Theorem 2 (Bass)

Let $A$ be a noetherian local ring; put $d = \dim(A)$. Let $x = (x_1, \ldots, x_d)$ be a maximal secant sequence of elements of $m_A$, and $x$ the ideal which it generates. The following conditions are equivalent:

(i) $A$ is a Gorenstein ring; (ii) one has $\operatorname{di}_A(A) = d$; (iii) there exists an integer $i > d$ such that $\operatorname{Ext}^i_A(\kappa_A, A) = 0$; (iv) one has $\operatorname{Ext}^i_A(\kappa_A, A) = 0$ for $i < d$ and the $\kappa_A$-vector space $\operatorname{Ext}^d_A(\kappa_A, A)$ is of dimension 1; (v) the ring $A$ is Macaulay and the $\kappa_A$-vector space $\operatorname{Hom}_A(\kappa_A, A/x)$ is of dimension 1; (vi) the sequence $x$ is $A$-regular and the $\kappa_A$-vector space $\operatorname{Hom}_A(\kappa_A, A/x)$ is of dimension 1.

The equivalence of (i), (ii) and (iii) follows from Lemma 2 and Prop. 10. If $\operatorname{Ext}^i_A(\kappa_A, A)$ is zero for every integer $i < d$, the ring $A$ is Macaulay (§ 2, No. 3, Prop. 3); if the ring $A$ is Macaulay, the sequence $x$ is $A$-regular (loc. cit., Th. 1); if the sequence $x$ is $A$-regular, one has $\operatorname{Ext}^i_A(\kappa_A, A) = 0$ for $i < d$ and the $\kappa_A$-vector spaces $\operatorname{Ext}^d_A(\kappa_A, A)$ and $\operatorname{Hom}_A(\kappa_A, A/x)$ are isomorphic (A, X, p. 166, Prop. 9). This proves the equivalence of conditions (iv), (v) and (vi).

Let us prove that (i) implies (v): if $A$ is a Gorenstein ring, it is a Macaulay ring (prop. 10). The sequence $x$ is then $A$-regular (§ 2, No. 3, prop. 4), hence $A/x$ is an artinian Gorenstein ring (example 2), so that the $\kappa_A$-vector space $\operatorname{Hom}_A(\kappa_A, A/x)$ is of dimension 1.

Lastly let us prove that (vi) implies (i): under assumption (vi), the ring $A/x$ is a Gorenstein ring (lemma 1), hence $A$ is a Gorenstein ring (example 3).