I am reading the proof of Proposition 2.19 in Liu's book on Algebraic Geometry and Arithmetic Curves. The proposition is as follows:
Let $X$ be an algebraic variety over $k$, and let $K / k$ be a Galois extension with group $G$. Then the set of classes $G \backslash X(K)$ injects in $X$, and the set of fixed points $X(K)^G$ can be identified with $X(k)$.
Here, $X(F)$ denotes the $F$-valued points of $X$, i.e. the morphisms of $k$-schemes from $\operatorname{Spec} F$ to $X$. In the couse of the proof, the author says that the natural action of $G$ on $\operatorname{Hom}_{k-algebras}(k(x), K)$ is transitive, where $k(x)$ is the residue field of some point $x \in X$ that is the image of an element of $X(K)$.
This seems like it should just be basic Galois theory, but I have a nagging doubt, since $k(x)$ will generally not be an algebraic extension of $k$. Can anyone give a proof that the action of the Galois group of $K / k$ on $k$-algebra homomorphisms from $A \to K$ is transitive for an arbitrary (maybe only finitely-generated?) $k$-algebra $A$, or a reference (a quick look through the algebra books on my shelf did not find this statement, but it must be somewhere if it's true)?
A $k$-map $\operatorname{Spec} F \to X$ is equivalent to a choice of point $x\in X$ and a map of residue fields $k(x)\to F$ (ref), so we get a composition of extensions $k\subset k(x) \subset F$. So if $F$ is algebraic over $k$ then $k(x)$ is also algebraic over $k$.