On \(\Sigma\)-Hilbertian fields
Let \(\Sigma\) be a family of finite simple group. A finite group \(G\) is said to be \(\Sigma\)-Hilbertian if each composition factor of \(G\) belongs to \(\Sigma\). Let \(K\) be a field of characteristic \(0\). Then \(K\) is \(\Sigma\)-Hilbertian if for every finite Galois extension \(F/K(t)\) with a \(\Sigma\)-group as Galois group, there are infinitely many \(a\in K\) such that each decomposition subgroup of the Galois group over the specialization \(t\mapsto a\) coincides with the whole group. For a positive integer \(g\), we say that \(K\) is \(g\) -Hilbertian if \(K\) is not the union of finitely many sets of the form \(\phi(X(K))\) where \(X\) is an algebraic curve of genus \(\leq g\) and \(f\) is a rational function of degree at least \(2\). Classical Hilbertian fields are those which are \(g\)-Hilbertian for every \(g\); moreover they can be defined as being \(\Sigma\)-Hilbertian, for \(\Sigma\) being the class of all simple groups. \textit{U. Zannier} and the reviewer [Isr. J. Math. 105, 323-335 (1998; Zbl 0912.12002)] gave an elementary construction of \(0\)-Hilbertian fields which are not Hilbertian. In the paper under review it is proved that under some technical hypothesis on the family \(\Sigma\), every countable Hilbertian field \(K_{0}\) admits an algebraic extension which is \(\Sigma\)-Hilbertian but not Hilbertian. This result implies the existence, for every \(g\), of an algebraic extension of \(K_{0}\) which is \(g\)-Hilbertian, but not Hilbertian. The link between the two results arises from considering the family \(\Sigma\) of composition factor of Galois groups obtain from Galois closures of covering of genus at most \(g\) of the line.
scientific article; zbMATH DE number 1344271