(i) the sequence $x$ is completely secant for $A$ ; (ii) (resp. (ii')) for every prime ideal (resp. maximal ideal) $p \in V(J)$, the image of $x$ in $A_p$ is completely secant for $A_p$ ; (iii) (resp. (iii')) for every prime ideal (resp. maximal ideal) $p \in V(J)$, the ideal $J$ is completely secant at $p$ and the image of $x$ in the $\kappa(p)$-vector space $\kappa(p) \otimes_A J$ forms a basis of it.

When $A$ is noetherian, these conditions are also equivalent to:

(iv) for every integer $n \geq 0$, the $A/J$-module $J^n / J^{n+1}$ is free, and the images of the monomials $x_1^{\alpha_1} \ldots x_r^{\alpha_r}$ of total degree $n$ form a basis of it.

Let $p$ be a prime ideal of $A$; let us denote by $x_p$ the image of the sequence $x$ in $A_p$. For every integer $i \geq 0$, the $A_p$-module $H_i(x_p, A_p)$ is isomorphic to $(H_i(x, A))_p$ (A, X, p. 151, 2)) ; it is zero if $p$ does not contain $J$, since then one has $J_p = A_p$ (A, X, p. 148, cor. 2). The equivalence of (i), (ii) and (ii') follows from this and from II, § 3, No. 3, cor. 2 of th. 1. On the other hand, the equivalence of (ii) and (iii), and that of (ii') and (iii'), follow from § 1, No. 3, cor. 2 of th. 1.

The implication (i) $\Rightarrow$ (iv) is a consequence of th. 1 of A, X, p. 160. Finally, condition (iv) implies by localization the analogous condition for the $(A_p/J_p)$-modules $J_p^n / J_p^{n+1}$ for every $p \in V(J)$, and the latter entails (ii) by cor. 1 of A, X, p. 160.

(i) the ideal J is completely secant; (ii) the ideal J is completely secant at every maximal ideal $m \in V(J)$; (iii) the $A/J$-module $J/J^2$ is projective and the canonical homomorphism $\alpha_J : S_A(J) \longrightarrow \bigoplus_{n \in \mathbf{N}} J^n$ is bijective; (iv) the $A/J$-module $J/J^2$ is projective and the canonical homomorphism $\beta_J : S_{A/J}(J/J^2) \longrightarrow \operatorname{gr}_J(A)$ is bijective.

(i) ⇒ (ii): this is trivial.

(ii) ⇒ (iii): suppose condition (ii) satisfied. It is enough to prove that for every maximal ideal m of A, the $A_m/J_m$-module $J_m/J_m^2$ is free and the homomorphism $\alpha_{J_m} : S_{A_m}(J_m) \longrightarrow \bigoplus_n J_m^n$ is bijective (II, § 5, No. 2, Th. 1 and § 3, No. 3, Th. 1). But these assertions are trivial when m does not belong to V(J), since one then has J_m = A_m, and they result from A, X, p. 160, Th. 1 and p. 161, Remark, when m belongs to V(J).

(iii) ⇒ (iv): this is clear.

(iv) ⇒ (i): suppose condition (iv) satisfied; let p be a prime ideal of A containing J. Then the $A_p/J_p$-module $J_p/J_p^2$ is free. Let x = (x_1, ..., x_r) be a sequence of elements of J_p lifting a basis of $J_p/J_p^2$. The sequence x generates J_p (Nakayama's Lemma), and satisfies by construction condition (iv) of Prop. 1. It follows that the ideal J_p of A_p is completely secant, and J is completely secant at p.

In fact, the canonical mapping gr_J(A) → gr_J(\hat{A}) is an isomorphism of graded rings; it is therefore enough to apply criterion (iv).

More generally:

a) If J is completely secant, the ideal JB of B is completely secant.

b) Suppose that the ideal JB of B is completely secant and that every maximal ideal m ∈ V(J) is the inverse image of a maximal ideal of B. Then the ideal J is completely secant. This is the case for example if B is a faithfully flat A-module.

First observe that, since the A-module B is flat, $J^n \otimes_A B$ is identified with $J^nB$ and $(J^n/J^{n+1}) \otimes_{A/J} (B/JB)$ with $J^nB/J^{n+1}B$ for every integer $n \geq 0$. Assertion a) then results from criterion (iii). Under the hypotheses of b), the $A/J$-module B/JB is faithfully flat (I, § 2, No. 7, Cor. 2 of Prop. 8 and § 3, No. 5, Prop. 9) and J is completely secant by criterion (iv), taking account of I, § 3, No. 1, Prop. 2 and No. 6, Prop. 12. The last assertion results from Prop. 8 of I, § 3, No. 5.

Let in fact $m$ be a maximal ideal of $A$ containing $J$. The ideal $J_m$ of $A_m$ is generated by an $A_m$-regular sequence; hence $(A/J)_m$ is a Macaulay ring (resp. a Gorenstein ring) by Example 6 of § 2, No. 5 (resp. Example 2 of § 3, No. 9).

(i) $A$ is regular; (ii) every maximal ideal of $A$ is completely secant; (iii) every ideal $J$ of $A$ such that $A/J$ is regular is completely secant.

(i) $\Rightarrow$ (iii): suppose the ring $A$ regular; let $J$ be an ideal of $A$ such that $A/J$ is regular, and let $p$ be a prime ideal of $A$ containing $J$. Then the local rings $A_p$ and $A_p/J_p$ are regular, hence $J_p$ is generated by a completely secant sequence for $A_p$ (VIII, § 5, No. 3, Cor. 2 and Prop. 2), which means that $J$ is completely secant at $p$.

(iii) $\Rightarrow$ (ii): this is clear since a field is a regular ring.

(ii) $\Rightarrow$ (i): let $m$ be a maximal ideal of $A$; under assumption (ii), the maximal ideal $mA_m$ of $A_m$ is generated by a completely secant sequence for $A_m$, hence $A_m$ is regular (VIII, § 5, No. 2, Th. 1). Consequently $A$ is regular.

(i) the ideal $J$ is completely secant; (ii) for every prime (resp. maximal) ideal $p$ of $A$ containing $J$, the ring $A_p$ is regular; (iii) the separated completion of $A$ for the $J$-adic topology is regular.

According to Th. 1, condition (i) means that for every prime (resp. maximal) ideal $p$ of $A$ containing $J$, the ideal $J_p$ of the local ring $A_p$ is generated by a completely secant sequence for $A_p$. Since $A_p/J_p$ is regular by hypothesis, this latter condition is equivalent to the regularity of $A_p$ (VIII, § 5, No. 3, Prop. 2 and its Cor. 2); this proves the equivalence of (i) and (ii). The equivalence of (ii) and (iii) follows from § 4, No. 2, Cor. 3 of Prop. 4.

a) The ideal $J$ is completely secant, the $A_0$-module $J/J^2$ is finitely generated projective, and the ring $A_0$ is regular.

b) Let $\varphi : J/J^2 \to J$ be an $A_0$-linear section of the canonical surjection $J \to J/J^2$. The homomorphism of $A_0$-algebras $S_{A_0}(J/J^2) \longrightarrow A$ extending $\varphi$ extends to an isomorphism of the completion of the graded ring $S_{A_0}(J/J^2)$ onto the separated completion of $A$ for the $J$-adic topology.

a) Let $\mathfrak{p}$ be a prime ideal of $A$ containing $J$. We have $\mathfrak{p} = (\mathfrak{p} \cap A_0) \oplus J$ and hence $\mathfrak{p}^2 = (\mathfrak{p} \cap A_0)^2 \oplus \mathfrak{p}J$, therefore $\mathfrak{p}^2 \cap J = \mathfrak{p}J$. Let $i$ denote the canonical injection of $J$ into $\mathfrak{p}$. From the preceding, the mapping $i \otimes 1_{A/\mathfrak{p}} : J \otimes_A A/\mathfrak{p} \longrightarrow \mathfrak{p} \otimes_A A/\mathfrak{p}$ is injective, and the same is true of $i_p \otimes 1_{\kappa(\mathfrak{p})} : J_p \otimes_{A_p} \kappa(\mathfrak{p}) \longrightarrow \mathfrak{p}A_p \otimes_{A_p} \kappa(\mathfrak{p})$. Nakayama's lemma implies that the ideal $J_p$ of $A_p$ is generated by a subset of a system of coordinates of the regular local ring $A_p$. By Prop. 2 of VIII, § 5, No. 3, the ring $A_p/J_p$ is regular and the ideal $J_p$ is completely secant. Thus $J$ is completely secant, the ring $A_0$, isomorphic to $A/J$, is regular, and the $A_0$-module $J/J^2$ is finitely generated projective by Theorem 1 (No. 2).

b) It follows from Theorem 1 that the canonical homomorphism $\beta_J : S_{A_0}(J/J^2) \longrightarrow \mathrm{gr}_J(A)$ is bijective. Let $f : S_{A_0}(J/J^2) \longrightarrow A$ be the homomorphism of $A_0$-algebras extending the $A_0$-linear mapping $\varphi : J/J^2 \longrightarrow J$. If $A$ is endowed with the $J$-adic filtration and $S_{A_0}(J/J^2)$ with the filtration associated with its graduation, $\beta_J$ is identified with the homomorphism deduced from $f$ by passing to the associated graded rings. Assertion b) then follows from III, § 2, No. 8, Corollary 3 to Theorem 1.

Let us denote by $P$ the graded $A_0$-module $A_+/A_+^2$. By Proposition 4 of No. 3, the ring $A_0$ is regular and the $A_0$-module $P$ is projective and finitely generated. The homogeneous components of $P$ are therefore projective, and there exists an $A_0$-linear section $\varphi : P \to A_+$, graded of degree 0 , of the canonical surjection $A_+ \to P$. Let $f : S_{A_0}(P) \longrightarrow A$ be the homomorphism of graded $A_0$-algebras extending $\varphi$. By Proposition 4, $f$ extends to an isomorphism of the separated completion of $S_{A_0}(P)$ for the $S_{A_0}(P)_+$-adic topology onto the separated completion of $A$ for the $A_+$-adic topology. Consequently, $f$ is injective and its image is dense in $A$ for the $A_+$-adic topology. But since the topologies induced on the homogeneous components of $A$ are discrete and the image of $f$ is a graded submodule, this implies that $f$ is bijective.

By theorem 2, the ring $B_0$ is regular, hence a product of regular integral rings ($§ 4$, No. 2, example 2). For every idempotent element $e$ of $B_0$, the $B_0$-module $B_0e$ is projective, hence free, which implies that $e = 0$ or $1$; thus $B_0$ is integral. Assertion a) then follows from theorem 2 and from the fact that a graded finitely generated projective $B_0$-module is graded free.

Let us prove b).

(i) $\Leftrightarrow$ (ii): flat finitely generated graded $A_0$-modules are graded free (II, $§ 5$, No. 2, cor. 2 to theorem 1). The equivalence of (i) and (ii) therefore follows from A, X, p. 144, prop. 8.

(ii) $\Leftrightarrow$ (iii): since $B$ is integral over $A$ and is a finitely generated $A_0$-algebra, $A$ is a finitely generated $A_0$-algebra (V, $§ 1$, No. 9, lemma 5), hence a noetherian ring. The equivalence of (ii) and (iii) then follows from the corollary to proposition 8 of $§ 4$, No. 5 and from a) applied to $A$.

The implications (i) $\Rightarrow$ (ii) and (ii) $\Rightarrow$ (iii) are clear, and the implication (iii) $\Rightarrow$ (i) follows from A, X, p. 144, prop. 8, a).

(iv) ⇒ (iii): let us suppose condition (iv) satisfied, and let x be an algebraically free generating sequence of A consisting of homogeneous elements. The sequence x, being A-regular, is completely secant for A (A, X, p. 157, prop. 5), so that the A-module Tor_1^A(k, B) is isomorphic to H_1(x, B) (A, X, p. 159, remark 3); but the latter is zero, since the sequence x is B-regular.

Corollaries 1 and 2 imply lemma 5 of LIE, V, § 5, No. 5.

(i) the homomorphism u makes N into a flat A[T_1, ..., T_r]-module; (ii) the A-module N is flat and, for every A-module M, the sequence x is M ⊗_A N-regular; (iii) the A-module N is flat and the sequence x is κ_A ⊗_A N-regular; (iv) the A-module N/(x_1N + ... + x_rN) is flat and the sequence x is N-regular.

(i) ⇒ (ii): let us put T = (T_1, ..., T_r). Suppose that the A[T]-module N is flat. Since A[T] is flat over A, the A-module N is flat (I, § 2, No. 7, cor. 3 to prop. 8). Let M be an A-module. The sequence T is obviously M[T]-regular, hence M[T] ⊗_{A[T]} N-regular, since N is flat over A[T]. Now the A[T]-module M[T] ⊗_{A[T]} N is identified with M ⊗_A N, the homothety of ratio T_i corresponding to the endomorphism 1_M ⊗ (x_i)_N. Condition (ii) is therefore satisfied.

(ii) ⇒ (iii): this is trivial.

(iii) ⇒ (iv): this follows from proposition 10 of § 1, No. 6.

(iv) ⇒ (i): let t denote the ideal of A[T] generated by T. The (A[T]/t)-module N/tN is flat by assumption and N is ideally separated for t (III, § 5, No. 4, prop. 2). To prove that N is flat over A[T], it is enough to prove, in view of theorem 1 of loc. cit., No. 2, that the A-module Tor_1^{A[T]}(A[T]/t, N) is zero. But since the sequence T is A[T]-regular, this module is isomorphic to H_1(T, N) (A, X, p. 159, remark 3), which is zero since the sequence T is N-regular (loc. cit., p. 157, prop. 5).

(i) the sequence $x$ is $M$-regular; (ii) the homomorphism $u$ makes $M$ into a flat $k[T_1, \ldots, T_r]$-module; (iii) the homomorphism $\hat{u}$ makes $\hat{M}$ into a flat $k[[T_1, \ldots, T_r]]$-module. The equivalence of (i) and (ii) follows from the equivalence of conditions (i) and (iv) of proposition 5; the equivalence of (ii) and (iii) follows from III, § 5, No. 4, prop. 4.

These results make it possible to characterize macaulay modules in two important cases. Let $A$ denote a noetherian local ring, $M$ a finitely generated $A$-module. It is equivalent to say that the $A$-module $M$ is macaulay or that the $\widehat{A}$-module $\hat{M}$ is macaulay (§ 2, No. 7, cor. 4 to prop. 8). Henceforth we shall suppose that the noetherian local ring $A$ is complete.

1) Let us suppose first that $A$ possesses a subfield; then it admits a field of representatives $k$ (IX, § 3, No. 3, Theorem 1). Let $(x_1, \ldots, x_r)$ be a maximal secant sequence for $M$; let us denote by $u : k[[T_1, \ldots, T_r]] \longrightarrow A$ the unique continuous homomorphism such that $u(T_i) = x_i$ for $i = 1, \ldots, r$. By Lemma 4 b) of IX, § 2, No. 5 and Remark 1 of VIII, § 3, No. 2, $A/\mathrm{Ann}(M)$ is a finitely generated $k[[T_1, \ldots, T_r]]$-module and consequently $M$ is a finitely generated $k[[T_1, \ldots, T_r]]$-module. This being so, the following conditions are equivalent: (i) the $k[[T_1, \ldots, T_r]]$-module $M$ is free; (ii) the $A$-module $M$ is Macaulay.

In fact, it is equivalent to say that $M$ is a Macaulay $A$-module or that the sequence $(x_1, \ldots, x_r)$ is $M$-regular (§ 2, No. 3, Theorem 1). By the above Corollary, the latter condition means that the $k[[T_1, \ldots, T_r]]$-module $M$ is flat, or again that it is free since it is finitely generated.

2) Suppose that the residue field $\kappa_A$ of $A$ is of characteristic $p > 0$ and that one has $\dim(M/pM) < \dim(M)$. Let $(x_1, \ldots, x_r)$ be a maximal secant sequence for $M/pM$, so that $(p1_A, x_1, \ldots, x_r)$ is a maximal secant sequence for $M$. Let $C$ be a $p$-ring of length $+\infty$, with residue field $\kappa_A$ (IX, § 2, No. 3, Proposition 5). There exists a homomorphism $u_0$ of $C$ into $A$ which induces the identity on the residue fields; let $u : C[[T_1, \ldots, T_r]] \longrightarrow A$ be the unique homomorphism extending $u_0$ and sending $T_i$ to $x_i$ for every $i$. It follows as above from loc. cit., No. 5, Lemma 4 that $u$ makes $M$ into a finitely generated $C[[T_1, \ldots, T_r]]$-module. The following conditions are equivalent: (i) the $C[[T_1, \ldots, T_r]]$-module $M$ is free; (ii) the $A$-module $M$ is Macaulay.

In fact condition (ii) is equivalent to saying that the sequence $(x_1, \ldots, x_r)$ is $M$-regular and that the homothety of ratio $p$ in $M/(x_1M + \ldots + x_rM)$ is injective (§ 2, No. 3, Theorem 1). Now this latter condition means that the $C$-module $M/(x_1M + \ldots + x_rM)$ is torsion-free, hence flat ($A$, X, p. 9, Example 7). Thus, having regard to Proposition 5, (iv) $\Leftrightarrow$ (i), condition (ii) is equivalent to the fact that the $C[T_1, \ldots, T_r]$-module $M$ is flat, or again (III, § 5, No. 4, Proposition 4) that the $C[[T_1, \ldots, T_r]]$-module $M$ is flat, that is to say free since it is finitely generated.

(i) the $A$-module $B/J$ is flat and the ideal $J$ is completely secant;

(ii) for every $q \in V(J)$, the $A$-module $B_q$ is flat and, for every $A$-algebra $A'$ such that the ring $A' \otimes_A B$ is noetherian, the ideal $J(A' \otimes_A B)$ of $A' \otimes_A B$ is completely secant;

(iii) for every maximal ideal $n$ of $B$ containing $J$, the $A$-module $B_n$ is flat and the ideal $J(\kappa(\rho^{-1}(n)) \otimes_A B_n)$ of $\kappa(\rho^{-1}(n)) \otimes_A B_n$ is completely secant.

By Th. 1 of No. 2, condition (i) is equivalent to the following condition (i') (resp. (i'')):

(i') (resp. (i'')) for every prime ideal (resp. maximal ideal) $q$ of $B$ containing $J$, the $A_{\rho^{-1}(q)}$-module $B_q/J_q$ is flat and the ideal $J_q$ of $B_q$ is generated by a $B_q$-regular sequence.

(i') $\Rightarrow$ (ii): assume that (i') holds. Let $A'$ be an $A$-algebra such that the ring $B' = A' \otimes_A B$ is noetherian. Let $q$ be a prime ideal of $B$ containing $J$; put $p = \rho^{-1}(q)$. Since $B'_q$ is identified with $A'_p \otimes_{A_p} B_q$, it follows from the implication (iv) $\Rightarrow$ (ii) of Prop. 5 (No. 5) that the ideal $JB'_q$ of $B'_q$ is generated by a $B'_q$-regular sequence and that $B_q$ is flat over $A_p$, hence over $A$. Moreover, for every prime ideal $r$ of $B'$ containing $JB'$, the inverse image $q$ of $r$ in $B$ contains $J$, and $B'_r$ is identified with a ring of fractions of $B'_q$. Hence $JB'_r$ is a completely secant ideal of $B'_r$, and $JB'$ is a completely secant ideal of $B'$.

(ii) $\Rightarrow$ (iii): this is trivial.

(iii) $\Rightarrow$ (i''): suppose that (iii) is satisfied. Let $n$ be a maximal ideal of $B$ containing $J$; put $m = \rho^{-1}(n)$. Let $x$ be a sequence of elements of $J_n$ whose images in $J_n/nJ_n$ form a basis of this $\kappa(n)$-vector space. The sequence $x$ generates $J_n$; by Prop. 1 of No. 1, it is $(\kappa(m) \otimes_A B_n)$-regular. Since the $A_m$-module $B_n$ is flat by assumption, it follows from the implication (iii) $\Rightarrow$ (iv) of Prop. 5 that condition (i'') is satisfied.