$\def\spec{\mathop{\text{Spec}}}$
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 $X$.
$\newcommand{\Spec}{\mathrm{Spec}}$Let $X$ be a scheme over $k$. Note that $X_{\overline{k}}=X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k})$. So, to give a morphism $X_{\overline{k}}\to X_{\overline{k}}$ it suffices to give maps of $k$-schemes $X\to X$ and $\mathrm{Spec}(\overline{k})\to\mathrm{Spec}(\overline{k})$. If $\sigma\in\mathrm{Gal}(\overline{k}/k)$ then declare that the action of $\sigma$ on $X$ is trivial and that the action of $\sigma$ on $\mathrm{Spec}(\overline{k})$ is the one induced from the $k$-algebra map $\overline{k}\to\overline{k}$. The induced map $X_{\overline{k}}\to X_{\overline{k}}$ is the map $\sigma$. Since it's a map of schemes, it's also a map of topological space (read the remark at the end of the post to see a different convention).
If $X=\mathrm{Spec}(A)$ then $X_{\overline{k}}=\mathrm{Spec}(A\times_k \overline{k})$ and $\sigma$ is the map of schemes $X_{\overline{k}}\to X_{\overline{k}}$ corresponding to the ring map $A\otimes_k \overline{k}\to A\otimes_k \overline{k}$ given by $a\otimes \alpha\mapsto a\otimes \sigma(\alpha)$. Covering $X$ by affines with $k$-rational gluing data allows you to bootstrap this concrete understanding of the map to the general case.
How does this relate to the action on $X(\overline{k})$? Note that
$$X(\overline{k})=\mathrm{Hom}_k(\mathrm{Spec}(\overline{k}),X)=\mathrm{Hom}_{\mathrm{Spec}(\overline{k})}(\mathrm{Spec}(\overline{k}),X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k}))$$
The action on the first presentation of $X(\overline{k})$ is to take $x:\Spec(\overline{k})\to X$ and then $\sigma(x)$ is defined to be the composition
$$\mathrm{Spec}(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{x}X$$
where, again, the first map $\sigma$ is that induced by the ring map $\sigma:\overline{k}\to\overline{k}$. What is the action then on the second presentation of $X(\overline{k})$? It can't clearly be the same idea that for $y:\Spec(\overline{k})\to X\times_{\Spec(k)}\Spec(\overline{k})$ we just precompose by $\sigma$ since that won't be a morphism over $\Spec(\overline{k})$. What's true is that $\sigma(y)$ is the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{y}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\qquad (1)$$
To convince ourselves of this, we need to remind ourselves how the identification
$$\mathrm{Hom}_k(\mathrm{Spec}(\overline{k}),X)=\mathrm{Hom}_{\mathrm{Spec}(\overline{k})}(\mathrm{Spec}(\overline{k}),X\times_{\mathrm{Spec}(k)}\mathrm{Spec}(\overline{k}))$$
works. It takes an $x:\Spec(\overline{k})\to X$ and maps to $y:\Spec(\overline{k})\to X\times_{\Spec(k)}\Spec(\overline{k})$ which, written in coordinates, is the map $(x,\mathrm{id})$. So, we see that $\sigma(y)$ should just be $(x\circ\sigma,1)$. Let's now think about the composition in $(1)$ looks like when written in coordinates. The projection to $X$ is the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{(x,1)}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{p_1}X$$
Since $p_1\circ\sigma^{-1}$ is just the identity map we see that we get $x\circ\sigma$. What happens we do the second projection? We are looking at the composition
$$\Spec(\overline{k})\xrightarrow{\sigma}\Spec(\overline{k})\xrightarrow{(x,1)}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{\sigma^{-1}}X\times_{\Spec(k)}\Spec(\overline{k})\xrightarrow{p_2}\Spec(\overline{k})$$
which gives us $\sigma^{-1}\circ \sigma=1$. Thus, we see that the composition in $(1)$ is $(x\circ\sigma,1)$ as desired!
Let's do a concrete example. Let's take $X=\mathbb{A}^1_k$. Then, the map $\sigma:X_{\overline{k}}\to X_{\overline{k}}$ is the map corresponding to the ring map $\overline{k}[x]\to \overline{k}[x]$ defined by sending
$$\sum_i a_i x^i\mapsto \sum_i \sigma(a_i)x^i$$
The points of $X_{\overline{k}}$ come in two forms:
- The closed ideals $(x-\alpha)$ for $\alpha\in k$.
- The generic point $(0)$.
It's then clear that $\sigma$ sends $(x-\alpha)$ to $(x-\sigma^{-1}(\alpha))$ (recall that the induced map on $\Spec$ is by pullback! See the remark below) and sends the generic point to itself. That's what it looks like topologically. Let's think about what the action of $\sigma$ looks like on $\overline{k}$-points.
If we take an $\overline{k}$-point corresponding to the map (thinking of $X(\overline{k})$ as $\mathrm{Hom}_k(\Spec(\overline{k}),X)$)
$$x:k[x]\to \overline{k}:p(x)\mapsto p(\alpha)$$
then $\sigma(x)$, on the level of ring maps, is apply $\sigma$ as a poscomposition:
$$\sigma(x):k[x]\to \overline{k}:p(x)\mapsto \sigma(p(\alpha))=p(\sigma(\alpha)$$
where last commutation was because $p(x)\in k[x]$. Let's now think about the same situation in the second presentation $X(\overline{k})=\mathrm{Hom}_{\overline{k}}(\Spec(\overline{k})),X\times_{\Spec(k)}\Spec(\overline{k}))$. Our point $x$ above now corresponds to ring $\overline{k}$-algebra map
$$\overline{k}[x]\to \overline{k}:q(x)\mapsto q(\alpha)$$
If we just postcompose this ring map with $\sigma$ we get
$$\overline{k}[x]\to \overline{k}:q(x)\mapsto \sigma(q(\alpha))$$
which is NOT the desired map sending $q(x)\mapsto q(\sigma(\alpha))$ since $q$ doesn't have rational coefficients. But, if we apply the map $\sigma_X^{-1}$ (where I'm using the subscript $X$ to not confuse it with $\sigma$ on $\overline{k}$) on $X_{\overline{k}}$ this has the effect of applying $\sigma^{-1}$ to the coefficients of $q(x)$. So then, you can see that
$$\sigma(\sigma_X^{-1}(q)(\alpha))=q(\sigma(\alpha))$$
yay!
Hopefully this all makes sense.
NB: Depending on the author one can take the action on $X\times_{\Spec(k)}\Spec(\overline{k})$ to be what I have called $\sigma^{-1}$. This has a couple nice effects:
- It's action geometrically is more intuitive (e.g. instead of sending $(x-\alpha)$ to $(x-\sigma^{-1}(\alpha))$ it sends $(x-\alpha)$ to $(x-\sigma(\alpha))$).
- It's a left action (opposed to a right action).
- In $(1)$ above the inclusion of $\sigma^{-1}$ on $X_{\overline{k}}$ might be jarring, and so if we use this alternative convention we'd just have $\sigma$. I actually like it the way it is because it reminds me of representation theory where if $V$ and $W$ are representations of some $G$ then $\mathrm{Hom}(V,W)$ is a representation of $G$ by $(g\cdot f)(v)=g(f(g^{-1}(v))$.
It's one downside is that ring theoretically it's less natural since then its action on $a\otimes\alpha\in A\otimes_k\overline{k}$ is $a\otimes\sigma^{-1}(\alpha)$. Either convention is OK--nothing changes theory wise--it's just something you have to pay attention/be consistent about.
Best Answer
a) Note that $\operatorname{Spec}(L \otimes_k K)$ is a finite discrete topological space. Let $e \in L \otimes_k K$ be a non-invertible idempotent. Then the product $f$ of the $G$-conjugates of $e$ is in $(L \otimes_k K)^G=L$, and is idempotent as well. Since $e$ is not invertible, $f$ cannot be invertible, hence $f=0$.
In particular, suppose that $e$ is chosen equal to $1$ on every connected component but one (let’s call it $C$) where it is zero. For for any connected component $D$ of $X_K$, $f$ vanishes on $D$, so $e$ has a $G$-conjugate vanishing on $D$, so $D$ is in the $G$-orbit of $C$.
c) for affine $X$. The argument above carries over when you replace $L$ by a connected $k$-algebra $A$, and proves that for every closed open subset $C \subset Y:=\operatorname{Spec}(A \otimes_k K)$, the $G$-translates of $C$ cover $Y$.
In particular, let $x \in Y$, then every closed open subspace $C \subset Y$ must contain a point in the $G$-orbit of $x$ (which has cardinality at most $|G|$). It follows that every finite partition of $Y$ into closed open subspaces has cardinality at most $|G|$.
Considering such a partition of maximal cardinality, we find that $T$ is the finite disjoint union of its open connected components. Hence, if $C$ is a connected component of $Y$, then $Y$ is covered by the $G$-translates of $C$, which are also connected components. QED.
c) for general $X$. It’s well-known that the set of connected components (with Galois action) of $X_K$ is the same as that of $\operatorname{Spec}{\mathcal{O}(X_K)}=\operatorname{Spec}(\mathcal{O}(X)\otimes_k K)$ (since $K/k$ is flat). But $\operatorname{Spec}{\mathcal{O}(X)}$ is an affine connected $k$-scheme, so we’re reduced to the case where $X$ is an affine connected $k$-scheme.
b) we can assume $X$ integral. I claim that the irreducible components of $X_K$ are in canonical (in particular, preserving the Galois action) bijection with $\operatorname{Spec}(k(X) \otimes_k K)$.
This can be rephrased as “the generic points of $X_K$ are exactly those in the generic fibre of $X_K \rightarrow X$”.
Now, let $y \in X_K$ be a generic point, lying above $x \in X$. Because $K/k$ is flat, so is $X_K \rightarrow X$, hence $\mathcal{O}_{X,x} \rightarrow \mathcal{O}_{X_K,y}$ is a flat local morphism. Thus it is faithfully flat, so it induces a surjection of spectra; since $\mathcal{O}_{X_K,y}$ has a single maximal ideal, so does $\mathcal{O}_{X,x}$, so that $x$ is the generic point of $X$.
Hence the generic fibre of $X_K \rightarrow X$ (a finite morphism) is a subspace of $X_K$ stable under generization, containing all the generic points, and is the spectrum of a finite $k(X)$-algebra: it is a finite discrete space! So said generic fibre must be exactly the spectrum of $k(X) \otimes_k K$.