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 $\mathbb{Q}$.



Are all solvable groups *regularly* realizable over Q(x)?
It is known that all finite abelian groups are regularly realisable over $\mathbb{Q}(x)$. See e.g. B.H. Matzat, Konstruktive Galoistheorie, p. 224, M. Fried and M. Jarden, Field Arithmetic, Lemma 24.46, or J.P Serre, Topics in Galois Theory, p. 36. It is also known that, e.g., the set of regularly realisable groups is closed under wreath products, which gives you some non-abelian soluble groups.


But according to Jack Sonn, [Brauer groups, embedding problems, and nilpotent groups as Galois groups](http://www.springerlink.com/index/Y0874238L4X4R417.pdf), Israel Journal of Mathematics 85, which, alas, is 16 years old, it is not even known whether all finite nilpotent groups are regularly realisable over $\mathbb{Q}(x)$. I don't know of any more recent developments (there has been quite a lot of work over "large" fields instead of $\mathbb{Q}$).


