# Irreducibility of quotient stacks.

Let $[X/G]$ be a quotient stack such that $X$ is irreducible and $G$ acts trivially on $X$ (I am just adding automorphisms to every point). Under which hypothesis is $[X/G]$ irreducible as an Artin stack?

• I guess $[X/G]$ is always irreducible, as long as $X$ is (namely one doesn't even need to assume the $G$ action is trivial). If $[X/G]$ is the union of two proper closed substacks, or equivalently, there exist two nonempty open substacks that do not meet, then their inverse images in $X$ do not meet, too, contradicting the assumption. Jun 20 '11 at 21:37
• Supplementing shenghao's comment, you can show that for any quotient stack $X/G$, the open and closed substacks of $X/G$ are in bijection with the open and closed subschemes of $X$ (i.e., they're all of the form $Y/G$ for Y open or closed in $X$). Jun 20 '11 at 23:24
• Mike, I assume that you still mean that $G$ acts trivially on $X$, since $[Y/G]$ isn't defined unless $Y$ is $G$-invariant. Jun 21 '11 at 10:41

If you have a presentation $s, t:R \to U$ of a stack $[U/R]$, the set of points of $[U/R]$ is just the equivalence classes of points in $|U|$ determined by the equivalence relation given by the image of the map $|R| \to |U|\times |U|$. In particular $|U| \to |[U/R]|$ is always surjective.
The topological properties of algebraic stacks therefore behave as expected. It is a purely topological fact that if you have surjective continuous map $U \to V$ of topological spaces, then $V$ is irreducible if $U$ is. The corresponding statements hold for quasi-compactness and connectedness. As commented above, this applies in your situation with the stack quotient $X \to [X/G]$, regardless of the action of $G$ being trivial or not.
If the action, as in your case, is trivial, the equivalence relation on $|X|$ becomes trivial as well. Hence we see that the map $|X| \to |[X/G]|$ is a bijection. Assuming that $G$ is flat and locally of finite presentation (this is required if we want $[X/G]$ to be algebraic), we see that $|X| \to |[X/G]|$ is even a homeomorphism. This illustrates that stackiness is invisible to the Zariski topology. The stackiness may be explored pointwise by considering the residual gerbes.