Let $A$ be a commutative noetherian local ring, and let $D$ be a dualizing complex over $A$. Let $i$ be the minimal integer such that $H^i(D) \ne 0$ (I am assuming cohomological grading, so the differential is of degree $+1$).

Must the $A$-module $H^i(D)$ have full support? that is, is it true that its support is equal to $\operatorname{Spec}(A)$?

This is true if $A$ is Gorenstein (obvious) or Cohen-Macaulay (the canonical module has full support). But is it true in general?


If $(A,\mathfrak m, \kappa)$ is a Noetherian local ring admitting a dualising complex $\omega_A^\bullet$, then there exists a unique $n \in \mathbb Z$ such that $\omega_A^\bullet[n]$ is a normalised dualising complex [Stacks, Tag 0A7M]. Thus, it is harmless to assume that $\omega_A^\bullet$ is itself normalised.

Then $\omega_A^\bullet$ has injective amplitude in $[-d,0]$, where $d = \dim(A)$ [Stacks, Tags 0A7U and 0BUJ]. In other words, $i = -d$. Finally, the support of $\omega_A := H^{-d}(\omega_A^\bullet)$ is the union of the irreducible components of $\operatorname{Spec} A$ of dimension $d$ [Stacks, Tag 0AWE].

Thus, the answer to your question is positive if $A$ is equidimensional, but fails for stupid reasons if $A$ has components of different dimensions.


[Stacks] A.J. de Jong et al, The stacks project.


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