Let $k\newcommand{\Spec}{\operatorname{Spec}}\newcommand{\id}{\operatorname{id}}$ be a field, let $K/k$ be a Galois extension, let $G = \operatorname{Gal}(K/k)$, and let $X$ be a $k$-scheme. I want to understand the connections between two different actions of $G$ on objects relating to $X$. I will begin by stating their definitions and what I know about them.
There are two natural ways to define a left action of $G$ on $X(K)$. The first is to let $\sigma\in G$ act on a point $x\in X(K)$ by post-composition: to $\sigma$, there is an associated morphism $\sigma^\flat\!:\Spec{K}\to\Spec{K}$, and we set $\sigma x := x\circ\sigma^\flat$. It is then not hard to check that this is a left group action. A different approach is to use the identification of $X(K)$ with pairs $(x,j)$, where $x\in X$ and $j\!:\kappa(x)\hookrightarrow K$ is an embedding of the residue field at $x$ into $K$. One can then define the action by $(x,j)\mapsto (x,j\circ\sigma)$. Now, it is not too hard to check that these two approaches give the same action (for example by inspecting the proof of the identification used in the second approach).
One can also define a right action of $G$ on the $k$–scheme $X_K := X\times_k\Spec{K}$ (with the composition $X_K\to\Spec{K}\to\Spec{k}$ as the structure map) by assigning to $\sigma\in G$ the map $\id_X\times_k\sigma^\flat\!:X_K\to X_K$ which acts trivially on $X$ and applies $\sigma^\flat$ to $\Spec{K}$. That this is a right action is not unexpected, since $\Spec$ is contravariant. Essentially what we now have is a map $\newcommand{\op}{\text{op}}\newcommand{\Aut}{\operatorname{Aut}}G^\op\to\Aut_k(X_K) = \{k\text{-isom. }X_K\to X_K\}$.
My problem is in trying to relate these two approaches. I'm aware that one can get a left action on $X_K$ by switching from $\sigma$ to $\sigma^{-1}$, but this seems like the wrong direction to go. On the other hand, I have seen no other ways to get a left action out of the right action on $X_K$.
Basically, my question is: is it actually possible to relate the right action on $X_K$ to the left action on $X(K)$, and if so, how? Furthermore, as this comes up in identifying $X(k)$ as the fixed points of the left action on $X(K)$, is there an analogous such result for the action on $X_K$?
Best Answer
I think it is easiest to understand the relationship between the two actions by looking at the case $X = \mathbb{A}^1_k = \operatorname{Spec} k [t]$. Then $X (K)$ can canonically be identified with $K$ (very explicitly, by the image of $t$ in $K$), and the left action of $\textrm{Gal} (K \mid k)$ is then identified with the tautological action on $K$. On the other hand, if we identify $X_K$ with $\operatorname{Spec} K [t]$, then the right action of $\sigma \in \textrm{Gal} (K \mid k)$ is identified with the automorphism of $\operatorname{Spec} K [t]$ induced by applying $\sigma$ to the coefficients of the elements of $K [t]$. Thus, it should now be clear that the two actions are essentially an inverse pair: the root of $t - \sigma (a)$ is $\sigma^{-1} (a)$.
More generally, consider the following commutative diagram: $$\require{AMScd} \begin{CD} \operatorname{Spec} K @>>> X_K @>>> X \\ @| @VVV @VVV \\ \operatorname{Spec} K @>>{\textrm{id}}> \operatorname{Spec} K @>>> \operatorname{Spec} k \end{CD}$$ The projection $X_K \to X$ therefore induces a bijection $X_K (K) \to X (K)$, provided we understand that $X_K (K)$ denotes the set of $K$-rational points of $X_K$ as a $K$-scheme. The left action on $x \in X (K)$ is defined by precomposing $\sigma^\flat$, but if we precompose $\sigma^\flat$ onto a morphism $\operatorname{Spec} K \to X_K$ the result is not a morphism of $K$-schemes! Instead, we have to consider this commutative diagram to get an element of $X_K (K)$: $$\begin{CD} \operatorname{Spec} K @>{\sigma^\flat}>> \operatorname{Spec} K @= \operatorname{Spec} K \\ @VVV @VV{x_K}V @VV{x}V \\ X_K @>{\textrm{id} \times_k \sigma^\flat}>> X_K @>>> X \\ @VVV @VVV @VVV \\ \operatorname{Spec} K @>>{\sigma^\flat}> \operatorname{Spec} K @>>> \operatorname{Spec} k \end{CD}$$ That is, the element of $X_K (K)$ corresponding to $x \circ \sigma^\flat$ has the property that postcomposing with $\textrm{id} \times_k \sigma^\flat$ is the element of $X_K (K)$ corresponding to $x$; put it another way, the left action on $X_K (K)$ transported from $X (K)$ is the inverse of the right action on $X_K (K)$ induced by the right action on $X_K$, exactly as we saw in the concrete example.