In an article I am reading I found a claim that I do not know how to prove, and reading a solution to it would probably teach me a lot. The context is the following:
$f:\mathcal{X}\to Spec(k)$ is a connected smooth $0$-dimensional Artin stack over an algebraically closed field $k$. We work on a (suitable) derived category with $\ell$-torsion coefficients $D_{et,c}=D_{et,c}(\mathcal{X},\mathbb{F}_\ell)$. And we are trying to compute the dualizing complex $K_\mathcal{X}$.
The claim is that $K_\mathcal{X}\cong \mathbb{F}_\ell[0]$.
The article uses the following definition of dualizing complex for an Artin stack:
- A dualizing complex in $D_{et,c}$ is an object $K$ equipped with isomorphisms $f^!K\cong K_X$ for any locally finitely presented morphism $f:X\to \mathcal{X}$ where $X$ is itself locally of finite presentation over $Spec(k)$. The isomorphisms are required to be functorial.
The wording of the article makes it seem like this is a general fact about connected smooth 0-dimensional Artin stacks, but it might end up being only a fact of the specific stack they work with.
If you're working in a context that satisfies the six functor formalism (I think there's a six functor formalism in the context you're in, have a look at the papers of Liu and Zheng), then for any smooth map $f:X\to Y$ of dimension $d$ we have $f^!\simeq f^*[2d](d)$. In particular, if $X$ is smooth, i.e. $\pi_X:X\to\text{pt}$ is smooth, then the dualising sheaf $\pi_X^!k=k[2d_X](d_X)$ is a shift of $k$, where $d_X$ is the dimension of $X$.