Action of a Galois Group on an Algebraic Variety I've to solve the following exercise.
Let's be $X$ a connected algebraic variety over $k$ and let $K$ be a finite Galois Extension of $k$ with Gaolis Group $G$.
Now I have to prove that $G$ acts transitively on the connected components of $X_K$ but I have no ideas. Some hints? 
Thank you.
 A: $\def\spec{\mathop{\text{Spec}}}$
Let $\{X_i\}$ be the set of connected components of $X_K$, on which $G$ acts in the "obvious" way (via by pullback along $\sigma:K\to K$). Let $Y = \bigcup_{\sigma\in G} \sigma(X_1)$ and $Z = X_K - Y$, so $X_K = Y\cup Z$ is a disjoint union of closed-and-open $G$-stable subschemes. One can then use the general theory of descent ("only" Galois descent is needed) to disconnect $X$ as $Y_k\cup Z_k$, where $Y_k, Z_k\subseteq X$ are closed and open subschemes such that $(Y_k)_K = Y$ and $(Z_k)_K = Z$. Connectedness of $X$ then implies that $Y = X_K$, so the action on components is transitive.
While I think this is a complete answer, it may be instructive to outline the descent argument in this special case. First note that $X_K$ is finite, in particular affine, over $X$, say $X_K = \spec \mathscr{A}$ for $\mathscr{A}$ a (coherent) sheaf of $\mathcal{O}_X$-algebras -- indeed, we could define $\mathscr{A}$ explicitly as $\mathscr{A}(U) = \mathcal{O}_X(U)\otimes_k K$ for open $U\subseteq X$. The decomposition $X_K = Y\cup Z$ gives a ($\mathcal{O}_X$-linear) decomposition $\mathscr{A} = \mathscr{A}_Y\times \mathscr{A}_Z$, where $Y = \spec \mathscr{A}_Y$.
The action of $G$ on $X_K$ translates into an $\mathcal{O}_X$-linear action of $G$ on $\mathscr{A}$ preserving the decomposition $\mathscr{A} = \mathscr{A}_Y \times \mathscr{A}_Z$. For a sheaf $\mathscr{B}$ of $\mathcal{O}_X$-modules with $\mathcal{O}_X$-linear $G$-action, denote by $\mathscr{B}^G\subseteq \mathscr{B}$ the subsheaf of $G$-invariants, defined by $\mathscr{B}^G(U) = \mathscr{B}(U)^G$ (check this is a sheaf). From the description $\mathscr{A}(U) = \mathcal{O}_X(U)\otimes_k K$, one checks that $\mathscr{A}^G = \mathcal{O}_X$. As the action of $G$ preserves the decomposition of $\mathscr{A}$, this immediately gives $\mathcal{O}_X = (\mathscr{A}_Y)^G\times (\mathscr{A}_Z)^G$. Setting $\mathscr{I}_{Y_k} = \{0\}\times (\mathscr{A}_Z)^G$ and $Y_k = V(\mathscr{I}_{Y_k})$ and similarly for $Z$, one checks that this gives the claimed decomposition of $X$.
