Counting Subrings of Z n and Related Questions

Transcription

1 Counting Subrings Advisor: Francisco Munoz Yale University 9/4/2015

2 Introduction Goal of Project Given a ring R which is a free Z-module of finite rank, define f R (k) = #{subrings (with identity) of index k of R}. Compute f R (k).

3 Introduction Goal of Project Given a ring R which is a free Z-module of finite rank, define f R (k) = #{subrings (with identity) of index k of R}. Compute f R (k). Example For R = Z, we have f R (k) = { 1 k = 1 0 otherwise. Useful fact (Liu 2006): If gcd(k 1, k 2 ) = 1, then f R (k 1 k 2 ) = f R (k 1 )f R (k 2 ). So, we need only consider the case k = p e for a prime p.

4 Subrings of R = Z n We have focused most of our analysis on the case R = Z n, building off of work by Liu There are two general approaches to determining f Z n(k), hereafter written as f n (k):

5 Subrings of R = Z n We have focused most of our analysis on the case R = Z n, building off of work by Liu There are two general approaches to determining f Z n(k), hereafter written as f n (k): Fix the value of n and vary the index k.

6 Subrings of R = Z n We have focused most of our analysis on the case R = Z n, building off of work by Liu There are two general approaches to determining f Z n(k), hereafter written as f n (k): Fix the value of n and vary the index k. Fix the index k and vary the choice of n.

7 Subrings of R = Z n We have focused most of our analysis on the case R = Z n, building off of work by Liu There are two general approaches to determining f Z n(k), hereafter written as f n (k): Fix the value of n and vary the index k. Fix the index k and vary the choice of n. Although the first approach is more natural, the second is easier.

8 Fixing k and varying n: Subring Matrices Proposition (Liu 2006) Let R be a ring which is a free Z-module of rank n. Then there is a bijection between subrings of R with index k = p e and n n matrices in the following form: a 11 a 12 a a 1n 0 a 22 a a 2n 0 0 a a 3n a nn 1 j > i, a ii > a ij 0 2 determinant = k ( a ii = p α i for some α i s) 3 the columns form a multiplicatively closed lattice.

9 A Discussion of Irreducible Subrings Definition A subring matrix A is irreducible if its last column is (1, 1,..., 1) T (the multiplicative identity) and all other entries are divisible by p

10 A Discussion of Irreducible Subrings Definition A subring matrix A is irreducible if its last column is (1, 1,..., 1) T (the multiplicative identity) and all other entries are divisible by p Every subring of Z n of prime power index can be expressed uniquely as a direct product of irreducibles

11 A Discussion of Irreducible Subrings Definition A subring matrix A is irreducible if its last column is (1, 1,..., 1) T (the multiplicative identity) and all other entries are divisible by p Every subring of Z n of prime power index can be expressed uniquely as a direct product of irreducibles Every irreducible subring has rank at least two and index at least p

12 A Discussion of Irreducible Subrings Definition A subring matrix A is irreducible if its last column is (1, 1,..., 1) T (the multiplicative identity) and all other entries are divisible by p Every subring of Z n of prime power index can be expressed uniquely as a direct product of irreducibles Every irreducible subring has rank at least two and index at least p Hence, we define g n (p e ) = #{irreducible subrings of Z n of index p e }. By computing g n (p e ), we can compute f n (p e ). Thus we calculated f n (p 6 ), while previously f n (p e ) was known only for e 5.

13 Derivation of f n (p e ) Theorem (Grunewald, Segal, and Smith 1988) For a fixed n and e, the function g n (p e ) is one of finitely many polynomials depending on p.

14 Derivation of f n (p e ) Theorem (Grunewald, Segal, and Smith 1988) For a fixed n and e, the function g n (p e ) is one of finitely many polynomials depending on p. Using this result, we fix e and bound the degree of g n (p e ); then we calculate the value at g n (p e ) + 1 points, and interpolate to obtain the desired polynomial. With the polynomials g n (p e ) in hand, we apply the following equations:

15 Derivation of f n (p e ) Theorem (Grunewald, Segal, and Smith 1988) For a fixed n and e, the function g n (p e ) is one of finitely many polynomials depending on p. Using this result, we fix e and bound the degree of g n (p e ); then we calculate the value at g n (p e ) + 1 points, and interpolate to obtain the desired polynomial. With the polynomials g n (p e ) in hand, we apply the following equations: 2e n f n (p e ) = f i (p e ) i f i (p e ) = λ Pi i=2 S λ l(k) k Kλ e m=1 g λm (p km ).

16 Computation of f n (p 6 ) Using this formula, we derived the following entirely new result: Proposition (AKMM 2015) f n(p 6 ) = n + (p 2 n + 4p + 1) + (p p 2 n + p + 16) (p p 3 + 2p 2 n + 81p + 41) + (p 4 + p p 2 n + 111p + 226) (21p p 2 n + 616p + 743) + (210p 2 n p ) n n n n (1260p ) As long as the polynomial is unique, we can theoretically compute f n (p e ) for any fixed e, given a sufficiently powerful computer.

17 Computation of f n (p 7 ) Proposition (AKMM 2015) f n(p 7 ) = n + (3p 2 n + 4p + 1) + (10p p 2 n + p + 19) (15p p p 2 n + 121p + 51) + 5 (p 6 + p p p p 2 n + 206p + 326) + 6 (p p p p 2 n p ) + 7 (28p p p 2 n p ) + 8 (378p p 2 n p ) + (3150p 2 n p ) n n n n (17325p )

18 Fixing n and varying e Generating Functions Define the following generating functions: A n (p, x) = B n (p, x) = f n (p e )x e e=0 g n (p e )x e. e=0 The functions A n, B n are known only for n 4 (Nakagawa 1996, Liu 2006). Our goal is to determine A 5 and B 5. A related objective is to classify irreducible matrices of certain types and count them in terms of known generating functions.

19 Classifying Irreducible Matrices Definition For an irreducible subring L, denote by m L the unique maximal ideal. Then let ρ L = dim Fp (m L /m 2 L ) 1.

20 Classifying Irreducible Matrices Definition For an irreducible subring L, denote by m L the unique maximal ideal. Then let ρ L = dim Fp (m L /m 2 L ) 1. Nakayama s Lemma Given an irreducible subring L of Z n, then 1 ρ L n 1.

21 Classifying Irreducible Matrices Definition For an irreducible subring L, denote by m L the unique maximal ideal. Then let ρ L = dim Fp (m L /m 2 L ) 1. Nakayama s Lemma Given an irreducible subring L of Z n, then 1 ρ L n 1. Definition Let π : Z n Z n 1 be given by π(a 1,..., a n ) (a 2,..., a n ).

22 Classifying Irreducible Matrices Definition For an irreducible subring L, denote by m L the unique maximal ideal. Then let ρ L = dim Fp (m L /m 2 L ) 1. Nakayama s Lemma Given an irreducible subring L of Z n, then 1 ρ L n 1. Definition Let π : Z n Z n 1 be given by π(a 1,..., a n ) (a 2,..., a n ). Idea for computing A 4 : when ρ L = 1 or ρ L = n 1 there are surjective maps to count these subrings. For all L Z 4, 1 ρ π(l) 2, so these are the only cases. For A 5, we must consider the additional case when ρ π(l) = 2.

23 The Case of Z[x]/x n Additively isomorphic to Z n via n 1 i=0 a ix i (a n 1, a n 2,..., a 0 ) T

24 The Case of Z[x]/x n Additively isomorphic to Z n via n 1 i=0 a ix i (a n 1, a n 2,..., a 0 ) T Looking for subring matrices in Hermite normal form with last column (0, 0,..., 0, 1) T

25 The Case of Z[x]/x n Additively isomorphic to Z n via n 1 i=0 a ix i (a n 1, a n 2,..., a 0 ) T Looking for subring matrices in Hermite normal form with last column (0, 0,..., 0, 1) T Letting the diagonal entries be a ii = p b i, we see the condition of multiplicative closure b i + b j b i+j.

26 The Case of Z[x]/x n Additively isomorphic to Z n via n 1 i=0 a ix i (a n 1, a n 2,..., a 0 ) T Looking for subring matrices in Hermite normal form with last column (0, 0,..., 0, 1) T Letting the diagonal entries be a ii = p b i, we see the condition of multiplicative closure b i + b j b i+j. All subrings are already irreducible - can t reduce the problem Proposition (AKMM 2015) For fixed e, and n 2e, f n (p e ) = f 2e (p e ). In particular, f n (p e ) is not a polynomial in n.

27 The Case of Z[x]/x n Additionally, f n (p 3 ) and f n (p 4 ) are quasi-polynomial depending on if p is even; it is unknown whether or not there are more classes of primes. Proposition (AKMM 2015) { 0 n < i Let n i = Then it holds that 1 n i. f n (p 2 ) = n 2 + n 3 p + n 4 p 2 f{ n (p 3 ) = n 2 + n 3 (p 2 + p) + n 4 (p 3 + p 2 ) + n 5 (2p 3 1) + n 6 (3p 4 1) p 2 n 2 + n 3 (p 2 + p) + n 4 (p 3 + p 2 ) + n 5 (p 4 ) + n 6 (p 5 + 2p 4 1) p = 2.

28 Acknowledgements Francisco Munoz Nathan Kaplan, Sam Payne, Yale Math Department Audience