PUMaC 2009 · 加试 · 第 3 题
PUMaC 2009 — Power Round — Problem 3
题目详情
- If ~ a, b ∈ L then ~ a + b ∈ L . Figure 1: An example of a lattice. n Given a set S of vectors in Z , we can form a lattice in dimension n by taking the set of all integer linear combinations of vectors in S . This is called the lattice generated by S . For example, the lattice generated by S = { (1 , 1) , (1 , − 1) } is the set of all integer linear 2 combinations of (1 , 1) and (1 , − 1), i.e. the subset ( a + b, a − b ) of Z , where a ∈ Z and b ∈ Z . 2 This comes out to be simply the set of all ( x, y ) ∈ Z with the property that x + y is even. The diagram below shows the points representing the vectors of this lattice. Figure 2: The lattice generated by { (1 , 1) , (1 , − 1) } .
解析
Problem 3.5 (5pts) . Prove that if L is a full lattice in dimension n , then its determinant n is divisible by (div L ) . n n Let div L be denoted by d so that we have the series of inclusions L ⊂ d Z ⊂ Z . There n n n n are d colattices of d Z in Z explicitly given as A = { ( a , . . . , a ) + d Z | 0 ≤ a ≤ d − 1 } . 1 1 n i Consider the set of colattices of L which we will denote A . A function f : A → A can be 2 2 1 n ~ defined by f ( ~ a + L ) = ~ a + d Z . Note that it does not matter if we choose a different b to n ~ represent the same collatice because then ~ a − b ∈ L ⊂ d Z so they define the same collatice n − 1 n n of d Z . Let f ( ~ a + d Z ) denote the subset of A which maps to ~ a + d Z by f . This set 2 is non-empty and its size does not vary as we vary the collatice in A . Indeed, let c be 1 i − 1 n n such that { c + L | 1 ≤ i ≤ k } = f ( d Z ). For any ~ a + d Z ∈ A it is easy to check that i 1 − 1 n − 1 n { ~ a + c + L | 1 ≤ i ≤ k } = f ( ~ a + d Z ). Now A has been partitioned by the f ( ~ a + d Z ) i 1 n n n into d distinct sets each of size k . We conclude that kd = det L and so (div L ) divides the determinant of L . 4 Finite Generation (3 Problems; 13 Points)