Lemma 51.4.9. Let $(A, \mathfrak m)$ be a Noetherian local ring of dimension $d$. Then $H^ d_\mathfrak m(A)$ is nonzero and $\text{cd}(A, \mathfrak m) = d$.

Proof. By one of the characterizations of dimension, there exists an ideal of definition for $A$ generated by $d$ elements, see Algebra, Proposition 10.60.9. Hence $\text{cd}(A, \mathfrak m) \leq d$ by Lemma 51.4.3. Thus $H^ d_\mathfrak m(A)$ is nonzero if and only if $\text{cd}(A, \mathfrak m) = d$ if and only if $\text{cd}(A, \mathfrak m) \geq d$.

Let $A \to A^\wedge$ be the map from $A$ to its completion. Observe that $A^\wedge$ is a Noetherian local ring of the same dimension as $A$ with maximal ideal $\mathfrak m A^\wedge$. See Algebra, Lemmas 10.97.6, 10.97.4, and 10.97.3 and More on Algebra, Lemma 15.43.1. By Lemma 51.4.5 it suffices to prove the lemma for $A^\wedge$.

By the previous paragraph we may assume that $A$ is a complete local ring. Then $A$ has a normalized dualizing complex $\omega _ A^\bullet$ (Dualizing Complexes, Lemma 47.22.4). The local duality theorem (in the form of Dualizing Complexes, Lemma 47.18.4) tells us $H^ d_\mathfrak m(A)$ is Matlis dual to $\text{Ext}^{-d}(A, \omega _ A^\bullet ) = H^{-d}(\omega _ A^\bullet )$ which is nonzero for example by Dualizing Complexes, Lemma 47.16.11. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).