By van Oosten J.

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1(i», because LI( is regular over K. The same holds for i I( in place of L Hence by the Galois correspondence, L. 1(. (direct product) where G'=G(K/K'(X». We have G'~G by (a) and (b), hence G' has trivial center (by assumption). It follows that G(K/LI() = G(K/il() = the full centralizer of G' in G(K/K(X». Hence LI( = il(' This proves (d). (e) Let () be as in (a). 6). Set ()' = tJ((). Then tJ(K(X» = K(t), hence tJ(LI() = K(t)(()'). Thus the restriction of tJ yields an embedding J of LI( into ~I( := K((t»(()').

Gr is another system of generators of G with product 1, but with the classes C I and C 2 switched. Here we encounter the braiding action for the first time. It accounts for the nonuniqueness of the gl, ... , gr. See Chapters 9 and 10. C(x) with r < 2 branch points. And if r = 2 then G must be cyclic. ) Thus the first interesting case is that of r = 3 branch points. (e) Actually, RET remains true if we replace C by an algebraically closed subfield k. 10. The fact that there are usually several nonisomorphic FG-extensions of a given ramification type is the major obstacle in the present approach to the Inverse Galois Problem.

1 Branch Points and the Associated Conjugacy Classes We fix a compatible system (I;e )eeN of primitive eth roots of unity in k. That is, I;e is a primitive eth root of unity, and whenever e = e' e" then 1;:" = I;e" In the case k = C, the canonical choice is I;e = exp(2nRle). Let A = k«(t» as in the previous section. Let ~I A be an FG-extension of degree e. 4). Hence there is a unique element w E G(~I A) with w(8) = l;e8 (for each such 8). We call w the "distinguished generator" of G(~I A). ) For each 8' E ~ with (8')" = t for some integer e' 2: I we have w(8') = l;e,8'.