# 18.312: Algebraic Combinatorics Lionel Levine. Lecture 22. Smith normal form of an integer matrix (linear algebra over Z).

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1 18.312: Algebraic Combinatorics Lionel Levine Lecture date: May 3, 2011 Lecture 22 Notes by: Lou Odette This lecture: Smith normal form of an integer matrix (linear algebra over Z). 1 Review of Abelian Groups (= Z-modules) Recall that given a ring R with 1, an R-module is an Abelian group written additively with a map R M M ( scalar multiplication ) with R the scalars and M the vectors, satisfying r (m 1 + m 2 ) = rm 1 + rm 2 r (sm) = (rs) m 1m = m which is analogous to a vector space over R, with the difference that we may lack multiplicative inverses in R, by contrast with a vector space over a field. If G is an group, we have a map Z G G (n, g) g + + g }{{} n times (0, g) 0 if n > 0 (n, g) g + + g if n < 0 }{{} n times so we admit scalar multiplication by integers, but not anything else. abelian group has the structure of a Z-module. In particular, any Now, say G is an Abelian group, finitely generated from generators g 1,..., g n. Then there is a surjective group homomorphism f : Z n G taking basis elements to the generators f (e i ) = g i f c i e i = c i g i. i [n] i [n] 22-1

2 Let K be a kernel of f, the subgroup of Z n s.t. K = ker f = c i g i c i g i = 0 in G i [n] i [n] Definition 1 A group G is torsion-free if g G, g 0, and n Z, n 0 we have ng 0 Example 2 Z and Z n are torsion free, but Z/nZ is not torsion free, since ng = 0 for all g Z/nZ. Note that if G is torsion free then it is infinite or zero since with g G and g 0, then g, 2g, 3g,... are distinct, since otherwise ig = jg (i j) g = 0. By the Fundamental Theorem of Finitely Generated Abelian Groups (FTFGAG), any finitely generated abelian group G has the form G Z r (Z/n 1 Z) (Z/n k Z) (1) where r 0 is unique but the n 1,, n k > 1 are not necessarily unique. Example 3 Z 6 Z 4 Z 2 Z 3 Z 4 Z 2 Z 12, since Z m Z n = Z nm, for m n by the Chinese remainder theorem. There are two ways to get uniqueness: 1. require that all the n i are prime powers 2. or require n 1 n 2 n 3 n k. We consider the second of these today. Lemma 4 Any subgroup K Z n satisfies K Z r for some r n. Note that unlike subspaces of a vector space, is it possible to have r = n and K Z n. For example, 2Z n Z n while is it still the case that 2Z n Z n. In this sense abelian groups are more interesting than vector spaces. 22-2

3 Now, since K Z n K Z r for some r n, pick a basis x 1,..., x r K so that K = c i x i c i Z i [r] and define so that L : Z r Z n ; e i x i G Z n /K = Z n /Image (L) = Z n /LZ r We can think of L as an r n matrix and each x i = j [n] a i,je i for i [r], where the a i,j are the matrix entries of L. We can extend L to Z n, i.e. L : Z n Z n by setting e i 0 for i > r (add zero columns ). So far we have seen how defining an abelian group G via generators and relations leads to an n n matrix L such that G Z n /LZ n, where n is the number of generators. The question that Smith normal form address is: given a group in this form, Z n /LZ n, how do we express it in the factored form (1)? Example 5 Let and L = ( ) G = g 1, g 2 / 2g 1 g 2 = 0 g 1 + 2g 2 = 0 ( G = Z / 1 2 ) Z

4 3 2 a 0, b 1 1 a 1, b 1 a b a 1, b 0 2 Figure 1: This figure illustrates G = Z 2 /LZ 2, where LZ 2 consists of the points (2a + b, 2b a) for a, b Z, as marked by the symbol on the grid. The remaining symbols represent elements in the respective equivalence classes. The points enclosed by the box represent the members of all equivalence classes, illustrating that G = 5. So G Z/5Z. Now, consider the kinds of changes to L that don t change the isomorphism type of G. One approach is to change the generators. Example 6 Write the group G of example (5) using different generators G = h 1, h 2 ; h 1 = g 1, h 2 = 3g 1 + g 2 In general, if H GL n (Z), where GL n (Z) is the set of n n matrices U with det U = ±1, so the inverse also is an integer matrix, then since UZ n = Z n In addition, if V GL n (Z), since V Z n = Z n G Z n /LZ n = UZ n /LZ n Z n /U 1 LZ n Z n /U 1 LZ n = Z n /U 1 L (V Z n ) = Z n / ( U 1 LV ) Z n Example 7 Write the group G of example (5) with different relations 2g 1 g 2 = 0 g 1 + 2g 2 = 0 2g 1 g 2 = 0 3g 1 g 2 =

5 Definition 8 An n n integer matrix S is in Smith Normal Form (SNF) if S is a diagonal matrix and uniqueness condition (2) is satisfied with the diagonal elements (S) i,i d i, i.e. d 1 d 2 d 3 d n ; d i 0, i [n] Note that some d i may be zero, since any integer divides zero. Theorem 9 An integer matrix L = (a i,j ) i,j [n] can be written as L = USV where S is in SNF and U, V GL n Z (invertible over the integers). Moreover, the non-zero d i on the diagonal of S are unique (note, gcd ( ) is non-negative by definition): d 1 = gcd (a i,j ) d 1 d 2 = gcd (a i,j a k,l a i,l a j,k ) (i.e. the 2 2 determinants). d 1 d k = gcd (all k k minors of L) d 1 d n = det L. with G = det L. In terms of the group G, if L has SNF S then in particular G = Z n /LZ n Z n /SZ n g 1,, g n / (d i g i = 0, i [n]) g 1 / (d 1 g 1 ) g n / (d n g n ) Z/ (d 1 Z) Z/ (d n Z) the rank of G is # {i d i = 0} if G is finite (all d i > 0) then G = d 1 d n = det L Note: row and column operations don t change the gcd ( ) result since, if n 1,..., n k 0 (noting that if (m, n) = 1 then c, c Z s.t. cm + c n = 1) gcd (n 1,..., n k ) = min d > 0 d = c i n i for some c 1,, c k Z i [n] so L = USV. 22-5

6 Example 10 Consider L = Z 3 /LZ 3 Z 4 Z 8 since d 1 = gcd (a i,j ) = 1 d 1 d 2 = gcd ( 8, 0, 8, 0, 4, 12, 8, 12, 4) = 4 d 1 d 2 d 3 = det L = 32 2 Commutative Monoids Definition 11 A monoid M is a set with an associative operation µ : M M M and an identity element e M (e, m) m, m M M is commutative if µ (m 1, m 2 ) = µ (m 2, m 1 ). We are interested in how to get a group from this object Example 12 Consider m = g / (10g = 6g) ; (k + 4) g = kg, g 6 = {e, g, 2g,..., 9g} and so, writing µ as addition 8g + 8g = 16g = 12g = 8g Definition 13 An ideal of a monoid M is a subset satisfying i.e. x I, m M we have m + x I. I M s.t. I + M I 22-6

7 Theorem 14 Let M be a finite commutative monoid and let J be the minimal ideal of M J = I Then Jis an Abelian group. ideals I In example (12) e.g. I = {8g} + M 8g + g = 9g 8g + 2g = 6g 8g + 3g = 7g 8g + 4g = 8g = {8g + m m M} I = {6g, 7g, 8g, 9g} and in the table below, the second last column is the identity, while the last column is cyclic of order 4, with 9g the generator 6g 7g 8g 9g 6g 8g 9g 6g 7g 7g 9g 6g 7g 8g 8g 6g 7g 8g 9g 9g 7g 8g 9g 6g 22-7

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