By Henry B. Mann.
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Defects in the discrete case 31 We will now treat the general case. Write A = S −1 B where B is a finitely generated k-algebra and S a multiplicative set. 12] we obtain A = S −1 B. We have shown that B is a finite A-module and hence A is a finite A-module. 3. Let (L, w) ⊇ (K, v) be a finite extension of valued fields. Suppose one of the following hold: i. L/K is separable and ∆v ∼ = Z; ii. Ov contains a field k, Ov = K, K finitely generated over k, and trdegk (K) = 1. Then we have ∆v ∼ = Z and d(w/v) = 1.
Proofs. 1. Let (L, w) ⊇ (K, v) be a finite extension of valued fields. Let (M, x) be a finite normal extension of (K, v) containing (L, w). Then the following hold: i. the quantity n(w/v) is well-defined and one has n(w/v) = [Lh,x : Kh,x ]; Z ii. d(w/v) is well-defined and has values in pv ≥0 . gM,w gM,v · [L : K] = Furthermore, the quantities d, dw , e, et ew , f, f s , f i and n are multiplicative in towers. Proof. i. e. does not depend on the choice of M . Let M be another normal extension of K containing L with G = AutK (M ).
Note that LL = M , L ∩ L = Q. Furthermore, we know the splitting behavior of (2) in M : (2) = p21 p22 p23 where there is just one prime above p respectively p , which is totally wild, and there are two primes above q respectively q . Say that p1 lies above p . Then p1 /p is totally wild, but p1 |L /2Z is not even local. 16 do not hold in this case. 28 Chapter 1. 15 it is not necessarily true that L1 = L4 , L2 = L5 or L3 = L6 . Indeed, for the extension (Q(α), q)/(Q, 2Z) we have L1 = L2 = L3 = Q and L4 = L5 = L6 = Q(α).