# Why $g\in F[X]$ by Galois theorem?

Let $F$ a field anf $f\in F[X]$ a polynomial. Let $K_f$ the splitting field of $f$ and $G_f=\text{Gal}(K_f/F)$. Let $g\in K_f$ a polynomial that divide $f$ (i.e. $g\mid f$). Why the fact that for all $\sigma \in\text{Gal}(K_f/F)$, $\sigma (g)=g$, we have that $g\in F[X]$ by Galois correspondance theorem ?

What I see is the fact that $\sigma (g)=g$ then $$g\in K_f^{\text{Gal}(K_f/F)} =\{x\in K_f\mid \forall \sigma \in\text{Gal}(K_f/F), \sigma (x)=x\}.$$

By Galois theorem, $$\text{Gal}\left(K_f^{\text{Gal}(K_f/F)}\big/F\right)=\text{Gal}(K_f/F)\Big/K_f^{\text{Gal}(K_f/F)},$$

but how can I conclude that $g\in F$ ?

Galois correspondence implies (among other things) that if $G=Gal(K/F)$ then $K^G=F$, so if an element is invariant under all the Galois automorphisms, then it must be in the base field, which in your case means that the coefficients of $g$ are in $F$.