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?



Support of cohomology of a dualizing complex
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](https://stacks.math.columbia.edu/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](https://stacks.math.columbia.edu/tag/0A7U) and [0BUJ](https://stacks.math.columbia.edu/tag/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](https://stacks.math.columbia.edu/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.




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**References.**


[Stacks] A.J. de Jong et al, [*The stacks project*](http://stacks.math.columbia.edu).


