Number Theory

Algebra[Lecture notes] by K. Mathiak, Steffen Supra

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By K. Mathiak, Steffen Supra

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1 of§ 1, if N Mis a K-norm, M is compact. Conversely, assume that M is compact, and take v-=foO; then Mv is the subset of K corresponding to (Kv)nM under the isomorphism x ~ x v of K onto K v; therefore M v is compact and cannot be K. This completes our proof. COROLLARY 1. An open R-module M in V is a K-Iattice if it contains no subspace of V other than O. e. KvcM. Conversely, as every subspace of V, other than 0, is closed in V and not compact, no such subspace can be contained in M if M is compact.

8. PROPOSITION 10. Let K be a commutative p-field of characteristic p. Then 1 + P, as a Zp-module, is the direct product of a countably i1ifinite family of modules isomorphic to Zp. 35 Lattices over R § 4. By tho 8 of Chap. 1-4, we may regard K as the field of formal powerseries in one indeterminate 1t, with coefficients in the field F q with q = pI elements. Here it is easy to give explicitly a family of free generators for the Zp-module 1 +P. In fact, take a basis {1X 1 , ... ,IX/ } for Fq over the prime field F p" As generators of 1 + P, we take the elements 1 + IX/X", where 1 ~ i ~ f, n running through the set of all integers > 0, prime to p.

This follows at once from the inequality N(x' v' -xv) ~ sup(modK(x')N(v' - v), modK(x' - x)N(v») which is an immediate consequence of def. 1. Therefore N is continuous, and the sets Lr make up a fundamental system of closed neighborhoods Norms §l. 25 of 0; in particular, Lr must be compact for some r>O. Now, for any s>O, take aEK x such that modK(a) ~ r/s; then, as one sees at once, L. is contained in a-I Lr; therefore it is compact. COROLLARY 1. There is a compact subset A of V - {O} which contains some scalar multiple of every v in V - {O}.

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