Integer-Valued Polynomials Over Quaternion Rings

Home , Noncommutative ring, Quotient ring, Simple ring

INTEGER-VALUED POLYNOMIALS OVER QUATERNION RINGS

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for the Degree Doctor of Philosophy in the

Graduate School of The Ohio State University

Nicholas J. Werner, B.A., M.S.

Graduate Program in Mathematics

*****

The Ohio State University

2010

Dissertation Committee:

Professor K. Alan Loper, Advisor Professor S. Tariq Rizvi Professor Daniel Shapiro c Copyright by

Nicholas J. Werner

2010 ABSTRACT

When D is an integral domain with ﬁeld of fractions K, one may deﬁne the

ring Int(D) of integer-valued polynomials over D to be Int(D) = {f(x) ∈ K[x] |

f(a) ∈ D for all a ∈ D}. The goal of this dissertation is to extend the integer-valued polynomial construction to certain noncommutative rings. Speciﬁcally, for any ring

R, we deﬁne the R-algebra RQ to be RQ = {a + bi + cj + dk | a, b, c, d ∈ R}, where i,

j, and k are the standard quaternion units satisfying the relations i2 = j2 = −1 and

ij = k = −ji. When this is done with the integers Z, we obtain a noncommutative

ring ZQ; when this is done with the rational numbers Q, we obtain a division ring

QQ that contains ZQ. Our main focus will be on the construction and study of

integer-valued polynomials over a ring R such that ZQ ⊆ R ⊆ QQ. For such an

R, we deﬁne Int(R) := {f(x) ∈ QQ[x] | f(α) ∈ R for all α ∈ R}. In this treatise, we will prove that Int(R) always has a ring structure and will investigate elements,

generating sets, and prime ideals of Int(R). The ﬁnal chapter examines the idea of

integer-valued polynomials on subsets of rings. Throughout, particular attention will

be paid to the ring Int(ZQ).

ii For my mother, whose support and encouragement were always incomparable.

iii ACKNOWLEDGMENTS

Most grateful thanks are extended to my advisor, Alan Loper, for proposing this project in the ﬁrst place, and for his many helpful comments and suggestions during its completion. I would also like to thank the members of my dissertation committee,

Tariq Rizvi and Daniel Shapiro, for taking the time to evaluate my work.

Lastly, I want to thank all the friends I have made in the math department during my time here. In particular, I would like to express my appreciation to Liz, Jenn,

Jason, Eric, and Matt. Without all the fun times and diversions I’ve had with you, it is unlikely that I would have maintained my sanity long enough to ﬁnish this dissertation. Congratulations, guys. Your names are now oﬃcially attached to a thesis.

iv VITA

2004 ...... B.A. in Mathematics, The College of New Jersey 2006 ...... M.S. in Mathematics, The Ohio State University 2004-2009 ...... VIGRE Graduate Fellow, The Ohio State University 2009-Present ...... Graduate Teaching Associate, The Ohio State University

FIELDS OF STUDY

Major Field: Mathematics

Specialization: Algebra and Ring Theory

v TABLE OF CONTENTS

Page

Abstract ...... ii

Dedication ...... iii

Acknowledgments ...... iv

Vita...... v

Chapters:

1. Introduction ...... 1

1.1 Motivation ...... 1 1.2 Notation and Conventions: Quaternions ...... 3 1.3 Notation and Conventions: Polynomials ...... 7 1.4 When is a Quaternion Ring a Matrix Ring? ...... 10

2. Properties of Overrings ...... 14

2.1 Basic Properties ...... 14 2.2 Classifying Overrings of ZQ ...... 17 2.3 A Useful Theorem ...... 22

3. Rings of Integer-Valued Polynomials ...... 25

3.1 Proving Int(R) Is a Ring ...... 25 3.2 The Trouble with Bar Conjugation in Int(ZH) ...... 30 3.3 Localization Properties ...... 31

vi 4. Muﬃns and Generating Sets (I) ...... 34

4.1 What Are Muﬃns, and Why Do We Care? ...... 34

4.2 The Polynomials φR ...... 40 4.3 From Muﬃns to Generating Sets ...... 44 4.4 A Proof that Int(ZQ) is Non-Noetherian ...... 55

5. Muﬃns and Generating Sets (II) ...... 58

5.1 Quotient Rings of ZH ...... 58 5.2 Int(ZH) is Closed Under Bar Conjugation ...... 69 5.3 A Generating Set for Int(ZH)...... 79

6. Some Primes of Int(R) ...... 88

6.1 Prime Ideals in Noncommutative Rings ...... 88 6.2 Prime Ideals in Rings of Integer-Valued Polynomials ...... 89 6.3 Maximal Ideals in Int(ZQ)...... 96 6.4 Some Primes of Int(ZQ) above (1 + i, 1 + j) ...... 103

7. Integer-Valued Polynomials on Subsets ...... 107

7.1 Subsets and Muffins ...... 107 7.2 Some Sufficient Conditions ...... 112 7.3 Results for Reduced Subsets of ZQ ...... 117 7.4 The Case Γ(S) = 8 and the Classification of Reducible Subsets . . 131 7.5 Necessary and Sufficient Conditions for Finite Subsets ...... 139

Bibliography ...... 148

vii CHAPTER 1

INTRODUCTION

1.1 Motivation

The set of integer-valued polynomials is defined to be Int(Z) = {f(x) ∈ Q[x] | f(a) ∈ Z for all a ∈ Z}. It is not difficult to prove that Int(Z) is closed under the addition and multiplication of polynomials, and thus we may speak of the ring of integer-valued polynomials. Furthermore, this construction easily extends to an integral domain D with field of fractions K by defining Int(D) = {f(x) ∈ K[x] | f(a) ∈ D for all a ∈ D}, called the ring of integer-valued polynomials over D. In this more general context, integer-valued polynomials have inspired considerable research in recent decades; the book [2] by Cahen and Chabert is an excellent reference for the subject. However, one need look no further than Int(Z) to find interesting results.

For instance, it is easy to see that Z[x] ⊆ Int(Z) ⊆ Q[x], but no element of Q − Z

x(x−1) can lie in Int(Z), so Int(Z) 6= Q[x]. On the other hand, the polynomial 2 is

integer-valued without having integer coeﬃcients, so Z[x] 6= Int(Z). Thus, Int(Z) is

a ring that lies properly between Z[x] and Q[x]. Other notable properties of Int(Z) include the following (proofs may be found in the indicated chapter of [2]):

1 x x(x − 1) ··· (x − (n − 1)) • for each n ≥ 0, the binomial polynomial = lies n n! in Int(Z) (Chapter I of [2]).

nxo • the collection of all the binomial polynomials forms a basis for Int(Z) n n≥0 as a Z-module (Chapter I of [2]).

• the ring Int(Z) is non-Noetherian (Chapter VI of [2]), i.e. there exist ideals of

Int(Z) that are not ﬁnitely generated.

• for each prime number p of Z, there is a one-to-one correspondence between

the maximal ideals of Int(Z) containing p and elements of the ring Zp of p-adic integers (Chapter V of [2]).

• for each subset S ⊆ Z, the set Int(S, Z) = {f(x) ∈ Q[x] | f(a) ∈ Z for all a ∈ S} is closed under addition and multiplication of polynomials, and hence

is a ring (Chapter I of [2]).

The purpose of this treatise is to extend the integer-valued polynomial construction to a class of noncommutative rings (these rings, which we call quaternion rings, will be introduced in the next section). Our overriding strategy in this endeavour is to use what is known about Int(Z) as a roadmap and a source of inspiration. Most of the problems we will investigate regarding integer-valued polynomials over quaterion rings are fairly basic and are reminiscent of the properties of Int(Z) listed above. Among the issues we want to resolve are: do these sets of polynomials have a ring structure? Assuming that they do, what sorts of polynomials lie in these rings? What can we say about generators and prime ideals? Are the rings we get non-Noetherian?

2 And what happens when we consider integer-valued polynomials on subsets of quater-

nion rings? We will not completely answer all of these questions, although for each of

them we will give some resolution, at least in special cases. If nothing else, we have

posed enough questions to encourage further research.

1.2 Notation and Conventions: Quaternions

All rings under consideration are assumed to have a multiplicative identity, but

are not necessarily commutative. By an integral domain, we mean a commutative

ring with unity that has no non-zero zero divisors.

Given any ring R, we deﬁne the R-algebra RQ to be

RQ = {a + bi + cj + dk | a, b, c, d ∈ R},

where the symbols i, j, k satisfy the relations i2 = j2 = −1 and ij = k = −ji. We refer to i, j, and k as the quaternion units, and we shall call RQ the quaternion ring with coeﬃcients in R. Note that if char(R) 6= 2, then RQ is noncommutative, since ij 6= ji. Generally, we will work with RQ, where R is either an overring of Z in Q or

a quotient ring of Z, and our primary focus will be on the quaternion ring

ZQ := {a + bi + cj + dk | a, b, c, d ∈ Z} .

An element of ZQ is sometimes called a Lipschitz quaternion. Readers may be familiar

with a ring closely associated to ZQ and usually called the Hurwitz integers or Hurwitz

1+i+j+k quaternions. Speciﬁcally, let µ = 2 (which we refer to as the Hurwitz unit) and deﬁne the Hurwitz integers to be

ZH := ZQ[µ] .

3 The ring ZH was studied by Adolf Hurwitz in [5], where he determined that, among other interesting properties, ZH is a principal ideal ring. As we shall see, the

Hurwitz unit plays an important role in classifying quaternion rings over Z, and so, given any ring R, we deﬁne the R-algebra RH to be RH = RQ[µ]. By a quaternion ring, we mean a ring of form RQ or RH. The notations RQ and RH are meant to be suggestive. The Q in RQ indicates the presence of the quaternion units i, j, and k, and the H in RH indicates the presence of i, j, and k along with the Hurwitz unit

µ.

Whenever R is an overring of Z in Q, the quaternion rings RQ and RH lie inside the ring QQ, and QQ itself lives inside the Hamiltonians, the ring of quaternions with coeﬃcients in R, which we denote by H. At this point, we need to point out a slight conﬂict of notation. The quaternion unit i that we presume to be part of the algebra

RQ is distinct from the complex square root of −1 found in C. In our notation, the ring H of Hamiltonian quaternions may be written as H = RQ. However, while

H contains infinitely many subrings isomorphic to the complex numbers C, the C- algebra denoted by CQ is different than H. This notational conflict will not be a problem in this text, because the ring R as found in RQ will usually be a quotient ring of Z or an overring of Z in Q.

For any α = a + bi + cj + dk ∈ H, we refer to a, b, c, and d as the coefficients of α. In particular, we often call a the constant coefficient of α. Furthermore, we define the bar conjugate of α to be α = a − bi − cj − dk. We say that a subset S of H is closed under bar conjugation if α ∈ S for all α ∈ S. The reason that α is not called simply the conjugate of α is because we shall frequently look at quaternions of the form uαu−1, which are the multiplicative conjugates of α. Thus, the term conjugate,

4 if left unqualiﬁed, can be ambiguous. In this text, we will always qualify the term to

avoid confusion.

Another tool that is useful in dealing with quaternions is the norm. The norm of

α = a + bi + cj + dk ∈ H is denoted by N(α) and is deﬁned to be the real number

N(α) = a2 + b2 + c2 + d2 .

Elements α and β of H satisfy the following properties related to bar conjugation and the norm (these can be veriﬁed by direct calculation):

αα = N(α) = αα

N(αβ) = N(α)N(β) (i.e., the norm is multiplicative)

N(α) = 0 if and only if α = 0

−1 α for all α 6= 0, α = N(α) αβ = βα .

Note that the above expression for α−1 tells us QQ is actually a division ring, since for any α ∈ QQ we have α ∈ QQ and N(α) ∈ Q. Also, note that N(µ) = 1, so that

1−i−j−k whenever µ is in a quaternion ring R, its inverse µ = 2 = µ − i − j − k is also in R. Thus, we are justiﬁed in referring to µ as the Hurwitz unit.

Note that i−1 = −i, j−1 = −j, and k−1 = −k. Then, for any α = a+bi+cj +dk ∈

H, we have

−i(a + bi + cj + dk)i = a + bi − cj − dk

−j(a + bi + cj + dk)j = a − bi + cj − dk

−k(a + bi + cj + dk)k = a − bi − cj + dk;

5 thus, multiplicative conjugation by i, j, or k does nothing except negate some of the coeﬃcients of α. Similarly, multiplicative conjugation by µ merely permutes the coeﬃcients of i, j, and k:

µ(a + bi + cj + dk)µ−1 = a + di + bj + ck .

These relations will come up often in proofs and computations in the succeeding chapters, so it is recommended that the reader be familiar with them, or at least remember that they are given here.

One can verify that for any α ∈ H,

α2 = 2aα − N(α)

This identity implies that every α ∈ H satisfies a quadratic polynomial with real coefficients. Specifically, let a be the constant coefficient of α. Then, α solves x2 −

2ax + N(α) ∈ R[x]. For any α ∈ H, we deﬁne the minimal polynomial of α to be ( x2 − 2ax + N(α), α∈ / R minα(x) = x − a, α ∈ R .

It is straightforward to prove that for each α ∈ H, the minimal polynomial of α is unique, i.e. if f(x) is any monic polynomial in R[x] with deg(f) = deg(minα(x)), then f(α) = 0 if and only if f(x) = minα(x). An easily veriﬁable fact that we shall use frequently is the following:

for any α and β in H, minα(x) = minβ(x) if and only if α and β share the same norm and constant coeﬃcient.

Lastly, we explain the terminology we will use regarding ideals. Throughout this text, the term ideal refers to a two-sided ideal. If we have need to consider left or

6 right ideals, we will refer to them as such. As far as notation, given α ∈ R, we let

Rα denote the left ideal of R generated by α, and likewise αR denotes the right ideal generated by α. We use (α) to denote the two-sided ideal of R generated by α. Note that if R is noncommutative, then (α) contains not only triples βαγ with β, γ ∈ R (in general, the set of all such triples will not be closed under addition), but also ﬁnite sums of the form β1αγ1 + β2αγ2 + ··· + βnαγn. Finally, if S is some subset of R, then

(S) is the two-sided ideal of R generated by S; in particular, for α1, α2, ..., αn ∈ R, we have (α1, α2, ..., αn) = (α1) + (α2) + ··· + (αn).

1.3 Notation and Conventions: Polynomials

Most of the rest of this text will deal with polynomials that have quaternion coefficients, which means that we will be working with polynomials in noncommuting coefficients. Since one most often meets polynomials whose coefficients lie in commutative rings, some remarks are in order about the similarities and differences between the two cases (a detailed treatment of polynomials over division rings is given in §16 of [6]).

Assume for now that R is a noncommutative ring. We can form the polynomial ring R[x] just as in the commutative case. First, we deﬁne addition and multiplication of monomials. For any α, β ∈ R, we take

αxn + βxn = (α + β)xn and αxn · βxm = αβxn+m

Addition for arbitrary polynomials is performed componentwise (as usual) and multiplication is carried out by expanding products via the distributive laws and then

7 grouping terms with the same powers of x (also as usual). Note that when multiplying, the indeterminant x always commutes with the coeﬃcients. Thus, on the level of

R[x], nothing is amiss when we assume that R is noncommutative.

The big diﬀerence in dealing with polynomials over noncommutative rings has to do with evaluating products of polynomials. If f(x) and g(x) are polynomials with coeﬃcients in a ring R (whether commutative or noncommutative), then we let

(fg)(x) denote the product of the polynomials f(x) and g(x) in R[x]. That is,

(fg)(x) = f(x)g(x).

Now, if S is a commutative ring, then evaluating polynomials in S[x] at a ∈ S deﬁnes a ring homomorphism from S[x] to S. Explicitly, if a ∈ S, then the function

Ea : S[x] → S deﬁned by Ea(f) = f(a) is a ring homomorphism. Put slightly diﬀerently,

if f(x), g(x) ∈ S[x] and a ∈ S, then (fg)(a) = f(a)g(a).

However, this sort of relationship may not hold if we are working over a noncommutative ring. Consider the following example with polynomials in ZQ[x]. Let f(x) = x−i and g(x) = x − j. Then, (fg)(x) = f(x)g(x) = x2 − (i + j)x + k and

(fg)(i) = i2 − (i + j)i + k = 2k, but

f(i)g(i) = (i − i)(i − j) = 0 .

Thus,

(fg)(i) 6= f(i)g(i), and evaluation does not give a homomorphism in this case. In fact, if R is a noncommutative ring and α ∈ R, the function Eα : R[x] → R deﬁned by Eα(f) = f(α) is a ring homomorphism if and only if α lies in the center of R.

8 The important thing to take away from this discussion is that evaluating a product of polynomials in noncommuting coeﬃcients may be counterintuitive and must be dealt with carefully. In general, it is legal to evaluate a product (fg)(x) if the product is written in such a way that all the indeterminants are “on the right.” For

P r example, suppose that f(x) = r αrx (throughout this text, we use r as the index of summation in polynomials). Then, one way to write (fg)(x) would be

P r (fg)(x) = r αrg(x)x . (1.3.1)

It is then true that

P r (fg)(α) = r αrg(α)α

(in particular, this equation means that if g(α) = 0, then (fg)(α) = 0, a fact that will occasionally come in handy). Usually, we will evaluate products of polynomials by using formula (1.3.1) or something similar.

Because of the nature of the problems we are considering, the evaluation issue described above will be lurking behind the scenes in all of the major theorems in this manuscript. Recall that for an integral domain D with ﬁeld of fractions K, we deﬁne

Int(D) = {f(x) ∈ K[x] | f(a) ∈ D for all a ∈ D}. If R is an overring of ZQ in QQ, then R is not an integral domain, since R is not commutative. However, the fact that

R ⊆ QQ and QQ is a division ring means that R has no zero divisors. So, despite being noncommutative, overrings of ZQ make fair substitutes for integral domains. Accordingly, the definition of the set of integer-valued polynomial over a quaternion ring is similar to the above definition for Int(D). For any overring R of ZQ in QQ, we define the set of integer-valued polynomials over R to be

Int(R) = {f(x) ∈ QQ[x] | f(α) ∈ R for all α ∈ R} .

9 Notice that at this point, we refer to Int(R) only as a set. It is not hard to see that

Int(R) is closed under addition, but it is non-trivial to prove that Int(R) is closed

under multiplication (see Theorem 3.1.1). The reason such a proof is diﬃcult is that

the polynomials in Int(R) have noncommuting coeﬃcients and are deﬁned entirely

in terms of evaluation; but, as elucidated above, evaluating the product of two such

polynomials may lead to unexpected behavior. Once again, because we are working

over noncommutative rings, care must be taken when doing something that would be

easy in a commutative setting.

1.4 When is a Quaternion Ring a Matrix Ring?

Given a commutative ring R, the quaterion ring RQ is sometimes isomorphic to

M2(R), the ring of 2 × 2 matrices over R. Knowing when this occurs will occasionally

be useful in the chapters that follow, so this section is devoted to giving examples of

rings R with this property.

The theorems in this section are not new results. However, the author could not

ﬁnd proofs in the available literature. So, for the sake of completeness, proofs are

provided here. ∼ Suﬃcient conditions under which RQ = M2(R) are given in by the following

theorem.

Theorem 1.4.1. Let R be a commutative ring such that

(i) there exist A, B ∈ R such that A2 + B2 = −1, and

(ii) 2 is a unit in R.

Then, the R-algebra RQ is isomorphic to M2(R), the ring of 2 × 2 matrices over R.

10 Proof. We extend a constuction given in Chapter 3 of [4] that covers the case R = Fp, where p is an odd prime.

Given that A2 + B2 = −1, construct an R-module homomorphism φ : RQ →

M2(R) by deﬁning

1 0 0 1 AB B −A φ(1) = 0 1 , φ(i) = −1 0 , φ(j) = B −A , φ(k) = −A −B ,

2 2 −1 0 and extending linearly over R. It is easy to check that (φ(i)) = (φ(j)) = 0 −1 and φ(i)φ(j) = φ(k) = −φ(i)φ(j), so φ respects multiplication and thus is a ring

homomorphism.

Claim 1: φ is injective

Proof of Claim: Let a + bi + cj + dk ∈ Kern(φ). Then,

a+cA+dB b+cB−dA 0 0 −b+cB−dA a−cA−dB = 0 0 ,

so a + cA + dB and a − cA − dB are both 0 in R. This gives 2a = 0, and since 2 is a

unit in R, we have a = 0. Similarly, b = 0. So, we have

cA+dB cB−dA 0 0 cB−dA −cA−dB = 0 0 . (∗)

Then,

0 = B(cA + dB)

= (cB)A + dB2

= dA2 + dB2, using (∗)

= −d,

11 so d = 0. Similarly, we can show that c = 0 by considering A(cA + dB). Thus,

Kern(φ) = {0} and φ is injective. This proves Claim 1.

Claim 2: φ is surjective

Proof of Claim: Note that

1 1 0 1 0 0 φ( 2 (1 − Aj − Bk)) = 0 0 , φ( 2 (1 + Aj + Bk)) = 0 1 ,

1 0 1 1 0 0 φ( 2 (i − Bj + Ak)) = 0 0 , and φ(− 2 (i + Bj − Ak)) = 1 0 .

1 0 0 1 0 0 0 0 Since the matrices 0 0 , 0 0 , 1 0 , and 0 1 generate M2(R) over R, we have

Im(φ) = M2(R). Thus, φ is surjective. This proves Claim 2.

We have shown that φ is bijective, so φ : RQ → M2(R) is an isomorphism.

We point out that the previous result does not conﬂict with the theorem that H is

isomorphic to a proper subring of M2(C) (see [6] §1). The reason for this, as mentioned in Section 1.2, is that the quaternion unit i found in an R-algebra RQ is diﬀerent than the complex square root of −1 found in C. Thus, while we may view C as a subring of H in multiple ways (the obvious example would be to use {a + bi | a, b ∈ R}), the

C-algebra denoted by CQ is (by the above theorem) isomorphic to M2(C) and not to H. In any case, we will not employ this representation of H as a ring of complex matrices, so the diﬃculty mentioned here, while possibly confusing, will not be an issue in the rest of this text.

Corollary 1.4.2. The following rings satisfy the conditions of Theorem 1.4.1:

(i) any field F of odd characteristic; in particular, the finite field Fq, where q is a power of an odd prime.

12 (ii) Z/nZ, where n is any odd integer greater than 1.

(iii) Qp, the ﬁeld of p-adic numbers, where p is an odd prime.

∼ Thus, if R is any of the rings listed above, then RQ = M2(R).

Proof. (i) Let F be a ﬁeld of odd characteristic p. Then, 2 is invertible in F . Also, F

contains a copy of Fp, the finite field with p elements. Since every element in a finite

2 2 ﬁeld is the sum of two squares, there exist A, B ∈ Fp ⊆ F such that A + B = −1, so the conditions of Theorem 1.4.1 are met.

(ii) Let R = Z/nZ. Then, 2 is a unit in R because char(R) is odd. Assume that

e1 e2 et n has prime factorization n = p1 p2 ··· pt . Then, for each 1 ≤ ` ≤ t, there exist

2 2 x`, y` ∈ Z/p`Z such that x` + y` ≡ −1 mod p`. For each 1 ≤ ` ≤ t, we can apply

Hensel’s Lemma (see Chapter 10 of [1]) to the polynomial x2 + y2 + 1 ∈ / e` [x] ` Z p` Z e to obtain elements z , w ∈ / e` such that z2 + w2 ≡ −1 mod p ` . The existence ` ` Z p` Z ` ` ` of the required A and B in R now follows from the Chinese Remainder Theorem.

(iii) The rational numbers are contained in Qp, so 2 is invertible in Qp. As in part (ii), the necessary A and B can be obtained via Hensel’s Lemma (applied this time to the ring Zp of p-adic integers, which has maximal ideal pZp with quotient ﬁeld ∼ Zp/pZp = Fp).

13 CHAPTER 2

PROPERTIES OF OVERRINGS

2.1 Basic Properties

We begin by proving some results concerning general overrings of ZQ in QQ. The results of this chapter, in particular Theorem 2.2.3, were known to Hurwitz [5] in a

slightly diﬀerent form.

The ﬁrst lemma is elementary but will be extremely useful in the discussion to

follow.

x Lemma 2.1.1. Let S be any overring of Z in Q, and let y ∈ S with x and y relatively

1 prime. Then, y ∈ S.

Proof. Since x and y are relatively prime, x is invertible mod y. So, there exists

xm 1 m ∈ Z such that xm ≡ 1 mod y. Then, y = k + y for some k ∈ Z, which gives

1 y ∈ S.

In particular, we see that Lemma 2.1.1 holds for any overring R of ZQ in QQ, since the ring S = R ∩ Q is an overring of Z in Q.

Theorem 2.1.2. Let R be an overring of ZQ in QQ, and let a be a coeﬃcient of a

x element of R. If a∈ / R, then a = 2y for relatively prime odd integers x and y. 14 Proof. We prove the contrapositive. Assume that a is a coeﬃcent of α ∈ R. By

multiplying α by ±i, ±j, or ±k, we may assume that a is the constant coeﬃcient of

α. Furthermore, since 4a = α − iαi − jαj − kαk, we see that 4a ∈ R regardless of

the value of a.

x n Write a = 2ny where n ≥ 0, y is odd, and x is relatively prime to 2 y. We have three cases to consider.

Case 1: n = 0

x 4x In this case, a = y and y = 4a ∈ R. Since x is invertible mod y and y is odd, 4x

1 1 is also invertible mod y. Thus, y ∈ R by Lemma 2.1.1 and hence a = x( y ) ∈ R.

Note that Case 1 covers any instance in which x is even, so for the remaining

Cases we may assume that x is odd.

Case 2: n > 2

x x 1 In this case, a = 2ny and 4a = 2n−2y ∈ R. By Lemma 2.1.1, 2n−2y ∈ R. Then,

n−2 1 1 1 1 1 1 n−3 1 2 ( 2n−2y ) = y ∈ R and y( 2n−2y ) = 2n−2 ∈ R. Since 2n−2 ∈ R, so is 2 = 2 ( 2n−2 ).

1 1 n x 1 1 This means that 2n = ( 2 ) ∈ R, and ﬁnally a = 2ny = x( 2n )( y ) ∈ R.

Case 3: n = 2

x x 1 In this case, a = 4y with x odd. We know that y = 4a ∈ R, so we can get y ∈ R.

1 Thus, it suﬃces to show that 4 ∈ R. By multiplying α by odd integers, we can clear all the odd numbers from the

denominators of the coeﬃcients of α. So, WLOG we may assume that α = a + bi +

15 x cj + dk where a = 4 , x is odd and b, c, d are rational numbers whose denominators are all powers of 2 and whose numerators are all odd.

If any of the denominators of b, c, or d equals 2n with n > 2, then as in Case

1 1 1 2 2 we have 2 ∈ R, and hence 4 = ( 2 ) ∈ R. So, we may assume that each of the denominators of b, c, and d is 1, 2, or 4. Furthermore, by subtracting integer multiples

1 1 of 1, i, j, and k from α, we may assume that a, b, c, and d all lie in the interval (− 2 , 2 ]. Finally, by multiplying α by −1 we may guarantee that a > 0. Thus, WLOG we may

1 1 1 assume that α = 4 + bi + cj + dk with b, c, d ∈ {0, 2 , ± 4 }.

2 Recall that for any γ ∈ H with constant coeﬃcient γ0, we have γ = 2γ0 − N(γ).

1 2 1 Using this, we see that if any of b, c, or d equals ± 4 , then α = 2( 4 )α − N(α) =

1 1 2 α − N(α) has a coeﬃcient equal to ± 8 , and in this instance we are done by the

1 argument of the preceding paragraph. If all of b, c, and d are equal to 0 or 2 , then

k 2 N(α) = 16 where k is odd. This means that α has constant coeﬃcient equal to

1 k 2−k 1 1 1 8 − 16 = 16 . Since 2−k is odd, we can get 16 ∈ R by Lemma 2.1.1, so 4 = 4( 16 ) ∈ R, and once again we can get a ∈ R.

In each case, we see that a ∈ R. This completes the proof.

Note that the converse of Theorem 2.1.2 does not hold: if an overring R contains

x an element with a coeﬃcient a equal to 2y with x and y odd and relatively prime,

1+i+j+k 1 then a may or may not be in R. For example, 2 is in both Z[ 2 ]Q and ZH, but

1 1 1 2 ∈ Z[ 2 ]Q and 2 ∈/ ZH. However, the following results are always true regardless of the overring.

Corollary 2.1.3. Let R be any overring of ZQ in QQ. For any α = a+bi+cj +dk ∈ R, we have 2a, 2b, 2c, 2d ∈ R.

16 x Proof. The only case left to consider is if α has a coeﬃcient equal to 2y with x

x and y odd and relatively prime. But in this case, y is a coeﬃcient of 2α ∈ R, so

x x 2( 2y ) = y ∈ R.

Corollary 2.1.4. Let R be any overring of ZQ in QQ. Then, R is closed under bar conjugation.

Proof. Let α = a + bi + cj + dk ∈ R. Then, α = α − 2bi − 2cj − 2dk ∈ R.

Corollary 2.1.5. Let R be any overring of ZQ in QQ. Then, for any α ∈ R,N(α) ∈ R.

Proof. Let α = a + bi + cj + dk ∈ R. Then, α2 = 2aα − N(α), and both 2aα and α2

are in R, so N(α) ∈ R.

2.2 Classifying Overrings of ZQ

We can now prove that overrings of ZQ are closely related to overrings of Z. We begin with two lemmas.

Lemma 2.2.1. Let R be an overring of ZQ in QQ. Assume that there exists α ∈ R such that some coeﬃcient of α is not in R. Then, no coeﬃcient of α lies in R.

Proof. Suppose by way of contradiction that some coeﬃcient of α is in R, and write

α = a + bi + cj + dk. We have three cases to consider, and we will derive a contra-

diction in each case.

Case 1: Exactly three coeﬃcients of α are in R

In this case, by multiplying α by ±i, ±j, or ±k, we may assume that the constant

coeﬃcient a is the coeﬃcient not in R. But then, a = α − bi − cj − dk ∈ R, which is

17 a contradiction.

Case 2: Exactly two coeﬃcients of α are in R

In this case, assume WLOG that a, b∈ / R, and subtract cj and dk from α. So, we

x1 x2 may assume α = a + bi. By Theorem 2.1.2, we must have a = and b = with x` 2y1 2y2

x1 x2 and y` (` = 1, 2) odd and relatively prime. By Corollary 2.1.3, and (and hence y1 y2 1 and 1 ) are in R, but 1 ∈/ R because a∈ / R. y1 y2 2 Consider 2 2 2 2 2 2 2 2 x1 x2 x1y2 + x2y1 N(α) = a + b = 2 + 2 = 2 2 . 4y1 4y2 4y1y2

Since each of x1, x2, y1, and y2 is odd, the square of each of these integers is congruent

2 2 2 2 2 2 2 2 to 1 mod 4. So, x1y2 + x2y1 ≡ 2 mod 4 and we may write x1y2 + x2y1 = 2m for some odd integer m. This gives

2 2 2 2 x1y2 + x2y1 2m m N(α) = 2 2 = 2 2 = 2 2 . 4y1y2 4y1y2 2y1y2

2 2 Now, since m is odd, there exists n ∈ Z with m + 2n = y1y2. So, n m n m + 2n 1 N(α) + 2 2 = 2 2 + 2 2 = 2 2 = . y1y2 2y1y2 y1y2 2y1y2 2 By Corollary 2.1.5, N(α) ∈ R, and we know 1 , 1 ∈ R. Thus, the above equation y1 y2

1 implies that 2 ∈ R, which is a contradiction.

Case 3: Exactly one coeﬃcient of α is in R.

In this case, assume WLOG that a, b, c∈ / R and subtract dk from α. So, we may assume α = a + bi + cj. As in Case 2, we write a = x1 , b = x2 , and c = x3 , and 2y1 2y2 2y3

1 again 2 ∈/ R. We compute 2 2 2 2 2 2 2 2 2 x1y2y3 + x2y1y3 + x3y1y2 m N(α) = 2 2 2 = 4y1y2y3 4n 18 m 1 for some odd integers m and n. Multiplying N(α) by n gives 4 ∈ R, so 4 ∈ R by

1 Lemma 2.1.1. However, this implies 2 ∈ R, which is again a contradiction.

We arrive at a contradiction in each case, so we conclude that no coeﬃcient of α is in R.

1+i+j+k Recall that our convention is to let µ = 2 .

Lemma 2.2.2. Let R be an overring of ZQ in QQ. Assume that there exists α ∈ R such that some coeﬃcient of α is not in R. Then,

(i) µ ∈ R.

(ii) there exists α0 ∈ R such that α = µ + α0 and each coeﬃcient of α0 is in R.

Proof. We prove (i) and (ii) together. By Lemma 2.2.1 and Theorem 2.1.2, we may

x1 x2 x3 x4 write α = + i + j + k with x` and y` odd for all 1 ≤ ` ≤ 4 and with each 2y1 2y2 2y3 2y4

1 ∈ R. Since each x` is odd, for each ` there exists m` ∈ such that x` + 2m` = y`. y` Z Then, m` ∈ R for each ` and y` m m m m α + 1 + 2 i + 3 j + 4 k y1 y2 y3 y4 x + 2m x + 2m x + 2m x + 2m = 1 1 + 2 2 i + 3 3 j + 4 4 k 2y1 2y2 2y3 2y4 1 + i + j + k = 2

is also in R, proving (i). Taking α0 = −( m1 + m2 i + m3 j + m4 k) proves (ii). y1 y2 y3 y4

Theorem 2.2.3. Let R be any overring of ZQ in QQ and let S = R ∩ Q. Then, the following hold:

(i) R = SQ ⇐⇒ every coeﬃcient of every element of R is in R.

19 (ii) either R = SQ or R = SH.

(iii) if R = SH, then every α ∈ R may be written as α = α0 + eµ, where α0 ∈ SQ

and e is either 0 or 1.

Proof. First, note that SQ = {α ∈ R | every coeﬃcient of α is in R}. Let T = {q ∈

Q | q is a coeﬃcient of some α ∈ R}. Then, (i) is the claim that R = SQ ⇐⇒ T ⊆ R.

Also, T is a Z-module, and if we close T under multiplication we get the ring Z[T ]. Note the following relationships that exist between R,S, and T :

S = T ∩ R (∗)

SQ ⊆ R and R ⊆ Z[T ]Q. (∗∗)

Now, assume ﬁrst that R = SQ. Then, T ⊆ R, because R = (R ∩ Q)Q =

{q0 + q1i + q2j + q3k | q0, q1, q2, q3 ∈ R ∩ Q}. Conversely, assume that T ⊆ R, so that

T = T ∩ R. By (∗), we get S = T and consequently SQ = Z[T ]Q. Then, (∗∗) implies that R = SQ. Thus, (i) holds, and we may conﬁne ourselves to the case where S ⊂ T .

Assume that S is strictly contained in T . Then, there exists α ∈ R such that

at least one coeﬃcient of α is not in R. By Lemma 2.2.2, µ ∈ R, and there exists

α0 ∈ SQ such that α = µ + α0. Thus, α ∈ SH.

Now, since µ ∈ R and S ⊆ R, we have SH ⊆ R. Let β ∈ R. If all the coeﬃcients

of β are in R, then β ∈ SQ ⊆ SH. If some coeﬃcient of β is not in R, then the

previous paragraph shows that β ∈ SH. In either case, β ∈ SH, so R ⊆ SH and

therefore R = SH.

Part (iii) is essentialy a restatement of Lemma 2.2.2. Given α ∈ R = SH, if

α ∈ SQ, then take e = 0; if α∈ / SQ, then apply Lemma 2.2.2 part (ii) and take

e = 1.

20 Applying Theorem 2.2.3 to ZH gives us the following corollary.

Corollary 2.2.4.

(i) For each α ∈ ZH, either all the coeﬃcients of α are integers (in which case α ∈

ZQ), or all the coeﬃcients of α are half-integers (in which case α ∈ ZH − ZQ).

More explicitly, given α = a + bi + cj + dk ∈ ZH, either a, b, c, d ∈ Z, or

1 u a, b, c, d ∈ Z + 2 = { 2 | u is an odd integer}.

(ii) The unit group of ZH has order 24, and consists of the eight units in (ZQ)× =

u1+u2i+u3j+u4k {±1, ±i, ±j, ±k} and the sixteen elements in { 2 | u1, u2, u3, u4 ∈ {±1}}. Explicitly,

× (ZH) = {±1, ± i, ±j, ±k, ±µ, ±iµi,

± jµj, ±kµk, ±µ2, ±iµ2i, ±jµ2j, ±kµ2k}.

Proof. (i) Let α ∈ ZH, and write α in the form α = x + yi + zj + wk + eµ, where x, y, z, w ∈ ZH ∩ Q = Z and e ∈ {0, 1}. If e = 0, then α ∈ ZQ; if e = 1, then

(2x+1)+(2y+1)i+(2z+1)j+(2w+1)k 1 α = 2 ∈ Z + 2 .

× −1 α −1 −1 (ii) If α ∈ (ZH) , then the inverse of α is α = N(α) , and α = α ∈ ZH, so N(α) = αα is a unit in ZH. Since ZH ∩ Q = Z, we see that α ∈ (ZH)× if and only if N(α) = 1. So, let α ∈ (ZH)×, and write α = a + bi + cj + dk + eµ, where a, b, c, d ∈ Z and e ∈ {0, 1}. If e = 0, then α ∈ (ZQ)×, which equals {±1, ±i, ±j,

u1+u2i+u3j+u4k ±k}. If e = 1, then α ∈ ZH − ZQ, so α = 2 for some odd integers u1,

2 2 2 2 u2, u3 and u4. Since N(α) = 1, we must have u1 + u2 + u3 + u4 = 4, implying that

1+i+j+k 2 −1+i+j+k u1, u2, u3, u4 ∈ {±1}. Finally, using the fact that µ = 2 and µ = 2 , it

21 is not hard to see that

× (ZH) = {±1, ± i, ±j, ±k, ±µ, ±iµi,

± jµj, ±kµk, ±µ2, ±iµ2i, ±jµ2j, ±kµ2k}.

2.3 A Useful Theorem

The next result gives a way of expressing powers of elements of overrings of ZQ that will come in handy in our later work.

Theorem 2.3.1. Let R be any overring of ZQ in QQ, let S = R ∩ Q, and let α ∈ R. Assume that β ∈ R is such that α and β have the same constant coeﬃcient and

N(β) = N(α). Then,

n (i) For all n ≥ 0, there exist Cn,Dn ∈ S such that α = Cnα + Dn. Furthermore,

n β = Cnβ + Dn.

(ii) If f(x) ∈ S[x], then there exist C,D ∈ S such that f(α) = Cα + D. Further-

more, f(β) = Cβ + D.

(iii) If f(x) ∈ R[x], then there exist γ, δ ∈ R such that f(α) = γα + δ. Furthermore,

f(β) = γβ + δ.

Proof. (i) We will use induction on n and the fact that α2 = 2aα − N(α), where a is

the constant coeﬃcient of α.

22 0 For the base case of the induction, we have α = 1, so C0 = 0 and D0 = 1. So,

n assume that n ≥ 0 and α = Cnα + Dn for some Cn,Dn ∈ R ∩ Q. Then,

n+1 2 α = Cnα + Dnα

= Cn(2aα − N(α)) + Dnα

= (2aCn + Dn)α − CnN(α).

So, we can take

Cn+1 = 2aCn + Dn

Dn+1 = −CnN(α).

By assumption, Cn,Dn ∈ R ∩Q, and 2a, N(α) ∈ R ∩Q by Corollaries 2.1.3 and 2.1.5, so Cn+1,Dn+1 ∈ R ∩ Q.

The second assertion follows because Cn and Dn were deﬁned in terms of the constant coeﬃcient of α and N(α).

P r (ii) Assume that f(x) = r arx , where each ar ∈ S. Then,

X r f(α) = arα r X = ar(Crα + Dr) for some Cr,Dr ∈ S, by part (i) r X X = Cα + D, where C = arCr ∈ S and D = arDr ∈ S. r r

As above, the second assertion is true because α and β have the same norm and constant coeﬃcient.

(iii) This is essentially the same as the proof of (ii).

23 Remark. We proved the above theorem by using the relation α2 = 2aα − N(α), but we could also have proven parts (ii) and (iii) by using minα(x). Since minα(x) is monic, we may divide f(x) ∈ R[x] by minα(x) to get f(x) = q(x) minα(x)+γx+δ for some q(x), γx + δ ∈ R[x]. Then, f(α) = γα + δ, and we get f(β) = γβ + δ because

α and β have the same minimal polynomial.

24 CHAPTER 3

RINGS OF INTEGER-VALUED POLYNOMIALS

3.1 Proving Int(R) Is a Ring

Recall that for any overring R of ZQ in QQ, we deﬁne Int(R) = {f(x) ∈ QQ[x] | f(α) ∈ R for all α ∈ R}. The ﬁrst theorem of this chapter proves that Int(R) is a

ring for any overring R of ZQ in QQ.

Theorem 3.1.1. Let R be any overring of ZQ in QQ. Then, Int(R) is a ring.

Proof. It is easy to see that Int(R) is an additive subgroup of the polynomial ring

QQ[x], so it suﬃces to prove that Int(R) is closed under multiplication. Toward that end, let f(x), g(x) ∈ Int(R), and let α ∈ R. It suﬃces to show that (fg)(α) is in R.

P r Write f(x) = r αrx ∈ Int(R); then, each αr ∈ QQ. Notice that if u is any unit in R and β is an arbitrary element of R, then

X r (fu)(β) = αruβ r X r −1 = αruβ u u r X −1 r = αr(uβu ) u r = f(uβu−1)u .

25 Since f ∈ Int(R) and u ∈ R, we have f(uβu−1)u ∈ R. Thus, fu ∈ Int(R).

Next, since g(x) ∈ Int(R), g(α) ∈ R. Let S = R ∩ Q. Then, by Theorem 2.2.3, R ⊆ SH, and g(α) = a + bi + cj + dk + eµ where a, b, c, d ∈ S and e ∈ {0, 1}. Now,

P r P r we may write the polynomial (fg)(x) as (fg)(x) = ( r αrx ) g(x) = r αrg(x)x . We have

X r (fg)(α) = αrg(α)α r X r = αr(a + bi + cj + dk + eµ)α r X r r r r r = αr(aα + biα + cjα + dkα + eµα ) r X r X r X r = a αrα + b αriα + c αrjα r r r X r X r + d αrkα + e αrµα r r = af(α) + b(fi)(α) + c(fj)(α) + d(fk)(α) + e(fµ)(α) .

We know that i, j, k ∈ R, so af(α) + b(fi)(α) + c(fj)(α) + d(fk)(α) ∈ R. If

µ ∈ R, then e(fµ)(α) ∈ R; if µ∈ / R, then α ∈ SQ and e = 0. In either case, we

get (fg)(α) ∈ R and fg ∈ Int(R). Thus, Int(R) is closed under multiplication, and

hence is a ring.

It is worth noting that the technique in the proof of Theorem 3.1.1 extends to

diﬀerent kinds of rings. For example, let G be any ﬁnite group, and let ZG and QG

be the standard group rings. If we deﬁne Int(ZG) := {f ∈ QG[x] | f(α) ∈ ZG for

all α ∈ ZG}, then the steps in the above proof can be used to show that Int(ZG) is

a ring. The key to the proof is that ZG can be generated over Z by a set consisting entirely of units: the elements of G. A similar construction works over matrix rings

if we deﬁne Int(Mn(Z)) := {f ∈ Mn(Q)[x] | f(A) ∈ Mn(Z) for all A ∈ Mn(Z)}.

26 Furthermore, since reducing either ZG or Mn(Z) modulo an integer greater than 1 yields a ﬁnite quotient ring, the theorems in Chapter 4 can be used to construct

elements of Int(ZG) or Int(Mn(Z)). More generally, the proof of Theorem 3.1.1 and the techniques of Chapter 4 should

work as long as we have suitable algebras over Z and Q that have bases of units and as long as we have properly deﬁned what an integer-valued polynomial means for

such algebras (of course, we would need to determine exactly what “suitable” and

“properly deﬁned” mean). We shall not delve any further into this particular topic,

but it deﬁnitely warrants future research.

Recall that for α = a + bi + cj + dk ∈ H, the bar conjugate of α is deﬁned to be

α = a − bi − cj − dk. We have a similar deﬁnition for polynomials in H[x]. Given

P r f(x) = r αrx with coeﬃcients in H, deﬁne the bar conjugate polynomial f to be

P r f(x) = r αrx . A subset S ⊆ H[x] is said to be closed under bar conjugation if f ∈ S for all f ∈ S. The rest of this section is devoted to showing that if R = SQ for some overring S of Z in Q, then Int(R) is closed under bar conjugation. Proving this is straightforward, if a bit technical. It turns out that when R = SH, Int(R) is also closed under bar conjugation, but in this case the proof requires some of the techniques used in Chapters 4 and 5. In light of this, we delay the proof that Int(SH) is closed under bar conjugation until Chapter 5.

Lemma 3.1.2. Let α, β ∈ H. Then,

1 1 αβ = 2 αβ + αβ + 2 α(−iβi)i + α(−iβi)i i

1 1 + 2 α(−jβj)j + α(−jβj)j j + 2 α(−kβk)k + α(−kβk)k k

27 Proof. Before diving into the proof, it is useful to consider what this lemma actually

1 says. The expression 2 αβ + αβ represents the constant coeﬃcient of αβ, so the lemma is indicating that αβ and αβ have the same constant coeﬃcient. Similarly,

1 2 α(−iβi)i + α(−iβi)i equals the constant coefficient of α(−iβi)i, which is the negative of the coefficient of i in α(−iβi). Thus, the lemma says that coefficient of

i in αβ is the negative of the coeﬃcient of i in α(−iβi). Analogous statements hold

for j and k.

Now, let α = a + bi + cj + dk and β = x + yi + zj + wk. Then,

αβ = (a − bi − cj − dk)(x + yi + zj + wk)

= (ax + by + cz + dk) + (ay − bx − cw + dz)i

+ (az + bw − cx − dy)j + (aw − bz + cy − dx)k .

We have β = x − yi − zj − wk, and one may compute that the constant coeﬃcient of

αβ equals ax + by + cz + dk. So,

1 ax + by + cz + dk = 2 αβ + αβ .

Next, since −iβi = x − yi + zj + wk, computation shows that the coeﬃcient of i in α(−iβi) equals −ay + bx + cw − dz. So, the constant coeﬃcient of α(−iβi)i is

−(−ay + bx + cw − dz) = ay − bx − cw + dz, and thus

1 (ay − bx − cw + dz)i = 2 α(−iβi)i + α(−iβi)i i .

Arguing in a similar manner, one can show that

1 (az + bw − cx − dy)j = 2 α(−jβj)j + α(−jβj)j j

1 (aw − bz + cy − dx)k = 2 α(−kβk)k + α(−kβk)k k,

which proves the lemma.

28 The next lemma presents a polynomial analogue of Lemma 3.1.2.

Lemma 3.1.3. Let f(x) ∈ H[x] and β ∈ H. Then,

1 1 f(β) = f(β) + f(β) + f(−iβi)i + f(−iβi)i i 2 2 1 1 + f(−jβj)j + f(−jβj)j j + f(−kβk)k + f(−kβk)k k 2 2

Proof. Note that taking the bar conjugate in H[x] is additive: if g, h ∈ H[x], then g + h = g + h. Thus, it suﬃces to prove the lemma for monomials. Assume that f(x) = αxr ∈ H[x]. By Lemma 3.1.2, we have

f(β) = αβr 1 1 = αβr + αβr + α(−iβri)i + α(−iβri)i i 2 2 1 1 + α(−jβrj)j + α(−jβrj)j j + α(−kβrk)k + α(−kβrk)k k . 2 2

We always have βr = (β)r, so

1 1 1 αβr + αβr = α(β)r + α(β)r = f(β) + f(β) . 2 2 2

Next, consider the coeﬃcient of i in the above expression for f(β). Since i is invertible with inverse −i, we get −iβri = (−iβi)r. So,

1 1 α(−iβri)i + α(−iβri)i = α(−iβi)r i + α(−iβi)r i 2 2 1 = f(−iβi)i + f(−iβi)i . 2

Analogous results hold for j and k, from which we conclude the desired result.

Proposition 3.1.4. Let R be an overring of ZQ in QQ such that R = SQ for some overring S of Z in Q. Then, Int(R) is closed under bar conjugation.

29 P r Proof. Let f(x) = r αrx ∈ Int(R). By Lemma 3.1.3, for each β ∈ R we have

1 1 f(β) = f(β) + f(β) + f(−iβi)i + f(−iβi)i i 2 2 1 1 + f(−jβj)j + f(−jβj)j j + f(−kβk)k + f(−kβk)k k . 2 2

1 Now, f(β) ∈ R and 2 f(β) + f(β) equals the constant coeﬃcient of f(β). Since R = SQ, Theorem 2.2.3 part (i) tells us that every coeﬃcient of every element of

1 R is in R; thus, 2 f(β) + f(β) ∈ R. Similarly, the other coeﬃcients in the above expression for f(β) are in R. Thus, f(β) ∈ R and f(x) ∈ Int(R).

3.2 The Trouble with Bar Conjugation in Int(ZH)

We would like to be able to extend Proposition 3.1.4 to all overrings of ZQ in QQ. The diﬃculty with accomplishing this latter goal comes from the expression

1 1 f(β) = f(β) + f(β) + f(−iβi)i + f(−iβi)i i (3.2.1) 2 2 1 1 + f(−jβj)j + f(−jβj)j j + f(−kβk)k + f(−kβk)k k . 2 2

In the proof of Proposition 3.1.4, we showed that f(β) ∈ R by arguing that each

coeﬃcient of f(β) in (3.2.1) lies in R, which relied on Theorem 2.2.3 part (i). However,

this last result does not apply to arbitrary overrings of ZQ. For example, let f ∈

1 Int(ZH) and β ∈ ZH, and consider 2 f(β) + f(β) as in (3.2.1). By assumption,

1 f(β) is in ZH, and 2 f(β) + f(β) is simply the constant coeﬃcient of f(β). Every

1 1 coeﬃcient of every element of ZH lies in either Z or Z+ 2 , so 2 f(β)+f(β) is either an integer or a half-integer. Similarly, the other coeﬃcients in (3.2.1) all lie in either

1 Z or Z+ 2 . This is good; it means that we will not get anything bizarre or unexpected

in (3.2.1). The problem is that for any element α ∈ ZH either every coeﬃcient of α

30 1 lies in Z, or every coeﬃcient of α lies in Z + 2 , and a priori the coeﬃcients in (3.2.1) have nothing to do with one another. For example, if

1 3 f(β) + f(β) = 2 2

and 1 f(−iβi)i + f(−iβi)i = 3, 2

then we would be forced to conclude that f∈ / Int(ZH), since f(β) ∈/ ZH.

We will show in Chapter 5 (Theorem 5.2.4) that Int(ZH) is in fact closed under bar conjugation, so the situation just described cannot occur. However, the proof of

Theorem 5.2.4 is attained only after an extensive study of the quotient rings of ZH.

Thus, while the result of Proposition 3.1.4 is true when R = ZH, the proof given in this chapter is not applicable.

3.3 Localization Properties

When D is a (commutative) Noetherian domain, the formation of integer-valued polynomials over D behaves well with respect to localization, as the following theorem

illustrates.

Theorem 3.3.1. Let D be a (commutative) Noetherian domain and let S be a multi-

plicatively closed subset of D that does not contain 0. Then, S−1Int(D) = Int(S−1D).

Proof. This is Theorem I.2.3 in [2].

It turns out that we can get similar behavior with quaternion rings if the subset

S consists of integers.

Theorem 3.3.2. Let R be an overring of ZQ in QQ, and let T be a multiplicatively

closed subset of Z that does not contain 0. Then, T −1Int(R) = Int(T −1R).

31 Proof. The proof is essentially the same as the proof of Theorem I.2.3 in [2], except

that some minor modiﬁcations are necessary because R is noncommutative.

First, we show that T −1Int(R) ⊆ Int(T −1R). Let f(x) ∈ T −1Int(R), and let

α −1 α −1 t ∈ T R, where α ∈ R and t ∈ T . We need to show that f( t ) ∈ T R. To do so, we use induction on the degree of f. Let n = deg(f). If n = 0, then f is a

constant and there is nothing to prove. So, assume n > 0 and that every polynomial in

T −1Int(R) of degree less than n is an element of Int(T −1R). Let g(x) = tnf(x)−f(tx).

Since t ∈ Z, f(tx) is a polynomial in T −1Int(R), so g(x) ∈ T −1Int(R). Furthermore, deg(g) < deg(f), so by induction g(x) ∈ Int(T −1R). Now, tnf(x) = g(x) − f(tx), so

n α α t f( t ) = g( t ) − f(α) .

α −1 Consider the right-hand side of this equation. We have g( t ) ∈ T R since g(x) ∈

−1 −1 F (x) Int(T R), and f(x) ∈ T Int(R), so f(x) = t0 for some F (x) ∈ Int(R) and some

0 F (α) −1 n α −1 −1 t ∈ T ; hence, f(α) = t0 ∈ T R. So, t f( t ) ∈ T R. But, t is a unit in T R, so

α −1 −1 −1 −1 f( t ) ∈ T R and f(x) ∈ Int(T R). This shows that T Int(R) ⊆ Int(T R).

Pd r −1 For the reverse inclusion, let h(x) = r=0 αrx ∈ Int(T R). Let C = α0R +

α1R + ··· + αdR be the right R-module generated by the coeﬃcients of h. Since R is

Noetherian as a right R-module over itself and C is ﬁnitely generated, C is Noetherian as a right R-module (this follows from Proposition 1.4 in Chapter X of [7]). Let M be the right R-module generated by {h(β)}β∈R. Then, M ⊆ C and C is Noetherian,

so M is ﬁnitely generated as a right R-module. Let γ1, γ2, . . . , γm be the generators

for M as a right R-module.

−1 −1 −1 Now, since h(x) ∈ Int(T R), M is contained in T R. So, γ1, γ2, . . . , γm ∈ T R,

and by ﬁnding a common denominator, we see that there exists u ∈ T such that

uγ` ∈ R for each 1 ≤ ` ≤ m. Thus, for each γ ∈ M, uγ ∈ R. In particular, we

32 have uh(β) ∈ R for all β ∈ R. Hence, uh(x) ∈ Int(R) and h(x) ∈ T −1Int(R), as required.

When combined with Theorem 2.2.3, the previous theorem says, in eﬀect, that to prove something is true for Int(R) with R an overring of ZQ in QQ, it suﬃces to consider what happens for Int(ZQ) and Int(ZH). Assume that R is an overring of

ZQ in QQ. Then, by Theorem 2.2.3, we know that R = SQ or R = SH, where S is an overring of Z in Q. Now, by taking T to be the set of integers that are invertible in S, it is straightforward to check that T is multiplicatively closed and S = T −1Z.

So, Theorem 3.3.2 tells us that to study Int(R) it is enough to consider T −1Int(ZQ) and T −1Int(ZH). Most of the rest of this text will focus on topics (such as generating sets and prime ideals) that behave well under localization at subsets of Z. So, it is no loss to stop considering general overrings of ZQ and instead focus on ZQ and ZH, and we will follow this tactic in the chapters to follow.

33 CHAPTER 4

MUFFINS AND GENERATING SETS (I)

4.1 What Are Muﬃns, and Why Do We Care?

The major goal in this chapter is to establish a generating set for Int(ZQ) over

ZQ. That is, we wish to describe a collection P of polynomials in Int(ZQ) such that

Int(ZQ) = ZQ[P]. Furthermore, we would like these generators to be “nice” in the sense that all the polynomials in P have a standard form or satisfy similar properties.

Currently, we do not have such a generating set, although we can classify many of the polynomials in Int(ZQ) (see Corollary 4.3.10). In Chapter 5, we will describe a full generating set (in the sense of the previous paragraph) for Int(ZH). As we shall see, the reason we will have more success with

Int(ZH) is because ZH is a principal ideal ring, a property that makes the ideal structure of ZH less complicated than the ideal structure of ZQ. However, the present chapter lays all of the groundwork for the theorems of Chapter 5, and in fact most of the results proven here will carry over to Int(ZH), sometimes without change and other times with minor modiﬁcations.

Our basic strategy in this chapter is to replace calculations in the inﬁnite ring

Int(ZQ) with calculations in quotient rings of ZQ, which (with the exception of ZQ

34 itself) are all ﬁnite rings. In fact, the idea of working in quotient rings of ZQ—

and later ZH—will be a recurring theme for the remainder of this manuscript. The ﬁrst result of this chapter shows why this technique is applicable to the study of integer-valued polynomials.

Correspondence 4.1.1.

(i) Let f(x) ∈ Int(ZQ). Then, there exist g(x) ∈ ZQ[x] and an integer n > 0 such

g(x) that f(x) = n .

g(x) (ii) Let g(x) ∈ ZQ[x] and let n > 0. Then, n ∈ Int(ZQ) if and only if g(α) ≡ 0

in ZQ/(n) for all α ∈ ZQ/(n) (here, (n) is the ideal of ZQ generated by n).

Proof. (i) Essentially, we just need to ﬁnd a common denominator for the coeﬃcients

P r of f(x). Write f(x) = r αrx , where each αr ∈ QQ. Now, for α ∈ QQ we may write q q q q α = 1 + 2 i + 3 j + 4 k n1 n2 n3 n4

for some q`, n` ∈ Z, 1 ≤ ` ≤ 4. But, by ﬁnding a common denominator d ∈ Z for the

0 q` , we can write α = α for some α0 ∈ Q. Applying this to each coeﬃcient of f(x), n` d Z 0 we have f(x) = P αr xr, where each d ∈ and each α0 ∈ Q. Letting n be the r dr r Z r Z

P βr r least common multiple of the dr, we may write f(x) = r n x for some βr ∈ ZQ.

P r g(x) Thus, taking g(x) = r βrx ∈ ZQ[x], we have f(x) = n , as required.

g(x) g(x) (ii) (⇒) Let f(x) = n and assume that n ∈ Int(ZQ). Since f(α) ∈ ZQ for all

α ∈ ZQ, n divides g(α) for all α ∈ ZQ. This means that g(α) ≡ 0 in ZQ/(n) for any

residue α ∈ ZQ/(n).

35 (⇐) Assume that g(x) ∈ ZQ[x] and n > 0 are such that g(α) ≡ 0 in ZQ/(n) for

all α ∈ ZQ/(n). Let β ∈ ZQ, and let β0 represent the residue of β in ZQ/(n). Then,

g(β) ≡ g(β0) ≡ 0 in ZQ/(n), so g(β) ∈ (n) in ZQ. Since β ∈ ZQ was arbitrary, we

g(x) conclude that n ∈ Int(ZQ).

The preceding theorem gives us two powerful methods for dealing with Int(ZQ).

g(x) First, given f(x) = n ∈ QQ[x] with g(x) ∈ ZQ[x] and n > 0, to show that f(x) ∈

Int(ZQ) it suffices to check the values of f(x) on a finite set: the set of representatives Q g(x) for the residues in Z /(n). Second, if we know that f(x) = n ∈ Int(ZQ), then g(x) mod n (the polynomial resulting from reducing each coefficient of g(x) modulo the integer n) is a polynomial in ZQ/(n) [x] that sends each element of ZQ/(n) to

0 in ZQ/(n). If we can classify each such g(x) for each integer n > 1, then we will be

able to establish a generating set for Int(ZQ).

In light of these observations, it makes sense to study quotient rings of ZQ, i.e.

rings of the form ZQ/I, where I is an ideal of ZQ. If R is such a quotient ring and f ∈ R[x], then we frequently abuse our notation and consider f to be a polynomial in

ZQ[x]. Furthermore, if I is an ideal of R, then we may also reduce the coeﬃcients of f modulo I to get f in R/I [x]; we refer to this polynomial as f mod I. In general,

the ring from which a polynomial takes its coeﬃcients should be clear from context.

As mentioned above, it will be helpful to look for polynomials in quotient rings of

ZQ that kill each element of the ring. This inspires the following deﬁnition.

Deﬁnition 4.1.2. For any ring R, let Muﬀ(R) = {f(x) ∈ R[x] | f(α) = 0 for all α ∈

R}, called (for want of a better term) the muﬃn of R.

In terms of this deﬁnition, Correspondence 4.1.1 can be rephrased as follows:

36 g(x) • f(x) ∈ Int(ZQ) can be written as n with g(x) ∈ ZQ[x] and n > 0 if and only if g(x) mod n ∈ Muﬀ ZQ/(n) .

Q g(x) • if g(x) ∈ Muﬀ Z /(n) , then n ∈ Int(ZQ).

If R is an inﬁnite ring, then there is no guarantee that Muﬀ(R) 6= {0}. For

our purposes, we shall focus on ﬁnite quotient rings of ZQ, and (as shown below in Proposition 4.1.5) if R is such a ring, then Muﬀ(R) always contains non-trivial

elements. For any n ∈ Z, we deﬁne Z/nZ Q = {a + bi + cj + dk | a, b, c, d ∈ Z/nZ}. ∼ Then, we have Z/nZ Q = ZQ/(n); in particular, ZQ/(n) is ﬁnite when n 6= 0. We

shall tend to write ZQ/(n) when discussing the quotient ring of ZQ modulo (n), but we shall always assume elements of this ring have the form a + bi + cj + dk where

a, b, c, d ∈ Z/nZ, and we shall view Z/nZ as a subring of ZQ/(n). More generally, if

R is a ﬁnite quotient ring of ZQ and n is the characteristic of R, then we may view

Z/nZ as a subring of R.

If I is any non-zero ideal of ZQ, then N(α) = αα is in I for any α ∈ I, so I always contains a non-zero integral ideal (n) and hence ZQ/I is a quotient of the finite ring ZQ/(n). In general, we will be obtaining information about Muff ZQ/I where I is a proper, non-zero ideal of ZQ. In the case where I = (n) for some n > 0, we let Muff(n) = Muff ZQ/I . Note that Muff(1) = {0}. Before long, we shall be delving into a protracted discussion on muffins, but we first describe some notation considering ideals and quotient rings of ZQ.

When n ∈ Z, we will usually denote the ideal of ZQ generated by n as (n), and

we may write “(n) in ZQ.” If, instead, we wish to work with the ideal I generated by

n in a quotient ring R of ZQ, then we will normally write I = nR or I = Rn; this is

37 done to avoid confusion about the ring in which I resides. Since n ∈ Z, the left ideal Rn, the right ideal nR, and the two-sided ideal (n) are all the same in R, so there is no loss of information with this notation. If α is not central in R, then αR, Rα, and

(α) might all be diﬀerent, so when considering the ideal generated by α, we will tend to write “(α) in R” or “(α) in ZQ” if the ring in which the ideal resides might not be clear from context.

As discussed in Chapter 6, the definition of a prime ideal in a noncommutative ring is different than the definition in a commutative ring. However, the definition of a maximal ideal remains unchanged. For any ring R, let Max(R) denote the collection of maximal ideals of R. Then, one can show (see [4], Chapter 3) that

Max(ZQ) = {(1 + i, 1 + j)} ∪ {(p) | p is an odd prime}. Given this, the following proposition is not surprising.

Proposition 4.1.3. Let R be any quotient ring of ZQ. Let n = char(R). Let I = (1 + i, 1 + j) in R. Then,

(i) Max(R) ⊆ {I} ∪ {pR | p is an odd prime}.

(ii) if n is odd, then Max(R) = {pR | p is an odd prime and p divides n}.

(iii) if n is even, then Max(R) = {I} ∪ {pR | p is an odd prime and p divides n}.

Proof. (i) Since R is a quotient of ZQ, each maximal ideal of R is the image of some ideal in Max(ZQ) under the quotient map ZQ → R. Thus, Max(R) ⊆ {I} ∪ {pR | p is an odd prime }.

(ii) Assume n is odd. Then, both 1 + i and 1 + j are invertible in R (because they have norm 2), so (1 + i, 1 + j) in R equals R. For an odd prime p, we have pR 6= R if

38 and only if p | n. Furthermore, if p | n, then pR is a maximal ideal of R because (p) is a maximal ideal in ZQ.

(iii) This is the same as (ii), except that (1 + i, 1 + j) in R is a proper ideal of R because 1 + i and 1 + j are not invertible in R.

We now proceed to the ﬁrst substantial result about muﬃns, which shows that

Muﬀ(R) is always an ideal of R[x].

Proposition 4.1.4. Let R be a ﬁnite, non-zero quotient ring of ZQ. Then, Muﬀ(R) is an ideal of R[x].

Proof. This is similar to the proof of Theorem 3.1.1. It is easy to see that Muff(R) is closed under addition. Let f ∈ Muff(R), let g ∈ R[x], and let α ∈ R. It suffices to show that (fg)(α) and (gf)(α) are both 0. Write g(α) ≡ a + bi + cj + dk. Then, as in Theorem 3.1.1,

(fg)(α) ≡ af(α) + bf(−iαi)i + cf(−jαj)j + df(−kαk)k and since f ∈ Muﬀ(R), each of f(α), f(−iαi), f(−jαj), and f(−kαk) is 0. Hence,

P r (fg)(α) ≡ 0. Let g(x) = r βrx , where each βr ∈ R. Then,

X r (gf)(α) ≡ βrf(α)α ≡ 0 . r

Thus, fg, gf ∈ Muﬀ(R), and Muﬀ(R) is in ideal of R[x].

As mentioned above, Muff(R) is non-zero for any non-zero finite ring R. When I is a proper, non-zero ideal of ZQ and R = ZQ/I, we can explicitly describe some of the elements in Muff(R).

39 Proposition 4.1.5. Let R be a finite, non-zero quotient ring of ZQ, and let n = char(R). Then, Muff(R) contains a monic polynomial with coefficients in Z/nZ.

Proof. Given any α ≡ a + bi + cj + dk ∈ R, we know that α2 ≡ 2aα − N(α).

2 This means that α is a root of the polynomial x − 2ax + N(α) in Z/nZ [x]. Let 2 Q S = {x + 2bx + c ∈ R[x] | b, c ∈ Z/nZ} and let f(x) = g(x). Then, f(x) is g(x)∈S monic and has coeﬃcients in Z/nZ, and since each g(x) ∈ S is central in R[x] we have f(α) ≡ Q g(α) ≡ 0 for all α ∈ R. Thus, f ∈ Muﬀ(R), as desired. g(x)∈S

4.2 The Polynomials φR

With R and n as in the preceding proposition, we have established that Muff(R) always contains a monic polynomial with coefficients in Z/nZ. So, we may pick such a monic polynomial in Muff(R) of least degree, and we denote this polynomial by

φR. When R = ZQ/(n), we let φn = φR; when necessary (as in Corollaries 4.3.8 and 4.3.10), we take φ1 = x. We make no claims about the uniqueness of any φR; in most cases, there could be numerous choices for a φR. However, in the results that follow, no uniqueness will be required. So, the reader may assume at this point that we have ﬁxed φR for each ﬁnite, non-zero quotient ring R of ZQ. As we will show later, the collection of φR will help comprise a generating set for Int(ZQ) over ZQ

(and selecting diﬀerent polynomials φR will give diﬀerent generating sets). At this point, the most important thing to recognize about φn is that

φ for each n > 1, n ∈ Int( Q). n Z

At the present time, we can say little about what φR is for a general quotient ring

R of ZQ, but we do have enough tools to determine φp for an odd prime p. We shall

40 do this below, after proving a proposition and a lemma. The proposition gives some

basic information about elements of ZQ/(p) when p is an odd prime.

Proposition 4.2.1. Let p be an odd prime. Let α ∈ ZQ/(p), and let a be the constant

coeﬃcient of α. Then,

(i) α is a unit ⇐⇒ N(α) 6≡ 0.

(ii) α is nilpotent ⇐⇒ N(α) ≡ 0 and a ≡ 0.

(iii) α is a non-nilpotent zero divisor ⇐⇒ N(α) ≡ 0 and a 6≡ 0.

(iv) if α is nilpotent, then α2 ≡ 0.

(v) for all β ∈ ZQ/(2), either β2 ≡ 0 or β2 ≡ 1; hence, ZQ/(2) has no non-nilpotent

zero divisors.

Proof. (i) Assume N(α) 6≡ 0. Then, N(α) is invertible mod p and N(α)p−1 ≡ 1.

Since α and α commute and αα ≡ N(α), we have α(αp−2αp−1) ≡ N(α)p−1 ≡ 1 and

α is a unit.

Conversely, if N(α) ≡ 0, then αα ≡ 0, so α is a zero divisor and hence cannot be

a unit.

(ii) If N(α) ≡ 0 and a ≡ 0, then α2 ≡ 2aα − N(α) ≡ 0 and α is nilpotent.

Converserly, assume αn ≡ 0 for some n > 0. Then, α is not a unit, so N(α) ≡ 0.

Since α2 ≡ 2aα, a quick induction shows that αm ≡ (2a)m−1α for all m > 0. So,

0 ≡ αn ≡ (2a)n−1α.

Let α ≡ a + bi + cj + dk. If α ≡ a, then since α is nilpotent and Z/pZ is a ﬁeld, we must have a ≡ 0. If α 6≡ a, then one of b, c, or d is not equivalent to 0. WLOG,

41 assume that b 6≡ 0. Then, the fact that 0 ≡ (2a)n−1α means that (2a)n−1b ≡ 0. This forces (2a)n−1 ≡ 0, and consequently a ≡ 0, as required.

(iii) This follows from (i), (ii), and the fact that every non-unit in a ﬁnite ring is

a zero divisor.

(iv) This follows from (ii) and the fact that α2 = 2aα − N(α).

2 (v) Let β ∈ Z/2Z Q with constant coeﬃcient x. Then, β ≡ 2xβ−N(β) ≡ N(β) and N(β) is either 0 or 1. Thus, every element of Z/2Z Q is either a unit or is nilpotent.

Remark. It turns out that when p is an odd prime, ZQ/(p) is isomorphic to M2(Fp), the ring of 2×2 matrices over the ﬁeld of order p (see Corollary 1.4.2). So, Proposition

4.2.1 could have been formulated and proven in terms of matrices. However, this representation will not be particularly useful in the discussion to follow, so we do not emphasize it. Also, ZQ/(2) is not isomorphic to M2(F2) because, among other

reasons, M2(F2) contains matrices that are neither units nor nilpotent.

Recall that any α ∈ ZQ satisﬁes a polynomial with integer coeﬃcients, called the

minimal polynomial of α and denoted by minα(x).

Lemma 4.2.2. Let p be an odd prime, and let R = ZQ/(p). Then, every monic quadratic polynomial in Z/pZ [x] is the minimal polynomial of some α ∈ R −Z/pZ.

2 Proof. Let x + Ax + B ∈ Z/pZ [x]. Then, A ≡ −2a for some a ∈ Z/pZ. It suﬃces

2 2 2 2 to show that we can ﬁnd b, c, d ∈ Z/pZ, not all zero, such that b + c + d ≡ B − a .

Then, α ≡ a + bi + cj + dk will be an element of R − Z/pZ with the desired minimal

42 polynomial. If B − a2 6≡ 0, then since every element of a ﬁnite ﬁeld is a sum of two

2 2 2 squares, we can ﬁnd b, c ∈ Z/pZ such that b + c ≡ B − a , and at least one of b or c is necessarily non-zero. If B − a2 ≡ 0, then ﬁnd b and c such that b2 + c2 ≡ −1, and

2 take d = 1. In either case, we have minα(x) = x + Ax + B, as required.

p2 p Theorem 4.2.3. Let p be an odd prime. Then, φp(x) = (x − x)(x − x) is the unique monic polynomial of minimal degree in Muﬀ(p) ∩ Z/pZ [x].

Proof. Let R = ZQ/(p) and let f(x) be any polynomial in Muﬀ(p) ∩ Z/pZ [x].

Then, f(x) kills 0, 1, . . . , p − 1, so deg(f) ≥ p > 2. So, for any α ∈ R − Z/pZ, we may write

f(x) = g(x)minα(x) + sx + t,

where g(x), sx + t ∈ Z/pZ [x]. Then, f(α) ≡ sα + t ≡ 0 and f(α) ≡ sα + t ≡ 0.

Thus, sα + t ≡ sα + t and sα ≡ sα. Since s ∈ Z/pZ, we have either s ≡ 0 or α ≡ α.

We assumed that α∈ / Z/pZ, so α 6≡ α. Hence, s ≡ 0. Since sα + t ≡ 0, we also have t ≡ 0. So,

f(x) = g(x)minα(x) .

Since α was an arbitrary element of R−Z/pZ, this means that every possible minimal polynomial over ZQ/(p) must divide f(x). By Lemma 4.2.2, we conclude that every monic quadratic polynomial in Z/pZ [x] divides f(x). Thus, for φp to be a monic polynomial in Muff(p) ∩ Z/pZ [x] of minimal degree, φp must be the least common multiple in Z/pZ [x] of every monic quadratic polynomial in Z/pZ [x]. It is a standard result in the theory of finite fields that the product of all irreducible

p2 p2 polynomials in Z/pZ [x] of degree 2 or less is equal to x −x. Thus, x −x accounts for all the irreducible quadratics in the least common multiple we desire. To handle

43 the reducible quadratics, we note that a reducible quadratic must factor into a product

of two linear polynomials. To deal with any such factorization, we need to multiply

xp2 − x by x(x − 1) ··· (x − (p − 1)) = xp − x. Hence, (xp2 − x)(xp − x) is the least

p2 p common multiple we seek, so we take φp(x) = (x − x)(x − x). Finally, φp is unique

because all of the calculations in this paragraph took place over the ﬁeld Z/pZ.

p2 p φp(x) (x −x)(x −x) Corollary 4.2.4. For each odd prime p, the polynomial p = p is an

element of Int(ZQ).

Proof. This follows from Correspondence 4.1.1.

2 (xp −x)(xp−x) Now that we know that the polynomials p are in Int(ZQ), we have

enough information to prove that Int(ZQ) is non-Noetherian. The proof is given in Section 4.4 at the end of this chapter.

4.3 From Muﬃns to Generating Sets

Before proceeding to the next result on muﬃns, we give a couple pieces of notation.

We deﬁne the annihilator of an ideal I in a ring R to be AnnR(I) = {α ∈ R |

αI = Iα = (0)}. If it is clear from context what ring we are working in, we write

simply Ann(I). Given any α ∈ R, we deﬁne Ann(α) to be the annihilator of the ideal

generated by α, i.e. Ann(α) = Ann((α)).

Deﬁnition 4.3.1. Let I be an ideal in a ring R, and let π : R → R/I be the quotient map. We deﬁne

Muﬀ(] R/I) = π−1(Muﬀ(R/I)) = {f(x) ∈ R[x] | f(α) ∈ I for all α ∈ R}.

In other words, Muﬀ(] R/I) is the inverse image of Muﬀ(R/I) in R[x].

44 Theorem 4.3.2. Let R be a ﬁnite, non-zero quotient ring of ZQ and let n = char(R). Then,

(i) for any ideal I of R, Muﬀ(R) mod I ⊆ Muﬀ(R/I).

(ii) for any γ ∈ R and any ideal I of R such that I ⊆ Ann(γ), we have

γMuﬀ](R/I) ⊆ Muﬀ(R). In particular, if n factors as n = `m for integers

` > 1 and m > 1, then `Muﬀ](R/mR) ⊆ Muﬀ(R).

(iii) Assume that γ is central in R. If f ∈ Muﬀ(R) and f = γg for some g ∈ R[x],

then f ∈ γMuﬀ](R/Ann(γ)).

(iv) for any ideal I of R, deg(φR) ≥ deg(φR/I ).

(v) Assume n factors as n = `m with ` > 1, m > 1, and gcd(`, m) = 1. Then

deg(φR) = max{deg(φR/`R), deg(φR/mR)}. In particular, if n has prime factor-

e1 e2 et ization n = p1 p2 ··· pt , then deg(φR) = max{deg(φR/per R)}. 1≤r≤t r

Proof. (i) Let g ∈ Muff(R). Then, g(α) ≡ 0 in R, so the reduction of g mod I kills every residue in R/I. Thus, g mod I ∈ Muff(R/I) and Muff(R) mod I ⊆ Muff(R/I).

(ii) Let γf ∈ γMuﬀ(] R/I), and let α ∈ R. Then, reducing both f and α mod

I, we have f(α) ≡ 0. So, f(α) ∈ I in R, and hence γf(α) ≡ 0 in R. This gives

γf ∈ Muff(R) and γMuff(] R/I) ⊆ Muff(R). The second assertion is true because

`m ≡ 0, and so mR ⊆ Ann(`).

(iii) Assume that f ∈ Muﬀ(R) and f = γg for some g ∈ R[x]. Then, for any α ∈ R, γg(α) ≡ 0. Since γ is central in R, this is enough to conclude that

45 g(α) ∈ Ann(γ). Thus, g mod Ann(γ) ∈ Muﬀ(R/Ann(γ)) and f ∈ γMuﬀ(] R/Ann(γ)).

(iv) By (i), φR mod I is a monic polynomial in Muﬀ(R/I), and deg(φR) =

deg(φR mod I) ≥ deg(φR/I ).

(v) Assume that gcd(`, m) = 1. Let ψ = φR/`R and τ = φR/mR. WLOG, assume

that deg(ψ) = max{deg(ψ), deg(τ)}. By (iv), deg(φR) ≥ deg(ψ).

Now, let d = deg(ψ) − deg(τ). Since gcd(`, m) = 1, there exist a, b ∈ Z such that a` + bm = 1. By (ii), both `τ and mψ are in Muﬀ(R), so a`xdτ + bmψ equals a monic

polynomial in Muﬀ(R) of degree deg(ψ). By the way we deﬁned φR, we must have

deg(φR) ≤ deg(ψ). Thus, deg(φR) = deg(ψ) = max{deg(φR/`R), deg(φR/mR)}.

e1 e2 et Finally, if n has prime factorization n = p1 p2 ··· pt , then the stated result follows from induction on t.

Our ultimate goal is to determine generators for the muﬃn ideal of any ﬁnite

quotient ring of ZQ. Then, the generators for Muﬀ(n) will give rise to generators

for Int(ZQ). It turns out that the nicest cases occur when the characteristic of the quotient ring is odd.

Theorem 4.3.3. Let R be a ﬁnite, non-zero quotient ring of ZQ of characterisic n.

(i) If n is odd, then Muff(R) is finitely generated by polynomials with coefficients

in Z/nZ.

(ii) If n = p for some odd prime p, then Muﬀ(R) = (φR). In particular, Muﬀ(p) =

(φp).

Before giving the proof, we deﬁne some more notation.

46 P r Deﬁnition 4.3.4. For any f(x) = r(ar + bri + crj + drk)x ∈ H[x], let

X r X r f1(x) = arx , fi(x) = brx , r r X r X r fj(x) = crx , and fk(x) = drx , r r so that f = f1 + fii + fjj + fkk. We call f1, fi, fj, and fk the component polynomials of f.

Proof. (i) The ideal Muff(R) is finitely generated because R[x] is Noetherian. Let f be any polynomial in Muff(R). Then, since Muff(R) is an ideal of R[x], we have

−ifi, −jfj, −kfk ∈ Muﬀ(R). So, 4f1 = f + −ifi + −jfj + −kfk ∈ Muﬀ(R).

Assume n is odd. Then, 4 is invertible in R, so f1 ∈ Muff(R). Similarly, the other component polynomials of f are in Muff(R). Hence, a polynomial is in Muff(R) if and only if its component polynomials are in Muff(R), and since these component polynomials have coefficients in Z/nZ, we achieve the stated result.

(ii) Assume now that n = p is an odd prime. Certainly, we have (φR) ⊆ Muff(R). Let I = Muff(R) ∩ Z/pZ [x]; then φR ∈ I. Since Z/pZ is a field, I is generated by a polynomial of minimal degree in I, i.e. by φR. Now, let f ∈ Muff(R), and let

f1, fi, fj, and fk be the component polynomials of f. As in part (i), each of these

component polynomials is in Muﬀ(R). But, f1, fi, fj, and fk all have coeﬃcients in

Z/pZ, so f1, fi, fj, fk ∈ I. So, φR divides each component polynomial; hence, φR

divides f. It follows that Muﬀ(R) ⊆ (φR) in R[x], and therefore Muﬀ(R) = (φR).

For a quotient ring R of ZQ of even characteristic, we do not have such nice results. As we shall see, the generators we are determining for Muﬀ(R) are related

to the ideals of R. When n = char(R) is odd, the same steps used in the proof

47 of Theorem 4.3.3 part (i) can be used to show that every ideal of R is generated

by an element of Z/nZ, and this ideal structure will allow us to fairly easily derive generators for Muﬀ(R). When n is even, the ideal structure of R is more complicated,

and this complexity is reﬂected in the muﬃn ideal of R. The odd case is dealt with

shortly in Theorem 4.3.7. At present time, we do not have a complete proof for the

even case, although a plan of attack is outlined at the end of this section.

Recall that for a commutative ring R and a polynomial f ∈ R[x], the content of f is the ideal of R generated by the coefficients of f. We adopt the same definition m P r for a noncommutative ring. Explicitly, given f(x) = αrx ∈ R[x], we define the r=0 content of f to be the ideal (α0, α1, ..., αm). We denote the content of f by con(f),

and we say that f has content 1 if con(f) = (1).

Lemma 4.3.5. Let R be a ﬁnite, non-zero quotient ring of ZQ, and let n = char(R). If f ∈ Muﬀ(R) ∩ Z/nZ [x] and con(f) = (1), then deg(f) ≥ deg(φR).

Proof. Since φR ∈ Muﬀ(R)∩ Z/nZ [x] and φR is monic, we have con(φR) = (1). So, there exist polynomials in Muﬀ(R) ∩ Z/nZ [x] of content 1; assume WLOG that f

is of minimal degree among all such polynomials. Then, deg(f) ≤ deg(φR), and to

prove the stated theorem it suﬃces to show that deg(f) = deg(φR).

From here, we break the proof into three cases, depending on n.

Case 1: n = p for some prime p

If p is odd, then the lemma holds by Theorem 4.3.3 part (ii). If p = 2, then Muﬀ(R) ∩ Z/2Z [x] = {g ∈ Z/2Z [x] | g(α) = 0 for all α ∈ R}, which is an ideal in Z/2Z [x]. Since Z/2Z [x] is a PID, we have Muﬀ(R) ∩ Z/2Z [x] = (φR), so

48 the lemma holds when n = 2. This completes Case 1.

Case 2: n = pe for some prime p and some e > 0

We use induction on e. The base case was handled in Case 1, so assume that e > 1 and that the result is true for all powers of p less than pe. Suppose by way

of contradiction that deg(f) < deg(φR). Let c be the leading coeﬃcient of f. Since

−1 deg(f) < deg(φR), c cannot be a unit in Z/nZ; if that were the case, then c f would be a monic polynomial in Muﬀ(R) ∩ Z/nZ [x] of degree less than deg(φR). So, p divides c; let m ∈ {1, 2, . . . , e − 1} be such that pm is the highest power of p dividing

m c. Let ψ = φR/pe−mR. Then, by Theorem 4.3.2 part (ii), p ψ ∈ Muﬀ(R), and by

m Theorem 4.3.2 part (iv) we have deg(p ψ) = deg(ψ) ≤ deg(φR). If deg(f) < deg(ψ),

e−m then f mod p R is a content 1 polynomial in Muﬀ R/pe−mR ∩ Z/pe−mZ [x] of degree less than deg(ψ), which contradicts the inductive hypothesis. So, deg(f) ≥

deg(ψ) = deg(pmψ).

Since pm is the highest power of p dividing c and we are assuming that n = pe,

m m there exists y ∈ Z/nZ such that c ≡ yp . Let d = deg(f) − deg(p ψ). Then, we

d m may write f = x yp ψ + g, where g ∈ Z/nZ [x] is either 0 or deg(g) < deg(f). If g = 0, then con(f) ⊆ (p), which is a contradiction. So, deg(g) < deg(f), and

g ∈ Muﬀ(R) because f, pmψ ∈ Muﬀ(R). We cannot have con(g) = (1) because it

would contradict the minimality of deg(f), so we must have con(g) ⊆ (p). But then,

d m d m−1 g = ph for some h ∈ Z/nZ [x], and we have f = yx p ψ + g = p(x yp ψ + h), which yields con(f) ⊆ (p). Thus, we arrive at a contradiction no matter what we do,

so we conclude that deg(f) = deg(φR). This proves Case 2.

49 e1 e2 et Case 3: n has prime factorization n = p1 p2 ··· pt , where the pr (1 ≤ r ≤ t) are

distinct primes and each er > 0

e er Let p and e be such that φp has maximal degree among all the φpr . Then,

e by Theorem 4.3.2 part (v), deg(φR) = deg(φR/peR). Now, f mod p R is in R e e e Muﬀ /p R ∩ Z/p Z [x], so by Case 2 we have deg(f) ≥ deg(φR/p R) = deg(φR). Thus, the lemma holds in this case as well.

There are no more cases, so the proof is complete.

Theorem 4.3.6. Let R be a ﬁnite, non-zero quotient ring of ZQ, and let n = char(R).

Assume that n is odd. If f ∈ Muﬀ(R) and con(f) = 1, then deg(f) ≥ deg(φR).

Proof. Let f be a content 1 polynomial in Muﬀ(R). Since R is ﬁnite and non-zero, n >

e1 e2 et 1. Let n factor as n = p1 p2 ··· pt , where the pr (1 ≤ r ≤ t) are distinct odd primes

and each er > 0. By Theorem 4.3.2 part (v), we have deg(φR) = max{deg(φR/per R)}. 1≤r≤t r

e er Let p and e be such that φR/p R has maximal degree among all the φR/pr R. Then,

deg(φR) = deg(φR/peR), so it suﬃces to show that deg(f) ≥ deg(φR/peR).

Consider the component polynomials f1, fi, fj, and fk of f. Since n is odd,

these component polynomials are all elements of Muﬀ(R). Furthermore, con(f) ⊆

con(f1)+con(fi)+con(fj)+con(fk). Since con(f) = (1), this means that at least one of the component polynomials has content not contained in the maximal ideal pR of

R (if not, then con(f) ⊆ pR, a contradiction). WLOG, assume that con(f1) 6⊆ pR.

Then, in R/peR, we must have

e e con(f1 mod p R) = R/p R .

50 e R e e Since f1 mod p R lies in Muﬀ /p R and has coeﬃcients in Z/p Z, by Lemma 4.3.5 we have

e deg(f1 mod p R) ≥ deg(φR/peR) .

Thus,

e deg(f) ≥ deg(f1) ≥ deg(f1 mod p R) ≥ deg(φR/peR)

and therefore deg(f) ≥ deg(φR).

Our next few results give an explicit generating set for Muﬀ(n) when n is odd.

Theorem 4.3.7. Let R be a ﬁnite, non-zero quotient ring of ZQ, and let n = char(R).

Assume that n is odd and that p1, . . . , pt are all the primes dividing n. Then, R R Muff(R) = (φn, p1Muff]( /Ann(p1)), . . . , ptMuff]( /Ann(pt))).

R R Proof. Let I = (φR, p1Muff(] /Ann(p1)), . . . , ptMuff(] /Ann(pt))). By Theorem 4.3.2 part (ii), we have Muff(R) ⊇ I, so we just need the other inclusion. Let f ∈ Muff(R).

Since φR is monic, there exist g, h ∈ R[x] such that f = gφR + h , where either h = 0

or deg(h) < deg(φR). If h = 0, then f ∈ (φR) and we are done. So, assume deg(h) < R R deg(φR). It suffices to show that h ∈ (p1Muff(] /Ann(p1)), . . . , ptMuff(] /Ann(pt))).

Since deg(h) < deg(φR), Theorem 4.3.6 tells us that con(h) 6= (1). By Proposition

4.1.3, there exists a prime divisor p` of n such that con(h) ⊆ p`R. Then, h ∈ R p`Muﬀ(] /Ann(p`)), so Muﬀ(R) ⊆ I and we are done.

Remark. With R and n as in the theorem, we know via Theorem 4.3.3 that Muﬀ(R) is generated by polynomials in Z/nZ [x], so we could have proven Theorem 4.3.7 using only Lemma 4.3.5.

Corollary 4.3.8. Let n > 1 be odd. Then, Muﬀ(n) = ({mφ n | m divides n}). m

51 Proof. For each n > 1, let I = ({mφ n | m divides n}). The fact that Muﬀ(n) ⊇ n m

In follows from Theorem 4.3.2 part (ii). To get the other inclusion, we will apply

e1 e2 et Theorem 4.3.7. Let n have prime factorization n = p1 p2 ··· pt , and let E = e1 + e2 + ··· + et. We will use induction on E. The base case, E = 1, holds by Theorem

4.3.3. Assume that E > 1 and that the result holds for any odd integer with prime

f1 f2 fs factorization q1 q2 ··· qs such that f1 + f2 + ··· + fs < E.

n n By Theorem 4.3.7, we have Muff(n) = (φn, p1Muff(] ), . . . , ptMuff(] )). By in- p1 pt

n duction, for each 1 ≤ ` ≤ t we have Muﬀ( ) = In/p . Thus, for each `, we have p` `

n n p`Muff(] ) = pÌn/p . Each divisor of is a divisor of n, so pÌn/p ⊆ In, and the p` ` p` ` stated corollary now follows.

Corollary 4.3.9. For any odd prime p and any e > 0, we have Muﬀ(pe) =

2 e−1 (φpe , pφpe−1 , p φpe−2 , . . . , p φp).

Proof. This follows from Corollary 4.3.8.

Translating our results back to Int(ZQ), we achieve the following interesting corollary.

Corollary 4.3.10.

(i) Let p be an odd prime and let

n g(x) o R = f(x) ∈ Int( Q) f(x) = for some g(x) ∈ Q[x] and e ≥ 0 . p Z pe Z

hnφp` o i Then, Rp is a ring and Rp = ZQ (recall that our convention is to p` `≥0

take φ1 = x).

(ii) Let

n g(x) o R = f(x) ∈ Int( Q) f(x) = for some g(x) ∈ Q[x] and some odd n , Z n Z 52 and let

nφ ` o P = p p is an odd prime and ` ≥ 0 , p`

where we again take φ1 = x. Then, R is a ring and R = ZQ[P]. In particular,

R can be generated over ZQ by monic polynomials with rational coeﬃcients.

Proof. (i) It easy to verify that Rp is closed under addition and multiplication. So,

Rp is a subring of Int(ZQ), and it suﬃces to establish the claim about the generators

φp` g(x) p` . Let f(x) = pe ∈ Rp, where g(x) ∈ ZQ[x] and e ≥ 0. If e = 0, then f(x) ∈

φ1 ZQ[x] = ZQ[ 1 ] and we are done. So, assume that e > 0. Since f(x) ∈ Int(ZQ), g(x)

e e mod p lies in Muﬀ(p ). By Corollary 4.3.9 there exist polynomials g0, g1, . . . , ge−1 ∈ ZQ/(pe) [x] such that

e−1 e g(x) ≡ g0(x)φpe (x) + pg1(x)φpe−1 (x) + ··· + p ge−1φp(x) mod p .

So, in ZQ[x], we may write

e−1 e g(x) = g0(x)φpe (x) + pg1(x)φpe−1 (x) + ··· + p ge−1φp(x) + p ge(x)

g(x) for some ge(x) ∈ ZQ[x]. Hence, we can express f(x) = pe as

g(x) φ e (x) φ e−1 (x) φ (x) = g (x) p + g (x) p + ··· + g p + g (x) . pe 0 pe 1 pe−1 e−1 p e

nφp` (x)o Thus, f(x) can be generated over ZQ by . It follows that all the p` 0≤`≤e nφp` (x)o hnφp` o i polynomials in Rp can be generated by , i.e. Rp = ZQ . p` `≥0 p` `≥0 (ii) As in part (i), it is straightforward to check that R is a subring of Int(ZQ).

g(x) Let f(x) = n ∈ R. If n = 1, then f(x) ∈ ZQ[x] and we are done. So, assume

e1 e2 et that n > 1 and has prime factorization n = p1 p2 ··· pt . We use induction on t to show that f(x) can be generated over ZQ by polynomials in P. If t = 1, then we are

53 φ ` p1 done by part (i), since f(x) ∈ Rp1 and each polynomial ` ∈ P. So, assume that p1 t > 1 and that the result holds whenever n has less than t distinct prime factors. For

e2 et e e convenience, let p = p1, e = e1, and q = p2 ··· pt . Then, n = p q. Since p and q are relatively prime, there exist a, b ∈ Z such that ape + bq = 1. Hence,

g(x) g(x) g(x) = a + b . n q pe

g(x) e g(x) Since q = p f(x) ∈ R and pe = qf(x) ∈ R, the inductive hypothesis holds for

g(x) g(x) both q and pe . Since both of these polynomials can be generated over ZQ by polynomials from P, so can f(x). Thus, the desired result holds.

At this point, all of our ﬁrm results end, and we move into the realm of conjecture.

Other than a few examples, we cannot prove much about Muﬀ(R) when R has even

characteristic. By applying a brute force approach and using some of the theorems

we have already proven, one can compute generators for Muﬀ(n) for small even values

of n:

Muﬀ(2) = x2(x2 + 1), (1 + i)(1 + j)(x2 + x),

Muﬀ(6) = (x3 − x)(x9 − x), 3x2(x2 + 1), 3(1 + i + j + k)(x2 + x), and

Muﬀ(4) = x4(x2 + 1)2, (1 + i)(x6 + x5 + x4 + 3x3 + 2x),

(1 + j)(x6 + x5 + x4 + 3x3 + 2x), 2x2(x2 + 1),

2(1 + i)(1 + j)(x2 + x) .

At present, the best conjecture about the generators for Muﬀ(R) when n =

char(R) is even is that Muff(R) = φR, (1 + i)Muff(] R/Ann(1 + i)), (1 + j)Muff(] R/Ann(1 + j)), (4.3.11) R R p1Muff(] /Ann(p1)), . . . , ptMuff(] /Ann(pt)) ,

54 where p1, . . . , pt are all the odd primes dividing n. In general, the ideals Ann(1 + i)

and Ann(1+j) are not generated by integers, so the quotient rings R/Ann(1 + i) and

R/Ann(1 + j) can be quite messy. However, there does not seem to be a way around

dealing with them, which explains why we have tried to consider all quotient rings of

ZQ rather than just those of the form ZQ/(n). While we do not have proofs for the proposed decomposition of Muﬀ(R), we do

have a plan of attack. We would like

1) to be able to extend Theorem 4.3.6 to rings of even characteristic, and

2) to prove that if f ∈ Muﬀ(R) and con(f) ⊆ (1 + i, 1 + j), then f is contained in

the ideal generated by (1 + i)Muﬀ(] R/Ann(1 + i)) and (1 + j)Muﬀ(] R/Ann(1 + j)).

If both of these conjectures are true, then we can prove (4.3.11) the same way

we proved Theorem 4.3.7. So far, no counterexamples have been found to disprove

either 1) or 2).

Finally, there is one other thing to note about the generators given above for

Muff(2), Muff(4), and Muff(6). All these generators have the form γf, where γ ∈ ZQ/(n) and f ∈ Z/nZ [x]. Assuming that (4.3.11) is true, we can prove that any

ﬁnite quotient ring R of ZQ is generated by polynomials of the form γf. A corollary

γf of this is that Int(ZQ) is generated over ZQ by polynomials of the form n , where

γ ∈ ZQ, n ∈ Z, and f ∈ Z[x].

4.4 A Proof that Int(ZQ) is Non-Noetherian

Theorem 4.4.1. The ring Int(ZQ) is non-Noetherian.

55 p2 p φp(x) (x −x)(x −x) Proof. By Corollary 4.2.4, for each odd prime p the polynomial p = p

is in Int(ZQ). Let p1, p2,... be all the (distinct) odd primes in Z. For each n > 0, φ (x) φp1 (x) φp2 (x) φpn (x) pn+1 let In = ( , , ..., ) in Int( Q). We will show that ∈/ In for p1 p2 pn Z pn+1 each n > 0, and this will allow us to produce a strictly increasing chain of ideals in

φpn+1 (x) Int( Q). Fix n > 0 and suppose by way of contradiction that ∈ In. Since Z pn+1 φ (x) φ (x) each p` is central in Int( Q) (because each p` has rational coeﬃcients), there p` Z p` exist f1(x) , f2(x) ,..., fn(x) ∈ Int( Q) such that m1 m2 mn Z

φ (x) f (x) φ (x) f (x) φ (x) f (x) φ (x) pn+1 = 1 p1 + 2 p2 + ··· + n pn , (∗) pn+1 m1 p1 m2 p2 mn pn

and for each 1 ≤ ` ≤ n, f`(x) ∈ ZQ[x] and m` is a positive integer. For each `, let α`

be the constant coeﬃcient of f`(x).

2 φpn+1 (x) 1 Now, the coeﬃcient of x in is . We know that φp (x) = pn+1 pn+1 `

p2 p 2 f`(x) φp (x) (x ` − x)(x ` − x) for each 1 ≤ ` ≤ n, so the coeﬃcient of x in ` equals m` p` α` . By equating the coeﬃcients of x2 in (∗) we get m`p` 1 α α α = 1 + 2 + ··· + n . (∗∗) pn+1 m1p1 m2p2 mnpn

f`(x) For each 1 ≤ ` ≤ n, we have α` = f`(0) and ∈ Int( Q), so α` ∈ (m`) in Q. m` Z Z

Thus, for each `, there exists β` ∈ ZQ such that α` = m`β`. So, (∗∗) becomes

1 β β β = 1 + 2 + ··· + n . pn+1 p1 p2 pn

However, this is impossible; among other things, it would imply that pn+1 is invertible

in Z(pn+1)Q (here, Z(pn+1) is the localization of Z at the prime ideal (pn+1)). We reached

φpn+1 (x) φpn+1 (x) this contradiction by supposing that ∈ In, so we conclude that ∈/ In. pn+1 pn+1

It now follows that In+1 6= In for all n > 0. Hence, I1 ⊂ I2 ⊂ I3 ⊂ · · · is a strictly

increasing chain of ideals in Int(ZQ), and therefore Int(ZQ) is non-Noetherian.

56 φp(x) As we shall see in Chapter 5, the polynomials p are also elements of Int(ZH), so the previous theorem shows that Int(ZH) is non-Noetherian. In fact, the theorem applies to Int(R) for any overring R of ZQ in which inﬁnitely many primes of Z remain prime.

57 CHAPTER 5

MUFFINS AND GENERATING SETS (II)

5.1 Quotient Rings of ZH

The material covered in Chapter 4 focused on ZQ and Int(ZQ). However, the majority of what we accomplished falls through to ZH and Int(ZH) without much change. The reason for this is that whenever n is an odd number, ZH/(n) ∼= ZQ/(n), a fact that we now prove.

Theorem 5.1.1.

(i) If n is an odd integer, then ZH/(n) ∼= ZQ/(n).

(ii) Let I be a non-zero ideal of ZH, let R = ZH/I, and assume that n = char(R)

is odd. Then, R is isomorphic to a quotient ring of ZQ/(n).

Proof. (i) Assume n is odd. Let ι : ZQ → ZH be the injection map, let π : ZH →

ZH/(n) be the canonical quotient map, and let φ = π ◦ ι. We will use φ to get the desired isomorphism. Note that (n) in ZQ is contained in Kern(φ), and if α ∈

Kern(φ), then ι(α) = nβ for some β ∈ ZH. By Theorem 2.2.3, β = β0 + eµ for some

β0 ∈ ZQ and e ∈ {0, 1}. So, nβ = nβ0 + neµ. If e = 1, then neµ∈ / ZQ, since n is

58 odd. Hence, e = 0 and α = nβ = nβ0 ∈ (n) in ZQ. This shows that Kern(φ) = (n), and so ZH/(n) contains ZQ/(n) as a subring.

It remains to show that φ is onto. Let γ ≡ a + bi + cj + dk + Eµ be a residue in

ZH/(n), where a, b, c, d ∈ Z and E ∈ {0, 1}. If E = 0, then γ ∈ Im(φ), so assume that E = 1. Since n is odd, 2 is invertible in ZH/(n); hence, µ ≡ y(1 + i + j + k) for some y ∈ Z, and

γ ≡ a + bi + cj + dk + Eµ

≡ (a + y) + (b + y)i + (c + y)j + (d + y)k, which is also in Im(φ). Thus, φ is onto and ZQ/(n) ∼= ZH/(n).

(ii) Since n ≡ 0 mod I, we have (n) ⊆ I in ZH. So, R is a quotient ring of

ZH/(n), which by (i) is isomorphic to ZQ/(n).

Because of this theorem, everything that was proven in Chapter 4 for quotient rings of ZQ of odd characteristic also holds for quotient rings of ZH of odd characteristic.

The other theorems from Chapter 4 either hold for ZH without change or need only minor modiﬁcations. Explicitly,

• Correspondence 4.1.1, Theorem 4.3.2, and Lemma 4.3.5 hold without change

when ZQ is replaced ZH in the statement and proof.

• Proposition 4.1.3 can be proved using the fact that the maximal ideals of ZH are (1 + i) and (p), where p is an odd prime.

• Proposition 4.1.4 requires modiﬁcation, but can be proved in the same way as

Theorem 3.1.1.

59 • Proposition 4.1.5 is true if we work with all the monic quadratic polynomials

2 in Z/nZ, not just those of the form x + 2bx + c.

All the other results in Chapter 4 are stated for quotient rings of ZQ of odd characteristic, and hence are valid for ZH because of Theorem 5.1.1. Finally, the conjecture

(4.3.11) about Muff(R) when R is a quotient ring of ZQ and char(R) is even also has its analogue. We conjecture that when R is a quotient ring of ZH of even characteristic n, then Muff(R) = φR, (1 + i)Muff(] R/Ann(1 + i)), (5.1.2) R R p1Muff(] /Ann(p1)), . . . , ptMuff(] /Ann(pt)) , where φR is a monic polynomial in Muff(R) ∩ Z/nZ [x] of minimal degree, and p1, . . . , pt are all the odd primes dividing n. The reason we do not have

(1 + j)Muﬀ(] R/Ann(1 + j)) in (5.1.2) is because in ZH, µ(1 + i)µ−1 = 1 + j, so the ideal generated by 1 + i and 1 + j is principal with generator 1 + i.

Just as in the ZQ case, for (5.1.2) to hold, it suﬃces

10) to have Theorem 4.3.6 hold for quotient rings of ZH of even characteristic, and

20) to prove that if R is a quotient ring of ZH, f ∈ Muﬀ(R), and con(f) ⊆ (1 + i), then f is contained in the ideal generated by (1 + i)Muﬀ(] R/Ann(1 + i)).

What is nice about the ZH situation is that we have proofs for 10) and 20) (given in Theorems 5.3.2 and 5.3.4, respectively). Ultimately, the reason we will be able to establish 10) and 20) is because ZH is a principal ideal ring. Since the maximal ideal of ZH above (2) is generated by 1 + i, we will be able to keep quotient rings

60 of ZH of characteristic 2m “under control,” and we will be able to give convenient characterizations of these rings. Another useful consequence of our work with quotient rings of ZH will be a proof that Int(ZH) is closed under bar conjugation. We begin with a number theoretic lemma. This lemma is elementary and is unlikely to be a new result, but the author could not ﬁnd a proof in the available literature.

Lemma 5.1.3. Let a, b, c, d ∈ Z, and let m > 0.

(i) If a2 + b2 + c2 + d2 ≡ 0 mod 22m, then all of a, b, c, and d are divisible by 2m−1.

(ii) If b2 + c2 + d2 ≡ 0 mod 22m, then all of b, c, and d are divisible by 2m.

Proof. (i) We use induction on m. There is nothing to prove if m = 1, so assume that m > 1 and that the result is true for m − 1.

n1 n2 n3 n4 Let a = 2 q1, b = 2 q2, c = 2 q3, and d = 2 q4, where n1, n2, n3, and n4

are all greater than or equal to 0, and q1, q2, q3, and q4 are all odd. Assume that

a2 + b2 + c2 + d2 ≡ 0 mod 22m. Then,

2n1 2 2n2 2 2n3 2 2n4 2 2m 2 q1 + 2 q2 + 2 q3 + 2 q4 ≡ 0 mod 2 .

2m Suppose that some n` = 0; WLOG, assume that n1 = 0. Since m > 1, 2 ≥ 16,

so the above equivalence is valid mod 8, i.e.

a2 + b2 + c2 + d2 ≡ 0 mod 8 .

2 2n1 2 But, n1 = 0 and q1 is odd, so a ≡ 2 q1 ≡ 1 mod 8. Hence,

1 + b2 + c2 + d2 ≡ 0 mod 8,

for integers b, c, and d, which is impossible. So, each n` > 0.

61 2n1 2 2n2 2 2n3 2 2n4 2 2m Now, since each n` > 0 and 2 q1 +2 q2 +2 q3 +2 q4 ≡ 0 mod 2 , we have

2(n1−1) 2 2(n2−1) 2 2(n3−1) 2 2(n4−1) 2 2(m−1) 2 q1 + 2 q2 + 2 q3 + 2 q4 ≡ 0 mod 2 .

By induction, all of n1 − 1, n2 − 1, n3 − 1, and n4 − 1 are greater than or equal to

m − 2. Hence, n1, n2, n3, and n4 are all greater than or equal to m − 1, and therefore

a, b, c, and d are all divisible by 2m−1.

(ii) Assume that b2 + c2 + d2 ≡ 0 mod 22m. By applying part (i) with a = 0,

m−1 m−1 m−1 we see that each of b, c, and d is divisible by 2 . Write b = 2 b1, c = 2 c1,

m−1 2 2 2 2m and d = 2 d1 for some b1, c1, d1 ∈ Z. Since b + c + d ≡ 0 mod 2 , we have

2m−2 2 2 2 2m 2 2 2 2 (b1 + c1 + d1) ≡ 0 mod 2 . This means that b1 + c1 + d1 ≡ 0 mod 4. Thus, b1,

m c1, and d1 must all be even, and therefore b, c, and d are all divisible by 2 .

Next, we have the following proposition, which establishes a standard form for the

residues in ZH/(2m).

Proposition 5.1.4. Let m > 1, let R = ZQ/(2m), and let I = (2m−1 + 2m−1i +

2m−1j + 2m−1k) in R. Then, ZH/(2m) contains a subring S that is isomorphic to

R/I, and ZH/(2m) = S[µ].

Proof. Let ι : ZQ → ZH be the injection map, let π : ZH → ZH/(2m) be the

canonical quotient map, and let φ = π ◦ ι. Let S = Im(φ). Then, (2m) in ZQ is

contained in Kern(φ), so S is isomorphic to a quotient ring of R = ZQ/(2m).

Note that if α ∈ Kern(φ), then ι(α) = 2mβ for some β ∈ ZH, so N(α) = N(2mβ)

is divisible by 22m. By Lemma 5.1.3 part (i), α lies in (2m−1) in ZQ. Now, a residue

in R is both equivalent to an element of (2m−1) in ZQ and has norm divisible by 22m

62 if and only if it lies in I. So, we conclude that Kern(φ) = I and S ∼= R/I. Finally, since S is the image of ZQ under φ, S consists of all the residues in ZH/(2m) that can written as a + bi + cj + dk + eµ with e = 0. It follows that we can obtain the

entire quotient ring ZH/(2m) by adjoining µ to S. Thus, ZH/(2m) = S[µ].

The eﬀect of Proposition 5.1.4 is best illustrated with an example. Consider

ZH/(2). Each residue in ZQ/(2) represents a residue in ZH/(2), but 1+i+j+k = 2µ

in ZH, so 1+i+j +k ≡ 0 in ZH/(2). Thus, 1+i ≡ j +k, 1+j ≡ i+k, 1+j +k ≡ i,

etc., so the image of ZQ in ZH/(2) consists of the eight residues in the following set:

{0, 1, i, j, k, 1 + 1, 1 + j, 1 + k} .

Hence, ZH/(2) is a ring of order 16 that may be expressed as the following set of

residues:

{0, 1, i, j, k, 1 + 1, 1 + j, 1 + k,

µ, 1 + µ, i + µ, j + µ, k + µ, 1 + i + µ, 1 + j + µ, 1 + k + µ} .

Analogous expressions hold for ZH/(2m) whenever m > 1. In general, whenever we

choose elements from ZH/(2m), we shall assume the elements come from a set of

residues like the one above.

We can now prove some interesting results about elements and quotient rings of

ZH. Our ﬁrst theorem gives a nice characterization of those elements of ZH that lie in (2m).

Theorem 5.1.5. Let α ∈ ZH and let m > 0. Then, α ∈ (2m) if and only if N(α) ≡ 0 mod 22m.

63 Proof. (⇒) Assume that α ≡ 0 mod 2m. Then, either each coeﬃcient of α is divisible by 2m, or α ≡ 2m−1 +2m−1i+2m−1j +2m−1k mod 2m. In the former case, it is easy to see that N(α) ≡ 0 mod 22m, and in the latter case we have α = 2m−1(1+i+j+k)+2mβ

for some β ∈ ZH. Let β = x + yi + zj + wk + eµ, where x, y, z, w ∈ Z and e ∈ {0, 1}. Then,

α = 2m−1(1 + i + j + k + 2(x + yi + zj + wk + eµ))

= 2m−1(2x + e + 1 + (2y + e + 1)i + (2z + e + 1)j + (2w + e + 1)k),

so we have

N(α) = 22m−2((2x + e + 1)2 + (2y + e + 1)2 + (2z + e + 1)2 + (2w + e + 1)2)

= 22m−2(4(x2 + y2 + z2 + w2) + 4(x + y + z + w)(e + 1) + 4(e + 1)2)

≡ 0 mod 22m .

(⇐) Assume that N(α) ≡ 0 mod 22m, and write α = a + bi + cj + dk + eµ, where

a, b, c, d ∈ Z and e ∈ {0, 1}. If e = 1, then

1 2 1 2 1 2 1 2 N(α) = (a + 2 ) + (b + 2 ) + (c + 2 ) + (d + 2 )

2 2 2 2 1 = a + b + c + d + a + b + c + d + 4( 4 )

≡ 1 mod 2, because a, b, c, d ∈ Z.

which contradicts the assumption that N(α) ≡ 0 mod 22m. So, we may assume that

e = 0. Then, α ∈ ZQ and a2 + b2 + c2 + d2 ≡ 0 mod 22m. By Lemma 5.1.3 part (i), all of a, b, c, and d are divisible by 2m−1.

Assume that some coeﬃcient of α is divisible by 2m. WLOG, assume that a ≡ 0

mod 2m. Then, a2 ≡ 0 mod 22m, so b2 + c2 + d2 ≡ 0 mod 22m. By Lemma 5.1.3

64 part (ii), b, c, and d are all congruent to 0 mod 2m. So, α ∈ (2m) in this case. If no

coeﬃcient of α is divisible by 2m, then each coeﬃcient of α is congruent to 2m−1 mod

2m, and so in ZH/(2m),

α ≡ 2m−1 + 2m−1i + 2m−1j + 2m−1k ≡ 2mµ ≡ 0.

Thus, α is once again in (2m).

Note that Theorem 5.1.5 does not hold in ZQ, since for any m > 0, the norms of 2m and 2m−1(1 + i + j + k) are both equal to 22m, but 2m−1(1 + i + j + k) ∈/ (2m).

The ﬁrst calculation in the proof of the reverse implication of Theorem 5.1.5 is useful enough that we record it separately:

Lemma 5.1.6. Let α ∈ ZH and write α = a + bi + cj + dk + eµ where a, b, c, d ∈ Z and e = 1. Then, the norm of α is odd.

Using Lemma 5.1.6, we can prove the following proposition.

Proposition 5.1.7. Let R = ZH/(2), and let I = (1+i) in R. Then, I is a maximal ∼ ∼ ideal of R, and R/I = F4, the ﬁeld with 4 elements. Furthermore, ZH/(1 + i) = F4.

Proof. Recall that after Proposition 5.1.4 we determined that R is a ring of order 16

consisting of the following residues:

{0, 1, i, j, k, 1 + 1, 1 + j, 1 + k,

µ, 1 + µ, i + µ, j + µ, k + µ, 1 + i + µ, 1 + j + µ, 1 + k + µ} .

By Lemma 5.1.6, those residues not in I = {0, 1 + i, 1 + j, 1 + k} all have odd

norm, and hence are units in R. So, I must be a maximal ideal of R, and R/I ∼ ∼ is a division ring of order 4; hence, R/I = F4. Finally, ZH/(1 + i) = F4 because

ZH/(1 + i) ∼= R/I.

65 We can get a result similar to Theorem 5.1.5 for the ideal of ZH generated by 2m + 2mi, but we need the help of a lemma.

Lemma 5.1.8.

(i) Let α ∈ ZH. Then, there exist β, γ ∈ ZH such that α(1 + i) = (1 + i)β and (1 + i)α = γ(1 + i).

(ii) Let m ≥ 0 and let α ∈ (2m +2mi) in ZH. Then, there exist β, γ ∈ ZH such that α = (2m + 2mi)β and α = γ(2m + 2mi). Thus, the two-sided ideal (2m + 2mi)

equals both the right ideal (2m + 2mi)ZH and the left ideal ZH(2m + 2mi).

(iii) For each α ∈ ZH, α ∈ (1 + i) if and only if N(α) is even.

Proof. (i) It suﬃces to prove this for each u ∈ {i, j, k, µ}. Then, the stated result follows by writing α in the standard form α = a + bi + cj + dk + eµ with a, b, c, d ∈ Z and e ∈ {0, 1}.

If u ∈ {i, j, k}, then since ZQ/(2) is commutative, we have u(1 + i) ≡ (1 + i)u mod 2. Hence, there exists δ ∈ ZQ such that

u(1 + i) = (1 + i)u + 2δ = (1 + i)(u + (1 − i)δ) and we can take β = (u + (1 − i)δ). The existence of γ in the case where u ∈ {i, j, k} can be proved in a similar manner.

If u = µ, then one need merely notice that µ(1 + i) = (1 + i)(µ − k) and

(1 + i)µ = (µ − j)(1 + i), so we may take β = µ − k and γ = µ − j.

n m m P m m (ii) Since α ∈ (2 + 2 i), we may write α = αr(2 + 2 i)βr for some αr, βr r=1 0 0 (1 ≤ r ≤ n) in ZH. By part (i), for each r there exist αr, βr ∈ ZH such that

66 0 0 αr(1 + i) = (1 + i)αr and (1 + i)βr = βr(1 + i). Hence,

n n n X m m X m 0 m m X 0 α = αr(2 + 2 i)βr = (2 )(1 + i)αrβr = (2 + 2 i) αrβr, and r=1 r=1 r=1 n n n X m m X 0 m X 0 m m α = αr(2 + 2 i)βr = αrβr(2 )(1 + i) = αrβr (2 + 2 i), r=1 r=1 r=1

n n P 0 P 0 so we can take β = αrβr and γ = αrβr. r=1 r=1

(iii) (⇒) Assume ﬁrst that α ∈ (1+i). By part (i), α = (1+i)β for some β ∈ ZH, so N(α) = N(1 + i)N(β) = 2N(β).

(⇐) Assume that α ∈ ZH has even norm. Then, the residue of α modulo 2 is not

invertible in ZH/(2), so α mod 2 lies in the maximal ideal of ZH/(2) that is generated

by (1 + i). Note that part (ii) also holds in ZH/(2), so we may write α ≡ (1 + i)β

mod 2 for some β ∈ ZH. Thus, there exists α0 ∈ ZH such that

α = (1 + i)β + 2α0 = (1 + i)(β + (1 − i)α0),

and hence α ∈ (1 + i).

Theorem 5.1.9. Let α ∈ ZH and let m ≥ 0. Then, α ∈ (2m + 2mi) if and only if N(α) ≡ 0 mod 22m+1.

Proof. (⇒) Assume that α ∈ (2m + 2mi). Then, by Lemma 5.1.8 part (ii), α =

(2m + 2mi)β for some β ∈ ZH, and hence N(α) = N(2m + 2mi)N(β) = 22m+1N(β).

(⇐) Assume that N(α) ≡ 0 mod 22m+1. Then, N(α) ≡ 0 mod 22m, so by Theorem

5.1.5, α ∈ (2m) in ZH and α = 2mβ for some β ∈ ZH. Since N(α) is divisible by 22m+1 and N(2m) = 22m, the norm of β must be even. By Lemma 5.1.8, β ∈ (1 + i)

67 and we may write β = (1 + i)γ for some γ ∈ ZH. Then, α = 2mβ = (2m + 2mi)γ ∈ (2m + 2mi).

The remainder of this section will be dedicated to classifying the quotient rings of ZH of characteristic 2m. Most of the work for the classiﬁcation is done in the following theorem.

Theorem 5.1.10. Let m ≥ 0. If I is an ideal of ZH such that (2m+1) ⊂ I ⊂ (2m), then I = (2m + 2mi).

Proof. Since (2m+1) ⊂ (2m + 2mi) ⊂ (2m), there exist ideals of ZH properly between

(2m+1) and (2m). So, let I be such an ideal of ZH.

Let R = ZH/(2m+1) and let π : ZH → R be the quotient map. We will prove that the only ideal of R properly between 2m+1R and 2mR is (2m + 2mi)R. Then, since there is an inclusion preserving bijection between ideals of R and ideals ZH containing (2m+1), we may conclude that I = π−1((2m + 2mi)R) = (2m + 2mi).

Let α ∈ 2mR. Since ZH is a principal ideal ring, the same is true for R, so it suﬃces to show that (α) in R equals one of 2mR, 2m+1R, or (2m + 2mi)R. Let

S = {0, 1, i, j, k, 1 + 1, 1 + j, 1 + k,

µ, 1 + µ, i + µ, j + µ, k + µ, 1 + i + µ, 1 + j + µ, 1 + k + µ}

(these are the residues for ZH/(2) that were given following Proposition 5.1.4). Then,

α ≡ 2mβ for some β ∈ S. If β∈ / {0, 1 + 1, 1 + j, 1 + k}, then by Lemma 5.1.6, β has odd norm, and hence is a unit in R. In this case, (α) = 2mR. If β = 0, then (α) = (0); that is, (α) = 2m+1R. Thus, the only way that (α) 6= 2mR and (α) 6= 2m+1R is if

β ∈ {1 + 1, 1 + j, 1 + k}, in which case (α) = (2m + 2mi)R.

68 Corollary 5.1.11. Let m > 0 and let R = ZH/(2m). Then, the ideals of R are

linearly ordered under inclusion.

Proof. First, let α ∈ R − {0} and let N(α) = 2nq, where 0 ≤ n ≤ 2m − 1 and q is

odd. If n is even, then n = 2` for some ` ∈ Z, ` ≤ m − 1, and by Theorems 5.1.5 and 5.1.9, α ∈ 2`R but α∈ / (2` + 2`i)R, so (α) = 2`R by Theorem 5.1.10. Similarly, if n is odd and n = 2` + 1, then (α) = (2` + 2`i)R.

Next, let I = (β) and J = (γ) be two ideals in R (here, we are using the fact that

n1 n2 R, like ZH, is a principal ideal ring). Let N(β) = 2 q1 and N(γ) = 2 q2, where n1, n2 ≥ 0 and q1 and q2 are odd. By the arguments of the preceding paragraph, we see that if n1 ≥ n2, then I ⊆ J, and if n2 ≥ n1, then J ⊆ I. We have containment in

either case, so the ideals of R are linearly ordered under inclusion.

Corollary 5.1.12. Let m > 0 and let R be quotient ring of ZH with char(R) = 2m.

Then, either R ∼= ZH/(2m) or R ∼= ZH/(2m−1 + 2m−1i).

Proof. Since char(R) = 2m, R is isomorphic to a quotient ring of S = ZH/(2m). Let

I be the ideal of S such that R ∼= S/I.

If 2m−1S ⊆ I, then char(R) ≤ 2m−1, which is a contradiction. By Corollary

5.1.11, I is an ideal of S properly contained in 2m−1S. By Theorem 5.1.10, either

I = (2m−1 + 2m−1i)S or I = 2mS. In the former case, R ∼= ZH/(2m−1 + 2m−1i), and

in the latter case R ∼= ZH/(2m).

5.2 Int(ZH) is Closed Under Bar Conjugation

Using the results of the previous section, we can prove that Int(ZH) is closed under bar conjugation. The gist of the proof is embodied in the following theorem.

69 Theorem 5.2.1. The ring Int(ZH) is closed under bar conjugation if and only if for each prime p and each m > 0, the ideal Muﬀ ZH/(pm) of ZH/(pm) [x] is closed

under bar conjugation.

Proof. (⇒) Assume that Int(ZH) is closed under bar conjugation. Let p be a

H m f(x) prime, let m > 0, and let f(x) ∈ Muﬀ Z /(p ) . Then, pm ∈ Int(ZH).

f(x) Since Int(ZH) is closed under bar conjugation, pm ∈ Int(ZH), which implies that f(x) ∈ Muﬀ ZH/(pm) . (⇐) Assume that for each prime p and each m > 0, Muﬀ ZH/(pm) is closed

g(x) under bar conjugation. Let f(x) ∈ Int(ZH). Then, we may write f(x) = n for

some g(x) ∈ ZH[x] and some n > 0. If n = 1, then f(x) ∈ ZH[x], so f(x) ∈ ZH[x] ⊆

Int(ZH) and we are done. So, assume that n > 1. Next, consider the case where n is a power of a prime. Assume that n = pe for

g(x) e some prime p and some e > 0. Then, f(x) = n ∈ Int(ZH), so g(x) mod p is in Muﬀ ZH/(pe) . Since Muﬀ ZH/(pe) is closed under bar conjugation, we have g(x)

e H e g(x) mod p ∈ Muﬀ Z /(p ) , and so f(x) = pe ∈ Int(ZH).

e1 et Finally, assume that n has prime factorization n = p1 ··· pt , where t > 1. Notice that if n factors as n = q1q2 for relatively prime integers q1 and q2, then there exist

g(x) bg(x) ag(x) g(x) a, b ∈ such that aq1 + bq2 = 1. In this case, = + , and both and Z n q1 q2 q1

g(x) e1 et are elements of Int( H). Now, since n = p ··· p , there exist a1, a2, . . . , at ∈ q2 Z 1 t Z t g(x) P arg(x) g(x) such that = er and each er ∈ Int( H). By the previous paragraph, n pr pr Z r=1 t g(x) g(x) P arg(x) er ∈ Int( H) for each r. Hence, f(x) = = er ∈ Int( H), and we pr Z n pr Z r=1 conclude that Int(ZH) is closed under bar conjugation.

70 Our strategy is now clear: to prove that Int(ZH) is closed under bar conjugation,

it suffices to work in ZH/(pm), where p is a prime. When p is odd, proving that Muff ZH/(pm) is closed under bar conjugation is easy given some of our previous results, but things are more difficult when p = 2, and we will rely heavily on the theorems in Section 5.1. The closure of the needed muffin ideals follows from Theorem

5.2.3, where we will actually achieve a stronger result that will be used when we establish a generating set for Int(ZH) in Section 5.3. Before dealing with Theorem 5.2.3, we prove one more result that will be useful.

Theorem 5.2.2. Let R be a quotient ring of ZH with char(R) = 2m. Let α ∈ R. Then, the following are equivalent:

(i) (1 + i)α ≡ 0 in R

(ii) α(1 + i) ≡ 0 in R

(iii) α ∈ AnnR(1 + i)

Proof. The implications (iii) ⇒ (i) and (iii) ⇒ (ii) are clear. We will ﬁrst prove that

(i) ⇔ (ii). By Theorem 5.1.12, either R ∼= ZH/(2m) or R ∼= ZH/(2m−1 + 2m−1i). In

the former case, we have

m (1 + i)α ≡ 0 in R ⇐⇒ (1 + i)α ∈ (2 ) in ZH

⇐⇒ N((1 + i)α) ≡ 0 mod 22m

⇐⇒ N(α(1 + i)) ≡ 0 mod 22m

⇐⇒ α(1 + i) ≡ 0 in R , and so (i) holds if and only if (ii) holds. Similarly, (i) ⇔ (ii) is true when R ∼=

ZH/(2m−1 + 2m−1i). This establishes that (i) ⇔ (ii).

71 To complete the proof, it suﬃces to show that (i) ⇒ (iii). Assume that (1+i)α ≡ 0 in R. Then, α(1 + i) is also equivalent to 0. Let β ∈ (1 + i). By Lemma 5.1.8 part

(ii), there exist γ, δ ∈ ZH such that β ≡ γ(1 + i) and β ≡ (1 + i)δ. Then,

βα ≡ γ(1 + i)α ≡ 0, and

αβ ≡ α(1 + i)δ ≡ 0 .

Thus, (iii) holds, and we are done.

We can now prove that the muﬃns ideals in ZH/(pm) are closed under bar conjugation. The proof of the following theorem is quite lengthy, and we break it up into several cases.

Theorem 5.2.3. Let p be a prime and let m > 0. Let R be a quotient ring of ZH with char(R) = pm, and let f ∈ Muﬀ(R). Then, f ∈ Muﬀ(R).

Proof. Assume ﬁrst that p is odd. Recall Lemma 3.1.3, where it was shown that for any g(x) ∈ H[x] and any β ∈ H, we have 1 1 g(β) = g(β) + g(β) + g(−iβi)i + g(−iβi)i i 2 2 1 1 + g(−jβj)j + g(−jβj)j j + g(−kβk)k + g(−kβk)k k . 2 2

Since 2 is invertible in R, this expression is also true over R. So, for each β ∈ R, we have

1 1 f(β) ≡ f(β) + f(β) + f(−iβi)i + f(−iβi)i i 2 2 1 1 + f(−jβj)j + f(−jβj)j j + f(−kβk)k + f(−kβk)k k . 2 2

Since f ∈ Muﬀ(R), each term on the right-hand side of the equivalence is congruent to 0 in R. Thus, f(β) ≡ 0 for each β ∈ R, and therefore f ∈ Muﬀ(R). This completes the proof in the case where p is odd.

72 Assume from now on that p = 2. Let α ∈ R. Since minα(x) is monic, we have

f(x) = q(x) min α(x) + γx + δ

for some q(x), γx + δ ∈ R[x]. Now, minα(x) has integer coeﬃcients, so minα(x) is central in R[x] and minα(x) = minα(x). This means that

q(x) min α(x) = min α(x) q(x) = min α(x)q(x) = q(x) min α(x) .

So, we have

f(x) = q(x) min α(x) + γx + δ .

We wish to show that f(α) ≡ 0. By the above, we have f(α) ≡ q(α) minα(α) +

γα + δ ≡ γα + δ, so it suﬃces to show that γα + δ ≡ 0. Notice that

γα + δ ≡ 0 ⇐⇒ γα + δ ≡ 0

⇐⇒ αγ + δ ≡ 0

⇐⇒ αγ + δ ≡ f(α) ≡ γα + δ

⇐⇒ γα − αγ ≡ 0 . (∗)

We will show that (∗) holds.

If minα(x) is linear, then γ ≡ 0, so (∗) holds. So, we may assume that minα(x) is quadratic, i.e. that α∈ / Z/2mZ in R.

Now, whenever β ∈ R and minβ(x) = minα(x), we have 0 ≡ f(β) ≡ γβ + δ.

Assume that α ≡ a + bi + cj + dk + eµ, where a, b, c, d ∈ Z and e ∈ {0, 1}. If e = 1, then

α − (−iαi) ≡ (a + bi + cj + dk + µ) − (a − bi + cj + dk + (µ − i))

≡ (a + bi + cj + dk + µ) − (a − (b + 1)i + cj + dk + µ)

≡ (2b + 1)i,

73 which is unit in R since 2b + 1 is odd. Hence,

0 ≡ f(α) − f(−iαi)

≡ (γα + δ) − (γ(−iαi) + δ)

≡ γ(α − (−iαi))

≡ (2b + 1)γi, implying that γ ≡ 0. Thus, (∗) holds when e = 1, and we may assume that e = 0.

Since e = 0, we have α ≡ a + bi + cj + dk, and we compute that

γα − αγ ≡ b(γi + iγ) + c(γj + jγ) + d(γk + kγ) .

So, it suﬃces to show that

b(γi + iγ) + c(γj + jγ) + d(γk + kγ) ≡ 0 . (∗∗)

∼ If R = ZH/(1 + i), then R = F4, the ﬁeld with 4 elements, in which case R is commutative and

b(γi + iγ) + c(γj + jγ) + d(γk + kγ) ≡ 2bγi + 2cγj + 2dγk ≡ 0,

so (∗∗) is true. So, assume that |R| > 4, and let I = AnnR(2) and J = AnnR(1 + i).

Since 0 ≡ f(α) − f(−iαi) ≡ 2bγ, we have bγ ∈ I. Similarly, cγ, dγ ∈ I. Furthermore,

0 let α ≡ a − ci − bj + dk. Then, minα0 (x) = minα(x), so

0 ≡ f(α) − f(α0)

≡ γ((b + c)i + (b + c)j)

≡ (b + c)γ(i + j) .

74 Since i+j is a generator for (1+i)R, Lemma 5.2.2 tells us that (b+c)γ ∈ J. Similarly,

(b + d)γ,(c + d)γ ∈ J.

From here, we have two broad cases to consider, depending on the isomorphism

type of R. By Corollary 5.1.12, either R ∼= ZH/(2m) or R ∼= ZH/(2m−1 + 2m−1i).

Case 1: R ∼= ZH/(2m)

In this case, I = 2m−1R and J = (2m−1 + 2m−1i)R. Then, J = {0, 2m−1 + 2m−1i,

2m−1 + 2m−1j, 2m−1 + 2m−1k}; in particular, note that J ⊆ I. Let

J 0 = J ∪ {2m−1, 2m−1i, 2m−1j, 2m−1k} .

Then, J ⊂ J 0 ⊂ I, and J 0 is not an ideal of R, but J 0 is closed under addition.

Furthermore, notice that each element of J 0 commutes (modulo 2m) with i, j, and k.

Now, we showed above that bγ, cγ, dγ ∈ I and (b + c)γ,(b + d)γ,(c + d)γ ∈ J.

If bγ ∈ J 0, then

cγ ≡ (b + c)γ − bγ ∈ J 0, and

dγ ≡ (b + d)γ − bγ ∈ J 0 .

Similarly, having cγ or dγ in J 0 forces all three of bγ, cγ, and dγ to be in J 0. So,

either all three of bγ, cγ, and dγ are in J 0, or all three are in I − J 0. Thus, we have

two subcases to consider.

Subcase 1a: bγ, cγ, dγ ∈ J 0

In this case, since each element of J 0 commutes with i, j, and k, we have

b(γi + iγ) + c(γj + jγ) + d(γk + kγ) ≡ 2bγi + 2cγj + 2dγk ≡ 0,

75 so (∗∗) holds and we are done.

Subcase 1b: bγ, cγ, dγ ∈ I − J 0

Let ε ∈ I − J 0. Then, ε ≡ 2m−1ε0, where ε0 is an element of

{µ, 1 + µ, i + µ, j + µ, k + µ, 1 + i + µ, 1 + j + µ, 1 + k + µ}, so we may write

m−2 m−2 m−2 m−2 ε ≡ u12 + u22 i + u32 j + u42 k,

m−1 m−1 where u1, u2, u3, u4 ∈ {±1}. Then, since 2 ≡ −2 ,

m−2 m−2 εi + iε ≡ 2 (−u2 + u1i + u4j − u3k) + 2 (−u2 + u1i − u4j + u3k)

≡ −2m−1 + 2m−1i

≡ 2m−1 + 2m−1i .

Similarly, εj + jε ≡ 2m−1 + 2m−1j and εk + kε ≡ 2m−1 + 2m−1k. Since bγ, cγ, and dγ are all in I − J 0, we get

b(γi + iγ) + c(γj + jγ) + d(γk + kγ)

≡ (2m−1 + 2m−1i) + (2m−1 + 2m−1j) + (2m−1 + 2m−1k)

≡ 2m−1 + 2m−1i + 2m−1j + 2m−1k

≡ 0 .

Thus, (∗∗) holds.

There are no more subcases, so Case 1 is complete.

76 Case 2: R ∼= ZH/(2m−1 + 2m−1i)

We use a strategy similar to that employed in Case 1. However, in this case it is

more diﬃcult to establish when two elements of R are equivalent. When in doubt,

the best way to check whether β1, β2 ∈ R are equivalent is to compute N(β1 − β2).

2m−1 m−1 m−1 If this norm is divisible by 2 = N(2 + 2 i), then β1 ≡ β2 in R. In light of this, it is not hard to see that

2m−1, 2m−1i, 2m−1j, and 2m−1k are all equivalent in R.

Keeping this relationship in mind will help expedite the calculations that follow.

Proceeding as in Case 1, we have I = (2m−2 + 2m−2i)R, J = 2m−1R, and J ⊆ I.

As in Case 1, |I| = 16 and |J| = 4. Considering I and J as additive groups, this

means that J is an index 4 subgroup of I. We will compute the cosets of J in I,

which will allow us to partition I into subsets that are amenable to the same sort of

calculations done in Case 1. Using the fact that 2m−1, 2m−1i, 2m−1j, and 2m−1k all

represent the same residue in R, we can compute that

J = {0, 2m−1, 2m−1u, 2m−1(1 + µ)} .

Then, the other cosets of J in I are given by

m−2 m−2 m−2 m−2 m−2 m−2 J1 = 2 + 2 i + J = {2 ± 2 i, 2 j ± 2 k},

m−2 m−2 m−2 m−2 m−2 m−2 J2 = 2 + 2 j + J = {2 ± 2 j, 2 i ± 2 k}, and

m−2 m−2 m−2 m−2 m−2 m−2 J3 = 2 + 2 k + J = {2 ± 2 k, 2 i ± 2 j} ,

so that I is the disjoint union of J, J1, J2, and J3.

Now, bγ lies in exactly one of the cosets J, J1, J2, or J3; assume that bγ ∈ β + J,

where β is an element of {0, 2m−2 + 2m−2i, 2m−2 + 2m−2j, 2m−2 + 2m−2k}. Since

(b + c)γ is in J, we must have cγ ∈ β + J. Similarly, dγ ∈ β + J.

77 Note that each element of J commutes (in R) with i, j, and k. Since bγ ∈ β + J, we have bγ ≡ β + ε for some ε ∈ J. Then,

b(γi + iγ) ≡ βi + iβ + εi + iε

≡ βi + iβ + 2εi

≡ βi + iβ,

where the last equivalence follows because ε ∈ I, so 2ε ≡ 0. Similarly, we can compute

that

c(γj + jγ) ≡ βj + jβ and d(γk + kγ) ≡ βk + kβ .

Thus, to establish that (∗∗), it suﬃces to verify that

(βi + iβ) + (βj + jβ) + (βk + kβ) ≡ 0

as β runs along the coset representatives {0, 2m−2 + 2m−2i, 2m−2 + 2m−2j, 2m−2 +

2m−2k}. This is certainly true when β ≡ 0. If β ≡ 2m−2 + 2m−2i, then βi + iβ ≡

2(−2m−2 + 2m−2i) ≡ 0, βj + jβ ≡ 2m−1j, and βk + kβ ≡ 2m−1k, so

(βi + iβ) + (βj + jβ) + (βk + kβ) ≡ 0 + 2m−1j + 2m−1k ≡ 0 .

Similarly, (βi + iβ) + (βj + jβ) + (βk + kβ) ≡ 0 when β ≡ 2m−2 + 2m−2j or

β ≡ 2m−2 + 2m−2k. Therefore, we conclude that (∗∗) holds in all instances. This

proves Case 2.

There are no more cases to consider, so the proof is complete.

Corollary 5.2.4. Int(ZH) is closed under conjugation; that is, if f ∈ Int(ZH), then

f ∈ Int(ZH).

Proof. This follows from Theorems 5.2.1 and 5.2.3.

78 5.3 A Generating Set for Int(ZH)

Having established that Int(ZH) is closed under bar conjugation, we now have enough tools to construct a generating set for Int(ZH). The first step in this process is to extend Theorem 4.3.6 to quotient rings of ZH. We shall do this in stages, with Theorems 5.3.1 through 5.3.3. Before stating Theorem 5.3.1, we recall two definitions from Chapter 4. By the content of a polynomial f, we mean the ideal generated by the coefficients of f (see the paragraph prior to Lemma 4.3.5). Also, when R is a quotient ring of ZH of characteristic n, the polynomial φR is a monic polynomial of minimal degree in Muff(R) ∩ Z/nZ [x] (see Section 4.2).

Theorem 5.3.1. Let R be a quotient ring of ZH with char(R) = 2m, where m > 0. Assume that f is a content 1 polynomial of minimal degree in Muﬀ(R). Then, the leading coeﬃcient of f is a unit, so we may assume that f is monic.

∼ ∼ Proof. We use induction on |R|. The base case occurs when R = ZH/(1 + i) = F4.

In this case, R[x] is a PID, so Muﬀ(R) = (φR) and we may take f = φR. Since φR is monic, we are done.

So, assume that |R| > 4 and that the result holds for any ﬁnite, non-zero quotient ring S of ZH such that char(S) is a power of 2 and |S| < |R|. Let γ be the leading coeﬃcient of f; then, γ 6≡ 0 in R. We want to show that γ is a unit in R. Let M = (1 + i)R. By Corollary 5.1.11, M is the unique maximal ideal

× of R and M = R − R . Suppose that γ ∈ M. Let I = AnnR(γ) and let S = R/I.

Since γ ∈ M, I 6= (0), so S 6= R.

Let g be a content 1 polynomial of minimial degree in Muﬀ(S). By the inductive hypothesis, we may assume that g is monic. Now, since con(f) = (1), we have

79 con(f mod I) = (1), so

deg(f) ≥ deg(f mod I) ≥ deg(g) .

Now, let G be a polynomial in R[x] such that G is monic, deg(G) = deg(g), and G ≡ g mod I (such a G must exist because we can always take G = g, where we abuse notation and consider the coefficients of g to be elements of R). Since g ∈ Muff(S), for all α ∈ R we have G(α) ∈ I = Ann(γ). Hence, γG(α) = 0 in R for all α ∈ R and so γG ∈ Muff(R). By the above, deg(γG) = deg(g) ≤ deg(f). Let d = deg(f) − deg(γG). Then, h = f − γxdG ∈ Muff(R), and since G is monic, either h = 0 or deg(h) < deg(f). If h = 0, then f = γxdG and con(f) ⊆ γR ⊆ M, which is a contradiction. So, deg(h) < deg(f).

Since con(f) = (1), there exists t ∈ {0, 1,..., deg(f)} such that the coeﬃcient

t αt of x in f is a unit (if all the coeﬃcients of f are non-units, then con(f) ⊆ M).

t d t Let βt be the coeﬃcient of x in x G. Then, αt − γβt is the coeﬃcient of x in h.

Since αt ∈/ M but γβt ∈ M, we cannot have αt − γβt in M. Thus, con(h) = (1).

This contradicts the minimality of deg(f). Hence, γ must be a unit, and γ−1f is a

monic, content 1 polynomial in Muﬀ(R) with the same degree as f. This completes

the proof.

Theorem 5.3.2. Let R be a quotient ring of ZH with char(R) = 2m, where m > 0.

If f is a content 1 polynomial in Muﬀ(R), then deg(f) ≥ deg(φR).

Proof. Since con(φR) = (1), Muﬀ(R) contains polynomials of content 1. Let f be

such a polynomial of minimal degree. We need to show that deg(f) ≥ deg(φR).

Assume ﬁrst that f has a coeﬃcient α of the form α ≡ a + bi + cj + dk + eµ,

where a, b, c, d ∈ Z and e = 1. By Theorem 5.2.3, f ∈ Muﬀ(R), so f + f ∈ Muﬀ(R)

80 has a coeﬃcient equivalent to α + α ≡ 2a + e. Since e = 1, α + α is a unit in R. So, con(f + f) = (1). Since f + f ∈ Muﬀ(R) ∩ Z/2mZ [x], we may apply Lemma 4.3.5

to f + f to conclude that deg(f + f) ≥ deg(φR). Thus,

deg(f) ≥ deg(f + f) ≥ deg(φR)

in this case. So, we may assume that f has coeﬃcients in ZQ/(2m).

By Theorem 5.3.1, we may assume that f is monic. Letting f1, fi, fj, and fk be the component polynomials of f (see Deﬁnition 4.3.4), we have deg(f) = deg(f1), and this degree is stricly larger than deg(fi), deg(fj), or deg(fk).

m Now, since f ∈ Muﬀ(R), 2f1 = f + f ∈ Muﬀ(R) ∩ Z/2 Z [x]. Similarly, since

m −if, −jf, and −kf are all in Muﬀ(R), we get 2fi, 2fj, 2fk ∈ Muﬀ(R) ∩ Z/2 Z [x]. Also,

−1 2 −2 f + µfµ + µ fµ = 3f1 + (i + j + k)(fi + fj + fk) ∈ Muﬀ(R) .

Let g = fi + fj + fk. Then,

−1 2 −2 f1 + (i + j + k)g ≡ f + µfµ + µ fµ − 2f1 ∈ Muﬀ(R) .

By Corollary 5.1.12, either R ∼= ZH/(2m) or R ∼= ZH/(2m−1 + 2m−1i), so we

have two cases to consider.

Case 1: R ∼= ZH/(2m)

Let β ∈ R. Then, since 2f1 ∈ Muﬀ(R), we have 2f1(β) ≡ 0 in R. This means that

m−1 m−1 m−1 f1(β) ∈ AnnR(2) = 2 R. Similarly, fi(β), fj(β), fk(β) ∈ 2 R. So, g(β) ∈ 2 R

and hence g(β) ≡ 2m−1γ for some γ ∈ R. Since 2m−1 + 2m−1i + 2m−1j + 2m−1k ≡ 0

81 in R,

(i + j + k)g(β) ≡ (i + j + k)2m−1γ

≡ (2m−1i + 2m−1j + 2m−1k)γ

≡ 2m−1γ

≡ g(β) .

Now, f1 + (i + j + k)g ∈ Muﬀ(R), so

0 ≡ (f1 + (i + j + k)g)(β)

≡ f1(β) + (i + j + k)g(β)

≡ f1(β) + g(β)

≡ (f1 + g)(β) .

Since β was an arbitrary element of R, we conclude that f1 + g ∈ Muﬀ(R). Further-

m m more, since f1 + g has coeﬃcients in Z/2 Z, we have f1 + g ∈ Muﬀ(R) ∩ Z/2 Z [x].

So, by Lemmma 4.3.5, deg(f1 + g) ≥ deg(φR). As mentioned above, the degrees of fi, fj, and fk are all strictly less than deg(f1), so deg(f1 + g) = deg(f1). Therefore,

deg(f) = deg(f1) = deg(f1 + g) ≥ deg(φR) .

This proves Case 1.

Case 2: R ∼= ZH/(2m−1 + 2m−1i)

As in Case 1, let β ∈ R. In this case, the condition 2f1(β) ≡ 0 in R forces

m−2 m−2 m−2 m−2 f1(β) ∈ (2 + 2 i)R. Similarly, fi(β), fj(β), fk(β) ∈ (2 + 2 i)R, so g(β) ∈

82 (2m−1 + 2m−1i)R and there exists γ ∈ R such that g(β) ≡ (2m−2 + 2m−2i)γ. Then,

(i + j + k)g(β) ≡ (i + j + k)(2m−2 + 2m−2i)γ

≡ (2m−2i + 2m−2j + 2m−2k − 2m−2 + 2m−2j − 2m−2k)γ

≡ (−2m−2 + 2m−2i + 2m−1j)γ

≡ (2m−2 + 2m−2i + (2m−1 + 2m−1j))γ

≡ g(β) .

At this point, proceeding as in Case 1 shows that deg(f) ≥ deg(φR), which proves

Case 2.

There are no more cases, so we are done.

Theorem 5.3.3. Let R be a ﬁnite, non-zero quotient ring of ZH. Let f ∈ Muﬀ(R), and assume that con(f) = (1). Then, deg(f) ≥ deg(φR).

e1 et Proof. Let n = char(R), and assume that n has prime factorization n = p1 ··· pt . By using Theorem 4.3.6 and Theorem 5.3.2, we see that, for all 1 ≤ ` ≤ t,

e` deg(f) ≥ deg(f mod p R) ≥ deg(φ e` ) . ` R/p` R

Thus, by Theorem 4.3.2,

e deg(f) ≥ max{deg(φR/p ` R)} = deg(φR) . 1≤`≤t `

Theorem 5.3.3 establishes point 10) in Section 5.1. Proving the second point, 20), is not nearly as arduous.

83 Theorem 5.3.4. Let R be a ﬁnite, non-zero quotient ring of ZH, and let n = char(R). Assume that n is even. If f ∈ Muﬀ(R) and con(f) ⊆ (1 + i)R, then f

R is contained in the ideal of R[x] generated by (1 + i)Muﬀ] /AnnR(1 + i) .

P r Proof. Assume that f ∈ Muﬀ(R) and con(f) ⊆ (1 + i)R. Write f(x) = r αrx .

Then, each αr ∈ (1 + i)R, so for each r there exists βr ∈ R such that αr ≡ (1 + i)βr.

P r P r Let g(x) = r βrx . Then, f(x) = r αrx = (1 + i)g(x).

We would like to show that g(α) ∈ AnnR(1 + i) for each α ∈ R. If that is the R case, then we can show that f ∈ (1 + i)Muff] /AnnR(1 + i) . So, fix α ∈ R. Then, f(α) ≡ 0, so (1 + i)g(α) ≡ 0. Let β ∈ (1 + i)R. It suffices to show that βg(α) ≡ 0 and g(α)β ≡ 0. By Lemma 5.1.8, there exist γ, δ ∈ R such that β ≡ γ(1 + i) and

β ≡ (1 + i)δ. Using β ≡ γ(1 + i) gives βg(α) ≡ γ(1 + i)g(α) ≡ 0. So, we just need to prove that g(α)β ≡ 0.

m m Factor n as n = 2 q, where m > 0 and q is odd. Let I1 = 2 R, I2 = qR,

R R m R1 = /I1 and R2 = /I2. In R1, (1 + i)g(α) ≡ 0 and char(R1) = 2 , so by

Theorem 5.2.2 we have g(α)(1 + i) ≡ 0 mod I1. So, g(α)(1 + i) ∈ I1 in R. Note that we are done if q = 1, so assume that q > 1. Then, R2 is a non-zero ring.

In R2, we have (1 + i)g(α) ≡ 0. Since char(R2) = q is odd, 1 + i is a unit in

R2. So, g(α) ≡ 0 in R2, which means that g(α) ∈ I2 in R. Thus, g(α)(1 + i) ∈ I2 in R, and therefore g(α)(1 + i) ∈ I1 ∩ I2 = (0). Now, using β ≡ (1 + i)δ, we have g(α)β ≡ g(α)(1 + i)δ ≡ 0. Hence, g(α) ∈ AnnR(1 + i).

We have shown that g(α) ∈ AnnR(1 + i) for each α ∈ R, so g(x) mod AnnR(1 + i) R R lies in Muﬀ /AnnR(1 + i) and f(x) = (1 + i)g(x) ∈ (1 + i)Muﬀ] /AnnR(1 + i) , as required.

84 We can now describe the generators of Muﬀ(R) for any ﬁnite, non-zero quotient ring of ZH.

Theorem 5.3.5. Let R be a ﬁnite, non-zero quotient ring of ZH, and let n =

char(R). Let p1, . . . , pt be all the odd primes dividing n.

R R (i) If n is odd, then Muff(R) = (φR, p1Muff]( /Ann(p1)), . . . , ptMuff]( /Ann(pt))).

(ii) If n is even, then

Muﬀ(R) = (φR,(1 + i)Muﬀ] R/Ann(1 + i) ,

R R p1Muﬀ]( /Ann(p1)), . . . , ptMuﬀ]( /Ann(pt))) .

Proof. (i) If n is odd, then ZH/(n) ∼= ZQ/(n), so this is just a restatement of

Theorem 4.3.7.

(ii) If n is even, then we can proceed just as in the proof of Theorem 4.3.7, noting

that the maximal ideals of R are exactly p1R, . . . , ptR, and (1 + i)R.

Corollary 5.3.6.

(i) For each n > 1, Muﬀ(ZH/(n)) can be generated by polynomials of the form γf, where γ ∈ ZH/(n) and f ∈ Z/nZ [x]. In fact, we may assume that γ ∈ {1, 1 + i}.

γf (ii) Int(ZH) can be generated over ZH by polynomials of the form n , where γ ∈

ZH, f ∈ Z[x], and n ≥ 1. In fact, we may assume that γ ∈ {1, 1 + i}.

∼ ∼ Proof. (i) If R is a simple quotient ring of ZH, then either R = ZH/(1 + i) = F4, or ∼ ∼ R = ZH/(p) = ZQ/(p) for some odd prime p. In the former case, Muﬀ(R) = (φR)

85 because R[x] is a PID, and in the latter case Muﬀ(R) = (φR) by Theorem 4.3.3. So,

the result holds for simple quotient rings of ZH.

e1 et Now, let R = ZH/(n) and let n have prime factorization n = p1 ··· pt . If t = 1, then for convenience let p = p1 and e = e1. We use induction on e. If p is odd and e = 1, then we are done by the previous paragraph. If p = 2 and e = 1, then Theorem

5.3.5 shows that Muﬀ(R) = (φR, (1 + i)φZH/(1+i)), so we are done in this case as well. The result for the case t = 1 now follows by Theorem 5.3.5 and induction on e.

If t > 1, then this can be proven using Theorem 5.3.5 and induction on t.

To prove that we may assume each γ ∈ {1, 1 + i}, let δf be a generator for Muﬀ(R), where f ∈ Z/nZ [x]. By Theorem 5.3.5, we may assume that δ is a product of integers and powers of 1 + i. Since (1 + i)2 = 2i, (1 + i)3 = −2(1 − i),

4 m and (1 + i) = −4, we have δ ≡ a(1 + i) , where a ∈ Z/nZ and 0 ≤ m ≤ 3. There is nothing to prove if m = 0 or m = 1. If m = 2, then there is no loss in using iδf to

generate Muﬀ(R). If m = 3, then we may use −jδfj = (−jδj)f in place of f. So, in each case we may write δf as γg, where γ ∈ {1, 1+i} and g ∈ Z/nZ [x], as required.

(ii) We know that ZH[x] ⊆ Int(ZH), so to prove (ii) it suﬃces to focus on

f(x) polynomials in Int(ZH) that may be written as n with n > 1. For each n > 1, let γ , γ , . . . , γ ∈ H/ and f , f , . . . , f ∈ / [x] be such n1 n2 nmn Z (n) n1 n2 nmn Z nZ H that Muﬀ(Z /(n)) = (γn1 fn1 , . . . , γnmn fnmn ). Furthermore, for each n > 1, let ∞ γn1 fn1 γnmn fnmn S Pn = { n ,..., n } ⊆ Int(ZH). Take P1 = {x} and let P = Pn. We wish n=1 to show that Int(ZH) = ZH[P]. The inclusion Int(ZH) ⊇ ZH[P] is easily veriﬁed,

so we just need to show that Int(ZH) ⊆ ZH[P].

86 g Let f ∈ Int(ZH), and write f = n for some g ∈ ZH[x] and some n ≥ 1. If n = 1, then f ∈ ZH[x], and we are done. If n > 1, then g mod n ∈ Muﬀ(ZH/(n)), so we can express g mod n as the ﬁnite sum

t X g mod n = FrhrGr, r=1 H where for each 1 ≤ r ≤ t, hr ∈ {γn1 fn1 , . . . , γnmn fnmn } and Fr,Gr ∈ Z /(n) [x].

This means that over ZH, t X g = nG0 + FrhrGr r=1 for some G0 ∈ ZH[x]. So,

t ! t g 1 X X hr f = = nG + F h G = G + F G , n n 0 r r r 0 r n r r=1 r=1 which is an element of ZH[P]. Thus, Int(ZH) can be generated over ZH by the set P, and each polynomial in P has the desired form. The last assertion, that we can take each γ ∈ {1, 1 + i}, can be proven just as in part (i).

87 CHAPTER 6

SOME PRIMES OF INT(R)

6.1 Prime Ideals in Noncommutative Rings

In this chapter, we will begin to describe the prime ideals of Int(ZQ). Because we are working in a noncommutative setting, the usual deﬁnition for a prime ideal over a

commutative ring no longer holds the same utility. Instead, we will use the following

standard deﬁnition for a prime ideal in a noncommutative ring.

Deﬁnition 6.1.1. A proper ideal P in a ring R is called a prime ideal if whenever

x, y ∈ R with xRy ⊆ P , then either x ∈ P or y ∈ P .

There are several equivalent characterizations of a prime ideal in a noncommutative ring, but the above deﬁnition seems to be the easiest one with which to work.

The other characterizations are summarized in the following proposition. The proof is given in §10 of [6].

Proposition 6.1.2. For a proper ideal P in a ring R, the following are equivalent:

(i) P is prime

(ii) whenever I and J are ideals of R such that IJ ⊆ P , then either I ⊆ P or J ⊆ P

88 (iii) whenever a, b ∈ R with (a)(b) ⊆ P , then either a ∈ P or b ∈ P

(iv) whenever I and J are left ideals of R such that IJ ⊆ P , then either I ⊆ P or

J ⊆ P

(v) whenever I and J are right ideals of R such that IJ ⊆ P , then either I ⊆ P or

J ⊆ P

Note that when R is commutative, the new deﬁnition for a prime ideal reduces to the customary one.

6.2 Prime Ideals in Rings of Integer-Valued Polynomials

We ﬁrst consider some of the prime ideals in Int(D), where D is a (commutative) integral domain. Given a (commutative) integral domain D, a prime ideal P of D, and a ∈ D, we deﬁne

PP,a := {f ∈ Int(D) | f(a) ∈ P } .

Then, it is not diﬃcult to show that PP,a is a prime ideal of Int(D) (although not

every prime ideal of Int(D) needs to have this form; see [2], Chapter V). We will use

this type of ideal as a model to construct prime ideals in Int(ZQ). To do this, we

ﬁrst need to extend the deﬁnition of PP,a to noncommutative rings.

Recall that in Corollary 2.2.4, we determined the unit groups of ZQ and ZH to be

× (ZQ) = {±1, ±i, ±j, ±k}

89 and

× (ZH) = {±1, ± i, ±j, ±k, ±µ, ±iµi,

± jµj, ±kµk, ±µ2, ±iµ2i, ±jµ2j, ±kµ2k}.

We use these facts in the following deﬁnition.

Deﬁnition 6.2.1. Given an overring R of ZQ in QQ and an element α ∈ R, we denote the (full) multiplicative conjugacy class of α in R by ConjR(α) (or simply Conj(α) if

−1 the ring R is clear from context). That is, ConjR(α) = {uαu | u ∈ R is a unit}.

Let S = R ∩ Q. If R = SQ, then we deﬁne the restricted multiplicative conjugacy

−1 × class KonjR(α) to be KonjR(α) = {uαu | u ∈ (ZQ) }. If R 6= SQ, then R = SH

−1 × and we deﬁne KonjR(α) = {uαu | u ∈ (ZH) }. As with the full multiplicative conjugacy class, we write Konj(α) if R is clear from context.

While ConjR(α) and KonjR(α) are the same when R = ZQ or ZH and α ∈ R, the importance of the restricted multiplicative conjugacy class arises when dealing with arbitrary overrings of ZQ. For an arbitrary overring of R of ZQ and α ∈ R, the full multiplicative conjugacy class ConjR(α) might be infinite, whereas KonjR(α) is always finite. This difference will be important if we are to extend some of our results, such as Theorem 6.2.3 below, to overrings of ZQ. The following fact regarding multiplicative conjugacy in quaternion rings will be used several times and is worth pointing out: if β ∈ ConjR(α), then α and β share the same norm and constant coefficient, and thus have the same minimal polynomial

(however, the converse is not true, since, for example, i and j have the same minimal polynomial but are not conjugate in ZQ).

90 Deﬁnition 6.2.2. Given an overring R of ZQ in QQ, an ideal I of R, and an element

α ∈ R, we deﬁne PI,α := {f ∈ Int(R) | f(β) ∈ I for all β ∈ KonjR(α)}.

If the ring R is a (commutative) integral domain, α ∈ R, and P is a prime ideal of R, then the deﬁnition of PP,α reduces to {f ∈ Int(R) | f(α) ∈ P }, which, as mentioned above, is a prime ideal of Int(R). Our next major goal is to determine to what extent this carries over to quaternion rings, i.e. we want to answer the question: if R is an overring of ZQ in QQ, α ∈ R, and P is a prime ideal of R, then when is the set PP,α a prime ideal of Int(R)? Currently, we have answers to this question only for the case when R = ZQ, although the techniques we use should also work for ZH and other overrings of ZQ.

In Chapter 3 of [4], we are told that the prime ideals of ZQ are (0), (1 + i, 1 + j),

and (p), where p is an odd prime of Z. Our ﬁrst result shows that if P 6= (1+i, 1+j),

then PP,α is at least an ideal of Int(ZQ).

Theorem 6.2.3. Let n ∈ Z and let I = (n) in ZQ. Let α ∈ ZQ. Then, PI,α is an Int( Q) ideal of Int(ZQ), and when n 6= 0, Z /PI,α is a ﬁnite ring.

Proof. This is similar to the proofs of Theorem 3.1.1 and Proposition 4.1.4. Since

(f + g)(γ) = f(γ) + g(γ) for any f, g ∈ Int(ZQ) and any γ ∈ ZQ, we see that PI,α is closed under addition.

91 P r Now, ﬁx f(x) = r αrx ∈ PI,α, and let g(x) ∈ Int(ZQ). Let β ∈ Konj(α).

Then, there exist s, t, u, v ∈ Z such that g(β) = s + ti + uj + vk. We have

X r (fg)(β) = αrg(β)β r X r = αr(s + ti + uj + vk)β r = sf(β) + t(fi)(β) + u(fj)(β) + v(fk)(β)

= sf(β) + tf(−iβi)i + uf(−jβj)j + vf(−kβk)k .

Since β ∈ Konj(α), so are −iβi, −jβj, and −kβk. Since f ∈ PI,α, each of f(β), f(−iβi), f(−jβj), and f(−kβk) is in I. So, (fg)(β) ∈ I and hence fg ∈ PI,α.

To prove the ﬁrst assertion in our theorem, it remains to show that gf ∈ PI,α.

Writing f(β) = a + bi + cj + dk for some integers a, b, c, and d and using the same steps as above, we get

(gf)(β) = ag(β) + bg(−iβi)i + cg(−jβj)j + dg(−kβk)k .

Since I = (n), the fact that f(β) ∈ I means that a, b, c, and d are all in (n). Since g ∈ Int(ZQ), it follows that (gf)(β) ∈ I, and hence gf ∈ PI,α. Thus, PI,α is an ideal of Int(ZQ). Int( Q) Next, let R = Z /PI,α, and assume that n 6= 0. Note that n ∈ PI,α, so that

R contains a subring isomorphic to ZQ/(n). For every F ∈ Int(ZQ), we deﬁne a map

F ∗ : Konj(α) → ZQ/(n)

β 7→ F (β) mod n .

92 Then, for all F,G ∈ Int(ZQ),

F ∗ = G∗ ⇐⇒ F ∗(β) = G∗(β) for all β ∈ Konj(α)

⇐⇒ F (β) ≡ G(β) mod n for all β ∈ Konj(α)

⇐⇒ F (β) − G(β) ≡ 0 mod n for all β ∈ Konj(α)

⇐⇒ F − G ∈ PI,α

⇐⇒ F ≡ G in R .

These equivalences show that for any F ∈ Int(ZQ), the residue of F modulo PI,α is determined by the associated map F ∗. Thus, the number of possible residues in

R is bounded by the number of possible maps from Konj(α) to ZQ/(n). Since both

Konj(α) and ZQ/(n) are finite sets, there are only finitely many maps between them. Int( Q) Therefore, we conclude that R = Z /PI,α is a finite ring.

The preceding result also holds when P = (1+i, 1+j), but since P is not principal

in this case, the proof given does not apply. In Section 6.4, we will prove the theorem

for (1 + i, 1 + j) by using some diﬀerent techniques. It is also likely that the proof of

Theorem 6.2.3 will extend to arbitrary overrings of ZQ, but for now we will remain

focused on ZQ.

Theorem 6.2.3 indicates that PI,α is an ideal of Int(ZQ) whenever I is generated

by an integer. A necessary condition for PI,α to be prime is that I is prime ideal of

ZQ.

Proposition 6.2.4. Let n > 1 and let I = (n) in ZQ. Assume that I is not a prime ideal of ZQ. Then, for any α ∈ ZQ, PI,α is not a prime ideal of Int(ZQ).

93 Proof. Since I is not a prime ideal of ZQ, n is either composite or n = 2. Assume ﬁrst that n is composite with factorization n = `m. Then, for any α, the set ` Int(ZQ) m ⊆

PI,α, but neither ` nor m is in PI,α. So, PI,α is not prime.

Next, assume that I = (2). We will show that (1 + i)Int(ZQ)(1 − i) is contained in PI,α, and since neither 1 + i nor 1 − i is in (2), this will be enough to conclude that

−1 1−i PI,α is not prime. Let ε = 1 + i. Then, the inverse of ε in QQ is ε = 2 . Now, if

β = a + bi + cj + dk ∈ ZQ, then direct computation shows that

εβε−1 = a + bi − dj + ck, and

ε−1βε = a + bi + dj − ck, so both εβε−1 and ε−1βε are elements of ZQ (in fact, since multiplicative conjugation by ε also respects the ring operations in ZQ, we have exhibited an automorphism of

ZQ).

We claim that whenever f ∈ Int(ZQ), the polynomial εfε−1 is also in Int(ZQ).

To see this, notice that whenever β ∈ ZQ, we have

(εfε−1)(β) = ε(f(ε−1βε))ε−1.

Since f ∈ Int(ZQ) and ε−1βε ∈ ZQ, we have f(ε−1βε) ∈ ZQ. Thus, ε(f(ε−1βε))ε−1 ∈

ZQ and εfε−1 ∈ Int(ZQ). Let g(x) = (1 + i)f(x)(1 − i) ∈ (1 + i)Int(ZQ)(1 − i).

Then, for any β ∈ ZQ, g(β) = 2(εfε−1)(β) ∈ (2) .

Thus, (1+i)Int(ZQ)(1−i) ⊆ PI,α regadless of the choice of α. However, neither 1+i nor 1 − i can be in PI,α, so PI,α is never a prime ideal. This complete the proof.

The question remains whether PI,α is prime if I is prime. Under certain conditions on α, we will prove that PP,α is in fact a prime ideal of Int(ZQ) when P = (p) and

94 p is an odd prime. The case P = (0) is true without restrictions on α, as we show in the next proposition.

Proposition 6.2.5. Let P = (0) in ZQ. Then, for any α ∈ ZQ, PP,α is a prime ideal of Int(ZQ).

Proof. Let f, g ∈ Int(ZQ) and assume that fInt(ZQ)g ⊆ PP,α. We need to show that either f ∈ PP,α or g ∈ PP,α. If g ∈ PP,α, then there is nothing to prove. So, assume g∈ / PP,α. Then, there exists β ∈ Konj(α) such that g(β) ∈/ P , i.e. such that g(β) 6= 0. Since g(β) is non-zero, N(g(β)) is a non-zero integer.

Let u be any unit in ZQ. Then, the constant polynomial ug(β) is an element of

P r Int(ZQ). Let f(x) = r αrx . Since fInt(ZQ)g ⊆ PP,α, we have

0 = (fug(β)g)(β)

X r = αr(ug(β)g)(β)β r X r = αrug(β)g(β)β r X r = αruN(g(β))β r X r = N(g(β)) αruβ r = N(g(β))(fu)(β)

= N(g(β))f(uβu−1)u .

We know that N(g(β)) 6= 0, so we must have f(uβu−1)u = 0. Since u is a unit, this implies that f(uβu−1) = 0. But u was an arbitrary unit, so f(γ) = 0 for all

γ ∈ Konj(α). Thus, f ∈ PP,α and therefore PP,α is a prime ideal.

95 When P 6= (0), not every PP,α ideal is prime; it depends on P and α. Furthermore,

showing that PP,α is prime is more diﬃcult. However, there is one instance that is

easy to deal with.

Proposition 6.2.6. Let P be any prime ideal of ZQ, and let α ∈ Z. Then, PP,α is

a prime ideal of Int(ZQ).

Proof. The key to the proof is the fact that since α is central in QQ, we have (fg)(α) =

f(α)g(α) for any f, g ∈ QQ[x]. We will use this fact freely.

Since P is an arbitrary prime ideal of ZQ, we ﬁrst need to check that PP,α is

an ideal. If f, g ∈ PP,α and h ∈ Int(ZQ), then (f + g)(α) = f(α) + g(α) ∈ P ,

(fh)(α) = f(α)h(α) ∈ P , and (hf)(α) = h(α)f(α) ∈ P , so PP,α is an ideal.

Next, assume that fInt(ZQ)g ⊆ PP,α. Since α ∈ Z, we have Konj(α) = {α}, so to verify the primality of PP,α it suﬃces to have either f(α) ∈ P or g(α) ∈ P . Now,

γ ∈ Int(ZQ) for any γ ∈ ZQ, so

(fγg)(α) = f(α)γg(α) ∈ P

and thus f(α)ZQg(α) ⊆ P . Since P is prime, either f(α) ∈ P or g(α) ∈ P , and we are done.

6.3 Maximal Ideals in Int(ZQ)

As mentioned in the previous section, when P = (p) for an odd prime p, the ideal

PP,α is not always prime. The goal of this section is to determine necessary and

suﬃcient conditions under which PP,α is a prime ideal of Int(ZQ). Before proceeding, we require two results from number theory regarding integers

that can be represented as sums of three squares. Both theorems are due to Gauss,

96 and a thorough treatment in terms of quadratic forms is given in Chapter 5 of [3]. The

ﬁrst theorem gives the well-known condition under which an integer can be written

as a sum of three integral squares.

Theorem 6.3.1. Let M ∈ Z be positive, and write M = 4mu, where m ≥ 0 and

4 - u. Then, there exist y, z, w ∈ Z such that M = y2 + z2 + w2 if and only if u 6≡ 7 mod 8.

The second result is a weakened form of Theorem 8.7 in Chapter 5 of [3].

Theorem 6.3.2. Let u be a positive integer. If u ≡ 1, 2, 3, 5, or 6 mod 8, then there

exist y, z, w ∈ Z such that u = y2 + z2 + w2 and gcd(y, z, w) = 1.

Remark. Calling Theorem 6.3.2 a weakening of Theorem 8.7 in Chapter 5 of [3] is

a bit of an understatement. The theorem in [3] gives a precise count of the number

of triples (y, z, w) ∈ Z3 such that u = y2 + z2 + w2 and gcd(y, z, w) = 1. However, these counts rely on the orders of class groups of quadratic forms, and a worthwhile

discussion of that topic would take us too far aﬁeld.

We can now begin our discussion of maximal ideals in Int(ZQ).

Theorem 6.3.3. Let p be an odd prime, let P = (p) in ZQ, and let α ∈ ZQ. Let

T = {a + bi + cj + dk ∈ ZQ | a, b, c, d ∈ {0, 1, . . . , p − 1}} and let L = {α1x +

α0 ∈ ZQ[x] | α1, α0 ∈ T }. Then, L represents a complete (though not necessarily Int( Q) irredundant) set of residues in Z /PP,α.

Int( Q) Proof. Let R = Z /PP,α and let π : Int(ZQ) → R be the quotient map. We

know that L ⊆ Int(ZQ). We wish to show that π(L) = R.

Let m(x) be a monic polynomial in Z[x] of minimal positive degree such that

m(α) ∈ P . Then, deg(m(x)) ≤ 2, because if α ≡ a mod p for some a ∈ Z, then we

97 can take m(x) = x − a, and if not, then we can take m(x) = minα(x). Note that m(x) ∈ PP,α, so (m(x), p) ⊆ PP,α. This means that whenever f ∈ ZQ[x], there exist g ∈ ZQ/(p) [x] and h ∈ L such that f(x) ≡ m(x)g(x) + h(x) mod p. It follows that

π(f) ∈ π(L).

Next, assume that f(x) is a generic element of Int(ZQ). Then, there exist F (x) ∈

F (x) e ZQ[x] and n ∈ Z such that f(x) = n . Write n = p q, where e ≥ 0 and gcd(p, q) = 1.

F (x) −1 F (x) Then, q is invertible mod p, so π n = π(q) π pe . Thus, it suﬃces to show F (x) that π pe lies in π(L).

If e = 0, then f(x) ∈ ZQ[x] and we are done. So, assume e > 0. By Corollary

4.3.10 part (i), there exist polynomials g0, g1, . . . , ge ∈ ZQ[x] such that

F (x) φ e (x) φ e−1 (x) φ (x) = g (x) p + g (x) p + ··· + g p + g (x), pe 0 pe 1 pe−1 e−1 p e

φpm (x) and for each 1 ≤ m ≤ e, we have pm ∈ Int(ZQ). Thus, it suﬃces to show that for

φpm (x) every m > 0, pm can be represented mod PP,α by a residue in L. Assuming this

F (x) to be the case, we can then represent the residue of f(x) = n in R by a polyomial

G(x) ∈ ZQ[x], i.e. there will exist G(x) ∈ ZQ[x] such that π(f) = π(G), and we know that π(G) ∈ π(L) for such a G.

φpm (x) So, let m > 0, and let φ(x) = pm ∈ Int(ZQ). Consider ﬁrst the case where

α ∈ Z. Since φ has rational coeﬃcients, φ(α) must lie in Z. So, α and φ(α) are both integers, and hence there exists A ∈ {0, 1, . . . , p − 1} such that φ(α) − (α − A) ∈ P .

Because α ∈ Z, this is enough to conclude that φ(x) − (x − A) ∈ PP,α. So, π(φ) = π(x − A) ∈ π(L).

98 From now on, assume that α∈ / Z. Since φpm ∈ Z[x], Theorem 2.3.1 tells us that

there exist C,D ∈ Z such that φpm (α) = Cα + D. Write α = a + bi + cj + dk. Then,

φ m (α) φ(α) = p pm Cα + D = pm (Ca + D) + Cbi + Ccj + Cdk = pm Ca + D Cb Cc Cd = + i + j + k . pm pm pm pm Claim: pm | C and pm | D

Proof of Claim: Assume ﬁrst that α 6≡ a mod p. Then, either p - b, p - c, or p - d.

Cb WLOG, assume that p - b. Since φ ∈ Int(ZQ), we must have φ(α) ∈ ZQ, so pm ∈ Z

m m Ca+D and p | Cb. The condition p - b now forces p | C, as desired. Since pm is also an integer, we get pm | D.

Next, assume that α ≡ a mod p. Let M = b2 + c2 + d2. Since α∈ / Z, M > 0. So,

we may write M = 4`u, where ` ≥ 0, u > 0, and 4 - u. Since M is a sum of three

squares, Theorem 6.3.1 shows that u 6≡ 7 mod 8. Since 4 - u, we see that u ≡ 1, 2, 3, 5,

or 6 mod 8. Thus, by Theorem 6.3.2 there exist y, z, w ∈ Z such that u = y2 +z2 +w2 and gcd(y, z, w) = 1. So,

M = 4`u = 22`(y2 + z2 + w2) = (2`y)2 + (2`z)2 + (2`w)2 .

Let β = a+2`yi+2`zj+2`wk ∈ ZQ. Then, β and α share the same constant coeﬃcient

2 and N(α) = a + M = N(β). By Theorem 2.3.1, this means that φpm (β) = Cβ + D.

Hence, Ca + D 2`Cy 2`Cz 2`Cw φ(β) = + i + j + k pm pm pm pm and this is an element of ZQ. Since gcd(y, z, w) = 1, either p - y, p - z, or p - w, and p cannot divide 2` because p is odd. Proceeding as in the case where α 6≡ a mod p

99 now gives pm | C and pm | D. This completes the proof of the Claim.

By the Claim, there exist C0,D0 ∈ Z such that φ(α) = (C0a + D0) + C0bi + C0cj +

C0dk = C0α + D0. Let ψ(x) = C0x + D0 ∈ ZQ[x]. Then, because both φ and ψ have

rational coeﬃcients, we have, for any unit v ∈ ZQ,

φ(vαv−1) = vφ(α)v−1 = vψ(α)v−1 = ψ(vαv−1) .

Thus,

φ(γ) = ψ(γ) for all γ ∈ Konj(α). (∗)

Recall that in the proof of Theorem 6.2.3, we introduced the following notation: for every τ ∈ Int(ZQ), let τ ∗ : Konj(α) → ZQ/(p) be the map deﬁned by τ ∗(γ) = τ(γ)

mod p for all γ ∈ Konj(α). In that same proof, we showed that for any τ, σ ∈ Int(ZQ),

∗ ∗ we have τ = σ if and only if τ and σ are equivalent modulo PP,α. Phrased in these

terms, (∗) indicates that φ∗ = ψ∗, and consequently φ ≡ ψ in R. But, ψ ∈ ZQ[x] and hence is equivalent to an element of L. Therefore, π(φ) ∈ π(L), and L represents a

complete set of residues in R.

Corollary 6.3.4. Let p be an odd prime, let P = (p) in ZQ, and let α ∈ ZQ. Then, Int( Q) 8 Z /PP,α ≤ p .

Int( Q) Proof. Let L be as in the statement of Theorem 6.3.3. Then, Z /PP,α ≤ |L| = p8.

Int( Q) Now that we know that any residue in Z /PP,α can be represented by a linear polynomial, we can begin to analyze the structure of these quotient rings.

100 Theorem 6.3.5. Let p be an odd prime, let P = (p) in ZQ, and let α ∈ ZQ. Let Int( Q) R = Z /PP,α and let m(x) be a monic polynomial in Z[x] of minimal positive degree such that m(α) ∈ P . Then, R is a simple ring if and only if m(x) is irreducible mod p.

Proof. (⇒) We prove the contrapositive. Assume that m(x) is reducible mod p.

Then, there exist non-constant, monic f, g ∈ Z[x] such that m ≡ fg mod p. Since m(α) ∈ P , the minimality of deg(m) implies that f(α) ∈/ P and g(α) ∈/ P . So, the residues of f and g are non-zero in R, but fg ≡ m ≡ 0 in R. Thus, the residue of f in R is a central, non-zero zero divisor. It follows that the ideal of

R generated by the residue of f is a proper, non-zero ideal of R. Thus, R is not simple.

(⇐) Assume that m(x) is irreducible mod p. If m(x) is linear, then by Theorem

6.3.3, the set T = {a+bi+cj +dk | a, b, c, d ∈ {0, 1, . . . , p−1}} represents a complete set of residues in R. So, R is a non-zero quotient ring of ZQ/(p); however, (p) is a ∼ maximal ideal of ZQ, so ZQ/(p) is a simple ring. Thus, R = ZQ/(p) is a simple ring ∼ (and in fact, appealing to Theorem 1.4.2 shows that R = M2(Fp)).

So, assume that deg(m(x)) > 1. Then, α∈ / Z, so we know that minα(x) is a monic quadratic polynomial in PP,α. Thus, we can assume that m(x) has degree 2.

Let I be the ideal of Z[x] generated by p and m(x). In Int(ZQ), both m(x) and p are ∼ contained in PP,α, so the ring R contains a subring F isomorphic to Z[x]/I = Fp2 . Furthermore, F can be represented by polynomials with integer coeﬃcients, and hence F is a central subring of R. Adjoining the quaternion units to this copy of

8 Fp2 shows that R contains a subring S isomorphic to Fp2 Q. However, |Fp2 Q| = p , and by Corollary 6.3.4, |R| ≤ p8. Thus, we must have equality between R and S,

101 ∼ and hence R = Fp2 Q. Furthermore, another application of Theorem 1.4.2 shows that ∼ R = M2(Fp2 ), a simple ring.

The known results about maximal ideals of Int(ZQ) are summarized in the following corollary.

Corollary 6.3.6. Let p be an odd prime, let P = (p) in ZQ, and let α ∈ ZQ. Let Int( Q) R = Z /PP,α and let m(x) be a monic polynomial in Z[x] of minimal positive degree such that m(α) ∈ P . Then, deg(m(x)) ≤ 2, and

(i) PP,α is a maximal ideal of Int(ZQ) if and only if m(x) is irreducible mod p.

∼ ∼ (ii) if m(x) is linear, then PP,α is a maximal ideal of Int(ZQ) and R = ZQ/(p) =

M2(Fp).

(iii) if m(x) is quadratic and irreducible mod p, then PP,α is a maximal ideal of ∼ ∼ Int(ZQ) and R = Fp2 Q = M2(Fp2 ).

(iv) if m(x) is quadratic and reducible mod p, then PP,α is not a prime ideal of

Int(ZQ). However, if x − A represents an irreducible factor of m(x) mod p, ∼ then M := (PP,α, x − A) is a maximal ideal of Int(ZQ), and Int(ZQ)/M = ∼ ZQ/(p) = M2(Fp).

Proof. Parts (i), (ii), and (iii) were all shown in Theorem 6.3.5.

(iv) Since m(x) is reducible mod p, there exist integers a and b such that minα(x) ≡

(x − a)(x − b) mod p, and (β − a)(β − b) ∈ P in ZQ for all β ∈ Konj(α). Consider

(x − a)Int(ZQ)(x − b). We wish to show that this set of polynomials lies in PP,α.

102 Toward that end, let h(x) ∈ Int(ZQ) and let H(x) = (x − a)h(x)(x − b). We have, for any β ∈ Konj(α),

H(x) = h(x)x2 − ah(x)x − bh(x)x + abh(x) and

H(β) = h(β)β2 − ah(β)β − bh(β)β + abh(β)

= h(β)(β − a)(β − b) ∈ P.

Since h was an arbitrary element of Int(ZQ), we have (x − a)Int(ZQ)(x − b) ⊆ PP,α.

However, neither x − a nor x − b is in PP,α because m(x) is not linear. Therefore,

PP,α cannot be a prime ideal.

Assuming that m(x) is quadratic and that x − A is an irreducible factor of m(x) mod p, the proof of the forward implication of Theorem 6.3.5 shows that the residue of x − A in R generates a non-zero, proper ideal I of R. Taking into account The- orem 6.3.3, we see that Int(ZQ)/M ∼= R/I is a non-zero ring that can be completely represented by residues in ZQ/(p). Since ZQ/(p) is simple, we must have ∼ ∼ Int(ZQ)/M = ZQ/(p) = M2(Fp).

6.4 Some Primes of Int(ZQ) above (1 + i, 1 + j)

In Section 6.3, we described when PP,α is a prime ideal of Int(ZQ) for the cases P = (0) and P = (p), where p is an odd prime. In this section, we will determine what happens when P = (1 + i, 1 + j). We begin with a lemma concerning elements of P and the quotient ring ZQ/P .

Lemma 6.4.1. Let P = (1 + i, 1 + j) in ZQ. Then,

103 (i) If α ∈ P , then there exist γ, δ, γ0, δ0 ∈ ZQ such that α = γ(1 + i) + δ(1 + j) and α = (1 + i)γ0 + (1 + j)δ0. Thus, the two-sided ideal (1 + i, 1 + j) equals both the

right ideal (1 + i, 1 + j)ZQ and the left ideal ZQ(1 + i, 1 + j).

∼ (ii) ZQ/P = F2, the ﬁeld with 2 elements.

P Proof. (i) Let α ∈ P . Then, we may write α = r γrβrδr, where each γr, δr ∈ ZQ and each βr ∈ {1 + i, 1 + j}. Let S = {r | βr = 1 + i} and T = {r | βr = 1 + j}.

Since ZQ/(2) is commutative, we have, working mod 2,

X X X α ≡ γrβrδr ≡ γrδr(1 + i) + γrδr(1 + j) . r r∈S r∈T P P So, in ZQ, α = r∈S γrδr(1 + i) + r∈T γrδr(1 + j) + 2ε, for some ε ∈ ZQ. Since 2 = (1 − i)(1 + i), we get

X X α = γrδr(1 + i) + γrδr(1 + j) + 2ε r∈S r∈T X X = γrδr(1 + i) + γrδr(1 + j) + ε(1 − i)(1 + i) r∈S r∈T X X = γrδr + ε(1 − i) (1 + i) + γrδr(1 + j) r∈S r∈T = γ(1 + i) + δ(1 + j),

P P 0 0 where γ = r∈S γrδr + ε(1 − i) and δ = r∈T γrδr. The existence of γ and δ can be proved similarly, so part (i) holds.

(ii) Let R = ZQ/(2). Then, it is not hard to see that M = (1+i, 1+j) in R is the ∼ ∼ unique maximal ideal of R and R/M = F2. But then, ZQ/P = ZQ/(2, 1 + i, 1 + j) = ∼ R/M = F2.

In the proof of Proposition 6.2.4, we showed that when ε = (1 + i) and α ∈ ZQ, we always have εαε−1 ∈ ZQ. We now record this formally as a lemma.

104 Lemma 6.4.2. Let ε ∈ {1 + i, 1 + j}. Then, for all α ∈ ZQ, εαε−1 ∈ ZQ.

−1 1−i Proof. Let α = a + bi + cj + dk ∈ ZQ. Assume ﬁrst that ε = 1 + i. Then, ε = 2 and direct computation shows that

−1 εαε = a + bi − dj + ck ∈ ZQ.

Similarly, when ε = 1 + j, we have

−1 εαε = a + di + cj − bk ∈ ZQ.

We can now prove that when P = (1 + i, 1 + j) and α ∈ ZQ, the set PP,α is not just an ideal of Int(ZQ), but is in fact a maximal ideal of Int(ZQ).

Theorem 6.4.3. Let P = (1 + i, 1 + j) in ZQ, and let α ∈ ZQ. Then,

(i) PP,α is an ideal of Int(ZQ).

Int( Q) ∼ (ii) PP,α is a maximal ideal of Int(ZQ), and Z /PP,α = F2.

Proof. (i) Let f ∈ PP,α and g ∈ Int(ZQ). We need to show that (fg)(β), (gf)(β) ∈ P for all β ∈ Konj(α). Let β ∈ Konj(α). Showing that (fg)(β) ∈ P may be done just

as in the proof of Theorem 6.2.3. To show that (gf)(β) ∈ P , we use Lemma 6.4.1.

P r Let g(x) = r βrx . Since f(β) ∈ P , there exist γ, δ ∈ ZQ such that f(β) =

105 γ(1 + i) + δ(1 + j). For convenience, let ε1 = 1 + i and ε2 = 1 + j. Then,

X r (gf)(β) = βrf(β)β r X r = βr(γε1 + δε2)β r X r X r = βrγε1β + βrδε2β r r X r −1 X r −1 = βrγε1β ε1 ε1 + βrδε2β ε2 ε2 r r −1 −1 = (gγ)(ε1βε1 )ε1 + (gδ)(ε2βε2 )ε2 .

−1 −1 By Lemma 6.4.2, both ε1βε1 and ε2βε2 are in ZQ. Since gγ, gδ ∈ Int(ZQ), we

−1 −1 have (gγ)(ε1βε1 ) ∈ ZQ and (gδ)(ε2βε2 ) ∈ ZQ. Since ε1, ε2 ∈ P , the above equation tells us that (gf)(β) ∈ P . It follows that PP,α is an ideal of Int(ZQ).

Int( Q) ∼ (ii) Let R = Z /PP,α. It suﬃces to show that R = F2, since this will force

PP,α to be a maximal ideal of Int(ZQ).

Let f ∈ Int(ZQ) and let ε = 1 + i. Then, ε ∈ PP,α, so fε = f + fi ∈ PP,α. In particular, this means that f(α) + f(−iαi)i = f(α) + (fi)(α) ∈ P . Thus,

f(α) ∈ P ⇐⇒ f(−iαi)i ∈ P ⇐⇒ f(−iαi) ∈ P.

Similar results hold when ε = 1 + j or ε = 1 + k, so we see that

f(α) ∈ P ⇐⇒ f(β) ∈ P for all β ∈ Konj(α)

⇐⇒ f ∈ PP,α .

∼ Now, since ZQ/P = F2, either f(α) ∈ P or f(α) + 1 ∈ P . Thus, either f ∈ PP,α or f +1 ∈ PP,α. Stated diﬀerently, this means that the residue of f in R is equivalent either to 0 or to 1. Since f was an arbitrary element of Int(ZQ), we conclude that ∼ R = F2, which completes the proof.

106 CHAPTER 7

INTEGER-VALUED POLYNOMIALS ON SUBSETS

7.1 Subsets and Muﬃns

When D is a (commutative) integral domain with ﬁeld of fractions K and E is any subset of D, one may deﬁne

Int(E,D) := {f(x) ∈ K[x] | f(a) ∈ D for all a ∈ E}, and it is not diﬃcult to show that Int(E,D) is a ring, which we call the ring of integer-valued polynomials on E. In this chapter, we investigate how the notion of integer-valued polynomials on subsets may be extended to quaternion rings. For now, we will work exclusively with ZQ. As mentioned above, Int(E,D) is always a ring when D is a (commutative) integral domain and E ⊆ D. However, this might not be true when working over ZQ. To illustrate the disparity with the commutative case, we begin with a simple negative example.

Example 7.1.1. Deﬁne Int({i}, ZQ) = {f(x) ∈ QQ[x] | f(i) ∈ ZQ}. Notice that x−i x−i x2−(i+j)x+k 3 , x − j ∈ Int({i}, ZQ). However, letting f(x) = 3 (x − j) = 3 , we have

107 2k f(i) = 3 ∈/ ZQ. Thus, Int({i}, ZQ) is not closed under multiplication, and therefore cannot be a ring.

As the example above shows, it is non-trivial to determine which subsets of ZQ give rise to rings of integer-valued polynomials. A satisfactory resolution to this problem is the goal of the present chapter. Toward that end, we begin with some deﬁnitions.

Deﬁnition 7.1.2. For any subset S ⊆ ZQ, deﬁne the set Int(S, ZQ) to be

Int(S, ZQ) = {f(x) ∈ QQ[x] | f(S) ⊆ ZQ}. For any quotient ring R of ZQ, define the set MuffR(S) to be MuffR(S) = {f(x) ∈ R[x] | f(α) = 0 in R, for all α ∈ S}.

We call MuffR(S) the muffin of S in R. Lastly, a ringset is defined to be a subset S of ZQ such that Int(S, ZQ) is a ring.

Recall Correspondence 4.1.1, which gave convenient ways to write and work with elements of Int(ZQ). A similar result exists for elements of Int(S, ZQ): any f(x) ∈

g(x) Int(S, ZQ) may be written as f(x) = n ∈ Int(S, ZQ) with g(x) ∈ ZQ[x] and n ∈ Z, and when f(x) is expressed in this way, g(x) mod n is in MuﬀR(S), where

R = ZQ/(n). The theorems in this section more carefully analyze the relationship between the sets Int(S, ZQ) and MuﬀR(S), where R is a quotient ring of ZQ.

Generally, we will be trying to show either that Int(S, ZQ) is a ring, or that

MuﬀR(S) is an ideal of R[x]. It is easy to verify that for any S ⊆ ZQ and any quotient ring R of ZQ, the sets Int(S, ZQ) and MuﬀR(S) are groups under addition.

Thus, to prove (or disprove) that Int(S, ZQ) is a ring or that MuﬀR(S) is an ideal of R[x], it is enough to consider the multiplicative aspects of the relevant structure.

This technique makes the proof of our ﬁrst result especially brief.

108 Proposition 7.1.3. Let S ⊆ Z. Then, S is a ringset.

Proof. Let f, g, ∈ Int(S, ZQ) and let a ∈ S. Then, f(a), g(a) ∈ Z and (fg)(a) = f(a)g(a) ∈ Z, so fg ∈ Int(S, ZQ). Thus, Int(S, ZQ) is closed under multiplication and therefore is a ring, i.e. S is a ringset.

Theorem 7.1.4. Let S ⊆ ZQ, and for all n > 1, let Rn = ZQ/(n). Then, the following are equivalent:

(i) Int(S, ZQ) is ring

(ii) for each n > 1, MuﬀRn (S) is an ideal of Rn[x]

(iii) for each prime p and for every e > 0, Muﬀ (S) is an ideal of R e [x] Rpe p

Proof. (i) ⇒ (ii) Assume Int(S, ZQ) is a ring. Fix n > 1, let f(x) ∈ MuﬀRn (S), and

let g(x) ∈ Rn[x]. It suffices to show that (fg)(x) and (gf)(x) are in MuffRn (S). By abusing notation, we may consider f(x) and g(x) to have coefficients in ZQ, in which

f(x) case n ∈ Int(S, ZQ) and g(x) ∈ ZQ[x] ⊆ Int(S, ZQ). Since Int(S, ZQ) is a ring,

(fg)(x) (gf)(x) both n and n are in Int(S, ZQ). It follows that mod n, both (fg)(x) and

(gf)(x) are in MuﬀRn (S), and therefore MuﬀRn (S) is an ideal of Rn[x].

(ii) ⇒ (i) Assume that for each n > 1, MuﬀRn (S) is an ideal of Rn[x]. It suﬃces to show that Int(S, Q) is closed under multiplication. Let f(x) , g(x) ∈ Int(S, Q), Z n1 n2 Z (fg)(x) where f(x), g(x) ∈ Q[x] and n1, n2 > 0. We want to show that ∈ Int(S, Q). Z n1n2 Z

Let α ∈ S and let g(α) = a + bi + cj + dk ∈ ZQ. Then, (fg)(α) = af(α) +

g(x) b(fi)(α) + c(fj)(α) + d(fk)(α). Since ∈ Int(S, Q), we have g(α) ∈ (n2) in Q, n2 Z Z so each of a, b, c, and d is divisible by n2.

Now, f(x) mod n is in Muff (S), and Muff (S) is an ideal of R [x], so 1 Rn1 Rn1 n1 (fi)(x), (fj)(x), and (fk)(x) are also in Muff (S). In Q, this means that each Rn1 Z 109 of f(α), (fi)(α), (fj)(α), and (fk)(α) are in (n1). Combining this with the above

paragraph, we see that each of af(α), b(fi)(α), c(fj)(α), and d(fk)(α) is in (n1n2)

in Q. Thus, (fg)(α) ∈ Q. Since α ∈ S was arbitrary, we have (fg)(x) ∈ Int(S, Q), Z n1n2 Z n1n2 Z as required.

(ii) ⇒ (iii) This is immediate.

(iii) ⇒ (ii) Let n > 1, let f ∈ MuﬀRn (S), and let g ∈ Rn[x]. Assume that n has

e1 et prime factorization n = p1 ··· pt . Fix m between 1 and t, and let p = pm and e = em.

e e e Let F = f mod p and let G = g mod p . Then, F ∈ MuﬀRpe (S) and G ∈ Rp [x],

and since MuﬀRpe (S) is an ideal, we see that FG ∈ MuﬀRpe (S). Fix α ∈ S. Then,

e e (FG)(α) ≡ 0 mod p for all α ∈ S, so in Rn, we have (fg)(α) ∈ p Rn. Since p was

an arbitrary prime dividing n, we conclude that

t \ em (fg)(α) ∈ pm Rn = {0}; m=1 thus, (fg)(α) ≡ 0 mod n. Similarly, (gf)(α) ≡ 0 mod n. Since α was an arbitrary

element of S, we get fg, gf ∈ MuﬀRn (S), and therefore MuﬀRn (S) is an ideal of

Rn[x].

Recall Deﬁnition 6.2.1: for any α ∈ ZQ, we deﬁne the restricted multiplicative

conjugacy class Konj(α) of α to be Konj(α) = {uαu−1 | u ∈ (ZQ)×}; here, (ZQ)× = {±1, ±i, ±j, ±k}.

Theorem 7.1.5. Let n > 1, let R = ZQ/(n), and let S ⊆ ZQ. Then, MuﬀR(S) is

an ideal of R[x] ⇐⇒ for all f ∈ MuﬀR(S), for all α ∈ S, and for all β ∈ Konj(α), we have f(β) ≡ 0 mod n.

Proof. (⇒) Assume that MuﬀR(S) is an ideal of R[x]. Let f ∈ MuﬀR(S), α ∈ S, and

β ∈ Konj(α). Then, there exists u ∈ {±1, ±i, ±j, ±k} such that β = uαu−1. Since

110 MuﬀR(S) is an ideal of R[x], (fu)(x) mod n ∈ MuﬀR(S) and (fu)(α) ≡ 0 mod n.

Thus,

f(β) ≡ f(uαu−1) ≡ (fu)(α)u−1 ≡ 0 mod n,

proving the forward implication.

(⇐) This proof is almost identical to that of Proposition 4.1.4. Assume that for

all f ∈ MuﬀR(S), for all α ∈ S, and all β ∈ Konj(α), we have f(β) ≡ 0 mod n. Let

P r P r f(x) ∈ MuﬀR(S), g(x) ∈ R[x], and α ∈ S. Let f(x) = r αrx , let g(x) = r βrx , and let a, b, c, d ∈ Z be such that g(α) ≡ a + bi + cj + dk. Then,

X r (gf)(α) ≡ βrf(α)α ≡ 0, r and

X r (fg)(α) ≡ αrg(α)α r X r ≡ αr(a + bi + cj + dk)α r ≡ af(α) + b(fi)(α) + c(fj)(α) + d(fk)(α)

≡ af(α) + bf(−iαi)i + cf(−jαj)j + df(−kαk)k .

Because α, −iαi, −jαj, and −kαk are all in Konj(α) we see that (fg)(α) ≡ 0 mod n.

Thus, both (fg)(x) and (gf)(x) are in MuﬀR(S), and therefore MuﬀR(S) is an ideal

of R[x].

Corollary 7.1.6. Let S ⊆ ZQ, and for all n > 1, let Rn = ZQ/(n). Then, the following are equivalent:

(i) Int(S, ZQ) is ring

(ii) for all n > 1, for all f ∈ MuﬀRn (S), for all α ∈ S, and for all β ∈ Konj(α), we have f(β) ≡ 0 mod n

111 (iii) for all primes p, for all e > 0, for all f ∈ Muﬀ (S), for all α ∈ S, and for all Rpe β ∈ Konj(α), we have f(β) ≡ 0 mod pe

Proof. This is just a combination of Theorems 7.1.4 and 7.1.5.

7.2 Some Suﬃcient Conditions

The following two theorems provide suﬃcient conditions for S to be a ringset.

Theorem 7.2.1. Let S ⊆ ZQ. Assume that for all u ∈ {i, j, k} we have uSu−1 ⊆ S. Then, S is a ringset.

Proof. This is a direct application of Corollary 7.1.6. Let n > 1, let R = ZQ/(n), let

f ∈ MuﬀR(S), let α ∈ S, and let β ∈ Konj(α). Since α ∈ S, we have Konj(α) ⊆ S.

Thus, f(β) ≡ 0 mod n. By Corollary 7.1.6, we conclude that Int(S, ZQ) is a ring.

By the preceding theorem, for S to be a ringset it is enough that S be closed

under multiplicative conjugation. However, this condition is not necessary.

Example 7.2.2. Consider S = {i, j}, and let our notation be as in Corollary 7.1.6.

Given f ∈ MuﬀRn (S), we know that f(i) ≡ f(j) ≡ 0 mod n, and it suﬃces to have

f(−i) ≡ 0 and f(−j) ≡ 0. Since ±i and ±j all share the same norm and constant coeﬃcient, Theorem 2.3.1 implies that there exist γ, δ ∈ Rn[x] such that f(α) ≡ γα+δ

for all α ∈ {±i, ±j}. So, 0 ≡ f(i) − f(j) ≡ γ(i − j). Multiplying this equivalence on

the right by −i + j yields 0 ≡ 2γ. But then, f(i) − f(−i) ≡ 2γi ≡ 0, from which we

conclude that f(−i) ≡ 0. Similarly, f(−j) ≡ 0, and thus S is a ringset in this case.

In the next theorem, we generalize the technique used in the above example.

Recall some notation from previous chapters: for any α = a + bi + cj + dk ∈ ZQ − Z,

2 we deﬁne the minimal polynomial minα(x) of α to be minα(x) = x − 2ax + N(α).

112 Theorem 7.2.3. Let S ⊆ ZQ. Assume that for all α ∈ S − Z and all u ∈ {i, j, k},

there exists β ∈ S and ε ∈ ZQ such that

(1) minα(x) = minβ(x), and

(2) (α − β)ε = α − uαu−1 .

Then, S is a ringset.

Proof. We ﬁrst deal with any integers that might lie in S. For any a ∈ S ∩ Z and any

f(x), g(x) ∈ Int(S, ZQ), we have (fg)(a) = f(a)g(a) ∈ ZQ and (gf)(a) = g(a)f(a) ∈

ZQ, so no obstacles will arise from S ∩ Z. So, assume that S − Z 6= ∅.

Let n > 1, let R = ZQ/(n), let f ∈ MuﬀR(S), let α ∈ S − Z, and let u ∈ {i, j, k}. By Theorem 2.3.1 part (iii) and condition (1) above, it follows that there exist γ, δ ∈ R

such that γα+δ ≡ 0 mod n and γβ+δ ≡ 0 mod n. Subtracting these two equivalences

and applying condition (2) above, we have

0 ≡ γ(α − β)

≡ γ(α − β)ε

≡ γ(α − uαu−1)

≡ γα + δ − (γuαu−1 + δ)

≡ −(γuαu−1 + δ)

≡ −f(uαu−1),

where the last equivalence follows because α and uαu−1 share the same minimal

polynomial. By Corollary 7.1.6, we conclude that S is a ringset.

As with Theorem 7.2.1, the conditions speciﬁed by Theorem 7.2.3 are not neces-

sary for S to be a ringset.

113 Example 7.2.4. Pick α ∈ ZQ − Z, and for each n > 1, let αn = α + n and

Sn = Konj(αn). Let ∞ [ S = Sn ∪ {α} . n=1

Then, no element of S has the same constant coeﬃcient as α, so minβ(x) 6= minα(x) whenever β ∈ S − {α}. So, the only way that condition (1) of Theorem 7.2.3 can be met for α is to take β = α. However, in that case condition (2) becomes

0 = α − uαu−1 for all u ∈ {i, j, k} .

Since α∈ / Z, there exists u ∈ {i, j, k} such that α and u do not commute, and for this u, we have α − uαu−1 6= 0. Thus, the conditions of Theorem 7.2.3 do not apply.

Nevertheless, S is a ringset, which we prove as follows. Given β ∈ Konj(α) and n > 1, there exists βn ∈ Konj(αn) ⊆ S such that β ≡ βn mod n. Hence, if f ∈ MuﬀZQ/(n)(S), then f(β) ≡ f(βn) ≡ 0 mod n. Therefore, we can meet the conditions of Corollary

7.1.6 and conclude that Int(S, ZQ) is a ring.

In Section 7.4, we shall give necessary and sufficient conditions for some subsets of ZQ to be ringsets; in Section 7.5, we will give necessary and sufficient conditions for any finite subset of ZQ to be a ringset. The hope is that the insight gained from these cases will allow us to determine necessary and sufficient conditions that work for more general S ⊆ ZQ. Before proceeding to those results, however, we prove two more useful theorems that hold for general subsets of ZQ.

Theorem 7.2.5.

(i) Let S,T ⊆ ZQ, let n > 1, and let R = ZQ/(n). Assume that S ⊆ T . Then,

Int(T, ZQ) ⊆ Int(S, ZQ) and MuﬀR(T ) ⊆ MuﬀR(S).

114 (ii) Let {S`}`∈I be a collection of subsets of ZQ such that S` is a ringset for each S ` ∈ I. Let S = `∈I S`. Then, S is a ringset.

Proof. (i) Let α ∈ S, f ∈ Int(T, ZQ), and g ∈ MuﬀR(T ). Then, α ∈ T , so f(α) ∈ ZQ and g(α) ≡ 0 mod n. Hence, f ∈ Int(S, ZQ) and g ∈ MuﬀR(S).

(ii) Let n > 1, let R = ZQ/(n), let f ∈ MuffR(S), and let α ∈ S. Then, there exists m ∈ I such that α ∈ Sm ⊆ S. Since f ∈ MuffR(S) ⊆ MuffR(Sm) and

Int(Sm, ZQ) is a ring, Corollary 7.1.6 tells us that f(β) ≡ 0 mod n for all β ∈ Konj(α).

Thus, MuﬀR(S) is an ideal of R[x] and S is a ringset.

As the previous theorem shows, the collection of all ringsets of ZQ is closed under unions. However, this collection is not closed under intersections, since, for instance,

Int({i, j}, ZQ) and Int({i, k}, ZQ) are both rings, but Int({i}, ZQ) is not. Thus, there is not any obvious way to give ZQ a topology that can help identify ringsets.

Deﬁnition 7.2.6. For any S ⊆ ZQ, any n ∈ Z, and any m ∈ Z − {0}, we deﬁne the sets S + n and mS to be S + n = {α + n | α ∈ S} and mS = {mα | α ∈ S}.

Theorem 7.2.7. Let S ⊆ ZQ, let n ∈ Z, and let m ∈ Z − {0}. Then, the following are equivalent:

(i) S is a ringset

(ii) S + n is a ringset

(iii) mS is a ringset

Proof. (i) ⇐⇒ (ii) Assume ﬁrst that S is a ringset. It suﬃces to show that

Int(S + n, ZQ) is closed under multiplication. Let f(x), g(x) ∈ Int(S + n, ZQ), and

115 deﬁne F (x) = f(x + n) and G(x) = g(x + n). Then, F (x) and G(x) are polynomials in QQ[x], and since n ∈ Z, for all α ∈ S we have F (α) = f(α + n) ∈ ZQ and G(α) = g(α + n) ∈ ZQ, so F (x),G(x) ∈ Int(S, ZQ). Furthermore, letting

P r f(x) = αrx , we have, for all α + n ∈ S + n,

X r (fg)(α + n) = αrg(α + n)(α + n) r X r = αrG(α)(α + n) r = (FG)(α) ∈ ZQ.

Thus, (fg)(x) ∈ Int(S + n, ZQ). Thus, Int(S + n, ZQ) is a ring. If we assume that S + n is a ringset, then the same proof works when we take f(x), g(x) ∈ Int(S, ZQ),F (x) = f(x − n), and G(x) = g(x − n). This proves the equivalence of (i) and (ii).

(i) ⇐⇒ (iii) This proof is similar to the one above. Assume ﬁrst that S is a ringset, let f(x), g(x) ∈ Int(mS, ZQ), and deﬁne F (x),G(x) ∈ Int(S, ZQ) by F (x) = f(mx) and G(x) = g(mx). Then, as in the proof that (i) and (ii) are equivalent, we can show that (fg)(mx) = (FG)(x), from which we conclude that Int(mS, ZQ) is closed under multiplication, and hence is a ring.

To prove that (iii) ⇒ (i), one would take f(x), g(x) ∈ Int(S, ZQ), deﬁne

x x F (x),G(x) ∈ Int(mS, ZQ) by F (x) = f( m ) and G(x) = g( m ), and then proceed as before.

116 7.3 Results for Reduced Subsets of ZQ

As the examples in the previous sections show, deciding whether a subset of ZQ is a ringset can be challenging. In this section, we begin to derive necessary and sufficient conditions for a certain type of finite subset of ZQ to be a ringset. Our first theorem examines what happens with one-element subsets of ZQ.

Theorem 7.3.1. Let α ∈ ZQ. Then, Int({α}, ZQ) is a ring ⇐⇒ α ∈ Z.

Proof. (⇐) This follows from Proposition 7.1.3.

(⇒) We prove the contrapositive. Assume that α∈ / Z. Then, there exists u ∈ {i, j, k} such that α and u do not commute, i.e. αu − uα 6= 0.

f(x) Now, let f(x) = x − α. Then, n ∈ Int({α}, ZQ) for all n > 0. However,

αu−uα αu−uα 6= 0, so there exists m > 0 such that m ∈/ ZQ. Let g(x) = (x−α)(x−u) =

2 x−α g(x) x − (α + u)x + αu. Then, g(α) = αu − uα, so m (x − u) = m ∈/ Int({α}, ZQ), even

x−α though both m and x − u are in Int({α}, ZQ). Thus, Int({α}, ZQ) is not closed under multiplication and therefore is not a ring.

The technique used in Theorem 7.3.1 of multiplying a polynomial by one of x − i, x − j, or x − k will be employed frequently enough that we shall record some of the relevant computations. Assume that α = a+bi+cj+dk and let g1(x) = (x−α)(x−i), g2(x) = (x − α)(x − j), and g3(x) = (x − α)(x − k). Then,

g1(α) = αi − iα = 2dj − 2ck

g2(α) = αj − jα = −2di + 2bk, and (7.3.2)

g3(α) = αk − kα = 2ci − 2bj .

117 When S is not a singleton subset of ZQ, the situation is more complicated. In

Theorem 7.2.3, we found a suﬃcient condition for a subset S of ZQ to be a ringset, and in that theorem it was important to consider the minimal polynomials of elements

of S. So, it seems reasonable to focus on subsets of ZQ in which every element has the same minimal polynomial. Assume that S is such a set. Then, every element of

S has the same constant coeﬃcient a, and by Theorem 7.2.6 part (ii), S is a ringset if and only if S − a is a ringset. Each element of S − a has constant coeﬃcient equal to

0, so there is no loss in assuming that each element of S has constant coeﬃcient equal to 0. Furthermore, by applying Theorem 7.2.6 part (iii), there is no loss in assuming that the elements of S have no common positive integer factor other than 1. This leads us to the following deﬁntion.

Deﬁnition 7.3.3. A subset S ⊆ ZQ − Z is called reduced if 1) all elements of S have the same minimal polynomial,

2) all elements of S have constant coeﬃcient equal to 0, and

3) for all n > 1, S 6⊆ (n) in ZQ.

A subset S ⊆ ZQ−Z is called reducible if there exists a reduced set T ⊆ ZQ−Z, a ∈ Z,

and m ∈ Z, m > 0 such that S = mT + a. If such a T exists, we call it the reduction of S.

To illustrate this deﬁntion, we give some examples. The set {4 + 5i, 4 − 5k} is

reducible with reduction {i, −k} and the set {7 + 3j, 7 − i + 2j + 2k, 7 − 3k} is

reducible with reduction {3j, −i+2j +2k, −3k}. On the other hand, the set {6−4k,

4j} is not reducible and {2i + 2k, 2j − 2k} is reducible but not reduced. As the next

proposition shows, a reducible set is precisely a set where every element has the same

minimal polynomial.

118 Proposition 7.3.4.

(i) A reduced subset of ZQ must be non-empty and ﬁnite.

(ii) Let S be a subset of ZQ − Z. Then, S is reducible if and only if every element of S has the same minimal polynomial.

(iii) Assume S is a reducible subset of ZQ, and let T be the reduction of S. Then, S is a ringset if and only if T is a ringset.

Proof. (i) By condition 3) in the deﬁnition above, a reduced subset of ZQ must be non-empty. Also, for any positive integer n, there are only ﬁnitely many elements of

ZQ with norm n; thus, given any monic quadratic polynomial f(x) ∈ Z[x], there are

only ﬁnitely many elements of ZQ that have f(x) as a minimal polynomial. Thus,

by condition 1), a reduced subset of ZQ must be ﬁnite.

(ii) (⇒) Assume that S is reducible with reduction T . Then, there exist a, m ∈ Z, m 6= 0, such that S = mT + a. Since T is reduced, every element of T has the

same minimal polynomial; equivalently, every element of T has the same norm n and

constant coeﬃcient A, and we know that A = 0. So, every element of S has norm

a2 + m2n and constant coeﬃcient a. Thus, every element of S solves the polynomial

x2 − 2ax + a2 + m2n.

(⇐) Since every element of S has the same minimal polynomial, there exists

a ∈ Z such that each element of S has constant coeﬃcient equal to a. Let m be the

α−a largest positive integer such that S − a ⊆ (m) in ZQ, and let T = { m | α ∈ S}. Then, T is reduced and S = mT + a, so we are done.

119 (iii) This follows from Theorem 7.2.6.

If we deﬁne a relation ∼ on ZQ by α ∼ β ⇐⇒ minα(x) = minβ(x), then it is not

hard to see that ∼ is an equivalence relation on ZQ. Thus, any subset of ZQ−Z may be expressed as the disjoint union of reducible subsets. It seems plausible that if we

can characterize which reducible subsets are ringsets, then we can obtain information

about arbitrary subsets of ZQ. Necessary and suﬃcient conditions for reducible sets to be ringsets are given by Theorem 7.4.6, and these conditions will be extended to

any finite subset of ZQ in Section 7.5. However, finding a theorem that works for general subsets of ZQ is an open problem. The classification given in Theorem 7.4.6 depends on the following numerical

quantity.

Definition 7.3.5. For any subset S ⊆ ZQ, we define the integer Γ(S) to be Γ(S) = gcd {N(α − β)}α,β∈S . To avoid confusion with the defintion of gcd({0}), when S is

a singleton set we take Γ(S) = 0.

Remark. When S is reduced and |S| > 1, taking distinct elements α = bi+cj+dk and

β = yi+zj+wk in S and computing N(α−β) gives N(α−β) = 2N(α)−2(by+cz+dw).

Thus, when S is reduced, Γ(S) is always an even number.

As we shall see, when S is reduced we can often use Γ(S) to determine whether or

not S is a ringset (although Γ(S) falls just short of giving us a complete classiﬁcation).

The most interesting result is the following.

Theorem 7.3.6. Let S be a reduced subset of ZQ. If S is a ringset, then Γ(S) equals 2, 4, or 8.

120 The proof of Theorem 7.3.6 is involved and requires several intermediate theorems

and lemmas. It is given after Lemma 7.3.12. To illustrate the ﬂavor of the proof and

to motivate our subsequent results, we work through an example with a reduced

three-element subset of ZQ.

Example 7.3.7. Let S = {3i + 4j + 12k, −12i + 5k, −12i + 5j}; for convenience, let

α = 3i + 4j + 12k, β = −12i + 5k, and γ = −12i + 5j. Then, S is reduced, and we

compute that

α − β = 15i + 4j + 7k, of norm 290,

α − γ = 15i − j + 12k, of norm 370, and

β − γ = −5j + 5k, of norm 50,

so Γ(S) = 10. Furthermore, notice that

β − α ≡ γ − α ≡ −4j − 2k mod 5 .

So, letting f(x) = (4j + 2k)(x − α), we have

f(α) ≡ 0 mod 5,

f(β) ≡ N(β − α) ≡ 0 mod 5, and

f(γ) ≡ N(γ − α) ≡ 0 mod 5 .

f(x) Thus, 5 is a non-trivial element of Int(S, ZQ). Now, if Int(S, ZQ) were a ring,

f(x) then g(x) = 5 (x − j) would be in Int(S, ZQ). However, we can compute that

−12+24i−48j+96k g(α) = 5 , which is not in ZQ. Therefore, we conclude that Int(S, ZQ) is not a ring.

Two things power this example. The ﬁrst is the fact that 4j + 2k will kill all

three of α − β, α − γ, and β − γ modulo 5; this allows us to determine that the

121 f(x) polynomial 5 is in Int(S, ZQ). Secondly, to demonstrate that Int(S, ZQ) is not a

f(x) ring, it suﬃced to multiply 5 by the “nice” polynomial x − j. As we shall see, neither of these developments was an accident. Given a reduced subset S and an odd prime p that divides Γ(S), we will prove (essentially) that there always exists a linear

f(x) polynomial f(x) ∈ ZQ[x] such that p ∈ Int(S, ZQ), and furthermore that at least

f(x) f(x) f(x) one of p (x − i), p (x − j), or p (x − k) is not in Int(S, ZQ). A similar technique will work when Γ(S) is a power of 2 greater than 8.

The process of proving Theorem 7.3.6 begins with the following proposition.

Proposition 7.3.8. Let S ⊆ ZQ be a reduced set. Assume that S is a ringset. Then, for any odd prime p, Muﬀ (S) contains no linear polynomials. ZQ/(p)

Proof. Let p be an odd prime, and let R = ZQ/(p). Let γ, δ ∈ R be such that

f(x) = γx + δ ∈ MuﬀR(S). It suﬃces to show that γ ≡ 0 mod p, since this will force

δ ≡ 0 and consequently f(x) = 0.

Now, for all α, β ∈ S, we have 0 ≡ f(α) − f(β) ≡ γ(α − β) mod p. This means

that for all α ∈ S, the polynomial γ(x − α) ∈ MuﬀR(S). Since S is reduced, there

exists β = yi+zj +wk ∈ S such that β∈ / (p) in ZQ. Then, γ(x−β) ∈ MuﬀR(S), and

since S is a ringset, we know that MuﬀR(S) is an ideal of R[x]. Hence, the following

polynomials are in MuﬀR(S):

g1(x) = γ(x − β)(x − i),

g2(x) = γ(x − β)(x − j), and

g3(x) = γ(x − β)(x − k) .

122 Evaluating these polynomials at β yields

g1(β) ≡ γ(βi − iβ) ≡ 2γ(wj − zk),

g2(β) ≡ γ(βj − jβ) ≡ 2γ(−wi + yk), and

g3(β) ≡ γ(βk − kβ) ≡ 2γ(zi − yj) .

Since p is odd and g1(x), g2(x), and g3(x) are all in MuﬀR(S), the above equivalences imply that

γ(wj − zk) ≡ 0, γ(−wi + yk) ≡ 0, and γ(zi − yj) ≡ 0 .

Thus, 0 ≡ γ(wj − zk)(wj − zk) ≡ γ(z2 + w2),

0 ≡ γ(−wi + yk)(−wi + yk) ≡ γ(y2 + w2), and (∗)

0 ≡ γ(zi − yj)(zi − yj) ≡ γ(y2 + z2) .

Now, suppose that y2 + z2, y2 + w2, and z2 + w2 are all congruent to 0 mod p.

Then,

0 ≡ (y2 + z2) − (y2 + w2) ≡ z2 − w2,

so we get

0 ≡ z2 + w2 + z2 − w2 ≡ 2z2,

from which we conclude that z ≡ 0. Similarly, y ≡ w ≡ 0. But then, β = yi + zj +

wk ∈ (p) in ZQ, a contradiction. Thus, at least one of y2 + z2, y2 + w2, or z2 + w2 is non-zero, and hence a unit, mod p. By the equivalences in (∗), we must have γ ≡ 0

mod p, and we are done.

Proposition 7.3.8 generalizes the second major step—that of causing a problem by

multiplying a suitable polynomial by x − i, x − j, or x − k—in Example 7.3.7. What

123 remains is to determine what the “suitable” polynomial is, and doing that requires a

couple of lemmas.

Lemma 7.3.9. Let p be an odd prime, and let R = ZQ/(p). Let N = {α ∈ R |

α is nilpotent}. Then,

(i) |N| = p2

(ii) N is not closed under addition

∼ Proof. (i) We exploit the isomorphism R = M2(Fp) (see Corollary 1.4.2). A matrix

in M2(Fp) is nilpotent if and only if both its trace and determinant are 0. Assume

a b that A = c d ∈ M2(Fp) is nilpotent. Then, we must have d = −a and 0 = det(A) = a2 − bc.

From here, we have two cases. If b 6= 0, then to ensure that a2 − bc = 0 we must have c = b−1a2. In this situation, we have p − 1 choices for b and p choices for a, giving (p − 1)p diﬀerent nilpotent matrices. If b = 0, then a = 0. In this case, we

have p choices for c, yielding an additional p nilpotent matrices. These two cases

exhaust all the posibilities for A, so we have a total of (p − 1)p + p = p2 nilpotent

2 matrices in M2(Fp). It follows that |N| = p .

(ii) Let α ∈ N − {0}. Then, by Proposition 4.2.1, α ≡ bi + cj + dk for some b, c, d ∈ Z/pZ. Since α 6≡ 0, at least one of b, c, or d is not congruent to 0 mod p. WLOG, assume that b 6≡ 0. Now, −iαi is also in N, but α + (−iαi) ≡ 2bi is a unit mod p. Hence, α + (−iαi) 6∈ N, and N is not closed under addition.

124 In the next lemma and associated theorem, we will use the following notation: for

any α ∈ ZQ/(p), we let hαi = {nα | n ∈ Z/pZ}. That is, hαi is the cyclic subgroup generated by α within the additive group ZQ/(p).

Lemma 7.3.10. Let p be an odd prime, and let α and β be nilpotent elements of

ZQ/(p). Assume that α − β is also nilpotent. Then, either α ∈ hβi or β ∈ hαi.

Proof. Write α ≡ bi + cj + dk and β ≡ yi + zj + wk. Then, since α, β and α − β are all nilpotent, we have N(α) ≡ b2 + c2 + d2 ≡ 0, N(β) ≡ y2 + z2 + w2 ≡ 0, and

0 ≡ N(α − β)

≡ (b − y)2 + (c − z)2 + (d − w)2

≡ b2 + c2 + d2 + y2 + z2 + w2 − 2(by + cz + dw)

≡ −2(by + cz + dw) .

Now, for all n, m ∈ Z/pZ, nα + mβ has constant coeﬃcient equivalent to 0, and

N(nα + mβ) ≡ (nb + my)2 + (nc + mz)2 + (nd + mw)2

≡ n2(b2 + c2 + d2) + m2(y2 + z2 + w2) + 2nm(by + cz + dw)

≡ 0,

so nα + mβ is also nilpotent.

Let G = {nα + mβ | n, m ∈ Z/pZ} be the additive subgroup of ZQ/(p) generated by α and β. Then, by the above, each element of G is nilpotent. By Lemma 7.3.9 part

(i), |G| ≤ p2; furthermore, |G| divides ZQ/(p) = p4. However, G is closed under

addition, so by Lemma 7.3.9 part (ii), |G|= 6 p2. Thus, either G = {0} or |G| = p.

125 If G = {0}, then α ≡ β ≡ 0; if |G| = p, then G is cyclic and either G = hαi

or G = hβi. In any of these cases, we have reach the desired conclusion, so we are

done.

We can now prove that Γ(S) must be a power of 2 in order for Int(S, ZQ) to be a ring.

Theorem 7.3.11. Let S be a reduced subset of ZQ. If S is a ringset, then Γ(S) must be a power of 2.

Proof. We prove the contrapositive. The only way that Γ(S) can be 0 is if S is a singleton set, in which case Int(S, ZQ) is not a ring by Theorem 7.3.1. So, assume that |S| > 1. Then, by the remark following Deﬁnition 7.3.5, Γ(S) is a positive even number. We are assuming that Γ(S) is not a power of 2, so let p be an odd prime that divides Γ(S). We want to show that S is not a ringset. Let R = ZQ/(p). By

Proposition 7.3.8, it suﬃces to show that MuﬀR(S) contains a linear polynomial.

Let D = {α − β | α, β ∈ S}, the set of diﬀerences of elements of S. Since S is reduced and p | Γ(S), each element of D is nilpotent in R. Next, ﬁx β0 ∈ S and let

D0 = {α − β0 | α ∈ S} ⊆ D. Note that since |S| > 1, D0 6= {0}.

Next, let π : ZQ → R be the canonical quotient map. We will use the following:

Claim: for each non-empty subset T ⊆ π(D0), there exists γ ∈ T such that T ⊆ hγi

Proof of Claim: We proceed by induction on |T |. If |T | = 1, then T = {γ} ⊆ hγi for some γ ∈ π(D0) and we are done. So, assume that |T | > 1 and that the Claim is true for any subset of π(D0) of cardinality less than |T |. Let δ ∈ T . By induction, there exists ε ∈ T − {δ} such that T − {δ} ⊆ hεi. Now, δ and ε are elements of

126 π(D), so both δ and ε are nilpotent in R. Furthermore, there exist δ0, ε0 ∈ S such

that δ ≡ δ0 − β0 and ε ≡ ε0 − β0. So, δ − ε ≡ δ0 − ε0 and δ0 − ε0 ∈ D, so δ − ε is

also nilpotent. By Lemma 7.3.10, either δ ∈ hεi or ε ∈ hδi. Thus, either T ⊆ hδi or

T ⊆ hεi. This proves the Claim.

Now, if π(D0) = {0} (i.e. every element of D0 is congruent to 0 mod p), then the

polynomial x − β0 is in MuﬀR(S), and we are done. So, assume that π(D0) 6= {0},

i.e. some element of D0 is non-zero in R, and consequently |π(D0)| > 1. Applying

the Claim with T = π(D0), we have π(D0) ⊆ hγi for some γ ∈ π(D0). Let f(x) =

γ(x − β0) ∈ R[x]. Note that γ is non-zero (since π(D0) 6= {0}) and nilpotent (since

γ ∈ π(D0) ⊆ π(D)), so f(x) 6= 0 in R[x] and N(γ) ≡ N(γ) ≡ 0.

We know that, for each α ∈ S, the residue of α − β0 modulo p is in π(D0). But,

π(D0) ⊆ hγi, so for each α ∈ S there exists nα ∈ Z/pZ such that α − β0 ≡ nαγ. Thus, for each α ∈ S,

f(α) ≡ γ(α − β0) ≡ γnαγ ≡ nαN(γ) ≡ 0,

so f(x) is a linear polynomial in MuﬀR(S). Therefore, by Proposition 7.3.8, S is not a ringset.

Proving Theorem 7.3.11 is the most diﬃcult part in showing that Theorem 7.3.6

is true. From here, things are fairly straightforward.

Lemma 7.3.12. Let S be a reduced subset of ZQ. Then, for all α ∈ S, some coeﬃcient of α is odd.

Proof. Suppose there exists α ∈ S such that all coeﬃcients of α are even. Then,

N(α) ≡ 0 mod 4. Let β ∈ S. Since S is reduced, N(β) ≡ N(α) ≡ 0 mod 4. But

127 then, N(β) ≡ 0 mod 4 is a sum of three squares, implying that all coeﬃcients of β

are even. Since β was an arbitrary element of S, we conclude that S ⊆ (2) in ZQ, a contradiction. Therefore, every element of S has some coeﬃcient that is odd.

We are now ready for the proof of Theorem 7.3.6.

Proof of Thm. 7.3.6:

Assume that S is a ringset. By Theorem 7.3.11, Γ(S) must be a power of 2. Suppose by way of contradiction that Γ(S) is greater than 8. Then, for all α, β ∈ S, the norm of α − β is divisible by 16. Since N(α − β) is a sum of three squares, Lemma 5.1.3 tells us that each coeﬃcient of α − β is divisible by 4. Thus, for any β ∈ S, the

x−β polynomial 4 ∈ Int(S, ZQ).

x−β Now, ﬁx β = yi + zj + wk ∈ S and let f(x) = 4 . By Lemma 7.3.12, some coeﬃcient of β must be odd. WLOG, assume that y is odd, and let g(x) = f(x)(x−j)

(if either z or w were odd, we would take g(x) = f(x)(x−i)). Then, employing (7.3.2)

−2wi+2yk shows that g(β) = 4 , which is not an element of ZQ because y is odd. Thus,

g(x) ∈/ Int(S, ZQ), which contradicts the fact that Int(S, ZQ) is a ring. Therefore,

Γ(S) must be 2, 4, or 8. So far, we have only given negative results about whether S is a ringset. The

next theorem shows that if S is reduced and Γ(S) = 2 or 4, then S is a ringset.

Unfortunately, the case where Γ(S) = 8 turns out to be ambiguous. We will study

that situation in more detail in Section 7.4.

Theorem 7.3.13.

128 (i) Let S ⊆ ZQ be such that all elements of S have the same minimal polynomial. If Γ(S) = 2, then S is a ringset.

(ii) Let S be a reduced subset of ZQ. If Γ(S) = 4, then S is a ringset.

Q Proof. (i) Let n > 1, let Rn = Z /(n), and let f ∈ MuﬀRn (S). Since all elements of

S share the same minimal polynomial, there exist γ, δ ∈ ZQ such that for all α ∈ S, f(α) ≡ γα + δ mod n. So,

for all α, β ∈ S, 0 ≡ f(α) − f(β) ≡ γ(α − β) mod n . (∗)

Next, since Γ(S) = 2, there exist α1, α2, . . . , αt, β1, β2, . . . , βt ∈ S such that

gcd(N(α1 − β1),N(α2 − β2),...,N(αt − βt)) = 2 .

So, there exist A1,A2,...,At ∈ Z such that

2 = A1N(α1 − β1) + A2N(α2 − β2) + ··· + AtN(αt − βt) .

Hence,

2γ ≡ A1γN(α1 − β1) + A2γN(α2 − β2) + ··· + AtγN(αt − βt)

≡ A1γ(α1 − β1)(α1 − β1) + A2γ(α2 − β2)(α2 − β2) + ··· + At(αt − βt)(αt − βt)

≡ 0, by (∗).

Now, for any α ∈ S and any β ∈ Konj(α), there exists ε ∈ ZQ such that α−β = 2ε. Thus, modulo n we have

−f(β) ≡ f(α) − f(β) ≡ γ(α − β) ≡ 2γε ≡ 0 .

Therefore, S is a ringset by Corollary 7.1.6.

129 (ii) Assume that Γ(S) = 4. By Corollary 7.1.6, to show that S is a ringset, it e Q e suffices to show that MuffZQ/(p )(S) is an ideal of Z /(p ) [x] for all prime powers e e p . So, let p be any prime, and let e > 0. Let R = ZQ/(p ), and let f ∈ MuffR(S).

Since S is reduced, there exist γ, δ ∈ ZQ such that f(α) ≡ γα + δ mod pe for all α ∈ S. We have two cases to consider.

Case 1: p is odd

Since Γ(S) = 4, there exist α, β ∈ S such that N(α − β) is not divisible by p. So,

α − β is a unit mod pe. Hence, having 0 ≡ f(α) − f(β) ≡ γ(α − β) mod pe implies that γ ≡ 0 mod pe. Thus, for all α0 ∈ S and all β0 ∈ Konj(α0),

−f(β0) ≡ f(α0) − f(β0) ≡ γ(α0 − β0) ≡ 0 mod pe .

By Theorem 7.1.5, it follows that MuﬀR(S) is an ideal of R[x].

Case 2: p = 2

Since Γ(S) = 4, there exist α, β ∈ S such that N(α − β) = 4n, where n is odd.

Since S is reduced, N(α − β) is a sum of three squares. But, N(α − β) ≡ 0 mod

4, so each coeﬃcient of α − β must be even. Thus, there exists ε ∈ ZQ such that α − β = 2ε, and we must have N(ε) = n. Hence, ε is invertible mod 2e.

Now, f(α)−f(β) ≡ γ(α −β) ≡ 2γε mod 2e, implying that 2γ ≡ 0. Taking α0 ∈ S and β0 ∈ Konj(α0), there exists ε0 ∈ ZQ such that α0 − β0 = 2ε0, and working mod 2e we get

−f(β0) ≡ f(α0) − f(β0) ≡ γ(α0 − β0) ≡ 2γε0 ≡ 0 .

130 Hence, we once again conclude that MuﬀR(S) is an ideal of R[x].

There are no more cases to consider, so the proof of part (ii) of the theorem is complete.

7.4 The Case Γ(S) = 8 and the Classiﬁcation of Reducible Subsets

As the next example demonstrates, when the reduced set S satisﬁes Γ(S) = 8, S may or may not be a ringset.

Example 7.4.1. Let S = {i + j + k, i − j − k, −i + j − k, −i − j + k}. Then, it is easy to see that S is reduced and Γ(S) = 8. Furthermore, S = Konj (i + j + k), so ZQ S is a ringset by Theorem 7.2.1.

Next, let T = {2i + 3j + 4k, −5j − 2k, −2i + 3j + 4k}. Then, T is reduced, and one may verify that Γ(T ) = 8. Letting α = 2i + 3j + 4k, β = −5j − 2k, and

γ = −2i + 3j + 4k, we have

α − β ≡ 2i + 2k ≡ γ − β mod 4,

(i+k)(x−β) so f(x) = 4 ∈ Int(S, ZQ). Let g(x) = f(x)(x − i). Then, evaluation at β

−10+4i−10j−4k yields g(β) = 4 ∈/ ZQ, so Int(S, ZQ) is not a ring.

We shall give necessary and suﬃcient conditions under which a reduced set S with

Γ(S) = 8 is a ringset, but the conditions are somewhat obtuse and unenlightening.

Nevertheless, the ﬁrst step is to investigate properties of S when Γ(S) = 8.

Lemma 7.4.2. Let S be a reduced subset of ZQ such that Γ(S) = 8. Let D = {α−β | α, β ∈ S}. Then,

131 (i) for all δ ∈ D, every coeﬃcient of δ must be even

(ii) there exists δ ∈ D such that some coeﬃcient of δ is congruent to 2 mod 4

(iii) for all δ ∈ D, at least two coeﬃcients of δ are congruent to 0 mod 4

(iv) for all δ ∈ D, either exactly two coeﬃcients of δ are congruent to 0 mod 4, or

every coeﬃcient of δ is congruent to 0 mod 4; that is, for all δ ∈ D, the residue

of δ mod 4 lies in {0, 2i + 2j, 2i + 2k, 2j + 2k}.

(v) there exists δ ∈ D such that exactly two coeﬃcients of δ are congruent to

0 mod 4; that is, the residue of δ mod 4 lies in {2i + 2j, 2i + 2k, 2j + 2k}.

Proof. First, note that by the way we deﬁned D, we have Γ(S) = gcd {N(δ) | δ ∈

D}. Second, since S is reduced, the constant coefficient of every element of D is equal to 0, so for all of the following proofs, it suffices to focus on the coefficients of i, j, and k.

(i) Since S is reduced and Γ(S) = 8, the norm of any element of D is a sum of three squares that is divisible by 4. Hence, each coeﬃcient of every element of D must be even.

(ii) Suppose not. Then, each δ ∈ D would lie in (4) in ZQ, and thus 16 would divide N(δ) for all δ ∈ D. But then, 16 | Γ(S), a contradiction.

(iii) Suppose not. Then, there exists δ ∈ D such that N(δ) ≡ 4 mod 8, implying that Γ(S) 6= 8.

132 (iv) By (iii), for each δ ∈ D at least two coeﬃcients of δ are congruent to 0 mod

4. If there exists δ ∈ D such that exactly three coeﬃcients of δ are congruent to 0 mod 4, then N(δ) ≡ 4 mod 8, which forces Γ(S) 6= 8.

(v) Apply (ii) and (iv).

The next lemma gives one ﬁnal property of a reduced set S with Γ(S) = 8. It was not included in the previous lemma because the statement and proof are longer.

Lemma 7.4.3. Let S be a reduced subset of ZQ such that Γ(S) = 8. Let p be a

e prime, let e > 1, let R = ZQ/(p ), let f ∈ MuﬀR(S), and let γ1, γ0 ∈ ZQ be such that

e e for all α ∈ S, f(α) ≡ γ1α + γ0 mod p . Then, 4γ1 ≡ 0 mod p .

Proof. First, note that γ1 and γ0 as given in the statement of the lemma exist because

S is reduced. Let D = {α − β | α, β ∈ S}. Note that for any δ = α − β ∈ D, we have

e 0 ≡ f(α) − f(β) ≡ γ1δ mod p .

Now, if p is odd, then p - Γ(S), so there exists δ ∈ D such that p - N(δ). Hence,

0 ≡ γ1δ ≡ γ1δδ ≡ γ1N(δ),

e and N(δ) is invertible mod p , so we must have γ1 ≡ 0.

So, assume that p = 2. Since Γ(S) = 8, there exists ε ∈ D such that N(ε) = 8n, where n is an odd number (if no such ε exists, then 16 | Γ(S), which we know to be false). By part (i) of Lemma 7.4.2, there exists ε0 ∈ ZQ such that ε = 2ε0, and we must have N(ε0) = 2n. Thus, modulo pe, we have

0 0 0 0 ≡ γ1ε ≡ 2γ1ε ≡ 2γ1ε ε ≡ 4nγ1,

e and 4γ1 must be equivalent to 0 because n is invertible mod p .

133 We will prove one more lemma before proceeding to the classiﬁcation theorem of

those reduced sets S with Γ(S) = 8 that are ringsets. This lemma should really be part of the proof of that theorem, but the proof of the lemma is long enough that we deal with it separately.

Lemma 7.4.4. Let S be a reduced subset of ZQ such that Γ(S) = 8. Let D = {α−β | α, β ∈ S}. Assume that there exist δ, ε ∈ D such that the residues of δ and ε mod 4

lie in {2i+2j, 2i+2k, 2j +2k}, but δ 6≡ ε mod 4. Then, for each α = bi+cj +dk ∈ S,

the coeﬃcients b, c, and d are all odd.

Proof. Let π : ZQ → ZQ/(4) be the standard quotient map. By Lemma 7.4.2 part (iv), π(D) ⊆ {0, 2i + 2j, 2i + 2k, 2j + 2k}. Since 0 ∈ D, the residue of 0 mod 4 lies

in π(D); hence, the conditions on δ and ε imply that |π(D)| > 2.

Now, let α = bi + cj + dk ∈ S. By Lemma 7.3.12, at least one of b, c, or d is odd.

We have two cases to consider, and we will derive a contradiction in each one.

Case 1: exactly one of b, c or d is even

WLOG, assume that d is even and b and c are both odd. Then, d = 2E for some

E ∈ Z. Let β = yi + zj + wk ∈ S. Then, we have α − β ∈ D, so Lemma 7.4.2 part (i) tells us that b − y, c − z and d − k are all even. Hence, y and z must be odd and

w must be even; let W ∈ Z be such that w = 2W . Next, suppose that E 6≡ W mod 2. Since N(α) = N(β), we have b2 + c2 + d2 =

y2 + z2 + w2, and so b2 + c2 + 4E2 = y2 + z2 + 4W 2. The terms b2, c2, y2, and z2 are

all congruent to 1 mod 8 (because b, y, z, and w are all odd), so 4E2 ≡ 4W 2 mod 8.

However, one of E or W is odd and the other is even. Thus, one of 4E2 or 4W 2 is

134 congruent to 4 mod 8, and the other is congruent to 0 mod 8. This is a contradiction, so we must have E ≡ W mod 2.

Since E ≡ W mod 2, we have 2E ≡ 2W mod 4. So, the coeﬃcient of k in α − β is congruent to 0 mod 4. Hence, by Lemma 7.4.2 part (iv), the residue of α − β mod 4 lies in {0, 2i + 2j}. Since β ∈ S was arbitrary, for any β1, β2 ∈ S we have, modulo 4,

β1 − β2 ≡ (β1 − α) − (β2 − α) ∈ {0, 2i + 2j}.

But then, π(D) ⊆ {0, 2i + 2j}, which is a contradiction.

Case 2: exactly two of b, c or d are even

WLOG, assume that c and d are even and b is odd. Then, there exist C,E ∈ Z such that c = 2C and d = 2E. Let β = yi + zj + wk ∈ S. Applying Lemma 7.4.2 part (i) as in Case 1, we see that y is odd but z and w must be even; so we may write z = 2Z and w = 2W for some Z,W ∈ Z. Now, by Lemma 7.4.2 part (iv), α − β mod 4 lies in {0, 2i + 2j, 2i + 2k, 2j + 2k}.

We will show that α − β mod 4 lies in {0, 2j + 2k}.

Suppose ﬁrst that α − β ≡ 2i + 2j mod 4. Then, d ≡ w mod 4, so E ≡ W mod

2. Similarly, c 6≡ z mod 4, so C 6≡ Z mod 2. So, one of C or Z is odd and the other is even; WLOG, assume that C is odd and Z is even. Then,

b2 ≡ y2 mod 8 (since both b and y are odd),

c2 ≡ 4C2 ≡ 4 mod 8,

z2 ≡ 4Z2 ≡ 0 mod 8, and

d2 ≡ 4E2 ≡ 4W 2 ≡ w2 mod 8.

135 However, this implies that N(α) 6≡ N(β) mod 8, which is a contradiction because

N(α) = N(β). So, we conclude that α − β 6≡ 2i + 2j mod 4.

If α − β ≡ 2i + 2k mod 4, then we may derive a contradiction just as in the preceding paragraph. So, we are forced to conclude that α − β mod 4 lies in {0,

2j + 2k}. As in Case 1, it now follows that π(D) ⊆ {0, 2j + 2k}, which again contradicts the fact that |π(D)| > 2.

We get a contradiction in both Case 1 and Case 2, so all three of b, c, and d must be odd. This proves the lemma.

We can now prove the promised classiﬁcation theorem.

Theorem 7.4.5. Let S be a reduced subset of ZQ such that Γ(S) = 8. Let D = {α − β | α, β ∈ S}. Then, S is a ringset ⇐⇒ there exist δ, ε ∈ D such that the residues of δ and ε mod 4 lie in {2i + 2j, 2i + 2k, 2j + 2k}, but δ 6≡ ε mod 4.

Proof. (⇒) We prove the contrapositive. Assume for all δ, ε ∈ D that either δ ≡ 0 mod 4, ε ≡ 0 mod 4, or δ ≡ ε mod 4. Now, by Lemma 7.4.2 part (iv), the residue

modulo 4 of any element of D lies in {0, 2i+2j, 2i+2k, 2j +2k}, and by Lemma 7.4.2

part (v), there exists δ ∈ D such that the residue of δ lies in {2i+2j, 2i+2k, 2j +2k}.

Hence, for all ε ∈ D, either ε ≡ 0 mod 4 or ε ≡ δ mod 4.

Now, WLOG, assume that δ ≡ 2i + 2j mod 4. Let β = yi + zj + wk ∈ S and let

(i+j)(x−β) f(x) = 4 . By the preceding paragraph, for each α ∈ S either α − β ≡ 2i + 2j mod 4 or α − β ≡ 0 mod 4. In either case, (i + j)(α − β) ≡ 0 mod 4, so f(x) ∈

136 Int(S, ZQ). Next, we compute that

(i + j)(βi − iβ) = (i + j)(2wj − 2zk) = −2w − 2zi + 2zj + 2wk, and

(i + j)(βj − jβ) = (i + j)(−2wi + 2yk) = 2w + 2yi − 2yj + 2wk .

By Lemma 7.3.12, at least one of y, z, or w is odd. If y is odd, then letting

(i+j)(βj−jβ) 2w+2yi−2yj+2wk g(x) = f(x)(x−j) and evaluating g(x) at β yields g(β) = 4 = 4 ,

which is not in ZQ. If y is even, then take g(x) = f(x)(x − i); this gives g(β) =

(i+j)(βi−iβ) −2w−2zi+2zj+2wk 4 = 4 ∈/ ZQ. In either case, g(x) ∈/ Int(S, ZQ), so Int(S, ZQ) is not a ring.

(⇐) Assume that there exist δ, ε ∈ D as in the statement of the theorem. Then,

δ and ε are congruent mod 4 to diﬀerent elements of {2i + 2j, 2i + 2k, 2j + 2k}.

Now, WLOG, assume that δ ≡ 2i+2j and ε ≡ 2i+2k. Let p be a prime, let e > 1,

e let R = ZQ/(p ), and let f ∈ MuﬀR(S). Since S is reduced, there exist γ, γ0 ∈ ZQ

e e such that f(α) ≡ γα + γ0 mod p for all α ∈ S. Then, γζ ≡ 0 mod p for all ζ ∈ D,

and 4γ ≡ 0 mod pe by Lemma 7.4.3.

Let α = bi + cj + dk ∈ S. To show that S is a ringset, it suﬃces to show that

f(β) ≡ 0 for all β ∈ Konj(α), and to establish this latter condition it suﬃces to show

that γ(α − β) ≡ 0 for all β ∈ Konj(α). By Lemma 7.4.4, all of b, c, and d are odd, so

α − (−jαj) = 2bi + 2dk ≡ ε mod 4, and

α − (−kαk) = 2bi + 2cj ≡ δ mod 4 .

Thus, there exist δ0, ε0 ∈ ZQ such that

α − (−jαj) = ε + 4ε0, and

α − (−kαk) = δ + 4δ0 .

137 Hence, in R,

γ(α − (−jαj)) ≡ γε + 4γε0 ≡ 0, and

γ(α − (−kαk)) ≡ γδ + 4γδ0 ≡ 0, and so f(−jαj) ≡ f(−kαk) ≡ f(α) ≡ 0.

It remains to show that γ(α−(−iαi)) ≡ 0 in R. We know that α−(−iαi) = 2cj +

2dk, and it is easy to see that −jαj − (−kαk) = 2cj − 2dk. Since f(−jαj), f(−kαk),

and 4γ are all equivalent to 0 in R, we have

0 ≡ f(−jαj) − f(−kαk)

≡ γ(−jαj − (−kαk))

≡ γ(2cj − 2dk)

≡ γ(2cj − 2dk) + 4γdk

≡ γ(2cj + 2dk)

≡ γ(α − (−iαi)) .

Thus, f(−iαi) ≡ 0. It now follows by Corollary 7.1.6 that S is a ringset.

Remark. An examination of the proof of Theorem 7.4.5 shows that if S is reduced,

Γ(S) = 8, and S is a ringset, then it must be true that every α = bi + cj + dk ∈ S has

b, c, and d all odd. However, this condition is not suﬃcient; a simple counterexample

is S = {i + j + k, i − j − k}. Thus, we must use the more cumbersome condition

given in the theorem.

Combining Theorem 7.3.6, Theorem 7.3.13, and Theorem 7.4.5 gives a complete

classiﬁcation of those reducible subsets S of ZQ that are ringsets.

138 Theorem 7.4.6. Let S be a reducible subset of ZQ with reduction T and let D = {α − β | α, β ∈ T }. Then, S is a ringset if and only if exactly one of the following holds:

(i) Γ(T ) = 2,

(ii) Γ(T ) = 4, or

(iii) Γ(T ) = 8 and there exist δ, ε ∈ D such that the residues of δ and ε mod 4 lie in

{2i + 2j, 2i + 2k, 2j + 2k}, but δ 6≡ ε mod 4.

Theorem 7.4.6 is particularly nice when applied to two-element reducible subsets.

Corollary 7.4.7. Let S be a two-element reducible subset of ZQ, and let T = {α, β} be the reduction of S. Then, S is a ringset if and only if N(α − β) = 2 or 4.

Proof. Let D be as in the statement of Theorem 7.4.6. Then, D = {±(α − β)}, so

condition (iii) cannot hold for T . Also, Γ(T ) = N(α − β), so S is a ringset if and only

if N(α − β) = 2 or 4.

7.5 Necessary and Suﬃcient Conditions for Finite Subsets

In the previous section, we derived necessary and suﬃcient conditions for a reduced

subset of ZQ to be a ringset. It turns out that we can extend these conditions to

work for any finite subset of ZQ. The idea is to first take a finite subset S of ZQ and partition S into reducible subsets. There is more than one way to achieve such a

partition, but when it is done in a certain way, we will show that S is a ringset if and

only if each of the reducible subsets in the partition of S is a ringset. The relevant

partition is described in the following deﬁnition.

139 Deﬁnition 7.5.1. Let S be a ﬁnite subset of ZQ. Let S0 = S ∩ Z and let m1(x),

m2(x), . . . , mt(x) be all the distinct minimal polynomials of elements of S − S0. That

is, if α ∈ S − S0, then minα(x) = m`(x) for some 1 ≤ ` ≤ t; m`1 (x) 6= m`2 (x) unless

`1 = `2; and for each 1 ≤ ` ≤ t, there exists α` ∈ S − S0 such that m`(x) = minα` (x).

For each 1 ≤ ` ≤ t, let S` = {α ∈ S | minα(x) = m`(x)}. Then, S can be expressed t S as the disjoint union S = S`. When S is written in this way, we call it the minimal `=0 polynomial partition of S.

t S When S is a ﬁnite subset of ZQ with minimal polynomial partition S = S`, `=0 each S` with 1 ≤ ` ≤ t is a reducible subset of ZQ, so we can use Theorem 7.4.6

to decide whether S` is a ringset. As we will show below in Corollary 7.5.5, S is a

ringset if and only if S` is a ringset for each 1 ≤ ` ≤ t (S0 is always a ringset because

it lies in Z). With this goal in mind, we ﬁrst more closely examine how a reduced subset can fail to be a ringset. In the proofs of Theorems 7.3.1, 7.3.11, 7.3.6, and

7.4.5, our strategy for showing that a reduced subset S was not a ringset was always

the same: we exhibited two linear polynomials in Int(S, ZQ) whose product was not

in Int(S, ZQ). It will be useful to recall what these linear polynomials were. In the following lemma, we summarize the diﬀerent cases that occurred.

Lemma 7.5.2. Let T be a reduced subset of ZQ that is not a ringset. Then, there

γ(x−β) exist β ∈ T , γ ∈ ZQ, n > 0, and u ∈ {i, j, k} such that n , x − u ∈ Int(T, ZQ),

γ(x−β)(x−u) but n ∈/ Int(T, ZQ).

Proof. We have already veriﬁed this result, but the proof is spread out over several

theorems, so for convenience we collect everything here.

140 Since T is not a ringset, Γ(T ) must be a non-negative integer diﬀerent than 2 or

4. This gives us several cases to consider.

Case 1: Γ(T ) = 0

The only way this case can occur is if T is a singleton set. As in Theorem 7.3.1,

we take β to be the lone element of T , γ = 1, u ∈ {i, j, k} such that βu − uβ 6= 0, and n such that βu − uβ is not divisible by n.

Case 2: p | Γ(T ) for some odd prime p

In this case, we refer to Theorem 7.3.11, where we showed that we can take n = p, any β ∈ T − (p), and (depending on the situation) either γ = 1 or a particular

γ ∈ ZQ such that N(γ) ≡ 0 mod p. Also, the u that we choose will depend on β, but will be an element of {i, j, k}, as shown in Proposition 7.3.8.

Case 3: 16 | Γ(T )

For this case, Theorem 7.3.6 shows that we can use n = 4, any β ∈ T , and γ = 1.

As in Case 2, u will depend on β.

Case 4: Γ(T ) = 8

Finally, since we are assuming that T is not a ringset, Theorem 7.4.5 demonstrates that we can take n = 4 and any β ∈ T . Then, there exist γ ∈ {i + j, i + k, j + k} and u ∈ {i, j, k} (both depending on β) such that the Lemma holds.

There are no more cases to consider, so we are done.

141 We need one more lemma that will be useful in the theorem that follows. The lemma is a straightforward result about polynomials over noncommutative rings. The result is well-known, but a proof could not be found in the available literature, so one is provided below.

Lemma 7.5.3. Let R be a noncommutative ring and let f(x), g(x) ∈ R[x]. Assume that f(x) is central in R[x]. Then, for any α ∈ R, we have (fg)(α) = (gf)(α) = g(α)f(α).

P r P s Proof. Let f(x) = r arx and g(x) = s βsx . Since f(x) is central in R[x], each ar is central in R. Let α ∈ R. We have

X r (fg)(α) = arg(α)α r X r = g(α)arα because each ar is central in R r X r = g(α) arα r = g(α)f(α) .

Similarly,

X s (gf)(α) = βsf(α)α s X s s = βsα f(α) because f(α) and α commute for each s s = g(α)f(α) .

Thus, the lemma holds.

We now show what goes wrong when one of the reducible subsets in the minimal polynomial partition of a ﬁnite set fails to be a ringset.

142 Theorem 7.5.4. Let S be a ﬁnite subset of ZQ with minimal polynomial partition t t S S S = S`. Let T be a reducible subset of ZQ such that S` ∪ T is the minimal `=0 `=0 polynomial partition of S ∪T . That is, if m(x) is the minimal polynomial correspond-

ing to T and ε is any element of S, then minε(x) does not equal m(x). Assume that

T is not a ringset. Then, S ∪ T is not a ringset.

Proof. First, since T is reducible, there exists a reduced subset T 0 ⊆ ZQ, a ∈ Z, and b ∈ Z−{0} such that T = a+bT 0. Notice that (S −a)∪(T −a) ⊆ ZQ, so by Theorem 7.2.7, S ∪ T is a ringset if and only if (S − a) ∪ (T − a) is a ringset. Thus, from here

0 1 on we may assume WLOG that a = 0, i.e. that T = bT . Now, while b T ⊆ ZQ, there

1 0 is no guarantee that b S ⊆ ZQ. So, the best we can assume is that T = bT . Under this assumption, for each α ∈ T there exists α0 ∈ T 0 such that α = bα0.

Next, for each 1 ≤ ` ≤ t, let f`(x) be the minimal polynomial corresponding to t Q Q S`. Additionally, let f0(x) = (x − A), and let f(x) = f`(x). Then, f(ε) = A∈S0 `=0 0 for all ε ∈ S. Let m(x) be the minimal polynomial corresponding to T . By assumption, m(x) 6= f`(x) for all 1 ≤ ` ≤ t, so f`(α) 6= 0 for all α ∈ T and all

1 ≤ ` ≤ t. Furthermore, f0(α) 6= 0 for all α ∈ T (if f0(α) = 0 for some α ∈ T , then Q (α − A) = 0, which is impossible because ZQ has no zero divisors and α∈ / Z), so A∈S0 f(α) 6= 0 for all α ∈ T .

Note that f(x) has coeﬃcients in Z, hence is central in ZQ[x]. Thus, Lemma 7.5.3 applies to f(x). We will freely use this fact in the calculations that follow.

As in Lemma 7.5.2, we have several cases to consider, depending on the value of

Γ(T 0).

143 Case 1: Γ(T 0) = 0

We proceed as in Theorem 7.3.1. Since Γ(T 0) = 0, both T 0 and T are singleton sets, so there exists β ∈ ZQ − Z such that T = {β}. Let g(x) = (x − β)f(x). Then,

g(x) g(ε) = 0 for all ε ∈ S ∪ T , so n ∈ Int(S ∪ T , ZQ) for all n > 0. However, β∈ / Z, so there exists u ∈ {i, j, k} such that βu − uβ 6= 0. Letting h(x) = g(x)(x − u), we have h(ε) = 0 for all ε ∈ S, but h(β) = (βu − uβ)f(β) 6= 0, so there exists n > 1 such

h(β) g(x) h(x) g(x)(x−u) that n ∈/ ZQ. Thus, n and x − u are in Int(S ∪ T , ZQ) while n = n is not. Hence, S ∪ T is not a ringset.

Case 2: p | Γ(T 0) for some odd prime p

Let β0 ∈ T 0, γ ∈ ZQ, and u ∈ {i, j, k} be as in the second case in Lemma 7.5.2, when applied to the reduced set T 0. Let β ∈ T be such that β = bβ0. Then, for all

0 0 0 0 e1 α ∈ T , we have γ(α − β ) ≡ 0 mod p. Factor b as b = p q1, where e1 ≥ 0 and p - q1. Then, we have

γ(α − β) = bγ(α0 − β0) ≡ 0 mod pe1+1 for all α ∈ T. (∗)

Next, consider f(x). Since T is reducible, by Theorem 2.3.1 there exist C,D ∈ Z such that f(α) = Cα + D for all α ∈ T . Let F (x) = f(x)(−Cx + D). Then, since

F (x) has integer coeﬃcients and each α ∈ T has constant coeﬃcient equal to 0, we have

F (α) = f(α)(−Cα + D) = (Cα + D)(Cα + D) = N(Cα + D) ∈ Z .

Furthermore, since each element of T has the same norm, it is not diﬃcult to see that for any α, δ ∈ T we have N(Cα + D) = D2 + C2N(α) = D2 + C2N(δ) = N(Cδ + D).

Lastly, we know that f(α) 6= 0 for all α ∈ T , so N(Cα + D) 6= 0. Thus, there exists

144 e2 a positive integer y such that F (α) = y for all α ∈ T . Factor y as y = p q2, where

e2 ≥ 0 and p - q2. Then,

F (α) ≡ 0 mod pe2 for all α ∈ T. (∗∗)

Now, let R = ZQ/(pe1+e2+1) and let g(x) = γ(x − β)F (x). Then, F (ε) = 0 for all

ε ∈ S, so g(ε) = 0 for all ε ∈ S. By (∗) and (∗∗), we see that g(α) ≡ 0 mod pe1+e2+1 for all α ∈ T . Hence, g(x) ∈ MuﬀR(S ∪ T ).

Finally, let h(x) = γ(x − β)(x − u) and H(x) = g(x)(x − u). Then,

H(x) = g(x)(x − u) = γ(x − β)F (x)(x − u) = γ(x − β)(x − u)F (x) = h(x)F (x).

By construction, we know that γ(β0u − uβ0) 6≡ 0 mod p, so

h(β) = γ(βu − uβ)

= bγ(β0u − uβ0)

e1 0 0 = p q1γ(β u − uβ )

6≡ 0 mod pe1+1 .

Thus,

H(β) = h(β)F (β)

e1 0 0 = p q1γ(β u − uβ )y

e1+e2 0 0 = p q1q2γ(β u − uβ )

6≡ 0 mod pe1+e2+1,

and therefore H(x) ∈/ MuﬀR(S ∪ T ). However, H(x) = g(x)(x − u) with

g(x) ∈ MuﬀR(S ∪ T ) and x − u ∈ R[x]; thus, MuﬀR(S ∪ T ) is not an ideal of R[x]. It

145 follows that S ∪ T is not a ringset.

Case 3: 4 | Γ(T 0)

We handle both Case 3 and Case 4 from Lemma 7.5.2 at once. The proof for this

case is similar to the one where p | Γ(T 0).

As above, let β0 ∈ T 0, γ ∈ ZQ, and u ∈ {i, j, k} be as in Lemma 7.5.2, when

0 e1 applied to the reduced set T . First, factor b as b = 2 q1, where e1 ≥ 0 and 2 - q1. Then, by construction, γ(α0 − β0) ≡ 0 mod 4 for all α0 ∈ T 0, so γ(α − β) ≡ 0 mod

2e1+2 for all α ∈ T . Next, let F (x) and y be as in the previous case of this proof,

e2 e +e +2 and factor y as y = 2 q2, where e2 ≥ 0 and 2 - q2. Let R = ZQ/(2 1 2 ) and let

g(x) = γ(x − β)F (x). Then, g(x) ∈ MuﬀR(S ∪ T ), but following the same steps used

in the previous case will show that g(x)(x − u) ∈/ MuﬀR(S ∪ T ). Hence, S ∪ T is not

a ringset in this case.

There are no more cases to consider, so the proof is complete.

We end this section with the necessary and suﬃcient conditions for a ﬁnite subset

of ZQ to be a ringset.

Corollary 7.5.5. Let S be a ﬁnite subset of ZQ with minimal polynomial partition t S S = S`. Then, S is a ringset if and only if S` is a ringset for all 1 ≤ ` ≤ t. `=0

Proof. (⇐) This is true because of Proposition 7.1.3 and Theorem 7.2.5.

(⇒) There is nothing to prove if t = 0, so assume that t > 1. We prove the contrapositive. Assume that S` is not a ringset for some ` ∈ {1, 2, . . . , t}. WLOG,

146 t−1 S assume that St is not a ringset. Applying Theorem 7.5.4 to S` ∪ St shows that `=0 S is not a ringset, as required.

147 BIBLIOGRAPHY

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[3] D. E. Flath. Introduction to Number Theory. Wiley, New York, 1989.

[4] K. R. Goodearl and Warﬁeld R. B. Jr. An Introduction to Noncommutative Noetherian Rings. Cambridge University Press, New York, 2004.

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