I often hear $\operatorname{Gal}(L/K)$ acts on $L$, but why (how)?

Question

$\DeclareMathOperator{\Gal}{Gal}$ I often hear $\Gal(L/K)$ acts on $L$, but why(how) ? $\Gal(L/K)×L\to L$ by $(σ,x)\mapsto σ(x)$, here, $σ$ is an element of $\Gal(L/K)$, which is a ring homomorphism from $L$ to $L$ and identity on $K$, and $σ(x)$ means the image of $x$ by $σ$. We can check the axioms of group action, so in this way, $\Gal(L/K)$ acts on $L$, is my understanding correct?

Answer 1

Yes, your understanding is correct. Sometimes this is also called the "natural action" of $\operatorname{Gal}(L/K)$ on $L$.

Why is this action "natural"? As all symmetric groups, the group $S_L$ of all permutations on $L$ acts on $L$ in the obvious "natural" way. The set $\operatorname{Aut}(L)$ of all field automorphisms of $L$ is a subgroup of $S_L$, which inherits the natural action on $L$. And by definition, $\operatorname{Gal}(L/K)$ is just the (element-wise) stabilizer of $K$ of the latter group action, which again inherits the natural action on $L$.

This approach has the additional benefit that it is clear that we get a group action. It is not necessary to check the axioms explicitly.

Taking the above approach a step further: If $X$ is a set and $G$ is a group of bijections $X \to X$ (with respect to the ordinary concatenation of mappings), you always have the "natural" action $f \bullet x = f(x)$ (where $f\in G$ and $x\in X$) of $G$ on $X$.