Reality check: irreducible over a non-algebraically closed field means replace closed sets with Galois-stable closed sets, right? I am requesting a reality check. I'm pretty sure of something but my algebraic geometry education is insufficiently comprehensive to give me full conviction, and I can't readily find the answer in texts.
Say $V=\operatorname{Spec}A$ is an affine variety (i.e. reduced, finite-type scheme) over a field $k$ which is not algebraically closed. Then $V$ is irreducible as a topological space if and only if $A$ is an integral domain (so that there is a unique minimal prime ideal, namely zero). I would like to understand irreducibility in terms of the Zariski topology on the $\overline k$-points, where $\overline k$ is $k$'s algebraic closure. It seems to me it's the following, and my question is if this is right:
Let $\Gamma=\operatorname{Gal}(\overline k/ k)$ be the absolute Galois group of $k$. Then $\Gamma$ acts in a natural way on the set $V(\overline k)$ of $\overline k$-points of $V$. The set $V(\overline k)$ gets a topology via its natural bijection with the closed points of $V\otimes_k \overline k$, which have a subspace topology induced from the Zariski topology on $V\otimes_k \overline k$. I think $V$'s irreducibility should be equivalent to the statement that $V(\overline k)$ is not the union of two proper closed subsets each of which is stable under the $\Gamma$-action. Is this right?
I think this because my understanding is that stability under $\Gamma$-action is equivalent to "defined over $k$", and irreducibility of $V$ (since it is a $k$-scheme) ought to mean that it is not reducible into proper closed subsets that are defined over $k$. But I am not immediately seeing how to connect this with the definition in terms of the irreducibility of $V$ itself (i.e. $\operatorname{Spec} A$) as a topological space.
I will be satisfied with an explanation, but if you have recommendations of chapters or sections of texts where this issue is discussed, I would be delighted by that too. Thanks in advance.
 A: Yes, this is true, and your concept of why is basically accurate as well. Here's a rigorous accounting. Your condition on $V(\overline{k})$ can be phrased in terms of the map $V_{\overline{k}}\to V$ by a combination of the following four facts:

*

*$V(\overline{k}) = (V\times_k \overline{k})(\overline{k})$;

*the topology on $V_{\overline{k}}(\overline{k})$ is the induced topology from $V_\overline{k}$;

*the $\overline{k}$-points of $V_{\overline{k}}$ are dense in every closed subset;

*and the Galois action on $V_{\overline{k}}(\overline{k})$ is the restriction of the Galois action on $V_\overline{k}$.

From these we may conclude that irreducible closed subsets of $V(\overline{k})$ are in bijection with irreducible closed subsets of $V_\overline{k}$, and this bijection respects the Galois action.
Next, every irreducible component of $V_\overline{k}$ maps down to an irreducible component of $V$ and this map is surjective (Stacks 04KX). Further, the Galois group acts on the irreducible components of $V_\overline{k}$, and the fibers of the map from the set of irreducible components of $V_\overline{k}$ to the set of irreducible components of $V$ are exactly the Galois orbits (Stacks 04KY).
This proves your assertion. In one direction, if $V=W\cup Z$ is reducible, then $W_\overline{k}$ and $Z_\overline{k}$ are proper closed $G$-invariant subsets of $V_\overline{k}$ having union $V_\overline{k}$. On the other hand, if $V_\overline{k}$ can be written as a union of two proper closed subsets $X,Y$ which are stable under the $G$-action, then we can partition the irreducible components of $V_\overline{k}$ in to two disjoint nonempty collections, both of which are stable under the Galois action: those irreducible components that are only in $X$, and all the others. Since the fibers of the map from irreducible components of $V_\overline{k}$ to irreducible components of $V$ are exactly the Galois orbits, $V$ must have multiple irreducible components.
