Theorem 1

Let A be a local noetherian ring. The following conditions are equivalent: (i) A is regular. (ii) The ideal $m_A$ is generated by a secant subset for A (§ 3, No. 2, Def. 1). (iii) The ideal $m_A$ is generated by a completely secant sequence for A (A, X, p. 157, Def. 2). (iv) Let S be the symmetric algebra of the $\kappa_A$-vector space $m_A/m_A^2$, and let gr(A) = $\bigoplus_{n \geqslant 0} m_A^n/m_A^{n+1}$ be the graded ring associated with A. The canonical homomorphism $\gamma$ of S onto gr(A) is bijective. (v) There exists an integer $r \geqslant 0$ such that one has $H_{A,m_A} = (1 - T)^{-r}$, that is to say $m_A = 0$ if $r = 0$ and $[m_A^n/m_A^{n+1} : \kappa_A] = \binom{n + r - 1}{r - 1}$ for every integer $n \geqslant 0$ if $r > 0$. (vi) One has $H_{A,m_A} = (1 - T)^{-d}$ with $d = \dim(A)$.

If these conditions are fulfilled, every system of coordinates of A is a completely secant sequence for A.

(ii) \Rightarrow (i): for every secant subset has at most $\dim(A)$ elements (§ 3, No. 2, Theorem 1). (iii) \Rightarrow (ii): for every completely secant sequence is secant (§ 3, No. 2, corollary to Proposition 3). (iv) \Rightarrow (iii): let $(x_1, ..., x_r)$ be a sequence of elements of $m_A$ whose classes modulo $m_A^2$ form a basis $(\xi_1, ..., \xi_r)$ of $m_A/m_A^2$ over the field $\kappa_A$. If property (iv) is satisfied, $\mathrm{gr}(A)$ is the polynomial algebra $\kappa_A[\xi_1, ..., \xi_r]$ and the sequence $(x_1, ..., x_r)$ is completely secant (A, X, p. 160, Theorem 1). This also proves the last assertion of Theorem 1. (i) \Rightarrow (iv): put $r = [m_A/m_A^2 : \kappa_A]$. By formula (6) of § 4, No. 2, the Poincaré series of the graded vector space S over the field $\kappa_A$ is equal to

Suppose that the canonical homomorphism $\gamma : S \to \mathrm{gr}(A)$ is not bijective. Since $\gamma$ is surjective, there exists a homogeneous element $u$ of $S$, of degree $d > 0$, annihilated by $\gamma$. We then have

By Theorem 3 of § 4, No. 4 and Lemma 2 of § 4, No. 1, therefore $\dim(A) < r$, and A is not regular.

Finally, let us prove the equivalence of conditions (iv) to (vi). Now (iv) means that one has $H_{A,m_A} = (1 - T)^{-s}$ with $s = [m_A/m_A^2 : \kappa_A]$. Therefore conditions (iv), (v), (vi) mean that one has $H_{A,m_A} = (1 - T)^{-m}$, with respectively $m = [m_A/m_A^2 : \kappa_A]$, $m \geq 0$, $m = \dim(A)$. But if one has $H_{A,m_A} = (1 - T)^{-m}$, then $\dim(A) = m$ by § 4, No. 4, Theorem 3 and $[m_A/m_A^2 : \kappa_A] = m$ (since $(1 - T)^{-m} = 1 + mT + \cdots$). The equivalence of conditions (iv) to (vi) follows immediately from this.