Lemma 2

Let $A$ be a local noetherian ring such that $\mathrm{di}_A(A) = +\infty$. Then $\mathrm{Ext}_A^i(\kappa_A, A) \neq 0$ for every integer $i \geqslant \dim(A)$.

Let us reason by induction on $\dim(A)$. When $\dim(A) = 0$, $m_A$ is the unique prime ideal of $A$ and the assertion follows from Prop. 6 of No. 3. Suppose therefore that $\dim(A) > 0$ and let $p$ be a prime ideal of $A$ distinct from $m_A$; then $\dim(A_p) < \dim(A)$. If $\mathrm{di}_{A_p}(A_p) = +\infty$, the induction hypothesis implies $\mathrm{Ext}_{A_p}^j(\kappa(p), A_p) \neq 0$ for every integer $j \geqslant \dim(A_p)$; by Lemma 3 of § 1, No. 7, this implies $\mathrm{Ext}_A^i(\kappa_A, A) \neq 0$ for every integer $i \geqslant \dim(A_p) + \dim(A/p)$, and in particular for $i \geqslant \dim(A)$ (VIII, § 1, No. 3, Prop. 8, b)).

It remains for us to treat the case where $\mathrm{di}_A(A)$ is infinite but where the injective dimension of $A_p$ is finite for every prime ideal $p$ of $A$ distinct from $m_A$. For such an ideal, in this case, by Prop. 10, $\mathrm{di}_{A_p}(A_p) = \dim(A_p) < \dim(A)$, hence $\mathrm{Ext}_{A_p}^i(\kappa(p), A_p) = 0$ for $i \geqslant \dim(A)$. Since $\mathrm{di}_A(A)$ is infinite, Prop. 6 of No. 3 then implies $\mathrm{Ext}_A^i(\kappa_A, A) \neq 0$ for every integer $i \geqslant \dim(A)$.