Arithmetic of Hyperbolic Three-Manifolds by Colin Maclachlan, Alan W. Reid

By Colin Maclachlan, Alan W. Reid

Lately there was enormous curiosity in constructing innovations in keeping with quantity conception to assault difficulties of 3-manifolds; comprises many examples and many difficulties; Brings jointly a lot of the present literature of Kleinian teams in a transparent and concise method; at the present no such textual content exists

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2 ,\ X ,\ E H,\ with respect to for almost all >. 33) 0. Number-Theoretic Menagerie 38 The group G is topologised by taking as a neighbourhood system of the identity, the sets f1 UA , where UA is an open neighbourhood of the identity in GA for all A and UA = HA for almost all A . By Tychonov's theorem, this defines G as a locally compact topological group and the group and its topology is independent of the choice of finite subset no. In all of the cases we will consider, the index set will be the set of all places of a number field k and no will always contain the finite subset of Archimedean places, usually denoted 000• Thus let k be a number field and f2 = {v a place of k} .

The formula {0. 1 6} thus takes the form 1. d = efg. (0 . ] 3. Show that if P1 , P2 are distinct prime ideals of Jik, , then Pf + P� = Rk for all integers a, b > 1 . 4 . k = 1 if and only if k = Q . This implies that for every number field k =I= Q , there is always a prime ideal which ramifies in k I Q . Show that this last statement is not true in general for relative extensions f I k by considering f = Q (v's, i) and k = Q (J=5) . 5. Let k1 , k2 be such that [k1 Q] = n1 and [k2 : Q] = 'n2 . Let K be the compositum of k1 , k2 .

Now, the discriminant of xq2 -1 - 1 is not divisible by p. 8 holds in any Dedekind domain and so we deduce that the extension L I K is unramified. 29) , [L : K] = 2. If L I K is a quadratic unramified extension, then L is the unique quadratic extension of lFq and L = RL/PR£ . Thus as earlier, L = K(a) , where a is a primitive root of xq2 -1 - 1. Let L = K ({3) where (32 E K. If 1r is a uniformiser of K, it can also be taken to be a uniformiser of L. 8, {32 = 1rr u, where u E Rp . In the discrete valuation ring RL , this implies that r is even.

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