# Galois action on $\overline{k}$-valued points extends to action on $k$-scheme $X$

Let $$X$$ be $$k$$ variety or more genrally a $$k$$-scheme. Denote the algebraic closure of $$k$$ by $$\overline{k}$$. it's a fact that $$X(\overline{k}):=Hom(\operatorname{Spec} \ \overline{k}, X)$$ as set is dense in the underlying topological space of $$X$$.

Obviously the Galois group $$Gal(\overline{k}/k)$$ acts on $$X(\overline{k})$$ by composition: let $$\sigma \in Gal(\overline{k}/k)$$ then $$\alpha: X(\overline{k}) \to X(\overline{k}), \alpha \mapsto \alpha \circ \operatorname{Spec} \ \sigma$$ where $$\alpha \in X(\overline{k})$$ and $$\operatorname{Spec} \ \sigma$$ is the spec morphism induced by $$\alpha$$.

Question: Is it true and if yes then why that the action of $$\alpha$$ as described above on $$\overline{k}$$-valued points $$X(\overline{k})$$ extends in appropriate way to an action on the whole scheme $$k$$-scheme $$X$$. By density $$X(\overline{k})$$ of it's clear that if such extension of action exists it's unique. That is the question is if such extension always or under "weak" assumptions on $$X$$ always exist and how does it look like. Is there a concrete description of it?

If we consider $$X$$ as contravariant functor $$X: (Sch/k) \to (Set), Y \mapsto Hom_k(Y,X)=X(Y)$$ then obviously it suffice to show that the action of $$Gal(\overline{k}/k)$$ on $$X(\overline{k})$$ extends to an action on $$X(Y)$$ for every $$k$$-scheme $$Y$$. That's also not obvious.

Does somebody know when under extra assumptions on $$X$$ such Galois action on $$X(\overline{k})$$ as described above extends to $$X$$? My hope is that the density condition could "somehow" allow a continuation (in what sense ever) on the action.

• If you're interested in only a single element of the Galois group, and not the full Galois group, you can do this in characteristic $p$ for any power of Frobenius. I suspect this is the only example but wasn't able to prove it (except for $\mathbb A^1$). – Will Sawin Mar 13 '20 at 16:07

No, the Galois action on $$X(\overline k)$$ corresponds to a trivial action on the scheme $$X$$. This you can already see if $$X=\mathbf A^1_k=\mathop{\rm Spec}(k[T])$$ is the affine line. Then $$X(\overline k)=\overline k$$, with its obvious Galois action. However, the scheme $$X$$ has two kind of points: the generic point, and closed points corresponding to maximal ideals of $$k[T]$$, that is, to irreducible monic polynomials $$P(T)$$. The $$\overline k$$-roots of such a polynomial are exchanged by the Galois group, but the maximal ideal is itself invariant.
You have, however, a Galois action on $$X_{\overline k}=X\otimes_k \overline k$$, which is inherited by the Galois action on $$\overline k$$.