Explicit Multiplicative Relations Between Gauss Sums Brian J

Louisiana State University LSU Digital Commons

LSU Doctoral Dissertations Graduate School

2002 Explicit multiplicative relations between Gauss sums Brian J. Murray Louisiana State University and Agricultural and Mechanical College, [email protected]

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Recommended Citation Murray, Brian J., "Explicit multiplicative relations between Gauss sums" (2002). LSU Doctoral Dissertations. 757. https://digitalcommons.lsu.edu/gradschool_dissertations/757

This Dissertation is brought to you for free and open access by the Graduate School at LSU Digital Commons. It has been accepted for inclusion in LSU Doctoral Dissertations by an authorized graduate school editor of LSU Digital Commons. For more information, please [email protected]. EXPLICIT MULTIPLICATIVE RELATIONS BETWEEN GAUSS SUMS

A Dissertation Submitted to the Graduate Faculty of the Louisiana State University and Agricultural and Mechanical College in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in The Department of Mathematics

by Brian J. Murray B.A., Mathematics, Washington University in St. Louis, 1997 M.S., Louisiana State University, 1999 August 2002 Acknowledgments

First and foremost, I would like to thank my advisor Dr. Paul van Wamelen for both his guidance and lack thereof. Throughout the development of this dissertation, I was involved in all decisions and was always oﬀered support and advice. Rarely, however, was I given answers; for this, I am truly grateful. I would also like to thank Dr. van Wamelen for the informal nature of our relationship–although the last three years were challenging, they were certainly enjoyable.

This dissertation would not have been possible without the support of the De- partment of Mathematics at Louisiana State University. Speciﬁcally, I would like to thank my committee members, Professors George Cochran, Richard Litherland,

Augusto Nobile, Louise Perkins, and Robert Perlis. I would like to thank Dan Co- hen for helpful mathematical discussions as well as extremely helpful racquetball games. I am also grateful to Anthony Picado for keeping Lockett Hall running smoothly.

I would like to thank Robert Osburn for continued friendship, valued mathematical advice, and most importantly, indispensable commiseration.

Finally, I would like to thank my entire family for their continued support and encouragement of my education. This dissertation is dedicated to my wife, Ashley, whose love and support made this possible.

ii Table of Contents

Acknowledgments ...... ii

Abstract ...... iv

Introduction ...... 1

1. Background ...... 4 1.1 Algebraic Number Theory ...... 4 1.2 Cyclotomic Number Fields ...... 7 1.3 Class Field Theory ...... 8

2. Gauss Sums ...... 10 2.1 Preliminaries ...... 10 2.2 Sign Ambiguities ...... 14

3. Results ...... 17 3.1 Methodology ...... 17 3.2 Biquadratic Coset Sums ...... 21 3.3 A Product Formula ...... 35 3.4 Resolution of Ambiguity ...... 44

References ...... 51

Vita ...... 53

iii Abstract

H. Hasse conjectured that all multiplicative relations between Gauss sums essentially follow from the Davenport-Hasse product formula and the norm relation for

Gauss sums. While this is known to be false, very few counterexamples, now known as sign ambiguities, have been given. Here, we provide an explicit product formula giving an inﬁnite class of new sign ambiguities and resolve the ambiguous sign in terms of the order of the ideal class of quadratic primes.

iv Introduction

C. F. Gauss ﬁrst introduced the sums bearing his name in 1801 in his Disquisitiones

k 1 mn2 Arithmeticae. This sum, − ζ , now known as a quadratic Gauss sum, was Pn=0 k notoriously diﬃcult to evaluate, even in the special case that m = 1 and k is an odd integer, [3]. Gauss was easily able to prove that this sum has the value

√k, or i√k, depending on whether k was congruent to 1 or 3 modulo 4, but ± ± the resolution of the sign proved to be much more diﬃcult. In May, 1801, Gauss conjectured that the plus sign held in each case and then, for the next four years, devoted time every week to proving the conjecture, [3, 10]. Finally, in August, 1805,

Gauss proved his conjecture, recording in his diary “Wie der Blitz einschlägt,hat sich das Räthselgelöst.. . ” (as lightning strikes was the puzzle solved), [10]. Several years later, Gauss published a complete evaluation of the quadratic Gauss sum for all positive integers k. Using this evaluation, Gauss was able to give a fourth proof of his Theorema Aureum, or golden theorem, now known as the law of quadratic reciprocity. Since the initial work of Gauss, the determination of Gauss sums and the resolution of ambiguous signs has been fundamental in the study of reciprocity.

With the introduction of the multiplicative character χ modulo k in his treatise on primes in arithmetic progressions, G. L. Dirichlet was able to generalize the

k 1 mn Gauss sum, [3]. This sum, G(χ) = − χ(n)ζ , is also called a Gauss sum as it Pn=0 k coincides with the quadratic Gauss sum when χ has order 2 and k is taken to be a prime p not dividing m. In addition to providing the foundation for higher reciprocity laws, this generalization of the Gauss sum arises naturally in the study of cyclotomy and has many important applications throughout mathematics. Build- ing on the initial work of Gauss, Dirichlet, and Jacobi, many well-known mathe-

1 maticians have made contributions to the theory of Gauss sums and the closely related Jacobi sums. These include L. Carlitz, A. Cauchy, S. Chowla, H. Daven- port, G. Eisenstein, B. Gross, H. Hasse, N. M. Katz, N. Koblitz, L. Kronecker,

E. E. Kummer, D. H. and E. Lehmer, L. J. Mordell, S. J. Patterson, C. L. Siegel,

L. Stickelberger, and A. Weil, [3].

One powerful approach to simplifying the evaluation of Gauss sums is to study the multiplicative relations between them. The most basic of the multiplicative relations is the norm relation for Gauss sums, which states that the product of a

Gauss sum and its complex conjugate is, up to a unit of absolute value one, equal to the prime p. From the norm relation, it is then clear that the Gauss sums divide p and thus generate ideals which factor only into prime ideals above p. Moreover, using Stickelberger’s congruence for Gauss sums, the factorization of all Gauss sums can be given explicitly. Using these results of Stickelberger, H. Davenport and

H. Hasse were able to formulate the second type of multiplicative relation between

Gauss sums, the beautiful Davenport-Hasse product formula. Originally appearing in their 1934 paper Die Nullstellen der Kongruenzzetafunktionen in gewissen zyk- lischen F¨allen, the Davenport-Hasse product formula provides an entire class of nontrivial multiplicative relations. In fact, in [8, pg.465], H. Hasse conjectured the norm relation and the Davenport-Hasse product formula were essentially the only multiplicative relations connecting Gauss sums over Fp. However, in K. Yamamoto’s 1966 paper On a conjecture of Hasse concerning multiplicative relations of Gaussian sums, Yamamoto provided a simple counterexample disproving the conjecture. This counterexample was a new type of multiplicative relation involving an ambiguous sign not connected to elementary properties of Gauss sums. Shortly thereafter, working in the context of Jacobi sums and with the aid of a computer, further sign ambiguities were discovered by Muskat,

2 Muskat-Whiteman, and Muskat-Zee, [13, 15, 16]. Based on his difficulty in finding sign ambiguities, in [14] Muskat proposed that “. . . perhaps there are no more than the seven sign ambiguities noted above.” To date only nine sign ambiguities have been given explicitly. While very few had been found, in [22], Yamamoto succeeded in not only proving that infinitely many exist, but also produced a formula giving the exact “number” of sign ambiguities to be expected in each case. Despite this guarantee of new sign ambiguities, very little has been done since 1975.

With the 1998 publication of Gauss and Jacobi sums by Berndt, Evans, and

Williams, [3], there has been renewed interest in the study of Gauss sums. Ad- ditionally, the rapid evolution of computing power has made the computational techniques of Muskat, [14], much more eﬀective. Building on these ideas, we provide an inﬁnite class of new sign ambiguities.

In chapter 1, we quickly summarize some essentials of Algebraic Number Theory.

Included are speciﬁcs about quadratic and cyclotomic number ﬁelds as well as a brief description of the ideal class group and Dirichlet’s class number formula.

Chapter 2 continues with the necessary background for Gauss sums. After giving the basic deﬁnitions and properties of Gauss sums, we then cover multiplicative relations, Jacobi sums, the factorization of Gauss sums over cyclotomic ﬁelds, and sign ambiguities.

We present our main results in chapter 3. We ﬁrst discuss the methodology behind the computer search for sign ambiguities. Then, after a digression into biquadratic coset sums, we present and prove our main theorem.

3 1. Background

1.1 Algebraic Number Theory

An (algebraic) number field, K, is a subfield of the complex numbers of finite degree over Q. Let [K : Q] denote the degree of K over Q. Within a number field K lies the (algebraic) integers K , which consists of all elements of K that are roots of a O monic polynomial with integral coefficients. Given a nonzero ideal a of K , we can O form the residue field K /a. As the residue field is finite, we define the norm of a, O to be N(a) = K /a , the number of elements in K /a. While rings of integers are |O | O not necessarily unique factorization domains, they are Dedekind domains. That is, any nonzero ideal of K has a unique factorization into prime ideals of K . O O

Let L be a ﬁnite extension of K. Then p L is an ideal of L lying over or above O O e1 eg p, and has unique factorization p L = P Pg , where the Pi’s are the distinct O 1 ··· prime ideals of L containing p. The exponents ei are called the ramiﬁcation indices of p in Pi. In this situation we will also say that Pi is an ideal of L dividing p, O or equivalently, Pi contains p.

Now, since for i = 1, . . . , g, Pi contains p, we have a residue ﬁeld extension

L/Pi over K /p. The degree of this extension, denoted fi, is called the inertial O O degree of p in Pi. The number, g, of prime ideals lying over p, the ramiﬁcation indices, ei, and the inertial degrees, fi, are related by the following, see [6], [10].

Theorem 1.1. Let K be a number ﬁeld with ﬁnite extension L and let p be a prime ideal of K . Then O g e f = [L : K]. X i i i=1

A prime ideal p K is said to be inert in L if p L is again a prime ideal in ⊆ O O

L. If the factorization of p L contains any ramiﬁcation index ei > 1, then we say O

4 that p is ramified in L. Finally, if p is ramified in L, g = 1, and e1 = [L : K], then p is said to be totally ramified in L.

If L is a Galois extension of K the situation is considerably nicer. In particular,

Theorem 1.1 simpliﬁes to the following.

Theorem 1.2. Let K be a number ﬁeld with Galois extension L and let p be a prime ideal of K . Then all of the prime ideals of L above p have the same O O ramiﬁcation index e, the same inertial degree f, and efg = [L : K].

Additionally, the action of the Galois group on the prime ideals is given in the following, [6].

Theorem 1.3. Let L be a Galois extension of K and p a prime ideal of K . Then O the Galois group Gal(L/K) acts transitively on the prime ideals of L containing O p, i.e. if P and P0 are primes ideals of L above p, then there is a σ Gal(L/K) O ∈ such that σ(P) = P0.

In the case of a Galois extension, K L, an ideal p of K is ramified in L ⊆ O if e > 1 and unramified if e = 1. If e = f = 1, we say that the prime p splits completely in L, and thus, by Theorem 1.1, p L splits into [L : K] primes above p. O Classifying which primes of a number field are ramified or which split completely is a fundamental problem in Algebraic Number Theory. We first consider this problem in quadratic number fields, or number fields of degree 2 over Q.

If K is a quadratic number ﬁeld, then K = Q(√N), where N = 0, 1 is a 6 squarefree integer. The discriminant of K,

N, if N 1 (mod 4),  ≡ dK =  4N, otherwise,  

5 completely determines the ring of integers of K. In particular,

Z[ 1+√N ] if N 1 (mod 4), 2 ≡ K =  O  Z[√N] otherwise.   To determine how rational primes decompose in these number fields, we first need some definitions.

Deﬁnition 1.4. Let m, n be positive integers and let a be an integer relatively prime to m. We say that a is an nth power residue modulo m if xn a (mod m) ≡ is solvable. In particular, if n = 2, we say a is a quadratic residue, and if n = 4, we say that a is a biquadratic residue.

Deﬁnition 1.5. If a is an integer and p an odd prime, then the Legendre symbol,

a , is given by: p

0, if a 0 (mod p),  ≡ a  =  p 1, if a is a quadratic residue modulo p,   1, if a is a quadratic nonresidue modulo p. −  We remark that the Legendre symbol is multiplicative and depends only on the congruence class of a mod p.

We can now state a fundamental law of number theory formulated by Euler and

Legendre, but ﬁrst proven by Gauss.

Theorem 1.6 (Law of Quadratic Reciprocity). Let p and q be odd primes.

Then

p q p 1 q 1 = ( 1) −2 −2 . q p − We can now give a complete classiﬁcation of the decomposition of rational primes in quadratic number ﬁelds.

6 Proposition 1.7. Let p be an odd prime and let K be a quadratic number ﬁeld of discriminant dK . Then

dK 2 (i) If = 0, then p K = p for some prime p of K , (i.e. p ramiﬁes in K). p O O dK (ii) If = 1, then p K = pp0 for some primes p = p0 of K , (i.e. p splits p O 6 O completely in K).

dK (iii) If = 1, then p K is again prime in K , (i.e. p is inert in K). p − O O 1.2 Cyclotomic Number Fields

2πi/m m n For a positive integer m, let ζm = e . Then ζm is a root of x 1, but not x 1 − − for any n < m. Such a ζm is called a primitive mth root of unity. Since all powers

m m m 1 of ζm are also roots of x 1, we have that x 1 = (x 1)(x ζm) (x ζ − ) − − − − ··· − m m and M = Q(ζm) is the splitting ﬁeld of the polynomial x 1. Hence, M is Galois − over Q, [10, 20].

The field M = Q(ζm) is called the cyclotomic field of mth roots of unity, or simply the mth cyclotomic field. The Galois group of cyclotomic number fields is particularly simple.

Theorem 1.8. Gal(Q(ζm)/Q) = (Z/mZ)× . In particular, every a (Z/mZ)× ∼ ∈ a corresponds to the automorphism σa Gal(Q(ζm)/Q) sending ζm ζ . ∈ 7→ m

Corollary 1.9. [Q(ζm): Q] = φ(m), where φ(m) denotes the Euler phi function.

Cyclotomic ﬁelds also have a beautiful algebraic structure.

Theorem 1.10. The ring of algebraic integers in Q(ζm) is Z[ζm].

Proposition 1.11. If p is a prime and p - m, then every prime ideal P Z[ζm] ⊆ containing p is unramiﬁed.

Combining these properties of cyclotomic number ﬁelds, we get a complete decomposition of rational primes in Z[ζm].

7 Theorem 1.12 ([10],Th.2, pg.196). Let p be a rational prime, p - m, and let f be the least positive integer such that pf 1 (mod m). Then ≡

pZ[ζm] = P1P2 Pg, ··· where each Pi has inertial degree f and g = φ(m)/f.

Remark 1.13. In particular, if p is a rational prime congruent to 1 modulo m, then f = 1 and p splits completely into φ(m) primes of Z[ζm].

For cyclotomic extensions of cyclotomic ﬁelds, the decomposition of rational primes is again surprisingly simple.

Theorem 1.14 ([10], Prop.13.2.9, pg.198). Let p be a rational prime such that p - m, E = Q(ζm, ζp), and let P1,...,Pg be as in the previous theorem. Then for

p 1 all i, Pi E = ℘ − , i.e. Pi ramiﬁes in E, and thus O i O

p 1 p E = (℘1℘2 ℘g) − . O ···

Finally, we have the following two results concerning units of absolute value 1 in cyclotomic ﬁelds, [3, 10, 20].

Theorem 1.15. If all the algebraic conjugates of an algebraic integer α over Q have absolute value 1, then α is a root of unity.

Theorem 1.16. Let M = Q(ζm). The only elements of absolute value 1 in M O i i are ζ , i = 1, 2, . . . , m. In particular, the only roots of unity in M are ζ . ± m O ± m 1.3 Class Field Theory

Given an ideal I K and an element α K, we can form αI = αx x I . ⊆ O ∈ { | ∈ }

This is an K -submodule of K and is called a fractional ideal of K. The set of all O fractional ideals of K, denoted IK , is closed under multiplication. Moreover, each fractional ideal is invertible and thus, IK is a group. The ideals of IK of the form

8 α K , for some α K×, form a subgroup of IK and are called principal fractional O ∈ ideals, denoted PK .

Deﬁnition 1.17. The ideal class group of K, C(K), is deﬁned to be the quotient

C(K) = IK /PK .

The order of C(K) is denoted hK and is called the class number of K.

It is well-known that for number fields K, C(K) is a finite abelian group, [12], and so hK is finite. Given an ideal I K , we will denote by [I] its class in C(K) ⊆ O and denote its order in the class group by o([I]).

The ideal class group is a measure of how close K is to being a principal ideal O domain, i.e. K is a principal ideal domain if and only if hK = 1. It is a classical O problem of Number Theory to investigate the structure and order of C(K).

As C(K) is a ﬁnite abelian group, it is a product of cyclic groups of prime-power order. To understand the structure of the ideal class group of a quadratic number

ﬁeld K, it is important to understand its 2-Sylow subgroup. Using quadratic forms,

Gauss proved that the number of cyclic factors of C(K) with even order is equal to t 1, where t is the number of distinct primes dividing dK . For speciﬁc classes − of discriminants, dK , much more can be said, see, for example, [4].

Concerning hK , let us restrict our attention to imaginary quadratic number fields as these fields are the main focus of Chapters 2 and 3. For these fields, the class number hK is easily obtained using Dirichlet’s class number formula, [6, 7].

Theorem 1.18. Let K be a quadratic number ﬁeld of discriminant dK < 0. Then

dK 1 | |− dK hK = n, X n n=1

dK dK r dK where is deﬁned for n = p1p2 . . . pr, pi prime, by = . n n Qi=1 pi

9 2. Gauss Sums

2.1 Preliminaries

Let e be an integer, e > 2, and let p be a prime, p 1 (mod e). Let Fp be the ≡

finite field of p elements with (cyclic) multiplicative group Fp× generated by γ. We can then define a multiplicative character

χ: F× Q(ζe) by χ(γ) = ζe, p −→ where ζe is a primitive eth root of unity. We extend the character to all of Fp by setting χ(0) = 0.

Deﬁnition 2.19. For a Z, deﬁne the Gauss sum τ(a) to be ∈

a α τ(a) = χ (α)ζ Q(ζep). X p ∈ α Fp ∈ Remark 2.20. Since χ is a character of order e, we need only consider a mod e, and thus have e distinct Gauss sums, τ(0), τ(1), . . . , τ(e 1). − Remark 2.21. By deﬁnition as sums of roots of unity, Gauss sums are algebraic integers in Q(ζep).

We now provide some elementary properties of Gauss sums. Further properties and proofs may be found in [3, 10].

Proposition 2.22. For a 0 (mod e), 6≡

1. τ(0) = 0

2. τ(a) = χa( 1)τ( a) − −

3. τ(a) = p1/2. | |

10 The action of Galois group elements on Gauss sums can also be given explicitly,

[20]. Let E = Q(ζep) = Q(ζe, ζp) and M = Q(ζe). Then Gal(E/Q) is given by the automorphisms σc, where gcd(c, ep) = 1. Now, σc is completely determined by

σc(ζe) and σc(ζp) and we have

1. σc ﬁxes Q(ζe) element-wise if and only if c 1 (mod e), ≡

2. σc ﬁxes Q(ζp) element-wise if and only if c 1 (mod p). ≡

c Therefore, for c 1 (mod p) with gcd(c, ep) = 1, σc(ζe) = ζ and σc(ζp) = ζp and ≡ e we conclude that Gal(E/Q(ζp)) ∼= Gal(M/Q). Applying these automorphisms to Gauss sums, we obtain the following proposition.

Proposition 2.23. For j (Z/eZ)×, σj (τ(a)) = τ(ja). ∈

Proof. For j (Z/eZ)×, σj Gal(M/Q) = Gal(E/Q(ζp)), and ∈ ∈ ∼

σ (τ(a)) = σ  χa(α)ζα = σ (χa(α))σ (ζα) = χja(α)ζα = τ(ja). j j X p X j j p X p α Fp α Fp α Fp  ∈  ∈ ∈

Two main types of multiplicative relations exist between Gauss sums. The ﬁrst such relation follows from properties two and three of Proposition 2.22 and is often referred to as the norm relation. Connecting a Gauss sum and its complex conjugate, it states that for a 0 (mod e), τ(a)τ(a) = χa( 1)p, or equivalently, 6≡ −

τ(a)τ( a) = χa( 1)p. (2.1) − −

The second type of relation is the beautiful Davenport-Hasse product formula for composite e. It states that for e = mn, with m, n > 1, and for 1 t m 1, ≤ ≤ − n 1 τ(t) − τ(km + t) χtn(n) = 1. (2.2) τ(tn) Y τ(km) k=1

11 Example 2.24. Let e = 15, p be a prime with p 1 (mod e), and let χ be a mul- ≡ tiplicative character χ: F× Q(ζ15). Then, upon splitting the quotients of (2.2), p → we have the following six Davenport-Hasse relations along with the corresponding values of m, n, and t.

1. m = 3, n = 5, t = 1; χ5(5)τ(1)τ(4)τ(7)τ(10)τ(13) = τ(3)τ(5)τ(6)τ(9)τ(12)

2. m = 3, n = 5, t = 2; χ10(5)τ(2)τ(5)τ(8)τ(11)τ(14) = τ(3)τ(6)τ(9)τ(10)τ(12)

3. m = 5, n = 3, t = 1; χ3(3)τ(1)τ(6)τ(11) = τ(3)τ(5)τ(10)

4. m = 5, n = 3, t = 2; χ6(3)τ(2)τ(7)τ(12) = τ(5)τ(6)τ(10)

5. m = 5, n = 3, t = 3; χ9(3)τ(3)τ(8)τ(13) = τ(5)τ(9)τ(10)

6. m = 5, n = 3, t = 4; χ12(3)τ(4)τ(9)τ(14) = τ(5)τ(10)τ(12).

Another type of character sum closely related to the Gauss sum is the Jacobi sum, which has the added advantage of being an integer of Q(ζe), rather than

Q(ζep).

Deﬁnition 2.25. Let e, p, and χ be as before, and let m, n Z. We deﬁne the ∈ Jacobi sum J(m, n) to be

m n J(m, n) = χ (α)χ (1 α) Q(ζe). X − ∈ α Fp ∈ Remark 2.26. We again remark that as χ has order e, we need only consider m and n modulo e.

τ(m)τ(n) Theorem 2.27. If m + n 0 (mod e), then J(m, n) = . 6≡ τ(m + n)

12 Proof. By the deﬁnition of τ(m),

τ(m)τ(n) = χm(α)χn(β)ζα+β X X p α β

= ζδ χm(α)χn(β) X p X δ α+β=δ

= χm(α)χn(β) + ζδ χm(α)χn(δ α) X X p X − α+β=0 δ=0 α 6 = χn( 1) χm+n(α) + ζδ χm(δα)χn(δ δα) − X X p X − α δ=0 α 6 = 0 + J(m, n) χm+n(δ)ζδ X p δ=0 6 = J(m, n)τ(m + n).

When possible, it is often desirable to express products of Gauss sums in terms of products of Jacobi sums as it is then clear that the product is an element of the smaller cyclotomic ﬁeld Q(ζe). It is also of interest to note that by Theorem 2.27, both the norm relation, (2.1), and the Davenport-Hasse product formula, (2.2), can be reformulated in terms of Jacobi sums.

We now consider the prime ideal factorization of Gauss sums. Again, let M =

Q(ζe),E = Q(ζep), and let M and E denote their respective rings of integers. O O From the norm relation, (2.1), the ideal generated by τ(a) divides p, and thus factors only into primes above p. Let P M be a prime ideal above the rational ⊆ O prime p. By Remark 1.13, since p 1 (mod e), p M will split completely into ≡ O

φ(e) primes. Then, since Gal(M/Q) acts transitively on the primes of M above O p, we have

p M = Pj, O Y j (Z/eZ) ∈ × where Pj = σj(P ), σj Gal(M/Q). Furthermore, by Theorem 1.14, each Pj ∈ ramiﬁes totally in E and thus, for some prime ideal ℘ E, we have P M = O ⊆ O O

13 p 1 ℘ − , ℘ M = P. Recalling that Gal(E/Q(ζp)) = Gal(M/Q) and letting ℘j = ∩ O ∼

σj(℘), we then have

p 1 − p 1   p E = ℘ − = ℘j . O Y j Y j (Z/eZ) j (Z/eZ) ∈ ×  ∈ × 

Therefore, the ideal generated by τ(a) factors only into powers of the ℘j. Speciﬁ- cally, we have the following factorization formula due to Stickelberger.

Theorem 2.28 (Stickelberger).

p 1 − aj  { e }  τ(a) E = ℘ j 1 , O Y − − j (Z/eZ)  ∈ ×  1 where j− is taken modulo e and x represents the fractional part of x. − { } Example 2.29. Let e = 15, p be a prime with p 1 (mod 15), and ℘ be a prime ≡ ideal of E dividing p. Then τ(1) factors into powers of the φ(15) = 8 distinct O primes of E above p O

14 7 11 2 13 4 8 1 p 1 15 15 15 15 15 15 15 15 − τ(1) E = ℘ ℘ ℘ ℘ ℘ ℘ ℘ ℘ . O 1 2 4 7 8 11 13 14

In particular, if p = 31, then

14 7 11 2 13 4 8 1 30 15 15 15 15 15 15 15 15 τ(1) E = ℘ ℘ ℘ ℘ ℘ ℘ ℘ ℘ O 1 2 4 7 8 11 13 14 28 14 22 4 26 8 16 2 = ℘1 ℘2 ℘4 ℘7℘8 ℘11℘13℘14. 2.2 Sign Ambiguities

In section 1, we saw two distinct types of multiplicative relations connecting Gauss sums: the norm relation, (2.1), and the Davenport-Hasse product formula, (2.2). It is natural to ask if any other multiplicative relations exist. In [8, p. 465], H. Hasse conjectured that the norm relation and the Davenport-Hasse product formula were essentially the only multiplicative relations connecting Gauss sums over Fp. In [21],

14 K. Yamamoto proved Hasse’s conjecture if the Gauss sums are considered as ideals, but provided a simple counterexample for e = 12 if the sums are instead considered as numbers. Further counterexamples were given for e = 15, 20, 21, 24, 28, 39, 55, and 56 by Muskat, Muskat-Whiteman, and Muskat-Zee in [13, 15, 16]. For each counterexample, an explicit multiplicative relation was found involving a sign that could not be determined using elementary properties of Gauss sums and relations

(2.1) and (2.2). Such relations have become known as sign ambiguities.

Deﬁnition 2.30. A multiplicative relation of the form

τ(i) = uζk, Y τ(j) e i,j where u = 1 for some primes p, u = 1 for others, such that the sign cannot − be connected to elementary properties of Gauss sums, the norm relation, nor the

Davenport-Hasse product formula, is known as a sign ambiguity.

Illustration 2.31 (Muskat, [13]). Let e = 39 and p be a prime, p 1 (mod 39). ≡ Then

τ(1)τ(16)τ(34) = ζ13 indγ 13τ(2)τ(17)τ(32) ± e

i is a sign ambiguity, where indγa is the unique integer i such that a γ (mod p) ≡ for a ﬁxed primitive root γ mod p.

In [22], Yamamoto further investigated Hasse’s conjecture and determined that sign ambiguities existed for all composite values of e. Moreover, Yamamoto was able to determine exactly how many multiplicatively independent relationships were “missing” in each case.

r 1 Theorem 2.32 (Yamamoto, [22]). For e > 2, there are exactly 2 − 1 multi- − plicatively independent Gauss sum relations that are not direct consequences of the norm relation and the Davenport-Hasse relations, where r is the number of distinct

15 prime divisors of e, or, if e 2 (mod 4), r is the number of distinct prime divisors ≡ of e/2.

While this shows that new multiplicative relations exist for all composite values of e, very few have been explicitly given.

We restrict attention to odd values of e with two distinct prime divisors, as sign ambiguities for many other values of e can be reduced to these cases. Furthermore, by Theorem 2.32, such values of e will have exactly one new multiplicatively independent relation. In this dissertation, we present a product formula explicitly producing the missing sign ambiguities for inﬁnitely many values of e.

16 3. Results

3.1 Methodology

Initially, we obtained computational evidence using PARI/GP to aid intuition.

This was accomplished by reformulating the problem in terms of linear algebra.

Consider the Davenport-Hasse product formula (2.2),

n 1 τ(t) − τ(km + t) χtn(n) = 1, τ(tn) Y τ(km) k=1 which, upon splitting the quotient becomes

n 1 n 1 − − χtn(n)τ(t) τ(km + t) = τ(tn) τ(km). Y Y k=1 k=1

Since Gauss sums are integers of E = Q(ζep), the products above are integers as well, and we can consider the ideals they generate, that is,

n 1 n 1 − − χtn(n) τ(t) τ(km + t) = τ(tn) τ(km) . Y Y unit k=1 k=1 | {z } ideal generator ideal generator | {z } | {z } Thus, each Davenport-Hasse relation corresponds to distinct algebraic integers gen- erating the same ideal. From this point of view, it is clear that ﬁnding multiplicative relations is equivalent to ﬁnding distinct generators of the same ideal.

This reformulation laid the foundation for the computer search for new multiplicative relations between Gauss sums. Using Stickelberger’s factorization formula

(Theorem 2.28), we ﬁrst factored each of the e 1 distinct, non-trivial, Gauss sums; − for example, recall the example for e = 15:

Example 3.33. Let e = 15, p be a prime with p 1 (mod 15), E = Q(ζep), and ≡

℘ be a prime ideal of E dividing p. Then τ(1) and τ(2) factor into powers of the O

φ(15) = 8 distinct primes of E above p. We have O 14 7 11 2 13 4 8 1 p 1 15 15 15 15 15 15 15 15 − τ(1) E = ℘ ℘ ℘ ℘ ℘ ℘ ℘ ℘ O 1 2 4 7 8 11 13 14

17 and

13 14 7 4 11 8 1 2 p 1 15 15 15 15 15 15 15 15 − τ(2) E = ℘ ℘ ℘ ℘ ℘ ℘ ℘ ℘ . O 1 2 4 7 8 11 13 14 Each factorization is independent of the choice of the rational prime p in the sense that diﬀerent p’s will not alter the ratios of the exponents. Thus, each Gauss sum factorization can be represented as a vector of the exponents. Using the factorizations from Example 2.29, we have the correspondences

14 7 11 2 13 4 8 1 τ(1) E , , , , , , , O ←→ 15 15 15 15 15 15 15 15 and 13 14 7 4 11 8 1 2 τ(2) E , , , , , , , . O ←→ 15 15 15 15 15 15 15 15 Representing all e 1 non-trivial Gauss sums in this way, we then formed a −

φ(e) (e 1) matrix, Me, with the “factorizations” as the columns. For e = 15, × −

14 13 4 11 2 3 8 7 2 1 4 1 2 1  15 15 5 15 3 5 15 15 5 3 15 5 15 15   7 14 2 13 1 4 4 11 1 2 2 3 1 8   15 15 5 15 3 5 15 15 5 3 15 5 15 15       11 7 1 14 2 2 2 13 3 1 1 4 8 4   15 15 5 15 3 5 15 15 5 3 15 5 15 15     2 4 2 8 2 4 14 1 1 1 7 3 11 13   15 15 5 15 3 5 15 15 5 3 15 5 15 15  M15 =   .    13 11 3 7 1 1 1 14 4 2 8 2 4 2   15 15 5 15 3 5 15 15 5 3 15 5 15 15     4 8 4 1 1 3 13 2 2 2 14 1 7 11   15 15 5 15 3 5 15 15 5 3 15 5 15 15       8 1 3 2 2 1 11 4 4 1 13 2 14 7   15 15 5 15 3 5 15 15 5 3 15 5 15 15     1 2 1 4 1 2 7 8 3 2 11 4 13 14   15 15 5 15 3 5 15 15 5 3 15 5 15 15  To factor the ideal generated by a product of Gauss sums, we need only to add the appropriate columns of Me and multiply by p 1. Moreover, to ﬁnd two distinct − products of Gauss sums that generate the same idea, we need only ﬁnd distinct sets of columns that have identical sums. Thus, multiplicative relations correspond exactly with elements of the nullspace of Me. Using PARI/GP to compute a basis

18 for the nullspace of M15, we obtain

 0 1 0 0 0 0 0 0 0   1 1 0 0 0 0 1 1 1     −     1 0 0 0 0 0 0 0 0  −     1 0 1 1 1 1 0 0 0    −     1 1 1 1 0 0 0 0 0   − − −     0 1 1 0 1 1 1 1 1  − − − − − −       0 0 1 0 0 0 0 0 0  N15 =   .    0 0 0 1 1 0 1 0 0   − − −       0 0 0 1 0 1 0 1 0   − −     0 0 0 0 1 0 0 0 1  −       0 0 0 0 0 1 0 0 0       0 0 0 0 0 0 1 0 0         0 0 0 0 0 0 0 1 0       0 0 0 0 0 0 0 0 1 

Looking at the second column of N15, for example, we see that τ(1)τ(6) and

τ(2)τ(5) generate the same ideal of E; therefore, τ(1)τ(6) = µ τ(2)τ(5), for O · some unit µ E. ∈ The nullspace, however, spans all multiplicative relations, not just sign ambiguities. To isolate the new relations, we then coded the norm and Davenport-Hasse relations into another matrix, e, and searched for relations from Ne that were M not in the image of e. This method produced a large number of equivalent new M relations for relatively small values of e,(e < 200). While Theorem (2.32) guaran- teed only one new relation for our choices of e, that one relation has many diﬀerent forms as it must be considered modulo all of the other relations. However, given

19 a multiplicative relation, this method provided a simple computational test to determine if it followed from the norm relation and Davenport-Hasse.

In an eﬀort to obtain a relatively simple representative of the new multiplicative relation, we restricted our search to values of e 3 (mod 4) and, following Muskat, ≡ [14], we further restricted to products of Gauss sums ﬁxed by a large number of automorphisms. In particular, we looked for Gauss sums that generated ideals which factored over K , where K = Q(√ e). Given such a product, ω Q(ζe) O − ∈ ⊆

Q(ζep), and an automorphism σ Gal(Q(ζe)/K), we have that σ(ω) M = ω M , ∈ O O and thus σ(ω) = µω, for some unit µ Q(ζe). Relations of this type that do not ∈ follow from the known relations are sign ambiguities. Also, since ω generates an ideal factoring over K , it will factor only into powers of the quadratic primes O α β above p, i.e. if p K = p1p2, then ω E = p p E, for some α, β Z. O O 1 2 O ∈ After analyzing large amounts of data, an interesting observation was made: for

α β many values of e, there was a product of Gauss sums as above with ω E = p p E, O 1 2 O such that β α = hK /4, where hK is the class number of K. Moreover, in these − cases the number of individual Gauss sums comprising ω was small among all of the equivalent forms of this new relation. Based on this computational evidence, we make the following conjecture.

Conjecture 3.34. Let e = q1q2 3 (mod 4) with q1 a quadratic residue modulo q2 ≡ and let σ be an automorphism of Q(ζe) sending both √q1 and √ q2 to conjugates. − α hK /4 k If ω is a product of Gauss sums such that ω E = p p , then σ(ω) = ζ ω is a O 1 ± e sign ambiguity.

The conjecture provided a more eﬀective method of gathering data. Rather than computing nullspaces, we instead looked for products of Gauss sums with the factorization from the conjecture and then quickly checked if each was in the image

20 of the matrix of known relations. Using the conjecture in this way, we were able to generate many more new sign ambiguities, (e < 1000). To prove the conjecture, we then looked for similarities among products of the above form for various values of e, and attempted to ﬁnd a formula producing them.

In the next sections, we present a product formula giving a partial proof of the conjecture. In particular, we prove the conjecture for all values of e such that e = q1q2, with q1 5 (mod 8), q2 3 (mod 4), and q2 a biquadratic residue ≡ ≡ modulo q1.

3.2 Biquadratic Coset Sums

For any positive integer n and for any a Z/nZ, let Ln(a) denote the least ∈ positive integer congruent to a modulo n. Furthermore, for a Z, we will also ∈ let Ln(a) = Ln (πn(a)), where πn : Z Z/nZ is the quotient map. This should → not cause confusion as Ln simply returns the least positive integer congruent to a mod n. Finally, if n = e we will suppress the subscript, i.e. L = Le.

For the remainder of the dissertation, let e = q1q2, with q1 5 (mod 8), q2 3 ≡ ≡

(mod 4), and q2 a biquadratic residue modulo q1. Recall from Deﬁnition 1.4 that a is a biquadratic residue modulo m if x4 a (mod m) is solvable. Let G = ≡

(Z/eZ)× and let H4 denote the biquadratic residues modulo e. The proof of our main theorem will depend on the evaluation of the eight biquadratic coset sums of

G/H4. For i = 0,..., 7, let ci denote these cosets. Then the biquadratic coset sums are the sums

L(a), for i = 0,..., 7. X a ci ∈

We now consider these sums.

21 Proposition 3.35. Let e = q1q2 and G be as stated. Then there is a g G such ∈ that

2 3 G/H4 = H4, gH4, g H4, g H4 . {± ± ± ± }

Proof. Let gq1 and gq2 denote primitive roots modulo q1 and q2 respectively. Then

(Z/eZ)× = (Z/q1Z)× (Z/q2Z)× = gq gq . ∼ × ∼ h 1 i × h 2 i

Let ψ : gq gq (Z/eZ)× be the isomorphism given by the Chinese Remain- h 1 i × h 2 i → 1 α β der Theorem. Now, a H4 is a biquadratic residue, so ψ− (a) = (g , g ) is a ∈ q1 q2 α β biquadratic residue. But (gq1 , gq2 ) is a biquadratic residue if and only if each coor-

α dinate is. Since q1 1 (mod 4), g is a biquadratic residue if and only if α 0 ≡ q1 ≡

(mod 4), and as q2 3 (mod 4), the biquadratic residues modulo q2 are exactly ≡ the quadratic residues, so gβ is a biquadratic residue if and only if β 0 (mod 2). q2 ≡ 4 2 Hence, H4 = g g , and ∼ h q1 i × h q2 i

4 2 q1 1 q2 1 φ(e) H4 = g g = − − = . | | |h q1 i| · |h q2 i| 4 2 8

Let g = ψ(gq , gq ) and consider gH4 G/H4. Computing powers of g, we see 1 2 ∈ 4 i that g H4, but g H4 for i = 1, 2, 3. Thus gH4 has order 4 in G/H4. In ∈ 6∈ particular, we have the following cosets of H4

1 ψ− 4 2 H4 g g −−→h q1 i × h q2 i 1 ψ− 4 2 gH4 (gq , gq ) g g −−→ 1 2 h q1 i × h q2 i 1 2 ψ− 2 2 4 2 g H4 (g , g ) g g −−→ q1 q2 h q1 i × h q2 i 1 3 ψ− 3 3 4 2 g H4 (g , g ) g g −−→ q1 q2 h q1 i × h q2 i

22 Now consider gH4. −

1 4 2 ψ− ( gH4) = ( gq , gq )( g )g − − 1 − 2 h q1 i × h q2 i 4 2 = ( 1, 1))(gq , gq )( g g ) − − 1 2 h q1 i × h q2 i q1 1 q2 1 2− 2− 4 2 = (gq , gq )(gq , gq )( g g ) 1 2 1 2 h q1 i × h q2 i q1+1 q2+1 2 2 4 2 = (gq , gq )( g g ) 1 2 h q1 i × h q2 i = (g2α+1, g2β)( g4 g2 ), q1 q2 h q1 i × h q2 i for some α, β Z, since q1 1 (mod 4) and q2 3 (mod 4). Since the components ∈ ≡ ≡ 1 of the coset ψ− ( gH4) have exponents of diﬀerent parity, it clearly cannot be a − 1 power of ψ− (gH4). Computing powers of gH4 we see that it also has order 4 in −

G/H4, with powers distinct from those of gH4. Therefore, since

φ(e) G/H4 = = 8, | | φ(e) 8 we have that

2 3 G/H4 = H4, gH4, g H4, g H4 . {± ± ± ± }

We will now prove three lemmas concerning coset sums. We will ﬁrst prove that the coset sums L(gja) are equal for j = 0,..., 3 . By symmetry, this will a H4 P ∈ imply that the other four coset sums, L( gja), are equal to each other as a H4 P ∈ − well. We proceed in three parts. We ﬁrst show that

L(a) = L(g2a). X X a H4 a H4 ∈ ∈ We then show that

L(ga) = L(g3a). X X a H4 a H4 ∈ ∈

23 And ﬁnally, we show that all four are equal by showing that

L(a) + L(g2a) = L(ga) + L(g3a). X X X X a H4 a H4 a H4 a H4 ∈ ∈ ∈ ∈

Lemma 3.36. Let e = q1q2,G, and H4 be as above. Then

L(a) = L(g2a). X X a H4 a H4 ∈ ∈ 4 2 Proof. From Proposition 3.35, we have that H4 = g g . Let T4 denote the ∼ h q1 i × h q2 i biquadratic residues modulo q1 and N2 denote the quadratic residues modulo q2.

Let a1 Z be such that a1q2 1 (mod q1) and a2 Z be such that a2q1 1 ∈ ≡ ∈ ≡

(mod q2). Then by the Chinese Remainder Theorem,

L(H4) = L(ta1q2 + na2q1) t Lq (T4), n Lq (N2) . (3.3) { | ∈ 1 ∈ 2 }

Setting 4 = L(H4), 4 = Lq (T4), and 2 = Lq (N2), Equation (3.3) becomes H T 1 N 2

4 = L(ta1q2 + na2q1) t 4, n 2 . (3.4) H { | ∈ T ∈ N }

But, since q2 is a biquadratic residue modulo q1, a1 4 and ta1 t0 (mod q1), ∈ T ≡ q1 q2 for some t0 4. And since q1 1 (mod 4), = = 1, which implies that ∈ T ≡ q2 q1 a2 2 and na2 n0, for some n0 2. Combining and reindexing, we have ∈ N ≡ ∈ N

4 = L(tq2 + nq1) t 4, n 2 H { | ∈ T ∈ N } and need only show that

L(tq + nq ) = L(L(g2)(tq + nq )). X 2 1 X 2 1 t 4 t 4 n∈T 2 n∈T 2 ∈N ∈N

Setting g0 = L(g), for the second summand we then have

2 2 2 L g (tq2 + nq1) = L (g t)q2 + (g n)q1 . 0 0 0

24 2 2 But g and n are both squares modulo q2, so g n n0 (mod q2) for some n0 2. 0 0 ≡ ∈ N 2 Furthermore, for any t 4, g t t0 (mod q1), for some t0 4, so ∈ T 0 ≡ − ∈ T

2 2 L (g t)q2 + (g n)q1 = L( t0q2 + n0q1) for some t0 4 and n0 2. 0 0 − ∈ T ∈ N

Upon reindexing, we are reduced to showing that

L(tq2 + nq1) = L( tq2 + nq1). (3.5) X X − t 4 t 4 n∈T 2 n∈T 2 ∈N ∈N

Since 0 t < q1 and 0 n < q2, we have that 0 tq2 + nq1 < 2e, and thus ≤ ≤ ≤

tq2 + nq1, if tq2 + nq1 < e  L(tq2 + nq1) =  tq2 + nq1 e, if tq2 + nq1 > e.  − 

For ﬁxed t 4, ∈ T

q1q2 tq2 tq2 tq2 + nq1 > e n > − n > q2 . ⇐⇒ q1 ⇐⇒ − q1

tq2 Let kt = # n 2 n > q2 . { ∈ N | − q1 }

Similarly, for the right summand of Equation (3.5) we have that e < tq2 + − − nq1 < e, and thus

 tq2 + nq1, if tq2 + nq1 > 0 − − L( tq2 + nq1) =  −  tq2 + nq1 + e, if tq2 + nq1 < 0. − − 

tq2 tq2 Again ﬁxing t 4, tq2 + nq1 < 0 n < . Let lt = # n 2 n < . ∈ T − ⇐⇒ q1 { ∈ N | q1 } Proof of the lemma is now reduced to proof of the equality of the integer sums

ekt + tq2 + nq1! = elt + tq2 + nq1!. X − X X X − t 4 n 2 t 4 n 2 ∈T ∈N ∈T ∈N

25 We have

L(a) L(g2a) = X − X a 4 a 4 ∈H ∈H

= ekt + tq2 + nq1! elt + tq2 + nq1! X − X − X X − t 4 n 2 t 4 n 2 ∈T ∈N ∈T ∈N

= e(kt + lt) + 2tq2! X − X t 4 n 2 ∈T ∈N

= e(kt + lt) + 2tq2 1! X − X t 4 n 2 ∈T ∈N q2 1 = e(kt + lt) + 2tq2 − X − 2 t 4 ∈T

= ( e(kt + lt) + tq2(q2 1)). X − − t 4 ∈T Now,

tq2 lt = # n 2 n < ∈ N | q1

tq2 = # n 2 n > − ∈ N | − q1

tq2 = # n 2 q2 n > q2 . ∈ N | − − q1

Since n is a quadratic residue modulo q2, q2 n is a nonresidue, and lt is then − tq2 equal to the number of quadratic nonresidues greater than q2 . But kt is the − q1 tq2 number of quadratic residues modulo q2 greater than q2 , and since the sets − q1 of residues and nonresidues are disjoint, kt + lt is the number of units modulo q2

tq2 greater than q2 . Thus for each t 4, we have, − q1 ∈ T

tq2 kt + lt = q2 q2 − − q1

tq2 = q2 q2 + − q1

tq2 = . q1

26 But, since both t and q2 are biquadratic residues modulo q1,

tq2 tq2 Lq (tq2) tq2 t0 = 1 = , q1 q1 − q1 q1 − q1

for some t0 4. Reindexing and summing over t, ∈ T

tq2 tq2 t t(q2 1) = = − . (3.6) X q1 X q1 − q1 X q1 t 4 t 4 t 4 ∈T ∈T ∈T Therefore,

( e(kt + lt) + tq2(q2 1)) = (tq2(q2 1) e(kt + lt)) X − − X − − t 4 t 4 ∈T ∈T tq2 = tq2(q2 1) q1q2 X − − q1 t 4 ∈T tq2 = q2(q2 1) t q1q2 − X − X q1 t 4 t 4 ∈T ∈T t(q2 1) = q2(q2 1) t q1q2 − − X − X q1 t 4 t 4 ∈T ∈T

= q2(q2 1) t q2(q2 1) t − X − − X t 4 t 4 ∈T ∈T = 0.

Hence,

L(a) = L(g2a), X X a H4 a H4 ∈ ∈ proving the lemma.

Lemma 3.37. With notation as before,

L(ga) = L(g3a). X X a H4 a H4 ∈ ∈

27 Proof. As in the previous lemma, for a H4, a = L(tq2 +nq1) for some t 4, n ∈ ∈ T ∈

2. Then, N

L(ga) = L(g0(tq2 + nq1))

= L (g0tq2 + (g0n)q1)

= L (g0tq2 n0q1) , for some n0 2, − ∈ N since g0 is a nonsquare modulo q2 and q2 3 (mod 4). ≡ On the other hand,

3 3 L(g a) = L(g0(tq2 + nq1))

2 = L g0(g (tq2 + nq1)) 0

= L ((g0( t0q2 + n0q1)) −

= L ( (g0t0)q2 + (g0n0)q1) , − where the third equality follows from the proof of the previous lemma. And again, since g0 is a quadratic nonresidue modulo q2, g0n0 = n00, for some n00 2. − ∈ N Combining and reindexing, we now need only show that

L (g0tq2 nq1) = L ( g0tq2 nq1), X − X − − t 4 t 4 n∈T 2 n∈T 2 ∈N ∈N or equivalently

( ) ( ) ∗ ∗∗ L (Lq (g0t)q2 nq1) = L ( Lq (g0t)q2 nq1) . X z 1 }| − { X z − 1 }| − { t 4 t 4 n∈T 2 n∈T 2 ∈N ∈N

Now, e < Lq (g0t)q2 nq1 < e, and thus − 1 −

Lq1 (g0t)q2 nq1, if Lq1 (g0t)q2 nq1 > 0 ( ) =  − − ∗  Lq (g0t)q2 nq1 + e, if Lq (g0t)q2 nq1 < 0.  1 − 1 − 

28 For ﬁxed t 4, ∈ T

Lq1 (g0t)q2 Lq1 (g0t)q2 nq1 < 0 n > . − ⇐⇒ q1

Lq1 (g0t)q2 Let kt = # n 2 n > . { ∈ N | q1 }

For the right summand, 2e < Lq (g0t)q2 nq1 < 0, and thus − − 1 −

 Lq1 (g0t)q2 nq1 + e, if Lq1 (g0t)q2 nq1 > e ( ) = − − − − − ∗∗  Lq (g0t)q2 nq1 + 2e, if Lq (g0t)q2 nq1 < e. − 1 − − 1 − −  For ﬁxed t 4, ∈ T

Lq (g0t)q2 nq1 < e Lq (g0t)q2 + nq1 > e − 1 − − ⇐⇒ 1 q2(q1 Lq (g0t)) n > − 1 ⇐⇒ q1

q2(q1 Lq1 (g0t)) q2 n < q2 − ⇐⇒ − − q1

Lq1 (g0t)q2 q2 n < . ⇐⇒ − q1

Lq1 (g0t)q2 Letting lt = # n 2 q2 n < , we again need only show equality of { ∈ N | − q1 } the integer sums

ekt + Lq (g0t)q2 nq1! = elt + Lq (g0t)q2 nq1 + e!. X X 1 − X X − 1 − t 4 n 2 t 4 n 2 ∈T ∈N ∈T ∈N Combining and simplifying, we have

L(ga) L(g3a) = X − X a H4 a H4 ∈ ∈

= e(kt lt) + (2Lq (g0t)q2 e)! X − X 1 − t 4 n 2 ∈T ∈N

= e (kt lt) + 2Lq (g0t)q2 e X − X 1 − X t 4 t 4 t 4 ∈T n∈T 2 n∈T 2 ∈N ∈N q2 1 q1 1 q2 1 = e (kt lt) + 2q2 − Lq1 (g0t) e − − X − 2 X − 4 2 t 4 t 4 ∈T ∈T φ(e) = e (kt lt) + q2(q2 1) Lq (g0t) e . X − − X 1 − 8 t 4 t 4 ∈T ∈T

29 But n 2 = q2 n / 2, so lt is the number of nonsquares modulo ∈ N ⇒ − ∈ N Lq1 (g0t)q2 Lq1 (g0t)q2 q2 less than . Thus, the number of squares less than is given by q1 q1

Lq1 (g0t)q2 q2 1 lt. And since there are exactly − squares modulo q2, the number j q1 k − 2 Lq1 (g0t)q2 of squares modulo q2 greater than 2 is given by

q2 1 Lq1 (g0t)q2 − lt = kt. 2 − q1 −

Hence, for each t,

q2 1 Lq1 (g0t)q2 q2 1 Lq1 (g0t)q2 kt lt = − + lt lt = − , − 2 − q1 − 2 − q1 and as in Equation (3.6),

Lq (g0t)q2 Lq (g0t)(q2 1) 1 = 1 − . X q1 X q1 t 4 t 4 ∈T ∈T Therefore,

eφ(e) e (kt lt) + q2(q2 1) Lq (g0t) = X − − X 1 − 8 t 4 t 4 ∈T ∈T

q2 1 Lq1 (g0t)q2 eφ(e) = e − + q2(q2 1) Lq1 (g0t) X 2 − q1 − X − 8 t 4 t 4 ∈T ∈T

q2 1 Lq1 (g0t)q2 eφ(e) = e − e + q2(q2 1) Lq1 (g0t) X 2 − X q1 − X − 8 t 4 t 4 t 4 ∈T ∈T ∈T

eφ(e) Lq1 (g0t)(q2 1) eφ(e) = e − + q2(q2 1) Lq1 (g0t) 8 − X q1 − X − 8 t 4 t 4 ∈T ∈T φ(e) eφ(e) = e q2(q2 1) Lq (g0t) + q2(q2 1) Lq (g0t) 8 − − X 1 − X 1 − 8 t 4 t 4 ∈T ∈T = 0, and thus

L(ga) = L(g3a). X X a H4 a H4 ∈ ∈

30 Lemma 3.38. With notation as above,

L(a) + L(g2a) = L(ga) + L(g3a). X X X X a H4 a H4 a H4 a H4 ∈ ∈ ∈ ∈ Proof. Let H2 denote the quadratic residues modulo e. Then, since the elements

2 of the cosets H4 and g H4 partition H2, we have that

L(a) + L(g2a) = L(a) X X X a H4 a H4 a H2 ∈ ∈ ∈ and

L(ga) + L(g3a) = L(ga). X X X a H4 a H4 a H2 ∈ ∈ ∈ Hence, we need only show

L(a) = L(ga). X X a H2 a H2 ∈ ∈ 2 2 From the proof of Proposition 3.35, we have that H2 = g g . Thus, if T2 ∼ h q1 i × h q2 i denotes the quadratic residues modulo q1 and 2 = Lq (T2), then we must show T 1 that

L(tq + nq ) = L(g (tq + nq )), X 2 1 X 0 2 1 t 2 t 2 n∈T 2 n∈T 2 ∈N ∈N Once more, since g0 / 2, ∈ N

L (g0(tq2 + nq1)) = L (g0tq2 + (g0n)q1) = L (g0tq2 n0q1) , − and upon reindexing we need only prove that

L(tq2 + nq1) = L(g0tq2 nq1). X X − t 2 t 2 n∈T 2 n∈T 2 ∈N ∈N We proceed as in the previous two lemmas. For the left summand, we have

0 < L(tq2 + nq1) < 2e, and thus

tq2 + nq1, if tq2 + nq1 < e  L(tq2 + nq1) =  tq2 + nq1 e, if tq2 + nq1 > e.  − 

31 Fixing n 2, ∈ N

nq1 nq1 tq2 + nq1 > e t > q1 q1 t < . ⇐⇒ − q2 ⇐⇒ − q2

nq1 Let kn = # t 2 q1 t < . { ∈ T | − q2 }

For the right summand, since L(g0tq2 nq1) = L(Lq (g0t)q2 nq1) and e < − 1 − −

Lq (g0t)q2 nq1) < e, 1 −

Lq (g0t)q2 nq1, if Lq (g0t)q2 nq1 > 0 1 − 1 − L(g0tq2 nq1) =  −  Lq (g0t)q2 nq1 + e, if Lq (g0t)q2 nq1 < 0.  1 − 1 − 

Fixing n 2, ∈ N nq1 Lq1 (g0t)q2 nq1 < 0 Lq1 (g0t) < . − ⇐⇒ q2

nq1 Let ln = # t 2 Lq (g0t) < . Then we are again reduced to proving the { ∈ T | 1 q2 } equality of the integer sums

ekn + tq2 + nq1! = eln + Lq (g0t)q2 nq1!. X − X X X 1 − n 2 t 2 n 2 t 2 ∈N ∈T ∈N ∈T Now,

L(a) L(ga) = X − X a H2 a H2 ∈ ∈

= ekn + tq2 + nq1! eln + Lq (g0t)q2 nq1! X − X − X X 1 − n 2 t 2 n 2 t 2 ∈N ∈T ∈N ∈T

= e(kn + ln) + tq2 + nq1 Lq (g0t)q2 + nq1! X − X − 1 n 2 t 2 ∈N ∈T   = e (kn + ln) + q2 t Lq (g0t) + 2q1 n. − X X − X 1 X t 2  t 2 t 2  t 2 ∈T n∈T 2 n∈T 2  n∈T 2 ∈N ∈N ∈N It is well-known that for primes congruent to 1 modulo 4, the sum of the squares is equal to the sum of the nonsquares, [9, Theorem 2.1]. Thus, since q1 1 (mod 4), ≡

32 t 2, and Lq (g0t) / 2, the middle two terms of the previous expression cancel, ∈ T 1 ∈ T leaving only

e (kn + ln) + 2q1 n. (3.7) − X X t 2 t 2 ∈T n∈T 2 ∈N nq1 Fixing n 2, kn + ln will be the number of units modulo q1 less than . So, ∈ N q2 for the ﬁrst summation of (3.7), we have

nq1 e (kn + ln) = e , − X − X q2 n 2 n 2 ∈N ∈N leaving it necessary to prove only that

nq1 e + 2q1 n = 0. − X q2 X n 2 t T2 ∈N n∈ 2 ∈N Applying an argument similar to that leading up to Equation (3.6),

nq1 n(q1 1) = − , X q2 X q2 n 2 n 2 ∈N ∈N and thus,

nq1 nq1 e + 2q1 n = 2q1 n e − X q2 X X − X q2 n 2 t T2 t T2 n 2 ∈N n∈ 2 n∈ 2 ∈N ∈N ∈N q1 1 n(q1 1) = 2q1 − n e − 2 X − X q2 n 2 n 2 ∈N ∈N n(q1 1) = q1(q1 1) n e − − X − X q2 n 2 n 2 ∈N ∈N q1 1 = q1(q1 1) n q1q2 − n − X − q2 X n 2 n 2 ∈N ∈N

= q1(q1 1) n q1(q1 1) n − X − − X n 2 n 2 ∈N ∈N = 0.

Hence,

L(a) + L(g2a) = L(ga) + L(g3a). X X X X a H4 a H4 a H4 a H4 ∈ ∈ ∈ ∈

33 Combining the results from this section, we now have the following.

Theorem 3.39. For i = 0, 1, 2, 3, the coset sums L(gia) are equal, the coset a H4 P ∈ sums L( gia) are equal, and a H4 P ∈ −

L(gia) < L( gia), i. (1) X X − ∀ a H4 a H4 ∈ ∈ Furthermore, for all i,

hK L( gia) L(gia) = e, (2) X − − X 4 a H4 a H4 ∈ ∈ where hK denotes the class number of K = Q(√ e). −

Proof. By Lemmas (3.36), (3.37), and (3.38), we have

L(a) = L(g1a) = L(g2a) = L(g3a). X X X X a H4 a H4 a H4 a H4 ∈ ∈ ∈ ∈ But, for i = 0,..., 3,

φ(e) L(gia) + L( gia) = e = e , X − X 8 a H4 a H4 ∈ ∈ and so φ(e) L( gia) = e L(gia). (3.8) X − 8 − X a H4 a H4 ∈ ∈ Therefore, since the L(gia) are all equal, it must be that the L( gia) are P P − equal as well.

h Let H = Gal(Q(ζe)/Q(√ e)) = h G = 1 . Then by Dirichlet’s class − { ∈ | e } number formula for imaginary quadratic ﬁelds, we have that

ehK = L( a) L(a). X − − X a H a H ∈ ∈

34 i But H is the union of the elements in g H4 for i = 0, 1, 2, 3, so we have

ehK = L( a) L(a) X − − X a H a H ∈ ∈ = 4 L( gia) 4 L(gia), i, X − − X ∀ a H4 a H4 ∈ ∈ = 4 L( gia) L(gia)! , i, X − − X ∀ a H4 a H4 ∈ ∈ proving (2). And, since both e and hK are positive, we have (1), completing the proof.

Corollary 3.40.

1. For i = 0, 1, 2, 3,

L(gia) L( gia) 0 (mod e). X ≡ X − ≡ a H4 a H4 ∈ ∈

2. hK 4 (mod 8). ≡

Proof. By Theorem 3.39 (2), L( gia) L(gia) 0 (mod e) and a H4 a H4 P ∈ − − P ∈ ≡ thus L( gia) L(gia) (mod e). But, by statement (1) of Theo- a H4 a H4 P ∈ − ≡ P ∈ rem 3.39, L( gia) = L(gia), so we must have that L(gia) a H4 a H4 a H4 P ∈ − 6 P ∈ P ∈ ≡ L( gia) 0 (mod e), proving the ﬁrst statement. a H4 P ∈ − ≡ Reducing Equation 3.8 modulo 2, we have L( gia) 1 L(gia) a H4 a H4 P ∈ − ≡ − P ∈ (mod 2) = L( gia) L(gia) 1 (mod 2), for all i. Thus, com- a H4 a H4 ⇒ P ∈ − − P ∈ ≡ paring with Theorem 3.39 (2), we conclude that hK 4 (mod 8), completing the ≡ proof.

3.3 A Product Formula

For values of e as in the previous section, we claim that ω = τ(t) has the t H4 Q ∈ desired properties from the conjecture. For these values of e, we have the following

35 ﬁeld diagram with corresponding Galois groups over Q.

Field Galois Group over Q

Q(ζep) = E o / G (Z/pZ)× × p 1 −

Q(ζe) = M o / G

φ(e) 8

(H4)0 = K o / G/H4 b 2

Q(√ e, √q1) = K o / G/H2 − e 2

Q(√ e) = K o / G/H − 2

Q o / 1 . { }

Using the results concerning the biquadratic coset sums from the previous section, we ﬁrst show that ω K and that ω E factors over K . Then, using a result of ∈ O O b k Yamamoto from [22], we conclude that σt(ω) = ζ ω is indeed a sign ambiguity. ± e

r Lemma 3.41. Let X = ai be a subset of the integers modulo e such that { }i=1

ai 0 (mod e), then Pi ≡ r τ(a ) Q(ζ ). Y i e i=1 ∈

Proof. Let c Z/eZ and consider the Jacobi sum product ∈

J(a1, c)J(a2, a1 + c)J(a3, a1 + a2 + c) J(ar, a1 + a2 + + ar 1 + c). ( ) ··· ··· − ∗

36 We have,

τ(a1)τ(c) τ(a2)τ(a1 + c) τ(ar)τ(a1 + a2 + ar 1 + c) ( ) = ··· − ∗ τ(a1 + c) · τ(a1 + a2 + c) ··· τ(a1 + a2 + ar + c) ··· τ(c)τ(a1)τ(a2) τ(ar) = ··· τ(a1 + a2 + ar + c) ··· = τ(a1)τ(a2) τ(ar), since a1 + a2 + + ar 0 (mod e). ··· ··· ≡

r But for any α, β Z, J(α, β) Q(ζe) = ( ) Q(ζe) = τ(ai) ∈ ∈ ⇒ ∗ ∈ ⇒ Qi=1 ∈

Q(ζe).

Proposition 3.42. ω K. ∈ b Proof. By Corollary 3.40, L(a) 0 (mod e), so ω Q(ζ ) by the lemma. a H4 e P ∈ ≡ ∈

Furthermore, j H4, σj will only permute the Gauss sums comprising ω. Thus, ∀ ∈

ω is ﬁxed by every automorphism from H4, that is, ω (H4)0 = K. ∈ b

Although ω K, we will ﬁrst consider the ideal generated by ω in M , i.e. ∈ O b ω M , so we can use the Stickelberger factorization formula. O Theorem 3.43.

α hK ω M = (p p 4 ) M , O O where α = L(t) and p is a prime ideal of dividing p. t H4 K P ∈ O

Proof. Let p be a prime, p 1 (mod e), and let P be a prime ideal of M dividing ≡ O p. Then by the (corollary to) Stickelberger factorization formula,

ω M = τ(t)! M O Y O t H4 ∈ 1 ta− σ a G n e o a = P P ∈ Y t H4 ∈ 1 ta− σ t H a G n e o a = P P ∈ 4 P ∈

1 ta− σ a G t H n e o a = P P ∈ P ∈ 4 .

37 For a G/H4, ∈ 1 ta− 1 1 = L(ta− ). X e e X t H4 t H4 ∈ ∈ But, by Theorem 3.39, there are only two possibilities for L(ta 1). If ta 1 t H4 − − P ∈ ∈ giH , for some i, then L(ta 1) = L(t). Let α = L(t). If, on the 4 t H4 − t H4 t H4 P ∈ P ∈ P ∈ other hand, ta 1 giH , for some i, then L(ta 1) = L( t). Let − 4 t H4 − t H4 ∈ − P ∈ P ∈ − β = L( t) and recall from Theorem 3.39 that α < β. But ta 1 giH t H4 − 4 P ∈ − ∈ ⇐⇒ a H, so we have ∈

1 ta− σ a G t H n e o a ω M = P P ∈ P ∈ 4 O

α a H σa+β a /H σa = P P ∈ P ∈

α (β α) a /H σa (3.9) = p P − P ∈

hK α 4 a /H σa = p P P ∈

hK α σa 4 = p P Pa /H . ∈

For a G, consider (σa) K . Since Gal(K/Q) = G/H,(σa) K is nontrivial if and ∈ | ∼ | only if a / H. Thus, if p K = p1p2, then ∈ O

σa a H σa σa a /H σa p1 M = P = P P ∈ and p2 M = P = P P ∈ . O Y O Y a H a/H ∈ ∈

Therefore, continuing (3.9),

h h K hK K α a /H σa 4 α 4 α 4 ω M = p P P ∈ = p (p2 M ) = p p M . O O 2 O

Corollary 3.44. t H, (σt(ω)) M = ω M . ∀ ∈ O O

38 Proof. Let t H. Then, ∈

α hK (σt(ω)) M = σt (ω M ) = σt p p 4 M O O O α hK = p σt p 4 M O h α K = p (σt(p)) 4 M O α hK = p p 4 M , O since p K and K is the ﬁxed ﬁeld of H. ⊆ O

Remark 3.45. Since ω and σt(ω) generate the same ideal in M and have the O same absolute value, it follows that they can only diﬀer by a unit of absolute value

1. But, by Theorems 1.15 and 1.16, the only units of absolute value 1 in Q(ζe) are ζk, k = 1, . . . , e, so we have that ± e

kt σt(ω) = ζ ω, for some kt Z. (3.10) ± e ∈

k0 Fix t0 H H2. Then σt (ω) = ζ ω. We now determine k0 explicitly. ∈ \ 0 ± e Theorem 3.46.

 q1 indγ(q1) (mod e), if e = q1 3 − ·  k  (3.11) 0 3(1 t0)q2 indγ(q2) (mod e), if e = 5 q2 ≡ − ·  0 (mod e), if gcd(e, 15) = 1.  

k0 Proof. As ω, σt (ω) K, we must have that ζ K as well. We ﬁrst determine 0 ∈ e ∈ b b which roots of unity lie in K. Assume ζn K for some integer n > 2. Since ∈ b b K Q(ζe), n 2e, and thus either n = q1, or n = q2, (note that Q(ζ2q ) = Q(ζq )). ⊆ | i i b Furthermore, since [K : Q] = 8, φ(n) 8. But the only odd primes n with φ(n) 8 | | b are n = 3, 5. Hence, if gcd(e, 15) = 1, then the only roots of unity in K are 1 and ± b q1 σt (ω) = ω, i.e. k0 0 (mod e). If e = q1 3, then powers of ζ3 = ζ are the 0 ± ≡ · ± ± e

39 only possible roots of unity in K and thus, k0 0 (mod q1). Finally, if e = 5 q2, ≡ · b q2 then the only possible roots of unity in K are powers of ζ5 = ζ and k0 0 ± ± e ≡ b (mod q2).

Assume e = q1 3. Using Equation 2.2 with m = 3, n = q1, and t = 2, we obtain · the following Davenport-Hasse relation

q1 1 q1 1 − − χ2q1 (q ) τ(3k + 2) = τ(2q ) τ(3k). 1 Y 1 Y k=0 k=1

Comparing with the eight cosets of G/H4, this relation can be rewritten as

q1 1 − τ(3k) 2q1 k=1 σt0 (ω) = χ− (q1) Q . σt3 (ω)σ t2 (ω)σ 1(ω) 0 − 0 −

k0 Substituting this relation into σt (ω) = ζ ω and using the norm relation to 0 ± e reduce, we then have p(q1 1)/2 − k0 2q1 = ζe χ (q1). σt3 (ω)σ t2 (ω) ± 0 − 0

Multiplying numerator and denominator by σ t3 (ω) and again using the norm − 0 relation, we obtain

σ 3 (ω) σ (ω) t0 t0 k0 2q1 − = σ 2 = ζ χ (q ). t0 e 1 σ t2 (ω) − ω ± − 0

2 k0 t0k0 k0 2q1 Substituting σt (ω)/ω = ζ , the above reduces to ζe− = ζ χ (q1), and hence 0 ± e e 2 k0(t0+1) 2q1 2q1 2q1 indγ (q1) 2q1 indγ (q1) ζe− = χ (q1). Now, χ(γ) = ζe, so χ (q1) = χ (γ ) = ζe ,

2 k0(t0+1) 2q1 indγ (q1) and therefore ζe− = ζe . But, as k0 0 (mod q1), we have ≡

2 k0(t +1) 2q1 indγ (q1) 2 ζ− 0 = ζ k0(t + 1) 2q1 indγ(q1) (mod q1 3) e e ⇐⇒ − 0 ≡ · 2 k0(t + 1) 2q1 indγ(q1) (mod 3) ⇐⇒ − 0 ≡

k0 q1 indγ(q1) (mod 3). ⇐⇒ ≡ −

Hence, k0 q1 indγ(q1) (mod q1 3). ≡ − ·

40 Now assume e = 5 q2. We proceed as in the previous case. Applying Equation · k0 2.2 with m = 5, n = q2, for t = 1 and t t0 (mod 5), to σt (ω) = ζ ω, we ≡ 0 ± e obtain σ (ω) t0 k0 q2(1 t0) σ t2 = ζe− χ − (q2). − 0 ω ±

k0 Substituting σt (ω)/ω = ζ and recalling that k0 0 (mod q2), we have 0 ± e ≡

2 2 t k0 k0+q2(1 t0) indγ (q2) k0(1 t ) q2(1 t0) indγ (q2) ζ− 0 = ζ− − ζ − 0 = ζ − e e ⇐⇒ e e 2 k0(1 t ) q2(1 t0) indγ(q2) (mod 5q2) ⇐⇒ − 0 ≡ − 2 k0(1 t ) q2(1 t0) indγ(q2) (mod 5) ⇐⇒ − 0 ≡ −

k0 3(1 t0)q2 indγ(q2) (mod 5). ⇐⇒ ≡ −

Therefore, k0 3(1 t0)q2 indγ(q2) (mod 5q2), completing the proof. ≡ −

Remark 3.47. Note that the congruences for k0 in the previous theorem can be rewritten as

L3( indγ(q1))q1 (mod e), if e = q1 3  − ·  k  (3.12) 0 L5(3(1 t0) indγ(q2))q2 (mod e), if e = 5 q2 ≡ − ·  0 (mod e), if gcd(e, 15) = 1.   k0 We claim that σt (ω) = ζ ω is a sign ambiguity. We will need the following 0 ± e weakened version of a result implied in [22].

Lemma 3.48. Let e = p1p2, for any primes p1, p2. Assume

τ(a ) = ζk τ(b ) (3.13) Y i e Y j i ± j for some ai, bj, k Z, and let Λ = L(p1 +p2),L(p1 p2),L(p2 p1),L( p1 p2) . ∈ { − − − − }

If # i ai Λ # j bj Λ 1 (mod 2), then equation 3.13 is a not direct { | ∈ } − { | ∈ } ≡ consequence of the norm relation and the Davenport-Hasse product formula.

41 Proof. We prove the contrapositive. A multiplicative relation (3.13) follows from the norm relation and Davenport-Hasse if and only if it can written as a linear combination of these relations. For any norm relation, τ(a)τ( a) = χa( 1)p, we − − have that a Λ a Λ. Thus, each use of a norm relation contributes an ∈ ⇐⇒ − ∈ even number of elements to the set i ai Λ . Now consider the Davenport-Hasse { | ∈ } relations. Splitting the quotient of (2.2) and reindexing the left side, we have that for e = mn, with m, n > 1, and for 1 t m 1, ≤ ≤ − n 1 n 1 − − χtn(n) τ(km + t) = τ(tn) τ(km). Y Y k=0 k=1

On the right hand side, none of the Gauss sums will contribute to i ai Λ , since { | ∈ } gcd(tn, e) and gcd(km, e) = 1. For the left side, ﬁx t and assume km + t Λ. 6 ∈ Then km + t m n (mod e) = either t = mx + n or t = mx n, for some ≡ ± ± ⇒ − (unique) x. Without loss of generality, assume t = mx + n. Then km + t Λ ∈ ⇐⇒ km+(mx+n) = m(k+x)+n Λ m(k+x) m (mod n) k+x 1 ∈ ⇐⇒ ≡ ± ⇐⇒ ≡ ± (mod n) k 1 x (mod n). Thus, for ﬁxed t, there are exactly two values ⇐⇒ ≡ ± − of k, 1 k n 1, such that km + t Λ. Therefore, each use of a Davenport- ≤ ≤ − ∈

Hasse relation contributes an even number of elements to the set i ai Λ . { | ∈ } Hence, if a multiplicative relation (3.13) is written as a linear combination of norm and Davenport-Hasse relations, then each use of a relation will contribute an even number or elements to i ai Λ and thus { | ∈ }

# i ai Λ # j bj Λ 0 (mod 2), { | ∈ } − { | ∈ } ≡ proving the lemma.

We are now able to state and prove our main theorem which gives a partial answer to Conjecture 3.34. In the next section, we will resolve the sign ambiguity.

42 Theorem 3.49. Let e = q1q2 with q1 5 (mod 8), q2 3 (mod 4), and q2 a ≡ ≡ biquadratic residue modulo q1. Let p be prime, p 1 (mod e). Let H4 be the group ≡ of biquadratic residues modulo e and let t0 be a quadratic nonresidue modulo q1 and q2, i.e. t0 H H2. Then ∈ \ τ(t) k0 = ζe− (3.14) Y τ(t0t) ± t H4 ∈ is a sign ambiguity, where k0 is as in Theorem 3.46.

Proof. Rewriting (3.14), we want to show that

ω σt0 (ω)

k0 τ(t) = ζ− τ(t0t) (3.15) zY}| { ± e zY }| { t H4 t H4 ∈ ∈

k0 is a sign ambiguity. By Remark 3.45, we have that σt (ω) = ζ ω, so we only 0 ± e verify that (3.15) does not follow from the norm relation or Davenport-Hasse.

As in Lemma 3.48, let Λ = L( q1 q2) . Let a H4. Then a = tq2 + nq1, for { ± ± } ∈ some t 4, n N2 and we have ∈ T ∈

a Λ t 1 (mod q1) and n 1 (mod q2). ∈ ⇐⇒ ≡ ± ≡ ±

But n is square modulo q2 and since q2 3 (mod 4), 1 is a nonsquare. Therefore, ≡ − n 1 (mod q2). Furthermore, 1 4 = 1 4, and thus 6≡ − ∈ T ⇒ − 6∈ T

a Λ t 1 (mod q1) and n 1 (mod q2). ∈ ⇐⇒ ≡ ≡

For a H4, consider t0a = t0(tq2 + nq1) = (t0t)q2 + (t0n)q1. We have ∈

t0a Λ t0t 1 (mod q1) and t0n 1 (mod q2). ∈ ⇐⇒ ≡ ± ≡ ±

But for t T4, t0t is a nonsquare and 1 are both squares, so t0a 1 (mod q1). ∈ ± 6≡ ±

Therefore t0a Λ, a H4. Combining these results, we have that 6∈ ∀ ∈

# a H4 a Λ # a H4 t0a Λ = 1 0 1 (mod 2), { ∈ | ∈ } − { ∈ | ∈ } − ≡

43 which, by the lemma, implies that (3.14) does not follow from Davenport-Hasse and the norm relation. Hence,

τ(t) k0 = ζe− Y τ(t0t) ± t H4 ∈ is a sign ambiguity.

Remark 3.50. We remark that this product formula does, indeed, give an inﬁnite set of new sign ambiguities. From the statement of the theorem, there are three restrictions on q1 and q2:

1. q1 5 (mod 8), ≡

2. q2 3 (mod 4), ≡

3. q2 a biquadratic residue modulo q1.

Replacing the third condition with the stronger

3 0. q2 1 (mod q1), ≡ and applying the Chinese Remainder Theorem, we then need only that q1 5 ≡

(mod 8) and q2 3 (mod 4q1). But by Dirichlet’s theorem for primes in an arith- ≡ metic progression [7], there are infinitely many such primes q1, and for fixed q1, there are also infinitely many primes q2. Thus, there are infinitely many values of e satisfying the theorem.

3.4 Resolution of Ambiguity

We now turn to the resolution of the ambiguous sign in Theorem 3.49. In previous cases, Muskat, Muskat-Whiteman, and Muskat-Zee, [13, 15, 16], have obtained resolution via binary quadratic form decomposition of the prime p. For example, we have the following from [13].

44 Example 3.51 (Muskat). Let e = 39 and let p be a prime such that p 1 ≡ (mod 39). Then

k τ(1)τ(16)τ(34) = uζe τ(2)τ(17)τ(32) is a sign ambiguity, where k = 13 indγ13 and

+1, if p = x2 + 39y2  u =  1, if p = 3x2 + 13y2. −  Here, we take a related, but diﬀerent approach. Whereas resolution via binary quadratic forms depends on the representation of p in the form class group, CF (K), our method instead relies on an equivalent condition on the primes of K above p O in the ideal class group, C(K).

To resolve the ambiguity, we again return to our diagram:

Q(ζe) o / G

ϕ(e) 8

ω (H4)0 = K o / G/H4 ∈ b 2

Q(√ e, √q1) = K o / G/H2 − e 2

Q(√ e) = K o / G/H − 2

Q o / 1 . { }

In the previous section, we exploited the fact that ω K to conclude that σt(ω) = ∈ b ζk0 ω. We now resolve the sign by determining whether the product of a certain ± e root of unity and ω is in K K or K, and then connecting this to the order of the \ e ideal classes above p in C(K).

δk0 Proposition 3.52. Let k0 be as in Theorem 3.46. Then ζ ω K, where δ = 1 e ∈ − e if gcd(e, 5) = 1 and δ = 3(t0 + 1) if e = 5 q2. ·

45 3 Proof. As t0 H H2, by Proposition 3.35, either [t0] = [g] in G/H4 or [t0] = [g ] ∈ \ in G/H4. Choose t1 H H2 such that [t1] = [t0] in G/H4. Then for all t H H2, ∈ \ 6 ∈ \ either σt(ω) = σt0 (ω) or σt(ω) = σt1 (ω).

Since K is Galois over Q, ω K = σt (ω) K. Therefore, ω = ∈ ⇒ i ∈ OK b b b b (σt (ω)) , and it follows that i OK b

λi ki σt (ω) = ( 1) ζ ω, for some λi, ki Z, i − e ∈ and thus

λi ki λi tiki tiki+ki σt2 (ω) = σt (( 1) ζ ω) = ( 1) ζ σt (ω) = ζ ω. (3.16) i i − e − e i e

δk0 Since K is the ﬁxed ﬁeld of H2, ζ ω K if and only if for all t H H2, e ∈ ∈ \ e e δk0 δk0 3 σt2 (ζe ω) = ζe ω. And, since [ti] = [t1 i], −

σt2 (ω) = σt6 (ω) = σt2 σt4 (ω) = σt2 (ω), (3.17) i 1 i 1 i 1 i 1 i − − − − so without loss of generality, we may assume that t = t0. Let e = q1 3. Then by · Equation (3.16),

2 2 2 k0 t k0 t k0 t0k0+k0 k0t +k0t0+k0 σ 2 (ζ− ω) = ζ− 0 σ 2 (ω) = ζ− 0 (ζ ω) = ζ− 0 ω. t0 e e t0 e e e

k0 2 Therefore, ζ− ω is ﬁxed by σt2 if and only if k0t + k0t0 + k0 k0 (mod e), e i − 0 ≡ − that is,

k0 2 ζ− ω K k0t + k0t0 + k0 k0 (mod q1 and 3). (3.18) e ∈ ⇐⇒ − 0 ≡ − e 2 But for e = q1 3, t0 H H2 = t0 2 (mod 3), and thus k0t + k0t0 + k0 · ∈ \ ⇒ ≡ − 0 ≡

4k0 + 2k0 + k0 k0 (mod 3). And, by Theorem 3.46, k0 is equivalent to 0 − ≡ − 2 k0 modulo q1, so 0 k0 k0t +k0t0 +k0 (mod q1). Hence, by (3.18), ζ− ω K. ≡ − ≡ − 0 e ∈ e If e = 5 q2, then again by (3.17), we may assume t = t0, and we have ·

2 2 2 2 δk0 δk0t δk0t δk0t +k0t0+k0 k0(δt +t0+1) σ 2 (ζ ω) = ζ 0 σ 2 (ω) = ζ 0 σ 2 (ω) = ζ 0 ω = ζ 0 ω. t0 e e t0 e t0 e e

46 δk0 2 Hence, ζ ω is ﬁxed by σt2 if and only if k0(δt + t0 + 1) δk0 (mod e), that is, e i 0 ≡

δk0 2 ζ ω K k0(δt + t0 + 1) δk0 (mod 5 and q2). (3.19) e ∈ ⇐⇒ 0 ≡ e

2 Since t0 is a nonsquare, t0 2 or 3 (mod 5) = t 4 (mod 5), and ≡ ⇒ 0 ≡

2 δt + t0 + 1 3(t0 + 1) 4 + t0 + 1 13(t0 + 1) 3(t0 + 1) δ (mod 5). 0 ≡ · ≡ ≡ ≡

2 Hence, k0(δt + t0 + 1) δk0 (mod 5). And by Theorem 3.46, k0 0 (mod q2), so 0 ≡ ≡ 2 δk0 k0(δt + t0 + 1) k0δ (mod q2) as well. Therefore, by (3.19), ζ ω K. 0 ≡ e ∈ e Finally, let e be such that gcd(e, 15) = 1. Then, again by Theorem 3.46, k0 0 ≡ (mod e), and

k0(t0+1) 0(t0+1) σt2 (ω) = ζe ω = ζe ω = ω, implying that ω K. ∈ e

Since ζδk0 ω K, we must have that either ζδk0 ω K K or ζδk0 ω K K. e ∈ e ∈ ⊆ e ∈ \ e e e δk0 First assume ζ ω K K. Since t0 H and K is the ﬁxed ﬁeld of H, it follows e ∈ ⊆ ∈ e δk0 δk0 that σt0 (ζe ω) = ζe ω. But

δk0 δk0 t0δk0 δk0(1 t0) ζ ω = σt (ζ ω) = ζ σt (ω) = σt (ω) = ζ − ω, (3.20) e 0 e e 0 ⇒ 0 e

k0 and for all e, δk0(1 t0) k0 (mod e). Therefore, σt (ω) = ζ ω. − ≡ 0 e δk0 δk0 On the other hand, if ζ ω K K, then it must be the case that σt (ζ ω) = e ∈ \ 0 e e δk0 k0 ζ ω, which implies σt (ω) = ζ ω. Therefore, to determine the correct sign in − e 0 − e τ(t) k0 = ζe− , Y τ(t0t) ± t H4 ∈ we need only determine whether or not ζδk0 ω K. e ∈

hK δk0 Lemma 3.53. ζ ω K p 4 K is principal. e ∈ ⇐⇒ ⊆ O

47 Proof. If ζδk0 ω K, then e ∈ hK δk0 α β α β α α 4 (ζ ω) K = ω E K = p p = p p − = p p , e O O ∩ O 1 2 2 2

α hK /4 hK /4 and p p2 is principal. Therefore, p2 is principal as well.

hK /4 If, on the other hand, p is principal, then there exists ω0 K such that ∈ α hK /4 ω0 K = p p . Now, O

α hK r ω0 K = p p 4 = ω0 = ω = ω0 = ζ ω, for some r Z. O ⇒ OK OK ⇒ ± e ∈ b b r r r Since ω0 K, it follows that ζ ω K and thus σt (ζ ω) = ζ ω. But ∈ e ∈ 0 e e

r r t0r r t0r r(1 t0) ζ ω = σt (ζ ω) = ζ σt (ω) = σt (ω) = ζ − ω = ζ − ω. (3.21) e 0 e e 0 ⇒ 0 e e

Comparing Equations (3.20) and (3.21), it must be the case that δk0(1 t0) − ≡ δk0 r(1 t0) (mod e) and thus r δk0 (mod e). Hence, ζ ω K. − ≡ e ∈ We now state our complete main theorem.

Theorem 3.54. Let e = q1q2 with q1 5 (mod 8), q2 3 (mod 4), and q2 a ≡ ≡ biquadratic residue modulo q1. Let p be prime, p 1 (mod e). Let H4 be the group ≡ of biquadratic residues modulo e and let t0 be a quadratic nonresidue modulo q1 and q2. Then τ(t) k0 = uζe− , Y τ(t0t) t H4 ∈ where

 q1 indγ(q1) (mod e), if e = q1 3 − ·  k  0 3(1 t0)q2 indγ(q2) (mod e), if e = 5 q2 ≡ − ·  0 (mod e), if gcd(e, 15) = 1.   and +1, if o([p]) 1 (mod 2),  ≡ u =  1, if o([p]) 0 (mod 2). − ≡ 

48 Proof. By Theorem 3.49, we need only verify the value of u. As in the preceding remarks, the value of u is dependent only on the ideal class of phK /4. We have

hK δk0 +1 ζe ω K [p 4 ] = 1, u =  ⇐⇒ ∈ ⇐⇒  h δk K 1 ζ 0 ω K K [p 4 ] = 1. − ⇐⇒ e ∈ \ ⇐⇒ 6  e

By Corollary 3.40, hK 4 (mod 8), and thus hK /4 1 (mod 2). Therefore, ≡ ≡ hK hK h [p 4 ] = 1 [p] 4 = 1 o([p]) K o([p]) 1 (mod 2). ⇐⇒ ⇐⇒ | 4 ⇐⇒ ≡

We conclude with an example of our main theorem as well as a demonstration of how resolution via binary quadratic forms is then quickly deduced.

Example 3.55. Let e = 155 = 5 31, p be a prime p 1 (mod e), and p K · ≡ ⊂ O be a prime ideal above p. Then

H4 = 1, 16, 36, 41, 51, 56, 66, 71, 76, 81, 101, 111, 121, 126, 131 { } and we can take t0 = 12. Thus, by Theorem 3.54 and Remark 3.47,

τ(t) (2)(31) indγ (31) = uζ155 , Y τ(t0t) t H4 ∈ where +1, if o([p]) 1 (mod 2),  ≡ u =  1, if o([p]) 0 (mod 2). − ≡  For instance, using PARI/GP we compute that for primes p < 20000,

+1 if p = 311, 5581, 11471, 12401, 19531, 19841,  u =  1 if p = 1861, 2791, 4651, 8681, 11161, 13331, 16741, 17981, 18911. −  Now, o([p]) 1 (mod 2) [p] C(K)4. Therefore, via the isomorphism be- ≡ ⇐⇒ ∈ tween the ideal class group and the form class group, we have o([p]) 1 (mod 2) ≡ ⇐⇒

49 4 [p] CF (K) . Again using PARI/GP, we compute the corresponding quadratic ∈ forms, giving

+1 if p = x2 + xy + 39y2,  u =  1 if p = 5x2 + 5xy + 9y2. −  Remark 3.56. We remark that for values of e with larger class numbers, using the quadratic form resolution alone becomes increasingly diﬃcult as the number of forms from which to choose will be hK /4. For example, if e = 327, then hK = 12, and there are three forms which give u = +1 and three which give u = 1. − However, using information about the ideal class of a quadratic prime as above, a computational resolution is quickly achieved and, if necessary, resolution criteria using quadratic forms can be easily deduced.

50 References

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52 Vita

Brian J. Murray was born on November 19, 1974, in St. Charles, Missouri. He

ﬁnished his undergraduate studies at Washington University in St. Louis in May

1997. In August 1997 he came to Louisiana State University to pursue graduate studies in mathematics and earned a master of science degree in mathematics in

May 1999. He is currently a candidate for the degree of Doctor of Philosophy in mathematics, which will be awarded in August 2002.