By G. Cherlin

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Pk Our theorem claims: r > I. Let ~ have characteristic that r ~ 0 (mod p), and hence observe that for i__~(1_Pi(~1) = Hence r a in p and order q. We are going to prove r ~ I, as desired. Fn: Pi(a) = 0 for 1 ~ i ~ k otherwise. ,jn); that for each = F~ ~ we define ~ in F and ~ varies over = T T a~ i. Furthermore notice (*) then c~ = 0, so the only ~: (*) ~Ji ! (q-1)~d i < n(q-1). ) = 0. ~n ~ E a ~I F O. Here we have use~ the fact that ~ aj = 0 F for j < q-l, an elementary fact which is easily verified.

1. One defect of the above f o r m u l a t i o n of Theorem 16 is that it does not apply to p-adic completions of general number fields, where ramification concerning ramification weakened 2. occurs. to cover this case An e x c e l l e n t We w i l l may be b r i e f l y That 1 6 . 2 embedded o f t h e p r o o f o f Theorem 16 w i l l attempt a meaningful implies can be ~0]" Theorem 16 also applies tarily equivalent" [30]. The hypothesis is a triviality. of the proof. in". be found i n The method For t h e c o n v e r s e , we may ~2 assume for simplicity saturated of cardinality and we must prove structure residue that that ~I field and value in K, choose power series quires suppose steps.

Pure over that subgroup K of it is e a s y every element Z' generated by to s h o w of Z'/Z o has finite order. Now of some z has over ord fix an element element finite of order ord~K]. K. z of Since over Zo, In p a r t i c u l a r Z. z and there We is will show in a n y thus is an z that case has an z element a finite element is a of the order of Z', order K k such that a = kz. Let hence KI be the field a I/k is in K. z E ordtK], To see K(al/k). Since We o r d ( a I/k) will show that = z, we w i l l as d e s i r e d .