1
$\begingroup$

I have a question on a certain property of morphisms between schemes endowed with Galois action. The motivation arises from a comment by Phil Tosteson on this question.

Phil wrote: "If the map factors through the projection, it factors uniquely (the projection is dominant) . So the factorization is automatically galois invariant..."

I want to understand how the Galois action on the morphism is concretly described in the given setting in order to be able to talk about "Galois invariant morphisms".

What we know. Let $X,Y$ be $k$-varieties or more generally $k$-schemes. Let $\overline{k}$ be algebraic closure of $k$ and $Gal(\overline{k}/k)$ the Galois group. We consider the fiber products $X \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}, Y \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}$.

We introduce the abbreviation $\overline{X}:=X \times_{\operatorname{Spec} \ k} \operatorname{Spec} \ \overline{k}$.

Let $f: \overline{X} \to \overline{Y}$ a morphisms between the $\overline{k}$-schemes.

QUESTION: I order to decide if $f$ is "Galois invariant" we need to decede which Galois action on $f$ we consider. Do we have in this context "the action" (ie a canonical one)?

I know a couple possibilities to define an action of $f$ in different ways (see below) but I'm not sure which one is the standard one when the literature talks about "Galois invariant morphism".

Two ways I know to define an action on $f$:

1. Let $ \sigma \in Gal(\overline{k}/k)$. Then $\sigma$ acts on $\overline{X}$ via morphism $id_X \times (\operatorname{Spec} \ \sigma)$ with $\operatorname{Spec} \ \sigma: \operatorname{Spec} \ \overline{k} \to \operatorname{Spec} \ \overline{k}$. We use notation $\overline{\sigma}:=id_X \times (\operatorname{Spec} \ \sigma)$.

Then we can define on action on $f$ by $\sigma$ via "conjugation" $\sigma(f):= \overline{\sigma^{-1}} \circ f \circ\overline{\sigma}$

2. Consider the set of $\overline{k}$-valued points $\overline{X}(\overline{k}):=Hom(\operatorname{Spec} \ \overline{k}, \overline{X})$.

We can say two things:

-Galois group acts on $\overline{X}(\overline{k})$ via composition $\alpha \mapsto \alpha \circ (\operatorname{Spec} \ \sigma)$ for $\alpha \in \overline{X}(\overline{k})$ and $\sigma \in Gal(\overline{k}/k)$

-$f$ induces map $f(\bar{k}):\overline{X}(\overline{k}) \to \overline{Y}(\overline{k})$ by compostion $\alpha \mapsto f \circ \alpha$

Thus the Galois Group also acts on $f(\bar{k})$ via precomposition $f(\bar{k}) \mapsto f(\bar{k}) \circ (\operatorname{Spec} \ \sigma)$

The funny thing is that $\overline{X}(\overline{k})$ is dense in $\overline{X}$ and thus the Galois action from 2 on $f(\bar{k})$ induces by continuity & density a unique action on $f$.

That is we have (at least) two possibilities how Galois group could act on $Hom(\overline{X}, \overline{Y})$ and we can say that a $f \in Hom(\overline{X}, \overline{Y})$ is Gaois invariant if for every $\sigma(f)=f$ is fixed.

Back to my question: if we talk about "the" Galois action on $f$(or calling $f$ Galois invariant) which action is generally proposed if it is not explicitly explained as in my MO question: the 1 or the 2?

$\endgroup$
3
$\begingroup$

The first action is the usual one.

Actually, your second action is not well-defined. By definition, $\bar{X}(\bar{k})$ is the set of morphisms $\alpha:\operatorname{Spec}\bar{k}\to \bar{X}$ such that $\pi_2\circ\alpha$ is the identity on $\operatorname{Spec}\bar{k}$, where $\pi_2:\bar{X}\to\operatorname{Spec}\bar{k}$ is projection onto the second factor (that is, the structure map). If $\pi_2\circ\alpha$ is the identity, then $\pi_2\circ(\alpha\circ \operatorname{Spec}\sigma)=\operatorname{Spec}\sigma$ will not be the identity unless $\sigma=1$, so $\alpha\circ \operatorname{Spec}\sigma$ will typically not be in $\bar{X}(\bar{k})$. So your proposal 2 does not give a well-defined Galois action on $\bar{X}(\bar{k})$, nor on $\operatorname{Hom}(\bar{X},\bar{Y})$.

$\endgroup$
2
  • $\begingroup$ Yes I see. One remark on a slightly different situation (which motivated me to impose the wrong proposal 2): If we start instead of $\overline{X}$ with $k$-scheme $X$ and consider the points $X(\bar{k})$ then the Galois action on $X(\bar{k})$ by precomposition $\alpha \mapsto \alpha \circ (\operatorname{Spec} \ \sigma)$ seems in this case to work since in this case for $X$ we haven't more the projection $\pi_2$ forcing $\pi_2\circ\alpha$ beeing idenity. So if we have this Galois action on $X(\bar{k})$, can this action be extended to Galois action on $X$? That is if certain $\sigma$ acts as $\endgroup$ – user7391733 Feb 28 '20 at 17:25
  • $\begingroup$ described on $X(\bar{k})$, then can (possibly by a certain continuity argument) the action extended to such one $X$? The problem is the existence, since if such extension of the action exist then it must be unique by density of $X(\bar{k})$ as I remarked. But I'm not sure if we can expect that for arbitrary $k$-scheme such extension realizable. Do you know if it always work or one need to impose extra conditions of $X$ (e.g. certain finiteness conditions...) $\endgroup$ – user7391733 Feb 28 '20 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.