. The implication (iii) $\Rightarrow$ (iv) follows from VIII, loc. cit.

It now remains for us to prove that when the equivalent conditions (i) and (iv) are satisfied, the $A$-module $B$ is flat. Let $x$ be a system of coordinates of $A$. Since $d\rho$ is injective, the image of this sequence under $\rho$ forms part of a system of coordinates of $B$. Thus the sequence $x$ is completely secant for $A$ and for $B$ (VIII, § 5, No. 3, Proposition 2), and generates the ideal $\mathfrak{m}_A$ of $A$. By Remark 3 of A, X, p. 159, the $A$-module $\mathrm{Tor}_1^A(\kappa_A, B)$ is isomorphic to $H_1(x, B)$, and therefore is zero; it follows that $B$ is flat over $A$ (III, § 5, No. 2, Theorem 1 and No. 4, Proposition 2).

For in fact, for every maximal ideal $\mathfrak{n}$ of $B$, the $A_{\rho^{-1}(\mathfrak{n})}$-module $B_n$ is flat (II, § 3, No. 4, Proposition 15), so that the ring $B_n$ is regular by Proposition 9.