DUAL REFLECTION GROUPS OF LOW ORDER KENT VASHAW. A Thesis Submitted to the Graduate Faculty of

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1 DUAL REFLECTION GROUPS OF LOW ORDER BY KENT VASHAW A Thesis Submitted to the Graduate Faculty of WAKE FOREST UNIVERSITY GRADUATE SCHOOL OF ARTS AND SCIENCES in Partial Fulfillment of the Requirements for the Degree of MASTER OF ARTS Mathematics May 2016 Winston-Salem, North Carolina Approved By: Ellen Kirkman, Ph.D., Advisor Jeremy Rouse, Ph.D., Chair W. Frank Moore, Ph.D.

2 Acknowledgments I would like to thank a number of people who helped make this thesis possible. First and foremost, my advisor Dr. Ellen Kirkman, who directed me in this research. She has always held me to the high standard I hope to hold myself to, while also having the patience to introduce me to noncommutative algebra. Her guidance, both with respect to this thesis and in the classes I took from her here, have been invaluable to my algebra education. I would also like to thank faculty at Wake Forest; in particular, Dr. Frank Moore and Dr. Jeremy Rouse for teaching me the Magma computer algebra system, which was critical to calculations performed in this thesis, as well as both agreeing to sit on my thesis committee. I would also like to thank Dr. Jason Gaddis for always having an open door to talk about anything algebraic, and for reading this thesis and providing comments. In addition, I would also like to thank Dr. Hugh Howards, Dr. Sarah Raynor, Dr. Stephen Robinson, and Dr. Sarah Mason, all of whom I had the pleasure of taking classes from, and who have made my experience at Wake Forest an overwhelmingly positive one. In the same vein, I would extend gratitude to my fellow graduate students here, many of whom I spent a lot of time studying and socializing with: math is a collaborative and communal experience, at its best, and I ve been very fortunate in my community here. ii

3 Table of Contents Acknowledgments Abstract ii v Chapter 1 Introduction Chapter 2 Context in Commutative Algebra Classical Invariant Theory and Shepherd-Todd-Chevalley The Poincaré Polynomial Chapter 3 Generalization to the Noncommutative Case Initial definitions Generalization of Shepherd-Todd-Chevalley A motivating example Process Chapter 4 Results Overview Wreath products The dihedral group of order 2p for p prime Quaternions Dihedral group of order The alternating group of order The dicyclic group of order Dihedral group of order The dicyclic group of order Central product of D 8 and Z SmallGroup(16,3) Non-trivial semi-direct product of Z 4 and Z M SD The direct product of D 8 and Z Direct product of Q 8 and Z The dicyclic group of order iii

4 4.18 The symmetric group of order Metacyclic groups of order pq The wreath product of Z 2 and S Chapter 5 Conclusion Bibliography Curriculum Vitae iv

5 Abstract Kent Vashaw We consider group-gradings of noncommutative algebras with the following question in mind: for what groups G does there exist an Artin-Schelter regular algebra A with G grading A such that A e, the identity-graded component of A, is itself Artin-Schelter regular? And in particular, restricting further, what if we require A to be a quadratic domain? Results exist regarding the Poincaré polynomial of a group with respect to an arbitrary generating set that grades an Artin-Schelter regular algebra in this way, which we use to narrow down possible grades of the generators of A. In this thesis we examine a number of groups of low order dihedral groups of order 8, 12, and 16; the quaternions; the alternating group of order 12; the dicyclic groups of order 12, 16, and 20; all other noncommutative groups of order 16; the symmetric group of order 24; the class of metacyclic groups; and the wreath product of Z 2 and S 3. v

6 Chapter 1: Introduction Suppose G is a finite group of n n matrices acting as graded automorphisms on a polynomial ring in n variables over a field of characteristic zero. Classical invariant theory is concerned with the elements of the polynomial ring that are fixed by all elements of G, the so-called invariant subring. One natural question is: for what groups G is the invariant subring again a polynomial ring? In this commutative context, the Shephard-Todd-Chevalley Theorem provides the answer, giving a full classification of such groups. This question has an analogue in the noncommutative case, where we have Hopf algebras acting on noncommutative Artin-Schelter regular algebras (a class of noncommutative rings that includes skew polynomial rings). We can ask, when is the invariant subalgebra of an Artin-Schelter regular algebra under a Hopf action itself Artin-Schelter regular? In particular, in this thesis we consider the Hopf action that corresponds to a grading by some group, and the invariant subalgebra corresponds to the identity-graded component. Throughout this thesis, we consider possible group-gradings of AS regular algebras such that the invariant subalgebra is itself AS regular using computational tools. This leads us to consider, more generally, certain properties of generating sets of groups that has application outside the context of noncommutative invariant theory. 1

7 Chapter 2: Context in Commutative Algebra 2.1 Classical Invariant Theory and Shepherd-Todd-Chevalley Some of the oldest problems in mathematics deal with solving, and more generally, understanding, polynomial equations. It is not suprising, then, that classically in commutative algebra polynomial rings have received much consideration. Intuitively, the idea of degrees of polynomials is very natural; the idea of grading is a formalization of this, and can be applied to more than just polynomial rings. Note that throughout this thesis, k = C. Definition 1. A semigroup is a set together with an associative binary operation. Groups are examples of semigroups. So are the natural numbers, under the operation of addition. Definition 2. A semigroup-graded ring is a ring R such that R = g G R g where G is a semigroup. Furthermore, we require that R g R h R gh. A polynomial ring is an example of an N-graded ring. If R = k[x, y] then R 1 contains all homogeneous polynomials of degree 1, and therefore is the set {λ 1 x+λ 2 y} for all λ i k; R 2 contains all homogeneous polynomials of degree 2, and therefore is the set {λ 1 x 2 + λ 2 xy + λ 3 y 2 } for all λ i k, et cetera. The condition R i R j = R i+j is satisfied, since multiplying a monomial of degree i by a monomial of degree j, gives a monomial of degree i + j; this then extends to homogeneous polynomials. During our discussion of commutative algebra, we will primarily discuss N-graded rings; when we move to noncommutative algebra, group-graded rings will be of primary interest. Once we impose a grading on a ring, we can check whether maps preserve the grading. 2

8 Definition 3. An automorphism σ of a graded ring R is called graded if σ(r) has the same degree as r for all r R. Any matrix M GL n (k) defines an automorphism σ of an N-graded ring generated in degree 1 by being defined on generators as σ(x i ) = M 1,i x 1 + M 2,i x M n,i x n. ( ) a b For example, in variables x and y, the matrix would send x to ax + cy and c d would send y to bx + dy. This way of representing automorphisms as matrices defines a group action for a matrix group on a polynomial ring: if G is a matrix group, g G, r k[x 1,..., x n ], then g.r = σ g (r) where σ g is the automorphism represented by the matrix g. Definition 4. Suppose G is a finite matrix group acting linearly on a ring R. The invariant subring of R under G, denoted R G, is defined as the set of r R for which g.r = r for all g G. The elementary symmetric polynomials in n variables, denoted e k (x 1,..., x n ) for 1 k n, are given by e k (x 1,..., x n ) = 1 j 1 <j 2 <...<j k n x j1 x jk. Theorem 2.1 (Gauss). Let A = k[x 1,..., x n ]. Let G = S n represented as permutation matrices. Then A G is a polynomial ring generated by the n elementary symmetric polynomials. For example, when n = 2, it is easy to check that x + y and xy are invariant under ( ) 0 1 by computing the group action, but it is perhaps marginally more difficult 1 0 to see that they generate the entire ring of invariants. 3

9 Definition 5. A reflection group is a finite matrix group generated by generators {g} with all but one eigenvalue of each g i equal to 1. Gauss Theorem above gives an example where the invariant subring of a polynomial ring under a group action is itself a polynomial ring. One natural question to ask is: when does this happen? [5], [25] provide an answer. Theorem 2.2 (Shephard-Todd-Chevalley). If G is a non-trivial finite group acting linearly on k[x 1,..., x n ], then k[x 1,..., x n ] G is a polynomial ring if and only if G is a reflection group. Note that R = k[x 1,..., x n ] is an infinite dimensional vector space over k: it has a basis of {1, x 1, x 2,..., x n, x 1 x 2, x 2 1, x 2 2,..., x 1 x 2,...}. But in any given degree, it is a finite dimensional vector space. For example, R 1 has basis {x 1, x 2,..., x n } over k. And R 2 has a basis of {x 2 1, x 2 2,..., x 2 n, x 1 x 2, x 1 x 3,..., x n 1 x n }. Definition 6. A graded algebra A is said to be of finite type if the dimension of A d over k is finite for all d. Definition 7. If A is a graded algebra of finite type, then the Hilbert series of A is the formal power series P (A, t) = dim k A d t d. 0 Definition 8. A N-graded algebra A is called connected graded if A 0 = k. Theorem 2.3 (Hilbert-Serre). Let R = k[f 1,..., f m ] be a nonnegatively connected graded finitely generated k-algebra with f 1,..., f m R a complete set of homogeneous algebra generators. Denote their respective degrees d 1,..., d m. Let M be a finitely generated graded R module. Then the Hilbert series of M has the form P (M, t) = p(t) (1 t d 1 ) (1 t d m) 4

10 (where we interpret this equality using the Taylor series of the right hand side) for some polynomial p(t) with integer coefficients. Lemma 1. Let 0 M M M 0 be a short exact sequence of graded vector spaces of finite type over a field k. Then P (M, t) + P (M, t) P (M, t) = 0. Corollary 1. If M and M are both nonnegatively graded vector spaces of finite type over the same field, then their direct sum M M is also a nonnegatively graded vector space of finite type of the same field with Hilbert series P (M M, t) = P (M, t) + P (M, t). 2.2 The Poincaré Polynomial Definition 9. Let S = {g 1,..., g k } be a generating set of a group G and g G. Then the length of g with respect to S, denoted n g, is defined as j where g = g i1 g i2...g ij is the shortest way to write g as a product of elements in S. We define the length of the identity element to be 0. Note that if we have a directed Cayley graph of G under a generating set S, then the shortest path from the identity element to g corresponds to the length of g with respect to S. Definition 10. Suppose G is a group and S a generating set of G. Then we define the Poincaré polynomial of G with respect to S to be p(t) = g G t ng m = {g : g has length n} t n n=0 5

11 where m is the highest length of any element of G with respect to S, and all lengths are calculated with respect to S. Example: Let G = D 8 be the dihedral group of order 8, presented as follows: G = {r, ρ r 2 = ρ 4 = e, rρ = ρ 3 r}, so ρ is a 90 degree rotation, and r a reflection. Suppose S = {r, rρ, rρ 2 }, and we wish to compute the Poincaré polynomial of G with respect to S. Then the identity has n g = 0 and our generators have n g = 1. By writing all words of two generators, we find all elements with n g = 2, and so on, giving us the following table: n g : g G : e r ρ rρ 3 rρ ρ 3 rρ 2 ρ 2 Therefore, the Poincaré polynomial of G under S is p(t) = t 3 + 3t 2 + 3t + 1. Note that this polynomial is the product of cyclotomic polynomials: in particular, it is equal to (1 + t) 3. Finding Poincaré polynomials that are the product of cyclotomic polynomials is of primary interest in certain computations in Chapter 4. Recall that the n th cyclotomic polynomial is the irreducible polynomial of the n th roots of unity, and is denoted Φ n. If n is a power of a prime p, then Φ n (1) = p, and otherwise Φ n (1) = 1. Note that if p(t) is a Poincaré polynomial of a group, then p(1) must be the order of the group. This gives us the following lemma. Lemma 2. Suppose p(t) is the Poincaré polynomial of a group G with respect to some generating set, and p(t) is the product of cyclotomic polynomials. If G = m, 6

12 which has prime factorization p 1...p l, then p(t) has factors Φ n1, Φ n2,..., Φ nl, where n i is some power of p i. Any other factors of p(t) must be of the form Φ k where k is not a power of any prime. One example to consider is the class of groups called Coxeter groups. Definition 11. Let S be a set. Then a matrix m : S S {1, 2,..., } is called a Coxeter matrix if it satisfies m(s, s ) = m(s, s), m(s, s ) = 1 s = s. Definition 12. Let m be a Coxeter matrix. Then m determines a unique group called a Coxeter group defined by having generators S and relations (ss ) m(s,s ) = e. For example, D 2n, the dihedral group of order 2n is the Coxeter group with generators r and rρ, where r corresponds to a reflection and ρ corresponds to a rotation of order n. It has Coxeter matrix ( ) 1 n. n 1 S n, the symmetric group of order n!, is also a Coxeter group with generators (12), (23), (34),..., ((n 1)n). It has Coxeter matrix The following result appears in [4]: 7

13 Theorem 2.4. The Poincaré polynomial of a Coxeter group, with respect to the Coxeter generators, is the product of cyclotomic polynomials. The study of the expansion of groups is a current subject of much research see [8] for a survey. In particular, the question is: for S a subset of a group G, with S k = {x 1 x 2...x k : x i S}, how does S k grow as k grows? We can observe that if S is a generating set, this is equivalent to asking how the coefficients of t k in the Poincaré polynomial grow as k grows. Another way of framing this discussion is in terms of the diameter of the Cayley graph of G with respect to some set of generators. There is a distinction between a directed Cayley graph and an undirected Cayley graph: a directed Cayley graph uses a specific set of generators S = {s 1,..., s n }, while the undirected Cayley graph uses S and all inverses of elements in S (or, in other words, uses a generating set such that if s S, s 1 S). Getting a bound on either diameter for a group G gives us a bound on the degree of the Poincaré polynomials of G. Suppose p and q are primes, p 1 mod 1, i 1, i q 1 mod p. Then define the metacyclic group of order pq as G = {a, b : a p = b q = e, ba = a i b}. Note then that G is the unique nonabelian group of order pq. Theorem 2.5 ([3]). Suppose G is a metacyclic group of order pq for p, q primes. Then the diameter of G under any set of generators is less than 3p. Much of the current research in growth of groups is in finding asymptotic bounds [2], [9], [11]. In general, to find bounds on the degree of all Poincaré polynomials of G, it is enough to compute the Poincaré polynomials of all minimal generating sets of G, 8

14 since the Poincaré polynomial of any generating set S has degree less than or equal to the degree of the Poincaré polynomial of any subset of S. Theorem 2.6 ([7]). Let m(g) be the maximum size of a minimal generating set of G. Then m(g) is bounded by the length of the longest chain of subgroups in G. Furthermore, [7] tells us that with respect to the symmetric group, m(s d ) 2d 1 p. p 2d p primes For a metacyclic group G as defined as the unique nonabelian group of order pq for p, q primes such that q divides p 1, we have that m(g) = 2 [7]. Definition 13. A polynomial p(t) = a m t m + a m 1 t m a 1 t + a 0, with a i k, is called unimodal if there exists an l N such that a m a m 1... a l+1 a l a l 1... a 0. For example, the Poincaré polynomial computed above for D 8, p(t) = t 3 + 3t 2 + 3t + 1, is unimodal: 3 is the least natural number such that a l < a l 1, and there are no higher terms. The polynomial p(t) = 3t 2 + t + 3, on the other hand, is not, since 1 is the lowest natural number with a l < a l 1, but a 2 a 1. Computationally, to arrive at the results presented in this thesis, a number of Poincaré polynomials are calculated. Every Poincaré polynomial we calculated was unimodal, which, in some cases, meant that for some groups every minimal generating set has a unimodal Poincaré polynomial. From our observations, then, we ask the following question: Question 1. For any group, under any generating set, is the Poincaré polynomial unimodal? 9

15 If this question had a positive answer, it would be applicable to our treatment of metacyclic groups of order pq later in the text. 10

16 Chapter 3: Generalization to the Noncommutative Case 3.1 Initial definitions Artin and Schelter defined a class of algebras by considering connected graded algebras which share many of the same homological and algebraic properties of the commutative polynomial ring. Recall that a N graded algebra A is called connected graded if A 0 = k. Definition 14. Let A be a connected graded algebra. The Gelfand-Kirillov dimension (or GK dimension for short) of A is defined to be GKdim A = lim sup n log( n i=0 dim A i). (3.1) log n Definition 15. Let A be a connected graded algebra. We call A Artin-Schelter regular (or AS regular) of dimension d if the following conditions hold: (a) A has finite global dimension, (b) Ext i A( A k, A A) = Ext i A(k A, A A ) = 0 for all i d, (c) Ext d A( A k, A A) = Ext d A(k A, A A ) = k( l) for some l (i.e. the grading is shifted by l), and (d) A has finite Gelfand-Kirillov dimension. A commutative algebra is AS regular if and only if it is isomorphic to a polynomial ring. The Hilbert series of a quadratic AS regular algebra has the form 1 p(t) where p(t) is the product of cyclotomic polynomials [27]. Here we will consider only AS regular algebras that have quadratic relations and are Noetherian domains meaning that they satisfy the ascending chain condition and have no zero divisors and therefore are finitely generated. 11

17 Definition 16. Let R be a ring and σ an automorphism of R. Then we define the Ore extension R[x; σ] to have elements of the form n i=1 r ix i where r i is in R, with the relation that xr = σ(r)x, i.e. R[x; σ] = R x xr σ(r)x. If R is an Ore extension, then R[x; σ] is called an iterated Ore extension. Recall the Hilbert series defined in Chapter 2. If A = k[x 1,..., x n ] is the commutative polynomial ring with degree x i = 1, the Hilbert series of A is H A (t) = 1 (1 t) n. This is also the Hilbert series of an iterated Ore extension A = k[x 1 ; σ 1 ][x 2 ; σ 2 ][... ][x n ; σ n ] when each x i has degree 1. If R is an AS regular algebra and a Noetherian domain, then the Ore extension R[x; σ] is also an AS regular Noetherian domain, and are one of the primary examples of AS regular algebras used in this thesis [1]. One example of an Ore extension is the skew polynomial ring k q [x, y] for some q k. k q [x, y] = k x, y yx qxy. In k q [x, y], every element is a polynomial in x and y with coefficients in k and can be uniquely written as a k linear combination of monomials of the form x i y j for i, j N; and yx = qxy. If q = 1, k q [x, y] is a commutative ring, since yx = xy. One tool we use later in this thesis is the Koszul dual [21], [23], [26]. 12

18 Definition 17. The Koszul dual of a quadratic algebra A with generators x 1,..., x n, denoted A!, is the algebra whose relations are generated by the subspace of k X 1,..., X n, where X i is the characteristic function of x i made up of all operators which vanish on the relations of A. Theorem 3.1 ([26]). Suppose A is a quadratic AS regular algebra. Then Ext A(k, k) = A!. Theorem 3.2. If A has finite global dimension, Ext A(k, k) is a finite dimensional vector space. Definition 18. A finite dimensional k-algebra R is Frobenius if, as left R-modules, R = R, where R = Hom k (R, k) is given the left module structure induced by the right regular action of R on itself. Definition 19. Suppose R is a finite dimensional k-algebra. Then if (, ) : R R k such that (ab, c) = (a, bc) for all a, b, c R is a non-degenerate bilinear pairing, then we call (, ) a Frobenius pairing. Theorem 3.3 ([26]). Let R be a finite dimensional, connected graded k-algebra such that R n 0 but R n+1 = 0. The following are equivalent: R is Frobenius; dim k (R n ) = 1 and the map (, ) : R R k, defined by (a, b) = the component of ab in R n, is a Frobenius pairing; R = R [ n] as graded left R-modules. Theorem 3.4 ([26]). A is AS regular if and only if A! is Frobenius. 13

19 One assumption that we make is that the algebras we work with are Koszul. Theorem 3.5 ([26]). A quadratic Noetherian AS regular algebra with n generators in degree 1 has the Hilbert series 1 (1 t) n, then A is Koszul. 3.2 Generalization of Shepherd-Todd-Chevalley In the context of noncommutative algebra, one natural question that arises is: what results from commutative algebra generalize to the noncommutative case? In particular, this question, when applied to the Shepherd-Todd-Chevalley Theorem, forms the main motivation for this thesis work. Recall that the Shepherd-Todd-Chevalley Theorem gives a full classification of groups acting on polynomial rings with the invariant subring also a polynomial ring. AS regular algebras are, in some sense, the proper noncommutative analogue to polynomial rings, and, for the noncommutative case, we replace group actions with Hopf algebra actions. So our question becomes: which Hopf algebras act on an Artin-Schelter regular algebra such that the invariant subalgebra is itself AS regular? In particular, in this thesis, we have considered the case where the Hopf action in question corresponds (by considering the coaction) to a grading of the AS regular algebra by some group G, and the invariant subalgebra corresponds to the identity-graded component. Recall that a group-graded ring is similar to an N-graded ring except that homogeneous elements are put in R g where g is an element of the group G. R = g G R g, R g R h = 0 for all g, h G not equal, and R g R h R gh for all g, h G. If r R is in R g for some g G, then we often say that r is graded by g. In this thesis, we consider the question: which groups G grade some AS regular algebra A such that the identity component is also 14

20 AS regular? Definition 20. If A is an AS regular algebra graded by a group G with A e also AS regular, then we call G a dual reflection group for A. Assume that A is an AS regular, and graded by a finite group G. Then assume B = A e is also AS regular. The following are consequences of this assumption from [19]. A is a graded free module over B. A decomposes as A g, g G and each A g is a B module. Since each A g is a direct summand of a free B- module, each A g is a projective B module. In fact, each A g is a free B-module. Each A g is a rank one free B module. The generators of A have the following properties: There can be no more than one generator of grade g for all g e. The grades of the generators must generate the group (not necessarily minimally). There can be any number of generators in grade e. Suppose A 1 = kx 1 kx 2... kx l ky 1... ky m where x i is in grade g i for some g i not equal to the identity and each y i has grade e. Then by the above we know g 1,..., g l are distinct and l G 1, and {g 1,..., g l } generates the group G. However, m is arbitrary. 15

21 Consider A g for some g 1. Since A g is rank one, A g = u g B and u g = x j1 x j2...x jng where g = g j1 g j2...g jng and n g is the length of g. Recall that an AS regular algebra A, generated in degree 1, with quadratic relations, has Hilbert series H A (t) = with n the number of generators. Let 1 (1 t) n p(t) = 1 + g e t ng be the corresponding Poincaré polynomial in k[t]. From the decomposition of A over B, H A (t) = p(t)h B (t) so that H B (t) = 1 (1 t) n p(t). This last fact gives us the following lemma. Lemma 3. If G is a dual reflection group of A with a set S grading the degree 1 generators of A, then the Poincaré polynomial of G with respect to S must be the product of cyclotomic polynomials. 3.3 A motivating example The following case serves as a first example of a dual reflection group. Let R = k 1 [u, w]. Recall that R is defined as k u, w uw + wu. 16

22 Let σ = ( ) 0 1. Then let A = R[v; σ]. Since A is an Ore extension of R, and R is a 1 0 skew polynomial ring, A is AS regular. A is defined by the following relations: wu = uw, vu = wv, vw = uv. These are all homogeneous relations with respect to u, w, v so this is an N graded ring. Let D 8 be the dihedral group of order 8, which will be represented using ρ a rotation of 90 degrees and r a reflection so that D 8 = {r i ρ j : 0 i 1, 0 j 3, r 2 = ρ 4 = 1, ρr = ρ 3 r}. We claim that A can be graded by D 8 by letting u A r ; w A rρ 2 and v A rρ. We first check that the relations of A remain homogeneous under this grading: wu A rρ 2 r = A ρ 2. uw A rrρ 2 = A ρ 2. vu A rρr = A ρ 3. wv A rρ 2 rρ = A ρ 3. vw A rρrρ 2 = A ρ. uv A rrρ = A ρ. We claim that A can be also graded by D 8 under the degrees : u A r, w A rρ 2, v A rρ 3. We first check that the relations of A are also homogeneous under this grading: wu A rρ 2 r = A ρ 2. 17

23 uw A rrρ 2 = A ρ 2. vu A rρ 3 r = A ρ. wv A rρ 2 rρ 3 = A ρ 3 = A ρ. vw A rρ 3 rρ 2 = A ρ 3. uv A rrρ 3 = A ρ 3 = A ρ 3. Now, under each grading of A by D 8, we claim that A e = k[u 2, w 2, v 2 ]. The first grading: First, we can note that u 2, w 2, and v 2 commute and are algebraically independent; therefore, they generate the commutative polynomial ring k[u 2, w 2, v 2 ]. Since v A rρ, v 2 A rρrρ = A e. Since u A r, u 2 A r 2 = A e. Since w A rρ 2, w 2 A rρ 2 rρ 2 = A e. Note that every element in A can be written uniquely as λu i w j v k. Suppose a monomial λu i w j v k is in the identity component. Then r i (rρ 2 ) j (rρ) k = e. But since reflections have order 2, this holds if i, j, k are even. If any of i, j, k are odd, this is either r(rρ 2 )rρ = rρ 3 or r(rρ 2 ) = ρ 2 or rrρ = ρ or rρ 2 rρ = ρ 3. None of these are the identity. So i, j, k are all even. Therefore, λu i w j v k is in k[u 2, v 2, w 2 ]. Therefore A e = k[u 2, v 2, w 2 ]. 18

24 Under the second grading, a similar argument shows that A e = k[u 2, v 2, w 2 ]. Therefore, under both of these gradings, since the identity component is a commutative polynomial ring, it is AS regular. Hence D 8 is a dual reflection group for A = k[u, w][v; σ]. 3.4 Process Computationally, we approach our problem as follows: Consider a group G. Our goal is to check whether G is a dual reflection group for an AS regular algebra A. For the purposes of this thesis, we focus on quadratic AS regular algebras that are Noetherian domains, and that no generators of A have grade e (this assumption is only necessary for a few of the groups we consider). We also assume that A is Koszul. This class of algebras include quadratic Ore extensions. Our standard first step is to find a bound n on the total possible degree of a Poincaré polynomial associated to G. We compute the minimal generating sets of G (up to automorphism) and then compute the Poincaré polynomials for these minimal sets. Any generating set must contain a minimal generating set, and the bound on the Poincaré polyomial associated to a generating set is less than or equal to the Poincaré polynomial associated to any generating set contained in the original generating set. By adding generators, we can only reduce the bound on the degree of the Poincaré polynomial. We find the generating sets of G that generate G with a Poincaré polynomial that is a product of cyclotomic polynomials. To do this, we calculate what possible products of cyclotomic polynomials have degree less than or equal to n. Recall that if the order of G has prime factors p 1,..., p l, then a Poincaré 19

25 polynomial of G that is the product of cyclotomic polynomials must have factors Φ n1,..., Φ nl, where n i is some power of p i. We then find all generating sets that have one of the possibly cyclotomic Poincaré polynomials. For groups of low order this is sometimes computationally easy. However, for groups of higher order, we use the computer algebra system Magma to simply calculate all generating sets and then match the Poincaré polynomial of each generating set against the set of possible cyclotomic Poincaré polynomials. Then for each generating set matching one of the possible cyclotomic Poincaré polynomials, we suppose we have an algebra A graded by G, and compute what relations A must satisfy, using the fact that each A g must be free of rank 1 over A e. Often, the relations on A give us a contradiction either showing that A is not AS regular, or that A is not a domain, or that A is not quadratic. 20

26 Chapter 4: Results 4.1 Overview The table below summarizes some of the relevant data, followed by more discussion of each case. As noted in Chapter 2, the upper bound on the degree of the Poincaré polynomials corresponds to an upper bound on the diameter of the directed Cayley graph of G, given any generating set. Group # of cyclotomic bound shown that ating sets are Upper Have we Minimal gener- Poincaré on degree of reflection G is a dual polynomials p(t) group unimodal Q no yes D no yes A no yes Dic no yes D no yes Dic no yes D 8 Z no yes Smallgroup(16,3) 3 6 yes yes Z 4 Z yes yes M yes yes SD yes yes D 8 Z yes yes Q 8 Z no yes Dic no yes S no yes Z 2 S no unknown Metacyclic varies unknown no unknown groups with subgroups of prime orders p and q D 2p (p prime) varies p no unknown 21

27 4.2 Wreath products The following is a generalization of our initial example of D 8 grading k 1 [x 1, x 2 ][x 3 ; σ] above to an infinite family of wreath product groups. Definition 21. Suppose some group H acts on a set Ω. Then the wreath product of H and G, denoted G H, is defined as Π ω Ω G H, defined by the action of H on Π ω Ω G with h.(a ω ) := (a h 1 ).ω for h H, a G in other words, if the set Ω is a list of natural numbers from 1 to n, then G H = {((g 1, g 2,..., g n ), h) : g i G, h H} defined by the operation ((g 1,1, g 2,1,..., g n,1 ), h 1 )((g 1,2, g 2,2,..., g n,2 ), h 2 ) = ((g 1 g h 1 1.1,2, g 2,1g h 1 1.2,2...g n,1g h 1.n,2), h 1 h 2 ). So we ve extended an action of H on the set Ω to an action of H on Π ω Ω G and then formed a semi-direct product. Now, in particular, let H be some cyclic subgroup of S n generated by a cycle of order n. This has a natural action on the set {1,..., n}. Therefore, for any group G we can form G H as Π i n G H. In particular, assume G is cyclic say Z m. Note that D 8 = Z2 (12), shown by considering the following map φ : D 8 Z 2 (12), φ(r) = ((1, 0), (e)), φ(ρ) = ((1, 0), (12)) where r denotes a reflection and ρ denotes a 90 degree rotation. Note that ((1, 0), (e))((1, 0), (12)) = ((0, 1), (12))((1, 0), (e)) = ((1, 0), (12)) 1 ((1, 0), (e)). 22

28 Therefore, we can see that this is an isomorphism, since ((1, 0), (e)) and ((1, 0), (12)) together generate Z 2 (12), and φ(r i ρ j r k ρ l ) = φ(r i+k ρ l+( 1)kj ) = ((1, 0), (e)) i+k ((1, 0), (12)) l+( 1)k j = ((1, 0), (e)) i ((1, 0), (12)) j ((1, 0), (e)) k ((1, 0), (12)) l = φ(r i ρ j )φ(r k ρ l ) Then we claim, in general, G H grades A = k q [x 1,..., x n ][x n+1 ; σ] where q = 1 or 1 and σ(x i ) = x h 1.i with A e AS regular. A is AS regular since it is an Ore extension. The grading we claim works is defined as follows: x n+1 is graded by ((0, 0,..., 0), h), and for all i < n + 1, x i graded by ((0,..., 0, 1, 0,..., 0), (e)) where the 1 appears in the i th entry. This is a grading, and to see this we just need to check relations. First, for i, j < n + 1, x i x j graded by the same thing as x j x i, since both are graded by ((0,.., 0, 1, 0,..., 1, 0,...), (e)) where the 1 s are in the i and j entries. Then note that for any i < n + 1, x n+1 x i is in the same grade as x h 1.ix n+1, by the definition of the wreath product. Therefore, this is a well-defined grading. Then it can be shown that the identity component is A e = k r [x m 1,..., x m n ][x n n+1] where r is -1 if q is -1 and m odd, and 1 otherwise. Clearly, x m 1,..., x m n, x p n+1 are all elements of A e, and they all either commute or skew-commute. Now take an arbitrary element of A, x r 1 1 x r x r n+1 n+1. Suppose that this element is graded by the identity. Then clearly r n+1 0 mod n, and for i < n + 1, r i 0 mod m. Therefore, this element is in k r [x m 1,..., x m n ][x p n+1]. Since A e is a (skew-)commutative polynomial ring, it is AS regular. Therefore, G H is a dual reflection group. 23

29 4.3 The dihedral group of order 2p for p prime Let G = D 2n be the dihedral group of order 2n given by D 2n = r, ρ : r 2 = e, ρ n = e, ρr = rρ n 1. In [20], the authors show that D 2p is not a dual reflection group for p 3 if A has quadratic relations. Suppose A is graded by G where A is an AS regular domain. Suppose A e is AS regular. We know that the grades of the generators of A must generate G. Therefore, without loss of generality, we can assume that we have either {r, ρ} or {r, rρ} as grades of two of our generators of A. First we examine the former case. Consider the rotation ρ i. ρ i = rρ n i r, so it has length either i or n i+2; however, if i > n/2 + 1, then n i + 2 < n/ In other words, if i > n then write ρ i as rρ n i r, and otherwise write ρ i just as itself. Therefore, for rotations, the least number of generators required is n For a reflection rρ i = ρ n i r, these take i + 1 and n i + 1 generators to write, respectively. Note that if i > n, or in other words if i+1 > n +1, then n i+1 < n Therefore, the maximum number of generators needed to write any reflection is n Now look at the latter case, using the generating set rρ and r. Note that rρ is its own inverse. We know that the minimal words we can form using rρ and r are just alternating them r(rρ)r(rρ)... because if they weren t alternating, then two r s would cancel each other out, and two (rρ) s would also cancel. So we only write these alternating words. Consider the rotation ρ i. ρ i = (rrρ) i = (rρr) n i. If i n 2 then we represent it by (r(rρ)) i, and if i > n we represent it by 2 ((rρ)r)n i note that if i > n, aka 2i > n, 2 then 2n 2i < n. Therefore, the least number of generators needed to generate ρ i is n. 24

30 Now take the reflection rρ i. rρ i = r(rrρ) i = r(rρr) n i. This gives us either 2i 1 or 2n 2i+1 generators, respectively. If i > n+1, aka 2i 1 > n, then 2n 2i+1 < n. 2 Therefore, the least number of generators we need is n. Therefore, in these two cases, the bounds on the degree of the Poincaré polynomial are n and n, respectively. As an example, we consider D 6, which is also S 3. Then the Poincaré polynomial is a polynomial of degree less than or equal to 2 in the former case. However, we know that p(1) must be equal to 2n, which is equal to 6; therefore, it is the product of cyclotomic polynomials Φ n1 Φ nk, where one of the n i is a power of 3; however, the third cyclotomic polynomial is degree 2, and all powers of 3 have cyclotomic polynomials of degree greater than or equal to 2; but since this must be multiplied by some power of 2 s cyclotomic polynomial, the degree is strictly greater than 2; therefore, there are no possible Poincaré polynomials the product of cyclotomic polynomials. In the latter case, we re using r and rρ as our generating set. Since p(1) = 6 we get that p(t) = Φ 2 (t) Φ 3 (t). Therefore, p(t) = (1 + t)(1 + t + t 2 ) = 1 + 2t + 2t 2 + t 3. Let x 1 be the generator of A of degree r and x 2 be the generator of degree rρ. From p(t) we know that there are only two generators of A with grade e. Now note that both x 2 1 and x 2 2 are graded by the identity, and x 1 x 2 A ρ and x 2 x 1 A ρ 2. This means that x 1 x 2 x 1 and x 2 x 1 x 2 are both in A rρ 2. Since A rρ consists of all multiples of some generating element as per p(t), this means that x 1 x 2 x 1 = λx 2 x 1 x 2. But since x 1 x 2 and x 2 x 1 are not graded by the same element, the relation x 1 x 2 x 1 = λx 2 x 1 x 2 is not a consequence of quadratic relations. This contradicts our assumption that A is quadratic. 25

31 4.4 Quaternions Let Q be the quaternion group Q = {±1, ±i, ±j, ±k} with operation ij = k, jk = i, ki = j, with i, j, k all skew commuting. Now suppose A, an AS regular quadratic domain, is graded by Q. Q is minimally generated by either {i, j} or {i, j} without loss of generality. Therefore, A must have at least two generators, say x 1 graded by i and x 2 graded by ±j. In either case, it can be shown that the bound on p(t) is 3. p(t) must be the product of three cyclotomic polynomials: the n th 1, n th 2, n th 3 cyclotomic polynomial where n i is 2 raised to some power. However, since the bound on p(t) is 3, each cyclotomic polynomial must be at most degree 1. Therefore, p(t) is the second cyclotomic polynomial raised to the third power, ie p(t) = (1 + t) 3 = 1 + 3t + 3t 2 + t 3. Therefore, A has three generators not of grade e. If the third generator x 3 is graded by k or -1 then any element of Q can be written as the product of two of the three grades of generators. This is a contradiction, since there must be one element g of Q with n g = 3. Therefore, x 3 must be graded by i or j. Without loss of generality, let s say i. Then p(t) is the Poincaré polynomial in this case, since ij = k, ji = k, ii = 1, 26

32 j 3 = j. Or, in other words, we have 3 generators, n g for ±k and 1 is 2, and n g for j is 3. In [20], the authors show that Q is not a dual reflection group for any AS regular algebra, even though this generating set has a Poincaré polynomial that is the product of cyclotomic polynomials. 4.5 Dihedral group of order 12 Let G be D 12, the dihedral group of order 12, presented as G = r, ρ r 2 = ρ 6 = e, ρr = rρ 5. Suppose A is graded by G with A e AS regular. Then note that p(1) = 12 = 2 2 3, so p(t) has some factors Φ n1 (t), Φ n2 (t), and Φ n3 (t), where n 1 and n 2 are both powers of 2 and n 3 is a power of 3. First, we note that no matter what generating set we pick for G, n g is going to be bounded above by 6 therefore, p(t) is must be a degree 6 or less polynomial. Our generating set must have at least one reflection. Without loss of generality, let that reflection be r. Then possible minimal generating sets are: {r, ρ}, {r, ρ 2, ρ 3 }, {r, ρ 3, ρ 4 }, {r, ρ 5 }, {r, rρ}, {r, rρ 2, rρ 3 }, {r, rρ 3, rρ 4 }, {r, rρ 5 }. In each case, n g is bounded above by 6 by computation. Since p(t) has three cyclotomic polynomials Φ n1, Φ n2, and Φ n3 as factors where n 1 and n 2 are powers of 2 and n 3 is a power of 3, and our bound on the degree of p(t) is 6, we know that Φ n1, Φ n2, and Φ n3 must all have degree less than or equal to 4. The only n 3, a power of 3, such that Φ n3 has degree less than or equal to 4 is n 3 = 3. We find, then, that n 1 and n 2 can be either 2 or 4, giving us three possibilities for p(t): p 1 (t) = (t + 1)(t + 1)(t 2 + t + 1) = t 4 + 3t 3 + 4t 2 + 3t + 1, p 2 (t) = (t + 1)(t 2 + 1)(t 2 + t + 1) = t 5 + 2t 4 + 3t 3 + 3t 2 + 2t + 1, 27

33 p 3 (t) = (t 2 + 1)(t 2 + 1)(t 2 + t + 1) = t 6 + t 5 + 3t 4 + 2t 3 + 3t 2 + t + 1. We can then have Φ n multiplied by any of these three polynomials for n not a power of a prime, keeping the degree of p(t) less than 6. Therefore, we also have the following possibility: p 4 (t) = p 1 (t) Φ 6 (t) = t 6 + 2t 5 + 2t 4 + 2t 3 + 2t 2 + 2t + 1, We can reject p 3 (t), as it would imply only one non-identity graded generator of A; but D 6 requires at least two generators. This leaves p 1 (t), p 2 (t), and p 4 (t). By checking possible minimal generating sets, it can be shown that p 2 (t) is also not possible. However, there are generating sets that generate G in accordance with p 4 (t). Without loss of generality, there is just one, and we also find four generating sets matching p 1 (t). We consider these cases in order. Case 1: the grades of the non-identity graded generators are {r, rρ}. Let x 1 be an element of A graded by r, x 2 be an element of A graded by rρ. Then the following table, generated by letting the i, j entry of the table to be the grade of x i multiplied by the grade of x j, calculates the grades of degree 2 elements of A: x 1 x 2 x 1 e ρ x 2 ρ 5 e Therefore, there are no quadratic relations involving x 1 x 2 and x 2 x 1, since these elements are graded by different elements of G, which are also different from the grades of x 2 1 and x 2 2. However, note that x 1 x 2 x 1 x 2 x 1 x 2 and x 2 x 1 x 2 x 1 x 2 x 1 are both graded by ρ 3. Therefore, we get a new relation, not a consequence of our quadratic relations. Since we re assuming that our minimal relations must be quadratic, this is a contradiction. 28

34 Case 2: the grades of the non-identity graded generators are {r, ρ, ρ 5 }: Let x 1, x 2, x 3 A such that x 1 A r, x 2 A ρ, x 3 A ρ 5. Then the following table represents all degree 2 grades for A: x 1 x 2 x 3 x 1 e rρ rρ 5 x 2 rρ 5 ρ 2 e x 3 rρ e ρ 4 From this, we see that the only quadratic relations we obtain for our algebra are the following, with λ i k: x 1 x 2 = λ 1 x 3 x 1, x 1 x 3 = λ 2 x 2 x 1. Now note that both x 3 2 and x 3 3 are in A ρ 3. Since A ρ 3 is generated by a single element, this implies that x 3 2 = λ 3 x 3 3. However, this relation is not a consequence of our quadratic relations therefore, this is a new relation. So if we assume that A only has quadratic relations, this gives us a contradiction. Case 3: the grades of the non-identity graded generators are {r, rρ, rρ 2 }: Let x 1, x 2, x 3 A such that x 1 A r, x 2 A rρ, x 3 A rρ 2. Then the following table represents all degree 2 grades for A: x 1 x 2 x 3 x 1 e ρ ρ 2 x 2 ρ 5 e ρ x 3 ρ 4 ρ 5 e From this, we see that the only quadratic relations we obtain for our algebra are the following, with λ i k: x 1 x 2 = λ 1 x 2 x 3, 29

35 x 2 x 1 = λ 2 x 3 x 2. Now note that both x 3 x 1 x 3 and x 1 x 3 x 1 are in A rρ 4. Therefore, x 3 x 1 x 3 = λ 3 x 1 x 3 x 1 for some λ 3 k. However, this relation is not a consequence of our quadratic relations therefore, this is a new relation. So if we assume that A only has quadratic relations, this gives us a contradiction. Case 4: the grades of the non-identity graded generators are {r, rρ, rρ 5 }: Let x 1, x 2, x 3 A such that x 1 A r, x 2 A rρ, x 3 A rρ 5. Then the following table calculates the grades of all degree 2 elements of A: x 1 x 2 x 3 x 1 e ρ ρ 5 x 2 ρ 5 e ρ 4 x 3 ρ ρ 2 e From this, we see that the only quadratic relations we obtain for our algebra are the following, with λ i k: x 1 x 3 = λ 1 x 2 x 1, x 1 x 2 = λ 2 x 3 x 2. Now note that both x 2 x 3 x 2 and x 3 x 2 x 3 are in A rρ 3. Therefore, x 2 x 3 x 2 = λ 3 x 3 x 2 x 3 for some λ 3 k. However, this relation is not a consequence of our quadratic relations therefore, this is a new relation, and therefore a contradiction. Case 5: the grades of the non-identity graded generators are {r, rρ 4, rρ 5 }: Let x 1, x 2, x 3 A such that x 1 A r, x 2 A rρ 4, x 3 A rρ 5. Then the following table calculates degree 2 grades for A: 30

36 x 1 x 2 x 3 x 1 e ρ 4 ρ 5 x 2 ρ 2 e ρ x 3 ρ ρ 5 e From this, we see that the only quadratic relations we obtain for our algebra are the following, for some λ i k: x 1 x 3 = λ 1 x 3 x 2, x 2 x 3 = λ 2 x 3 x 1. Now note that both x 1 x 2 x 1 and x 2 x 1 x 2 are in A rρ 2. Therefore, x 1 x 2 x 1 = λ 3 x 2 x 1 x 2 for some λ 3 k. However, this relation is not a consequence of our quadratic relations therefore, this is a new relation, a contradiction Therefore G is a not a dual reflection group for any AS regular Koszul Noetherian domain. 4.6 The alternating group of order 12 Suppose some AS regular algebra A is graded by A 4 such that A e is AS regular with A having quadratic relations. We will use the presentation of A 4 : A 4 = a, b a 2 = b 3 = (ab) 3 = 1. Therefore, by this presentation, the elements of A 4 are: {1, a, b, ab, ba, b 2, aba, ab 2, bab, b 2 a, bab 2, b 2 ab}. 31

37 Some of the relations in this presentation may not be immediately intuitive, so we write out a few of the ways which we reduce words in a and b: abab = b 2 a ab 2 a = bab baba = ab 2 bab 2 a = b 2 ab b 2 aba = bab 2 b 2 ab 2 = aba In this presentation, b corresponds to a 3-cycle. Without loss of generality, to generate A 4 we need at least one 3-cycle. Let that 3-cycle be b, and let s let b be the three-cycle (123). Adding any element not equal to b 2 as a generator ends up generating the group. Then to get an initial upper bound on p(t), we just compute n g under all these minimal generating sets, and find n g is bounded by 4. Therefore, p(t) = Φ 2 (t) Φ 2 (t) Φ 3 (t) = t 4 + 3t 3 + 4t 2 + 3t + 1. Then by computation we find the following generating sets (with additional help from [24], [6]) that match this Poincaré polynomial: Case 1: generating set {b, ab, ba}: under this set, we get quadratic relations λ 1 x 2 1 = x 3 x 2, λ 2 x 2 2 = x 1 x 3, λ 3 x 2 3 = x 2 x 1, λ 4 x 1 x 2 = x 2 x 3, λ 5 x 1 x 2 = x 3 x 1, 32

38 for some λ i k, which leads to zero divisors: ( λ4 λ 2 x 1 + x 2 )( λ 2 λ 4 x 2 + x 3 ) = λ 4 x 1 x 2 + λ 2 λ 4 x 2 2 = λ 4 x 1 x 2 + λ 2 λ 4 x 2 2 = 0 λ4 λ 2 x 1 x 3 + x 2 x 3 λ4 λ 2 λ 2 x λ 4 x 1 x 2 Case 2: generating set {b, ab, aba}: similar to the first case, this case leads to zero divisors. Case 3: generating set {b, ba, aba}: this also leads to zero divisors. Therefore G is a not a dual reflection group for any AS regular Koszul Noetherian domain. 4.7 The dicyclic group of order 12 Suppose G is Dic 3, the dicyclic group of order 12, and G grades an AS regular algebra with quadratic relations A such that A e is AS regular. We use the following presentation for G: {t, z : t 2 = z 3, z 6 = e, tzt 1 = z 1 }, which can be represented as the group generated by matrices t = where ψ 6 is the sixth root of unity. ( ) ( ) 0 1 ψ6 0, z = ψ6 1 Note that (tz i )(tz j ) = z 3+j i ; it follows that (tz i ) 2 = z 3. Our first step is to determine all minimal generating sets for G. It can be shown that there is an automorphism of G defined by sending t to tz i for any i and fixing 33

39 z, and an automorphism of G defined by sending z to z 5 and fixing t. Using these automorphisms, we can reduce the number of generating sets we actually have to check. After we find all minimal generating sets, we find that the upper bound on n g is 5. We get two possibilities for p(t): p 1 (t) = (t + 1)(t + 1)(t 2 + t + 1) = t 4 + 3t 3 + 4t 2 + 3t + 1, p 2 (t) = (t + 1)(t 2 + 1)(t 2 + t + 1) = t 5 + 2t 4 + 3t 3 + 3t 2 + 2t + 1. Without loss of generality, there is one generating set that generates in accordance with p 1 (t): generating set {t, tz 2, z 3 }. However, supposing that we have an AS regular algebra A with generators x 1, x 2, x 3 graded by these elements, we find that A has a cubic relation, not a consequence of its quadratic relations, giving us a contradiction. Without loss of generality, there is one generating set that generates in accordance with p 2 (t): generating set {t, tz}. Under this generating set, we also get that A has cubic relations not a consequence of its quadratic relations, giving us a contradiction. Therefore G is a not a dual reflection group for any AS regular Koszul Noetherian domain. 4.8 Dihedral group of order 16 Let G be D 16, the dihedral group of order 16. Suppose A is an AS regular algebra graded by G with A e AS regular. Possible minimal generating sets of D 8 are: {r, ρ}, {r, ρ 3 }, {r, ρ 5 }, {r, ρ 7 }, {r, rρ}, {r, rρ 3 }, {r, rρ 5 }, {r, rρ 7 }. By computing the Poincaré polynomial for each minimal generating set, we find that the bound on n g = 8. Therefore, our possible p(t): 34

40 p 1 (t) = Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 2 (t) = t 4 + 4t 3 + 6t 2 + 4t + 1, p 2 (t) = Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 4 (t) = t 5 + 3t 4 + 4t 3 + 4t 2 + 3t + 1, p 3 (t) = Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 8 (t) = t 7 + 3t 6 + 3t 5 + t 4 + t 3 + 3t 2 + 3t + 1, p 4 (t) = Φ 2 (t) Φ 2 (t) Φ 4 (t) Φ 4 (t) = t 6 + 2t 5 + 3t 4 + 4t 3 + 3t 2 + 2t + 1, p 5 (t) = Φ 2 (t) Φ 2 (t) Φ 4 (t) Φ 8 (t) = t 8 + 2t 7 + 2t 6 + 2t 5 + 2t 4 + 2t 3 + 2t 2 + 2t + 1, p 6 (t) = Φ 2 (t) Φ 4 (t) Φ 4 (t) Φ 4 (t) = t 7 + t 6 + 3t 5 + 3t 4 + 3t 3 + 3t 2 + t + 1, p 7 (t) = Φ 4 (t) Φ 4 (t) Φ 4 (t) Φ 4 (t) = t 8 + 4t 6 + 6t 4 + 4t 2 + 1, p 8 (t) = Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 6 (t) = t 6 + 3t 5 + 3t 4 + 2t 3 + 3t 2 + 3t + 1, p 9 (t) = Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 6 (t) Φ 6 (t) = t 8 + 2t 7 + t 6 + 2t 5 + 4t 4 + 2t 3 + t 2 + 2t + p 10 (t) = Φ 2 (t) Φ 2 (t) Φ 2 (t) Φ 4 (t) Φ 6 (t) = t 7 + 2t 6 + 2t 5 + 3t 4 + 3t 3 + 2t 2 + 2t + 1, p 11 (t) = Φ 2 (t) Φ 2 (t) Φ 4 (t) Φ 4 (t) Φ 6 (t) = t 8 + t 7 + 2t 6 + 3t 5 + 2t 4 + 3t 3 + 2t 2 + t + 1. p 4 (t), p 6 (t), p 7 (t), p 10 (t), and p 11 (t) can all be eliminated by looking at the data from our minimal generating sets. With just a little bit more work, it can be shown that p 3 (t) and p 8 (t) are also not possible. This leaves possible p(t) to be p 1 (t), p 2 (t), p 3 (t), p 5 (t), and p 8 (t). By computation we found four generating sets (without loss of generality) that match p 1 (t). Case 1: generating set {r, ρ, ρ 5, rρ 2 }. Under this generating set, we get zero divisors in A, contradicting our assumption that A is a domain. Case 2: generating set {r, ρ, ρ 5, rρ 6 }. Under this generating set, we also get zero divisors. Case 3: generating set {r, rρ, rρ 2, rρ 3 }. Under this generating set, we get a new 35