PUMaC 2009 · 加试 · 第 4 题
PUMaC 2009 — Power Round — Problem 4
题目详情
Problem 4.3 (5pts) . Prove that every full lattice is finitely generated. 5 Isomorphism Types of Lattices (7 Problems; 28 Points) n Definition 5.1. Two lattices L and L in Z are said to be isomorphic iff there exists a 1 2 n n n m linear bijection f : Z → Z which is also a bijection from L to L . [A map f : Z → Z is 1 2 ~ ~ ~ ~ said to be linear iff f ( 0) = 0, and f ( ~ a + b ) = f ( ~ a ) + f ( b )]. [A map f : X → Y is a bijection if it gives a one-to-one correspondence between elements of X and elements of Y ]. 2 2 For example, the linear bijection f : Z → Z defined by f ( x, y ) = (5 x + 2 y, 2 x + y ) is also a bijection from the lattice L generated by { (2 , 1) , (1 , 2) } to the lattice L generated 1 2 by { f (2 , 1) , f (1 , 2) } = { (12 , 5) , (9 , 4) } ; hence L and L are isomorphic. Note that L is also 1 2 2 the lattice generated by { (3 , 0) , (0 , 1) } .
解析
Problem 4.3 (5pts) . Prove that every full lattice if finitely generated. We combine the last two problems to solve this problem. Consider the set of finitely generated full sublattices of L . These lattices have finite determinant by Problem 3.4 and so by the well ordering principle, there must be a finitely generated full sublattice M with ~ ~ minimal determinant. Assume M does not equal to L so that we can take l ∈ L with l / ∈ M . ~ The generators of M together with l generate a finitely generated full sublattice of L which ′ ′ ′ ′ we will call M . Clearly M % M so by Problem 4.1 det M < det M . Therefore M violates the minimality of M so it must have been that M = L and L is finitely generated. 5 Isomorphism Types of Lattices (7 Problems; 28 Points) n Definition 5.1. Two lattices L and L in Z are said to be isomorphic iff there exists a 1 2 n n n m linear bijection f : Z → Z which is also a bijection from L to L . [A map f : Z → Z 1 2 ~ ~ ~ ~ is said to be linear iff f ( 0) = 0, and f ( ~ a + b ) = f ( ~ a ) + f ( b )]. 2 2 For example, the linear bijection f : Z → Z defined by f ( x, y ) = (5 x + 2 y, 2 x + y ) is also a bijection from the lattice L generated by { (2 , 1) , (1 , 2) } to the lattice L generated 1 2 by { f (2 , 1) , f (1 , 2) } = { (12 , 5) , (9 , 4) } ; hence L and L are isomorphic. Note that L is also 1 2 2 the lattice generated by { (3 , 0) , (0 , 1) } .