By Paul J. McCarthy

*Math.*reports. Over 2 hundred workouts. Bibliography.

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In fact, H = 1. we can show that n He9t Hnormal For, suppose that a belongs to every normal subgroup of G(K/k) which belongs to 91. If a~ 1 there is an a E K such that a(a) ~ a. Let L be a finite normal extension of k in K which contains a. There is such an extension by Theorem 19, Chapter 1. Then G(K/L) E 91 and is a normal subgroup of G(K/k) but a¢ G(K/L). Hence we must have a= 1. Now we shall give the example promised in Section 10. This example will illustrate two important facts. First, there may be relatively few subgroups of G(K/k) which are closed and, second, there may be subgroups of finite index in G(K/ k) which are not closed.

Proof We see from §1, no. 2 of Bourbaki's Chapter III that we must show that 91 is a filter base and that if HE 9l and a E G then aH a 1 E 91. It is certainly true that 91 is not empty and that no element of 91 is the empty set. If, for i = 1, 2, Hi = G(K/Fi) E 91 then 60 \ 11. The Krull topology Let K be a Galois extension of k and let G = G(K/k). Let 91 be the family of all subgroups G(K/F) of G where k £ F £ K and [F: k] is finite. 21. There is a topology on G which is compatible with the group structure of G and which has 91 as a fundamental system of neighborhoods of the identity.

Choose once and for all a real number dwith 0 < d < 1. lfj(x)fg(x) is an arbitrary nonzero element of k(x) we set for all positive integers t. If we let t ,. oo we obtain lml <; 1 for all positive m E J, and so for all m E J. It then follows from the lemma that I I is nonArchimedean. Let I= {mE J llml < 1}. lal t = (absolute value of n)c. + + + 1/t for some positive real number c. Thus for all n E J or 1 nllt 1 lml ~ lnl(log m) j(log n) Now lett be a positive integer and replace m by mt. This gives 1 log n By interchanging the roles of m and n we see that for all integers m and n which are greater than 1, JmJlf(log m) = JnJlf(log n) = ec i=O < lnl(log m')f(lOg n) • for all positive integers t.