By Froberg R.

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We give a full proof in Chapters 4 to 6, using topology and Riemann surfaces. 14 (a) We see in particular that each finite group occurs as Galois group over C(x) - the Inverse Galois Problem has a positive solution over this field. C(x)-isomorphism) of a given type. They can be parametrized by the Aut(G)-c1asses of the tuples (gl, ... , gr) - however, not canonically. 14 for the topological interpretation. Alternatively, this can be deduced purely algebraically from the above version of RET. 4.

This can be done in the present purely algebraic set-up, but things become clearer in the geometric translation of Chapter 5. 11 We have seen that there are not a lot of Galois extensions of the form k(y)/ k(x). The next step is to look at arbitrary (non-Galois) extensions k(y)/k(x), and see what extensions of k(x) we get as their Galois closure. This can be formulated as a purely group-theoretic problem, via RET and the Riemann-Hurwitz formula. The conjecture of Guralnick and Thompson (see [GT]) expects that only finitely many nonabelian simple groups other than An can occur as a composition factor of the Galois group of such an extension.

Galois extensions ofQ with abelian Galois group). 29 Consider the polynomial f(x, y) is irreducible over Q(x). = y2 - x E Q[x, y], which (i) Let Qsolv denote the composite of all finite solvable extensions of Q. The field Qsolv is closed under taking square roots (because it consists of all algebraic numbers that are expressible by iterated radicals - Galois' theorem). Thus feb, y) is reducible over Qsolv for each b E Qsolv' Thus Qsolv is not hilbertian. 3 Algebraic Extensions of Hilbertian Fields 25 (ii) Since the square root of any rational number lies in Qab (by the theorem of Kronecker and Weber), it follows that f (b, y) is reducible in Qab [y] for each b E Q.