It is known for Hilbertian fields that all groups that are abelian, solvable, $A_n$ or $S_n$ are realizable over them. $\mathbb{Q}(x)$ is one such field, but it's not obvious that the extensions that are guaranteed by this theorem will be regular extensions.

$A_n$ and $S_n$ are regularly realizable through the symmetric polynomials. Is it known whether all solvable (or even abelian) groups are *regularly* realizable over $\mathbb{Q}(x)$?

Remark: for those who aren't familiar with the terminology, by regularly realizable over $\mathbb{Q}(x)$ I mean that there's an extension of it, $L$, such that $L$ over $\mathbb{Q}(x)$ is $G$-Galois, and such that the algebraic closure of $\mathbb{Q}$ in $L$ is $\mathbb{Q}$.