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By Ivan Fesenko

Creation to algebraic quantity theory

This direction (36 hours) is a comparatively common path which calls for minimum necessities from commutative algebra for its knowing. Its first half (modules over vital perfect domain names, Noetherian modules) follows to a definite quantity the ebook of P. Samuel "Algebraic concept of Numbers". Then integrality over jewelry, algebraic extensions of fields, box isomorphisms, norms and strains are mentioned within the moment half. broadly speaking 3rd half Dedekind jewelry, factorization in Dedekind earrings, norms of beliefs, splitting of leading beliefs in box extensions, finiteness of the right type staff and Dirichlet's theorem on devices are handled. The exposition occasionally makes use of equipment of presentation from the publication of D. A. Marcus "Number Fields".

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Then there are m linearly independent over R vectors y1 , . . ym ∈ Y such that y1 , . . , ym is a basis of the Z -module Y . Proof. Let x1 , . . , xm be a set of linearly independent elements in Y over R with the maximal m. Denote m L = {x ∈ Rn : x = ci x i : 0 ci 1}. i=1 The set L is bounded and closed, so L ∩ Y is finite. For y ∈ Y write y = with bi ∈ R. Define z=y− [bi ]xi = (bi − [bi ])xi ∈ L ∩ Y. m i=1 bi xi 35 Hence the group Y is generated by the finite set L ∩ Y and {xi }, and Y is finitely generated as a Z -module.

So the number of different characteristic polynomials of g −1 (Z) is finite, and so is g −1 (Z). 4. Hence the kernel of g , being the preimage of 0, is a cyclic finite group. On the other hand, every root of unity belongs to the kernel of g , since mg(z) = g(z m ) = g(1) = 0 implies g(z) = 0 for the vector g(z). We conclude that the kernel of g consists of all roots of unity T in F . Since for u ∈ U the norm NF/Q (u) = σi (u), as the product of units, is a unit in Z, it is equal to ±1. Then |σi (u)| = 1 and log |σ1 (u)| + · · · + log |σr1 (u)| + 2 log(|σr1 +1 (u)| ) + · · · + log(|σr1 +r2 (u)|2 ) = 0.

Example: points of Rn with integer coordinates form a discrete subgroup. 2. Proposition. Let Y be a discrete subgroup of Rn . Then there are m linearly independent over R vectors y1 , . . ym ∈ Y such that y1 , . . , ym is a basis of the Z -module Y . Proof. Let x1 , . . , xm be a set of linearly independent elements in Y over R with the maximal m. Denote m L = {x ∈ Rn : x = ci x i : 0 ci 1}. i=1 The set L is bounded and closed, so L ∩ Y is finite. For y ∈ Y write y = with bi ∈ R. Define z=y− [bi ]xi = (bi − [bi ])xi ∈ L ∩ Y.

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