$\newcommand{\Spec}{\mathrm{Spec}}$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
Your attempt at writing things down more explicitly is good. The important thing to realize is that because the equivalence between affine schemes and rings is contravariant, the order of composition changes and swaps left/right actions.
If the authors do not specify whether an action is a left/right action, there is a good chance it is either inferrable from the context, or does not matter. If you can point to specific examples where you think it does matter and it's not clear, these would be good things to ask about as a separate question.
For your explicit example involving actions on $\operatorname{Spec} \Bbb Q[x]$, note that all automorphisms you mention fix $\Bbb Q[x]$ (because they fix $\Bbb Q$) and so they're the identity map. If you instead talk about $\operatorname{Spec} \Bbb Q(i)[x]$, then you do see some movement happening: $(x-i)$ is swapped with $(x+i)$, for instance. One classic fact to learn (mentioned early on in Vakil, for instance) is that if $k\subset K$ is a Galois extension, then $Gal(K/k)$ acts on $\Bbb A^n_K$ and the orbits are precisely the points of $\Bbb A^n_k$. Another good thing to know is that if we have an automorphism $\sigma:k\to k$, then the induced action of $\sigma$ on the $k$-rational points of $\Bbb A^n_k$ is $(a_1,\cdots,a_n)\mapsto (\sigma(a_1),\cdots,\sigma(a_n))$ (proof: just write down the action on the maximal ideal $(x_1-a_1,\cdots,x_n-a_n)$). So these two facts should give you a full understanding of what the Galois action does on affine space.
For subvarieties of affine space, you may need to be a little careful when you define actions. If you want to define an action on $V(I)\subset \Bbb A^n_k$, then you'll need the Galois action to fix $I$. For an example of why this is necessary, think about $V(x-i)\subset \operatorname{Spec} \Bbb Q(i)[x]$. The Galois action here doesn't fix this subvariety and thus does not define an automorphism of it.
As far as group actions on non-affine schemes, things can get a little hairy depending on the specific context. Fortunately, in the case of the Galois action on a scheme $X$ over a field $k$, everything is induced from what happens on $\operatorname{Spec} k$: we define the action of $\sigma \in Gal$ on $X$ to be the map $X\times_k \operatorname{Spec} k\cong X\to X$ induced by taking the fiber product of $X\to \operatorname{Spec} k$ with the automorphism $\sigma:\operatorname{Spec} k\to\operatorname{Spec} k$. If you're interested in studying this action on $X$ via an embedding $X\to Y$, then you need to make sure that the embedding respects the action (this is often described in the literature as "the morphism is an intertwiner" or the like - it means that the morphism commutes with the automorphism).
This brings us to your final point - when we have a morphism of varieties and we want to think about group actions on the source and target, it's usually important to us that the morphism respects the actions: if $f:X\to Y$ is our morphism, we want $g\cdot f(x)=f(g\cdot x)$. Sometimes we don't mean this, though, and if we have an action on $X$, then we can get an action on $\operatorname{Hom}(X,Y)$ by precomposing a map $f:X\to Y$ with an automorphism $\sigma:X\to X$, or if we have an action on $Y$, then we can get an action on $\operatorname{Hom}(X,Y)$ by postcomposing a map $f:X\to Y$ with an automorphism $\sigma:Y\to Y$. So the "action on a morphism" you mention in the comments is just a version of this. (Specifically, saying that $\sigma(F)=F$ is the second version of this - $\sigma$ acts on $Y$, which induces an action on maps as described, if you want to think of it this way. I usually don't, though this isn't my main area and I'm a mathematician, not a cop.)
As for the descent question, the goal is to have $\phi_{\sigma\tau}:(\sigma\tau)V\to V$ factor as $\sigma(\tau V)\stackrel{\sigma\circ\phi_\tau}{\longrightarrow}\sigma V\stackrel{\phi_\sigma}{\longrightarrow} V$, which is just the cocycle condition from (non-Galois) descent.