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On Gauss Sums, Cyclotomic Number Fields and Stickelberger’S Theorem ON GAUSS SUMS, CYCLOTOMIC NUMBER FIELDS AND STICKELBERGER'S THEOREM KONSTANTIN DRAGOMIRETSKIY 01 June 2010 Undergraduate Honors Thesis Advisor: Professor Harold Stark University of California, San Diego Department of Mathematics 0 Contents 1. Introduction 2 2. Classical Gauss Sum 3 3. Quadratic Reciprocity 7 4. Gauss Sums 11 5. Quadratic Gauss Sums 15 6. Stickelberger's Theorem 19 7. Brumer-Stark Conjecture 25 References 26 1. Introduction This undergraduate thesis will survey Gauss sums, starting from the classical Gauss sum analyzed by C. F. Gauss, and continuing to more general character sums. Along the way, the classical Gauss sum will be utilized to prove the famous law of quadratic reciprocity, which Gauss refered to as the "fundamental theorem" in the Disquisitiones Arithmeticae. Various decompositions and recursions of Gauss sums will be developed. Prime ideal factorizations will be developed through Stickelberg's theorem. Finally, a closing remarks on the Stickelberger element annihilating the cyclotomic class group and its generalization, the Brumer-Stark conjecture. Acknowledgement First and foremost, I would like to thank Dr. Harold Stark, the advisor for my undergraduate honors thesis and my concurrent algebraic num- ber theory course professor. I would have prefered meeting Professor Stark as a graduate or postdoctoral student to be able to have much deeper dialogues in mathematics, nevertheless I feel very fortunate that our paths crossed and that I had the privilege of working with him. I thank him for his time invested into this thesis and wish him good luck with his book. I thank Dr. Cristian Popescu for encouraging number theory to me during my undergraduate career and reaffiring my aspirations to be- come a mathematician. I finally thank Dr. Lance Small, Dr. Audrey Terras, and Dr. Cristian Popescu for teaching me mathematics and for recommending me for the mathematics doctorate program at the University of California, Los Angeles. Konstantin Vitalyevich Dragomiretskiy 2 2. Classical Gauss Sum The Classical Gauss Sum was studied extensively by Carl Friedrich Gauss. This section is series of exercises from Abstract Algebra by Dummit and Foote. 2πi Put Kp = Q(ζp), where ζp = exp( p ) and p is an odd prime. A finite dimensional Q-vector space is called a number field. Kp is the pth cyclotomic number field. This is because the monic, irreducible polynomial over Q is the pth cyclotomic polynomial: xp − 1 Φ (x) = = 1 + x + ::: + xp−1 and [x]=hΦ (x)i ∼= K p x − 1 Q p p Put Gp = Gal(Kp=Q) = fσ 2 Aut(Kp=Q)j σ(q) = q 8q 2 Qg φ(p) p−1 X i X i i p−1 8a 2 Kp; a = qiζp = qiζp for qi 2 Q where fζpgi=1 is a i=1 i=1 power basis for Kp over Q. Hence τ 2 Gp is completely determined by α τ(ζp) = ζp where (α; p) = 1. ∼ × This observation elucidates the well-known isomorphism Gp = Z=pZ . X X Definition 2.1. η0 = τ(ζp) and η1 = τ(ζp) τ2H τ2σH where H is the unique normal subgroup of index 2 and the coset σH = G n H. η0 and η1 are called the two periods of ζp with respect to H. By the aformentioned isomorphism, Gp is cyclic, and hence the follow- ing one-to-one correspondence can be observed: gi p−1 hτ1i = fτi 2 G jτi(ζp) = ζp g ! f1; g; : : : ; g g = hgi = Gp. Remark. hτ1i is cyclic by the action of composition. Lemma 2.1. τ1(η0) = η1 and τ1(η1) = η0 p−1 p−1 X X X2 X2 Proof. τ1(η0) = τ1( τ(ζp)) = τ1(τ(ζp)) = τ1(τ2i(ζp)) = τ2i−1(ζp) = τ2H τ2H i=1 i=1 X τ(ζp) = η1: τ2σH 3 p−1 p−1 X X X2 X2 τ1(η1) = τ1( τ(ζp)) = τ1(τ(ζp)) = τ1(τ2i−1(ζp)) = τ2i(ζp) = τ2σH τ2σH i=1 i=1 X τ(ζp) = η0: τ2H X a X b Lemma 2.2. η0 = ζp and η1 = ζp a≡square b6≡square 2k k 2 2 Proof. a 2 H , a = g for some k 2 N , a = (g ) = gk , a is a square, hence a is a square (modulo p) if and only if a 2 H. p−1 p−1 2 2 X a X g2i X X ζp = ζp = τ2i(ζp) = τ(ζp) = η0 a≡square i=1 i=1 τ2H If b is not a square then b 62 H and hence b = g2i−1 for some i 2 N p−1 p−1 p−1 p−1 2 2 2 2 X b X g2i−1 X X X ζp = ζp = τ2i−1(ζp) = τ1(τ2i(ζp)) = τ1( τ2i(ζp)) = b6≡square i=1 i=1 i=1 i=1 τ1(η0) = η1 Definition 2.2. Let k be a field and K a cyclic extension with Galois group G of order n, given α 2 K; the Lagrange resolvent is n−1 X (α; ζ) = α + ζσ(α) + ::: + ζn−1σn−1(α) = ζiσi(α) i=0 where ζ is an nth root of unity and σ is a generator of G. Lemma 2.3. η0 + η1 = (ζp; 1) = −1 and η0 − η1 = (ζp; −1) p−1 p−1 2 2 X a X b X X Proof. η0 + η1 = ζp + ζp = τ2i(ζp) + τ2i−1(ζp) = a≡square b6≡square i=1 i=1 p−1 X 2 τi(ζp) = τ1(ζp) + τ2(ζp) + ::: + τp−1(ζp) = τ1(ζp) + τ1 (ζp) + ::: + i=1 p−2 φ(p)−1 τ1 (ζp) + ζp = ζp + τ1(ζp) + ::: + τ1 (ζp) = (ζp; 1): p−1 p−1 X X i η0 + η1 = τi(ζp) = ζp via rearrangement, and hence η0 + η1 = i=1 i=1 Φp(ζp) − 1 = −1, where Φp(x) is the pth cyclotomic polynomial. 4 g2 g4 gp−1 g1 g3 gp−2 η0 − η1 = (ζp + ζp + ::: + ζp ) − (ζp + ζp + ::: + ζp ) = p−1 p−2 X i i X i i (−1) τ1(ζp) = (−1) τ1(ζp) = (ζp; −1): i=1 i=0 p−1 X i2 Lemma 2.4. ζp = (ζp; −1) i=1 Proof. Observe that (p − k)2 ≡ p2 − 2pk + k2 ≡ k2(mod p). p−1 X i2 0 1 22 (p−2)2 (p−1)2 Put γ = ζp . γ = ζp + ζp + ζp + ::: + ζp + ζp = i=0 p−1 2 2 1 2 ( 2 ) p−1 1 + 2(ζp + ζp + ::: + ζp ): Rewritting the residues 1, 2, :::; 2 a p−1 2 (ga1 )2 (ga2 )2 (g 2 ) as powers of g, γ = 1 + 2(ζp + ζp + ::: + ζp ) = 1 + 2a p−1 g2a1 g2a2 g 2 2(ζp + ζp + ::: + ζp ) = 1 + 2η0 since each exponent of g in the parenthesized sum is even and unique mod φ(p): γ = 1 + 2η0 = 2η0 − (−1) = 2η0 − (ζp; 1) = 2η0 − (η0 + η1) = η0 − η1 = (ζp; −1): Lemma 2.5. τ(γ) = γ if τ 2 H and τ(γ) = −γ if τ 62 H p−1 Proof. τ 2 H ) τ = τ2i for some i, 1 ≤ i ≤ 2 ) τ(γ) = τ(η0 − η1) = τ(η0) − τ(η1) = τ2i(η0) − τ2i(η1) = η0 − η1 = γ since by Lemma 2.1, τ2 = τ1 ◦ τ1 is the identity on both η0 and η1: p−1 τ 62 H ) τ = τ2i+1 for some i, 0 ≤ i < 2 ) τ(γ) = τ(η0 − η1) = τ(η0) − τ(η1) = τ2i+1(η0) − τ2i+1(η1) = η1 − η0 = −γ since τ2i+1 = τ ◦ τ2i: Lemma 2.6. γ¯ = γ when p ≡ 1 (mod 4) and γ¯ = −γ when p ≡ 3 (mod 4) where γ¯ is the complex conjugate of γ: Proof. By Lemma 2.5, γ is fixed by H, and thus Q ⊆ Q(γ) ⊆ KH . It H is easily verifiable that γ 62 Q and thus Q(γ) = Kp , the subfield of Kp fixed by H. ∼ Gal(Kp=Q) ∼ Gal(Q(γ)=Q) = = G=H = hσ−1i = Z=2Z where σ−1 is Gal(Kp=Q(γ)) complex conjugation. p−1 2 A simple observation of orders gives us σ−1 = τ1 = τ p−1 : 2 p−1 p ≡ 1 (mod 4) ) ≡ 0 (mod 2) ) τ p−1 2 H ) γ¯ = σ−1(γ) = 2 2 τ p−1 (γ) = γ: 2 5 p−1 p ≡ 3 (mod 4) ) ≡ 1 (mod 2) ) τ p−1 62 H ) γ¯ = σ−1(γ) = 2 2 τ p−1 (γ) = −γ: 2 Lemma 2.7. γγ¯ = p p−2 p−2 X i i X i −i Proof. γ¯ = (−1) τ1(ζp) = (−1) τ1 (ζp) ) i=0 i=0 p−2 p−2 p−2 p−2 i−j X X X X τ (ζp) γγ¯ = ( (−1)iτ i(ζ ))( (−1)jτ −j(ζ )) = (−1)i−jτ j( 1 ) 1 p 1 p 1 ζ i=0 j=0 i=0 j−0 p p−2 p−2 k X k X j τ1 (ζp) = (−1) τ1 ( ) where k = i − j. ζp k=0 j=0 k k τ1 (ζp) j τ1 (ζp) k = 0 ) = 1 ) τ1 ( ) = 1: ζp ζp k τ (ζ ) k 1 p g −1 αk αk k 6= 0 ) = ζp = ζp = τ1 (ζp) where gcd(αk; p) = 1: ζp p−2 p−2 k p−2 p−2 0 X k X j τ1 (ζp) X k X j gk−1 γγ¯ = (−1) (p−1)+ (−1) τ1 ( ) = (p−1)+ (−1) τ1 (ζp ) = ζp k=1 j=0 k=1 j=0 p−2 p−2 p−2 p−2 X k X j αk X k αk X j (p−1))+ (−1) τ1 (τ1 (ζp)) = (p−1)+ (−1) τ1 ( τ1 (ζp)) = k=1 j=0 k=1 j=0 p−2 p−2 X k αk X k αk (p − 1) + (−1) τ1 [(ζp; 1)] = (p − 1) + (−1) τ1 (−1) = (p − 1) + k=1 k=1 p−2 X k+1 (−1) = (p − 1) + 1 − 1 + ::: + 1 − 1 + 1 = (p − 1) + 1 = p: k=1 2 p−1 Theorem 2.1.
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