2.1 Reciprocity Laws and the Dirichlet’s Theorem
Let p be an odd prime. It is known that the multiplicative group G = (Z=(p)) of Z=(p) is cyclic, and there is a natural epimorphism from G onto f 1g with kernel the squares G2 in G. This map is denoted by (p) and for a 2 Z with gcd(a; p) = 1 the image of a 2 G under (p) is called the Legendre symbol of a (with respect to p). Note that we only de…ne the values on the integers relatively prime to p and what the Legendre symbol describes is the sovablility of the equation X2 a (mod p). More explicitly, this equation is solvable if and only if a lies in G2 if and only if (ap) = 1. Now we list the well-known result about the computation of the Legendre symbol.
Theorem (Gauss). Let p be an odd prime. Then 1. ( p1) = ( 1)p21,
2. (2p) = ( 1)p28 1, and
3. (qp)(pq) = ( 1)p21q21 for any odd prime q 6= p.
Proof. See Page 48 in [4].
The third equation stated in the theorem is usually called the Quadratic Reciprocity Law.
We give an interesting consequence of this theorem: there are in…nitely many primes of the form 4n + 1. We prove this by contradiction. Suppose that there is only a …nite number of primes of the form 4n + 1, say p1, p2, ..., pk. Thus the integer N = (2p1p2 pk)2+ 1 is never a prime and has no prime divisors of the form 4n + 1 (otherwise pj divides 1 for some j). Therefore N is divisible by a prime p of the form 4n + 3, that is, (2p1p2 pk)2 1 (mod p)and X2 1 (mod p)is solvable, which contradicts to the fact that p 3 (mod 4).
Remarkably, there is a more general phenomenon.
Theorem (Dirichlet’s Theorem). Let a and m be relatively prime integers. Then there are in…nitely many primes p with p a (mod m).
Proof. See Page 138 in [4].
One advantage of this nontrivial theorem is that we can always select a prime of particular type that avoids a given …nite set of prime numbers.
2.2 Facts on Algebraic Number Theory
Now we recall some facts in algebraic number theory. Let k be any …eld. A map j j de…ned on k into the real numbers is called a valuation if for each a; b 2 k, it satis…es the following conditions:
1. jaj > 0 and jaj = 0 if and only if a = 0.
2. jabj = jajjbj.
3. ja + bj 6 jaj + jbj.
We call a valuation j j nonarchimedean if it satis…es the stronger assumption: ja + bj 6 maxfjaj; jbjg for each a; b 2 k; and archimedean otherwise. Notice that, in particular, a valuation is a metric. Two valuations are equivalent if they induce the same (metric) topology and inequivalent otherwise. This de…nes an equivalence relation on the set of all valuations on k, and an equivalence class is called a place of k. If v is a place of k, we select any valuation in v and denotes the completion of k with respect to this valuation by kv, which is independent of the choice of the valuation since all valuations in the same place induce the same topology. We also call kv the completion of k at the place v.
If j j is a nonarchimedean valuation on k. Then A = f 2 k j j j 6 1g is a local ring with M = f 2 k j j j < 1g as its maximal ideal, U = f 2 k j j j = 1g as its unit group and k as its …eld of fractions. We call A the valuation ring of k.
We now come to a concrete example. Let A be a Dedekind domain and k its …eld of fractions. Fix a nonzero prime ideal P of A, we have seen that the localization of A at P is a discrete valuation ring (DVR), i.e., a local PID that is not a …eld, with maximal ideal PAP. If 2 A is a nonzero element, we let vP(a) be the power of PAP occurred in the factorization of the principal ideal (a) of AP; this map can be extended to k by setting vP( ) = vP( 1) if 2 k but 2 A= P. Hence j jP = cvP( ) (where c < 1 is any positive number) de…nes a nonarchimedean valuation on k. Let kP be the completion of k with respect to j jP, then j jP can be extended to a valuation on kPwhich is still nonarchimedean, we also denote this extended map by j jP. The valuation ring cAP of kP is still a DVR j jq are inequivalent. Ostrowski’s theorem states that every valuation on Q is equivalent to j jp for some prime p or j j1 (the usual absolute value). Hence the set of places of Q is just the set of all prime numbers together with the symbol 1 (which denotes the place containing j j1). Let p be a prime. If A = Fp[t] and k = Fp(t). Let P = (P ) where P is a monic irreducible polynomial, we denote vP( ) by vP( ) and j jP = (1=pdeg(P ))vP( ) by j jP. There is an additional (nonarchimedean) valuation: j j1 =j jP = (1=p)vP( ) where P = (t 1) Fp[t 1], we denote the uniformizer t 1 by . Similar consequence as for Q, every valuation on Fp(t) is equivalent to j jP for some monic irreducible polynomial P or j j1. Hence the set of places of Fp(t) is the set of monic irreducible polynomail P with the symbol 1.
2.3 Approximation Theorem and Hensel’s Lemma
We now present a useful tool that helps us to esimate the distance with respect to a …nite number of pairwise inequivalent valuations simultaneously.
Theorem (Approximation Theorem). Let j j1, ..., j jn be nontrivial pairwise inequivalent valuations on k and 1,..., n 2 k. Then for any " > 0, there is an element 2 k such that j jjj < " for each j.
Proof. See Page 109 in [4].
Let k be a complete …eld with respect to a nonarchimedean valuation with its valuation ring A a DVR. Let M be the unique maximal ideal of A. The following lemma gives
su¢ cient conditions under which we may determine a factorization of a polynomial f (X) 2 A[X] from a factorization of f (X) 2 (A=M)[X].
Lemma 2.1 (Hensel’s lemma). If f (X) is monic and f (X) = G(X)H(X) in (A=M )[X]
where G(X) and H(X) are monic and relatively prime in (A=M )[X]. Then f (X) = g(X)h(X) in A[X] for some monic g(X); h(X)2 A[X] with g(X) = G(X), h(X) = H(X) in (A=M )[X] and deg(g) = deg(G), deg(h) = deg(H).
Proof. See Page 38 in [4].
We can use Hensel’s lemma to characterize the nonzero squares in Qpwhere p is a prime.
First we let p be an odd prime. From the canonical epimorphism Zp Zp=pZp, the units in Zpare mapped into the unit group in Zp=pZpand (Zp=pZp) ' (Z=(p)) ' Fp. Hence this induces an injection i : Up=Up2 ,! Fp=Fp2. If 2 Up with i( ) is a square in Fp, then this implies that f (X) = X2 2 Zp[X]has a root in Zp=pZp. Since f0(X) = 2X and p is odd, we …nd that f has no multiple roots in Zp=pZp. Now apply Hensel’s lemma, f (X) has a root in Zp, in other words, is a square in Up, proving surjectivity of i. Since Up=Up2 ' Fp=Fp2
and Fp is a cyclic group, we have Up=Up2 ' Z=(2). Since every element 2 Qp is uniquely expressed as pnu where n = vp( ) 2 Z and u 2 Up, we see that Qp ' Z Up so that Qp=Qp2 ' Z=(2) Up=Up2 ' Z=(2) Z=(2). This also tells us that is a square in Qp if and only if vp( ) is even and u 1 mod pZp (or u 1 (mod p)).
Next we let p = 2. We claim that U2=U22 ' (Z=(8)) . If is a square in U2, then
= (1 + 2 )2 where 2 Z2 and 2+ 2 2Z2 so = 1 + 4( 2 + ) 2 1 + 8Z2. From U2 ,! Z2 Z2=8Z2 we have U2=U22 ,! (Z2=8Z2) . If 2 U2 with 2 1 + 8Z2. Write
= 1 + 8 where 2 Z2. In order to be a square in U2, we have to …nd 2 Z2 such that (1 + 2 )2 = 1 + 8 or 2+ = 2 . Consider this time f (X) = X2+ X 2 2 Z2[X], f (X)always has a root modulo 2Z2 since 0 + 0 1 + 1 0 (mod 2), and f0(X) = 1modulo 2Z2, can indeed be selected by Hensel’s lemma. Thus U2=U22 ' (Z2=8Z2) ' (Z=(8)) ' Z=(2) Z=(2). The same argument as above, Q2=Q22 ' Z=(2) Z=(2) Z=(2). And
= 2nu 2 Q2 is a square in Q2 if and only if v2( ) is even and u 1 mod 8Z2 (or u 1 (mod 8)). We summarize this.
Proposition 2.1. Qp=Qp2 ' Z=(2) Z=(2) for p 6= 2 and Q2=Q22 ' Z=(2) Z=(2) Z=(2).
Remark 2.1. Let ; 2 Q . If j j is any valuation on Q and the following condition holds:
j j < " for su¢ ciently small ".
Then j = 1j < "=j j = "0 for "0 small enough. If j j = j j1, then this condition implies that = is positive, in other words, = is a square in R . If j j = j jp for some odd prime p, then this condition implies that = 1 2 pZp or = 2 1 + pZp. Therefore = is a square in Qp. If j j = j j2, then this condition implies that = 12 8Z2 or = 2 1+8Z2, which shows that = is a square in Q2 as well. We conclude that: if v is a place of Q, then j jv < " for su¢ ciently small " implies that = is a square in Qv.