PUMaC 2009 · 加试 · 第 5 题
PUMaC 2009 — Power Round — Problem 5
题目详情
Problem 5.7 (7pts) . Prove that divisor and determinant do not characterize lattices in 3 dimension three. That is, construct two lattices L and L in Z which have the same 1 2 determinant and the same divisor but which are not isomorphic. 4 6 Canonical Form (2 Problems; 13 Points) The following theorem is true (proving it is not part of this test). Theorem 6.1. Every lattice in dimension n is isomorphic to the lattice generated by { d ~ e , . . . , d ~ e } (6.1) 1 1 n n for some integers d ≥ 0 where d divides d . Furthermore, the sequence of integers i i i +1 ( d ; . . . ; d ) is isomorphism invariant; it is called the signature of the lattice. 1 n Here ~ e is the vector (0 , . . . , 0 , 1 , 0 , . . . , 0) where the 1 appears in the i th entry. i You may assume it is true for any of your work on problems appearing after this point in the test.
解析
Problem 5.6 (7pts) . Prove that if two full lattices in Z have the same determinant and 2 same divisor then they are isomorphic. Conclude that all full lattices in Z are isomorphic to one of the lattices from problem 5.5 (don’t forget the result of problem 3.5). By rescaling, it suffices to treat the case that d = 1. Suppose we are given a full lattice L in dimension two with div L = 1 and det L = ∆. We would like to show that L is isomorphic to the lattice generated by { (1 , 0) , (0 , ∆) } . Since L is full, we know that ( E, 0) ∈ L for some large E . Pick the smallest such E , and let ~ a = (0 , E ) ∈ L . Now again since L is full, the set of y -coordinates of vectors in L ~ contains some elements other than zero. Thus we can find a vector b = ( C, D ) ∈ L where ~ C > 0 is as small as possible. From our choice of vectors, it is clear that ~ a and b generate L . Thus since div L = 1, we know that gcd( C, D, E ) = 1. Thus there exists an integer k such that gcd( C, D + kE ) = 1. WLOG, we may assume that k = 0. Thus gcd( C, D ) = 1, C D so there exist integers x, y with Cx + Dy = 1. Now consider the matrix M = ( ). Since − y x − 1 det M = Cx + Dy = 1, M is invertible and M has integer entries. Thus we may apply − 1 ′ the matrix M to the generators of L and get generators of an isomorphic lattice L . Since ′ ~ M (1 , 0) = b , we conclude that (1 , 0) ∈ L . Now we are done, since every lattice containing (1 , 0) is equal to the lattice generated by { (1 , 0) , (0 , P ) } for some integer P . Comparing determinants, we conclude that P = ∆, so the desired isomorphism is demonstrated.