# closed substacks of cartesian powers of a stack

Let $\mathbb{Z}/2\mathbb{Z}$ act on $\mathbb{A}^1$ as $x \mapsto -x$, and let $\mathscr{X}$ be the quotient stack. It has coarse moduli space $\mathbb{A}^1$ and a residual gerbe $B\mathbb{Z}/2\mathbb{Z}$ at 0. There is then a surjective map from cartesian powers $\mathscr{X}^n$ to $\mathbb{A}^n$. What are the closed substacks of $\mathscr{X}^n$? Will there be anything other than pullbacks of closed subschemes of $\mathbb{A}^n$?

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A substack of $\mathcal{X}$ is given by a $Z/2Z$ invariant subscheme of $\mathbb{A}^1$ and hence given by a $Z/2Z$ invariant ideal $I\subset k[x]$. So for example, the ideals $(x^n)$ give substacks, but they only pullback back from subschemes of the coarse space for $n$ even: the coarse space is given by $Spec(k[x^2])$. The substack given by $(x)$ here is just a copy of $B(Z/2Z)$ embedded at the origin.
When $G$ acts on $X$, then substacks of $[X/G]$ are the same as $G$-invariant subschemes of $X$, which are given by $G$-invariant ideal sheaves, which are the same as ideal sheaves on $[X/G]$. Sheaves on the coarse moduli space $X/G$ don't come into the equation, except for the fact that you can pull them back to $[X/G]$ and obtain some of the ideal sheaves on $[X/G]$. I use $X$ and $X/G$ because in your example, unfortunately both $X$ and $X/G$ are isomorphic to $A^1$, which adds some confusion. – Arend Bayer Nov 19 '10 at 17:35