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?



Irreducibility of quotient stacks.
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 topology on the underlying sets of points of stacks is characterised by the
following two properties:


* 1-morphisms of stacks give continuous maps
* flat morphisms locally of finite presentation give open maps


These statements are can be found [here](http://math.columbia.edu/algebraic_geometry/stacks-git/locate.php?tag=04XJ) and [here](http://math.columbia.edu/algebraic_geometry/stacks-git/locate.php?tag=04XL) in the Stacks Project.


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.


