Let $F$ be a finite Galois extension of the rational function field $\mathbb Q(x)$. Let $k$ be the field of constants of $F$, i.e., the algebraic closure of $\mathbb Q$ in $F$. Is $k$ necessarily a Galois extension of $\mathbb Q$?



Field of constants of a Galois extension of function fields
Fix an embedding of $F$ in $\overline{\mathbb Q(x)}$. Then $F$ is Galois over $\mathbb Q(x)$ if and only if every automorphism $\sigma$ of $\overline{\mathbb Q(x)}$ over $\mathbb Q(x)$ satisfies $\sigma(F)=F$. Since $\sigma$ sends $\overline{\mathbb Q}$ to $\overline{\mathbb Q}$, doesn't that immediately imply that $\sigma$ maps $F\cap\overline{\mathbb Q}$ to itself, and hence this intersection is a Galois extension of $\mathbb Q$?


