Let $X$ be a $k$ variety (i.e. a $k-$scheme of finite type). Let $k^{s} \subset \bar{k}$ be the separable closure of $k$. I will wright $X(k^{s})$ for the set of $k$ morphism from $Spec(k^{s})$ to $X$. The group $Gal(k^{s}/k)$ naturally acts on $X(k^{s})$ (by composition...).
I would like to show the orbits of this action are finite.
What I've tried : as the image $x$ of an element of $X(k^{s})$ is a closed point of $X$ (since $k \subset k(x) \subset k^{s} \subset \bar{k} $ hence $k(x)$ is algebraic over $k$ hence $x$ is a closed point) it's enough to show their is only a finite number of $k$ embedding from $K$ to $k^{s}$ if $K$ is a finite extension of $k$.
Do you have ideas?
This is standard field theory. The number of $k$-homomorphisms $K \to k^s$ is at most $[K:k]$.