Theorem 1

Let A be a noetherian ring and J an ideal of A. The following conditions are equivalent:

(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.