Quadratic and Hilbert Reciprocity Timothy Curry B.S., Mathematics B.S., Biological Sciences An Undergraduate Honors Thesis Submitted in Partial Fulfillment of the Requirements for the Degree of Bachelor of Science at the University of Connecticut May 2014 i Copyright by Timothy Curry May 2014 ii APPROVAL PAGE Bachelor of Science Honors Thesis Quadratic and Hilbert Reciprocity Presented by Timothy Curry, B.S. Math, B.S., Biology Honors Major Advisor William Abikoff Honors Thesis Advisor Keith Conrad Honors Thesis Advisor Alvaro´ Lozano-Robledo University of Connecticut May 2014 iii ACKNOWLEDGMENTS My most sincere appreciation and gratitude goes to both Keith Conrad and Alvaro´ Lozano-Robledo. Without their constant guidance and patience, this thesis would never have been completed. iv Quadratic and Hilbert Reciprocity Timothy Curry, B.S. University of Connecticut, May 2014 v Contents Ch. 1. Introduction 1 Ch. 2. The p-adic numbers 2 2.1 Useful Definitions and Properties . 2 Ch. 3. The Hilbert Symbol 6 3.1 Definition and Basic Properties . 6 3.2 Square Classes . 9 3.3 Bimultiplicativity of the Hilbert Symbol . 11 3.4 Formula for the Hilbert Symbol . 17 3.5 Hilbert Reciprocity Law on Q ...................... 24 Ch. 4. The Hilbert Symbol on Q(i)π 28 4.1 Primes in Z[i] and completions of Q(i) . 28 4.2 Hilbert Symbol on Q(i)π ......................... 32 4.3 Square Classes . 33 4.4 Bimultiplicativity of the Hilbert Symbol over Q(i)v . 39 4.5 Q(i)τ = Q2(i)............................... 47 4.6 Hilbert Reciprocity on Q(i) ....................... 52 Bibliography 63 vi Chapter 1 Introduction The law of quadratic reciprocity provides conditions that tell whether an integer is a quadratic residue modulo primes. However, it does not treat 2 in the same manner as other primes. There is a supplementary law for 2 which details the conditions needed for it to be a quadratic residue. In this thesis, we will recall a completion of the rational numbers Q, called the p- adic numbers Qp. After exploring Qp, we will consider the Hilbert symbol, a particular × × pairing on Qp ×Qp . The Hilbert symbol satisfies the Hilbert reciprocity law, which we will show is equivalent to the law of quadratic reciprocity. However, unlike quadratic reciprocity, the Hilbert reciprocity law puts all primes on an equal footing, including 2. For a Gaussian integer prime π, we will also discuss the π-adic completion of Q(i), × denoted Q(i)π. Then we will examine the Hilbert symbol on Q(i)π and show that the Hilbert reciprocity law on Q(i) is equivalent to quadratic reciprocity in the Gaussian integers. 1 Chapter 2 The p-adic numbers 2.1 Useful Definitions and Properties In this thesis, we will assume prior knowledge of the p-adic numbers Qp. However, in this section we will recall several important definitions and properties regarding Qp that will be used frequently in this thesis. Definition 2.1.1. Let p 2 Z be prime. Define the p-adic valuation on Z to be the function vp : Z − f0g −! R such that for each n 2 Z − f0g, vp(n) is the unique vp(n) 0 0 positive integer satisfying n = p n where p - n . Furthermore, we extend vp so a × that for x = b 2 Q with a; b 2 Z − f0g, we have vp(x) = vp(a) − vp(b). Lastly, we set vp(0) = 1. 2 3 Definition 2.1.2. Let x 2 Q. We define the p-adic absolute value of x by 8 > 1 ; if x 6= 0; < pvp(x) jxjp := > :>0; if x = 0: The field Qp is defined as the completion of Q with respect to the p-adic absolute value, and we have the following theorem about the form of each element in Qp. × Theorem 2.1.3. Let x 2 Qp . Then x can be written uniquely in the form −n0 2 n X n x = b−n0 p + ··· + b0 + b1p + b2p + ··· + bnp + ··· = bnp n≥−n0 with 0 ≤ bn ≤ p − 1 and −n0 = vp(x). Proof. For a proof, see [3, p. 68, Corollary 3.3.11]. Definition 2.1.4. The ring of p-adic integers is Zp = fx 2 Qp : jxjp ≤ 1g. In × particular, the units of Zp are Zp = fx 2 Qp : jxjp = 1g. Theorem 2.1.5. Let n 2 Z such that n ≥ 1. Then the inclusion Z ,! Zp induces a n n ring isomorphism Z=p Z ! Zp=p Zp. Proof. See [3, p. 63, Corollary 3.3.6]. n Theorem 2.1.6 (Hensel's Lemma). Let f(x) = a0+a1x+···+anx be a polynomial in Zp[x]. Suppose that there exists a p-adic integer α0 2 Zp such that f(α0) ≡ 0 mod pZp and 0 f (α0) 6≡ 0 mod pZp: 4 Then there exists a unique p-adic integer α 2 Zp such that α ≡ α0 mod pZp and f(α) = 0: Proof. See [3, p. 70, Theorem 3.4.1]. × × Corollary 2.1.7. Let p be an odd prime and let u 2 Zp . Then u = in Zp if and only if u ≡ mod p. × × Proof. Let u 2 Zp (in particular, u 6≡ 0 mod p). First, if u = in Zp , then obviously u ≡ mod p. On the other hand, let u ≡ mod p. Then there exists some a 2 (Z=pZ)× such that a2 ≡ u mod p. Now consider the polynomial f(x) = x2 − u. We have that f(a) ≡ a2 − u ≡ 0 mod p and f 0(a) = 2a 6≡ 0 mod p. Thus, by Hensel's 2 lemma there exists α 2 Zp such that f(α) = 0. So α = u, meaning u = in Zp and 2 × × jαjp = jujp = 1, so α 2 Zp . Thus, u = in Zp . × Lemma 2.1.8. Let p be an odd prime and let a; b; c 2 Zp . Then there exist x; y 2 Z=pZ such that ax2 + by2 ≡ c mod p. 2 2 p+1 Proof. Rewrite the congruence as ax ≡ c − by mod p. Since there are 2 squares 2 p+1 in Z=pZ (including 0 mod p here), ax mod p has 2 values as x varies mod p, and 2 p+1 p+1 p+1 likewise c − by mod p has 2 values as y varies mod p. Since 2 + 2 = p + 1 > p = jZ=pZj, by the pigeonhole principle ax2 mod p : x 2 Z=pZ \ c − by2 mod p : y 2 Z=pZ 6= ;: 2 2 So there exist x0; y0 2 Z=pZ such that ax0 ≡ c − by0 mod p. × Corollary 2.1.9. Let p be an odd prime and let a; b; c 2 Zp . Then the equation 2 2 ax + by = c has a solution with x; y 2 Zp. 5 2 2 Proof. By Lemma 2.1.8 there are x0; y0 2 Zp such that ax0 + by0 ≡ c mod p and either x0 6≡ 0 mod p or y0 6≡ 0 mod p. The congruence is symmetric in the roles of 2 c−by0 x0 and y0, so without loss of generality, let x0 6≡ 0 mod p. Then a is congruent 2 mod p to a nonzero square x0, so by Hensel's lemma there exists x 2 Zp such that 2 2 c−by0 x = a and x ≡ x0 mod p. Now let y = y0. Then (x; y) is a solution to the equation ax2 + by2 = c. Chapter 3 The Hilbert Symbol 3.1 Definition and Basic Properties The completions of Q are Q2; Q3; Q5; :::; and R. To describe these with a uniform notation, let v be a place, either a prime or the symbol 1, and define Q1 = R. × Definition 3.1.1. For any a; b 2 Qv , the Hilbert symbol of a and b relative to Qv is defined as 8 > 2 2 2 3 <> 1; if ax + by = z has a solution in (x; y; z) 2 Qv − f(0; 0; 0)g; (a; b)v := > :>−1; otherwise. × 2 2 2 For a; b; c 2 Qv , we will often refer to ax + by = cz having a solution when we mean having a solution besides (0; 0; 0). Remark 3.1.2. Since we can multiply the equation ax2 + by2 = z2 by any nonzero square without changing the existence of a solution, if v is a finite place and there is a 6 7 2 2 2 solution to ax + by = z with x; y; z 2 Qp, then there is a solution with x; y; z 2 Zp × and x, y, or z in Zp . Example 3.1.3. We will evaluate (2; 3)3. This means that we are trying to find out 2 2 2 whether there is a solution to 2x +3y = z with x; y; z 2 Q3 besides (0; 0; 0). If such a solution does exist, we know that one exists with x; y; z 2 Z3 where at least one of them is a unit. With this knowledge, we can now reduce 2x2 + 3y2 = z2 mod 3 and obtain 2x2 ≡ z2 mod 3. Here, if x ≡ 0 mod 3 then we would have that z ≡ 0 mod 3. Then 3jx and 3jz, so 32jx2 and 32jz2. This means that 32j(z2 − 2x2). So 32j3y2, meaning 3jy2, which implies that 3jy. Now we have that x; y, and z are not units, which is a contradiction as we had at least one of them being a unit. Thus it must be that x 6≡ 0 mod 3.