On free products of an absolute Galois group (Q1972173)

The author gives a partial answer to the question by \textit{M.~Jarden} [Infinite Galois theory, Handbook of algebra. Vol. 1, 271-319 (1996; Zbl 0864.12003)] whether the class of profinite groups that are absolute Galois groups of fields is closed under free profinite products. Denote by \(G(K)\) an absolute Galois group of a field \(K\). Recall that a separable group is a topological group that has a countable base of neighborhoods about identity. The author proves: Theorem A. Let the absolute Galois groups \(\mathfrak G(K_i)=G_i\) of fields \(K_i\) be separable groups, \(i=1,2,\dots,n\). Let \(G=G_1*G_2*\dots *G_n\) be the free profinite product. Then, for some field \(L\), the group \(G\) is an absolute Galois group, \(G=\mathfrak G(L)\). Besides, the author proves: Theorem B. A profinite group \(G\) is isomorphic to an absolute Galois group of some field, \(G=\mathfrak G(K)\), if and only if it is a subgroup of an absolute Galois group of some field of rational functions \(\Phi_p(X)\) over a prime field \(\Phi_p\) of \(p\) elements for some set of variables \(X\).