1

Appendices

We collect some results that might be covered in a first course in algebraic number theory.

A. Quadratic Reciprocity Via Gauss Sums A1. Introduction In this appendix, p is an odd prime unless otherwise specified. A quadratic equation 2 modulo p looks like ax + bx + c =0inFp. Multiplying by 4a, we have 2 2ax + b ≡ b2 − 4ac mod p

Thus in studying quadratic equations mod p, it suffices to consider equations of the form

x2 ≡ a mod p.

If p|a we have the uninteresting equation x2 ≡ 0, hence x ≡ 0, mod p. Thus assume that p does not divide a.

A2. Definition The Legendre symbol a χ(a)= p is given by 1ifa(p−1)/2 ≡ 1modp χ(a)= −1ifa(p−1)/2 ≡−1modp.

If b = a(p−1)/2 then b2 = ap−1 ≡ 1modp,sob ≡±1modp and χ is well-defined. Thus

χ(a) ≡ a(p−1)/2 mod p.

A3. Theorem a The Legendre symbol ( p ) is 1 if and only if a is a quadratic residue (from now on abbre- viated QR) mod p. Proof.Ifa ≡ x2 mod p then a(p−1)/2 ≡ xp−1 ≡ 1modp. (Note that if p divides x then p divides a, a contradiction.) Conversely, suppose a(p−1)/2 ≡ 1modp.Ifg is a primitive root mod p, then a ≡ gr mod p for some r. Therefore a(p−1)/2 ≡ gr(p−1)/2 ≡ 1modp, so p − 1 divides r(p − 1)/2, hence r/2 is an integer. But then (gr/2)2 = gr ≡ a mod p, and a isaQRmodp. ♣ 2

A4. Theorem

The mapping a → χ(a) is a homomorphism from Fp to {±1}. Proof. We compute ab a b ≡ (ab)(p−1)/2 = a(p−1)/2b(p−1)/2 ≡ p p p so χ(ab)=χ(a)χ(b). ♣

A5. Theorem If g is a primitive root mod p, then χ(g)=−1, so g is a quadratic nonresidue (from now on abbreviated QNR) mod p. Consequently, exactly half of the integers 1,2,...,p − 1 are QR’s and half are QNR’s. Proof.Ifh2 ≡ g mod p then h is a primitive root with twice the period of g, which is impossible. Thus by (A4), χ(ag)=−χ(a), so a → ag gives a bijection between QR’s and QNR’s. ♣

A6. The First Supplementary Law From the definition (A2) and the fact that (p − 1)/2isevenifp ≡ 1 mod 4 and odd if p ≡ 3 mod 4, we have −1 1ifp ≡ 1mod4 =(−1)(p−1)/2 = p −1ifp ≡ 3mod4.

A7. Definition Let K be a field of characteristic = p such that K contains the p-th roots of unity. Let ζ ∈ K be a primitive p-th root of unity. Define the Gauss sum by p−1 a τ = ζa. p p a=1

A8. Theorem

2 − (p−1)/2 τp =( 1) p.

Proof. From the definition of Gauss sum and (A4) we have p−1 ab τ 2 = ζa+b. p p a,b=1 3

For each a, we can sum over all c such that b ≡ ac mod p. (As c ranges over 1,... ,p− 1, ac also takes all values 1,... ,p− 1.) Thus p−1 p−1 a2c τ 2 = ζa+ac. p p a=1 c=1 a2 Since p = 1, this simplifies to p−1 p−1 c τ 2 = ζa(1+c) . p p c=1 a=1

If 1 + c ≡ 0modp then 1,ζ1+c,ζ2(1+c),... ,ζ(p−1)(1+c) runs through all the roots (zeros) of Xp − 1 (note that ζp = 1). But the coefficient of Xp−1 is 0, so the sum of the roots is 0. Therefore the sum of ζ1+c,ζ2(1+c),... ,ζ(p−1)(1+c) is -1. If 1 + c ≡ 0modp, then we are summing p − 1 ones. Consequently, p−2 c −1 τ 2 = − +(p − 1) . p p p c=1 ≡ ↔ − − −1 (Note that 1 + c 0modp c = p 1.) We can sum from 1 to p 1ifweadd p , hence p−1 c −1 τ 2 = − + p . p p p c=1 The sum is 0 by (A5), and the result follows from (A6). ♣

A9. The Law of Quadratic Reciprocity Let p and q be odd primes, with p = q. Then p q =(−1)(p−1)(q−1)/4. q p Thus if either p or q is congruent to 1 mod 4, then p is a QR mod q if and only if q is a QR mod p; and if both p and q are congruent to 3 mod 4, then p isaQRmodq if and only if q is a QNR mod p. Proof. Let K have characteristic q and contain the p-th roots of unity. For example, take p K to be the splitting field of X − 1 over Fq. Then

q 2 (q−1)/2 τp =(τp ) τp which by (A8) is (q−1)/2 (p−1)/2 (−1) p τp. 4

Thus q − (p−1)(q−1)/4 (q−1)/2 τp =( 1) p τp.

But by the binomial expansion applied to the definition of τp in (A7), p−1 a τ q = ζaq p p a=1 (Recall that K has characteristic q,soaq = a in K.) Let c ≡ aq mod p and note that q−1 q = p p

1 because the product of the two terms is ( p ) = 1. Thus p−1 c q q τ q = ζc = τ . p p p p p c=1

q We now have two expressions (not involving summations) for τp ,so q (−1)(p−1)(q−1)/4 p(q−1)/2 = . p

(q−1)/2 p F Since p =(q ) by (A2), the above equation holds not only in K but in q, hence p can be written as a congruence mod q. Finally, multiply both sides by ( q ) to complete the proof. ♣

A10. The Second Supplementary Law 2 2 =(−1)(p −1)/8 p so 2 1ifp ≡±1mod8 = p −1ifp ≡±3mod8.

Thus if p ≡ 1 or 7 mod 8, then 2 is a QR mod p, and if p ≡ 3or5mod8,then2isa QNR mod p. Proof. Let K be a field of characteristic p containing the 8th roots of unity, and let ζ be a primitive 8th root of unity. Define τ = ζ + ζ−1. Then τ 2 = ζ2 + ζ−2 +2. Now ζ2 and ζ−2 are distinct 4th roots of unity, not ±1, so they must be negatives of each other (analogous to i and −i in C). Therefore τ 2 = 2. Modulo p we have 2 2 τ p =(τ 2)(p−1)/2 τ =2(p−1)/2 τ = τ = (ζ + ζ−1). p p 5

But by definition of τ, τ p =(ζ + ζ−1)p = ζp + ζ−p. Now (again as in C) ζp + ζ−p and ζ + ζ−1 will coincide if p ≡±1 mod 8, and will be negatives of each other if p ≡±3mod 8. In other words,

2 ζp + ζ−p =(−1)(p −1)/8(ζ + ζ−1).

Equating the two expressions for τ p, we get the desired result. We can justify the appeal to the complex plane by requiring that K satisfy the constraints ζ2 + ζ−2 = 0 and 2 ζp + ζ−p =(−1)(p −1)/8(ζ + ζ−1). ♣

A11. Example We determine whether 113 is a QR mod 127: 113 127 ≡ ( 127 )=(113 ) because 113 1mod4; 127 14 a ( 113 )=(113 ) because ( p ) depends only on the residue class of a mod p; 14 2 7 7 ≡ ( 113 )=(113 )( 113 )=(113 ) by (A4) and the fact that 113 1mod8; 7 113 ≡ ( 113 )=( 7 ) because 113 1mod4; 113 1 ≡ ( 7 )=(7 ) because 113 1mod7 1 and since ( 7 ) = 1 by inspection, 113 is a QR mod 127. Explicitly, 113 +13(127) = 1764 = 42 2.

A12. The Jacobi Symbol

Let Q be an odd positive integer with prime factorization Q = q1 ···qs. The Jacobi symbol is defined by s a a = . Q q i=1 i

It follows directly from the definition that a a a a a aa = , = . Q Q QQ Q Q Q

Also, a a a ≡ a mod Q ⇒ = Q Q

  because if a ≡ a mod Q then a ≡ a mod qi for all i =1,... ,s. Quadratic reciprocity and the two supplementary laws can be extended to the Jacobi symbol, if we are careful. 6

A13. Theorem If Q is an odd positive integer then −1 =(−1)(Q−1)/2. Q

Proof. We compute − s − s 1 1 (q −1)/2 s (q −1)/2 = = (−1) j =(−1) j=1 j . Q q j=1 j j=1

Now if a and b are odd then ab − 1 a − 1 b − 1 (a − 1)(b − 1) − + = ≡ 0mod2, 2 2 2 2 hence a − 1 b − 1 ab − 1 + ≡ mod 2. 2 2 2

We can apply this result repeatedly (starting with a = q1,b = q2) to get the desired formula. ♣

A14. Theorem If Q is an odd positive integer, then 2 2 =(−1)(Q −1)/8. Q

Proof. As in (A13), s s 2 2 (q2−1)/8 = = (−1) j . Q q j=1 j j=1

But if a and b are odd then a2b2 − 1 a2 − 1 b2 − 1 (a2 − 1)(b2 − 1) − + = ≡ 0mod8. 8 8 8 8

a2−1 (a−1)(a+1) 2 − ≡ In fact 8 = 8 is an integer and b 1 0 mod 8. (Just plug in b =1, 3, 5, 7.) Thus a2 − 1 b2 − 1 a2b2 − 1 + ≡ mod 8 8 8 8 and we can apply this repeatedly to get the desired result. ♣ 7

A15. Theorem If P and Q are odd, relatively prime positive integers, then P Q =(−1)(P −1)(Q−1)/4. Q P r s Proof. Let the prime factorizations of P and Q be P = i=1 pi and Q = j=1 qj. Then P Q pi qj (p −1)(q −1)/4 = =(−1) i,j i j . Q P q p i,j j i

But as in (A13),

r s (pi − 1)/2 ≡ (P − 1)/2mod2, (qj − 1)/2 ≡ (Q − 1)/2mod2. i=1 j=1

Therefore P Q =(−1)[(P −1)/2][(Q−1)/2] Q P as desired. ♣

A16. Remarks Not every property of the Legendre symbol extends to the Jacobi symbol. for example, 2 2 2 − − ( 15 )=(3 )( 5 )=( 1)( 1) = 1, but 2 is a QNR mod 15. 8

B. Extension of Absolute Values B1. Theorem Let L/K be a finite extension of fields, with n =[L : K]. If ||is an absolute value on K and K is locally compact (hence complete) in the topology induced by ||, then there is exactly one extension of ||to an absolute value on L, namely 1/n |a| = |NL/K (a)| . We will need to do some preliminary work.

B2. Lemma Suppose we are trying to prove that ||is an absolute value on L. Assume that ||satisfies the first two requirements in the definition of absolute value in (9.1.1). If we find a real number C>0 such that for all a ∈ L, |a|≤1 ⇒|1+a|≤C. Then ||satisfies the triangle inequality (|a + b|≤|a| + |b|).

Proof.If|a1|≥|a2|, then a = a2/a1 satisfies |a|≤1. We can take C = 2 without loss of generality (because we can replace C by Cc = 2). Thus

|a1 + a2|≤2a1 = 2 max{|a1|, |a2|} so by induction, r |a1 + ···+ a2r |≤2 max |aj|. If n is any positive integer, choose r so that 2r−1 ≤ n ≤ 2r. Then r |a1 + ···+ an|≤2 max |aj|≤2n max |aj|. (Note that 2r−1 ≤ n ⇒ 2r ≤ 2n. Also, we can essentially regard n as 2r by introducing zeros.) Now n n − n − |a + b|n = ajbn j ≤ 2(n +1) max |a|j|b|n j . j j j j=0 But |m| = |1+1+···+1|≤2m for m ≥ 1, so n |a + b|n ≤ 4(n +1) max |a|j|b|n−j . j j The expression in braces is a single term in a binomial expansion, hence |a + b|n ≤ 4(n +1)( |a| + |b|)n. Taking n-th roots, we have |a + b|≤[4(n +1)] 1/n(|a| + |b|). The right hand side approaches (|a| + |b|)asn →∞(take logarithms), and the result follows. ♣ 9

B3. Uniqueness

Since L is a finite-dimensional vector space over K, any two extensions to L are equivalent as norms, and therefore induce the same topology. Thus (see Section 9.1, Problem 3) for c some c>0wehave||1 =(||2) . But |a|1 = |a|2 for every a ∈ K,soc must be 1.

B4. Proof of Theorem B1 | |≤ ⇒| |≤ By (B2), it suffices to findC>0 such that a 1 1+a C. Let b1,... ,bn be a n basis for L over K.Ifa = i=1 cibi, then the max norm on L is defined by

|a|0 = max |ci|. 1≤i≤n

The topology induced by ||0 is the product topology determined by n copies of K. With respect to this topology, NL/K is continuous (it is a polynomial). Thus a →|a| is a composition of continuous functions, hence continuous. Consequently, ||is a nonzero continuous function on the compact set S = {a ∈ L : |a|0 =1}. So there exist δ, ∆ > 0 such that for all a ∈ S we have

o<δ≤|a|≤∆. ∈ ∈ | | | | n | | If 0 = a L we can find c K such that a 0 = c . (We have a = i=1 cibi, and if ci is the maximum of the |cj|, 1 ≤ j ≤ n, take c = ci.) Then |a/c|0 =1,soa/c ∈ S and

|a/c| 0 <δ≤|a/c| = ≤ ∆. |a/c|0

Now

|a/c| |a| = |a/c|0 |a|0 because |a/c|0 = |a|0/|c|. Therefore

|a| 0 <δ≤ ≤ ∆. |a|0

−1 ( Now suppose |a|≤1, so |a|0 ≤|a|/δ ≤ δ .Thus

|1+a|≤∆|1+a|0

≤ ∆[|1|0 + |a|0] (1) −1 ≤ ∆[|1|0 + δ ]=C where step (1) follows because ||0 is a norm. ♣ 10

C. The Different C1. Definition

Let OK be the ring of algebraic integers in the number field K. Let ω1,... ,ωn be an integral basis for OK , so that the field discriminant dK is det T (ωiωj)), where T stands for trace. Define

−1 D = {x ∈ K : T (xOK ) ⊆ Z}.

C2. Theorem D−1 Z ∗ ∗ is a fractional ideal with -basis ω1 ,... ,ωn, the dual basis of ω1,... ,ωn referred to Q ∗ the vector space K over . [The dual basis is determined by T (ωiωj )=δij, see (2.2.9).] −1 Proof. In view of (3.2.5), if we can show that D is an OK -module, it will follow that D−1 is a fractional ideal. We have

−1 x ∈OK ,y ∈D ⇒ T (xyOK ) ⊆ T (yOK ) ⊆ Z so xy ∈D−1. ∗ Z ∗ ··· Z ∗ ⊆D−1 By (2.2.2), the trace of ωi ωj is an integer for all j.Thus ω1 + + ωn .We ∈D−1 n ∗ ∈ Q must prove the reverse inclusion. Let x ,sox = i=1 aiωi ,ai . Then n ∗ T (xωj)=T ( aiωi ωj)=aj. i=1

−1 But x ∈D implies that T (xωj) ∈ Z,soaj ∈ Z and

n D−1 Z ∗ ♣ = ωi . i=1

C3. Remarks O ∈O ∗ ∈O Since K is the fraction field of K , for each i there exists ai K such that aiωi K . n D−1 ⊆O By (2.2.8), we can take each ai to be an integer. If m = i=1 ai, then m K , which gives another proof that D−1 is a fractional ideal.

C4. Definition and Discussion The different of K, denoted by D, is the fractional ideal that is inverse to D−1; D−1 −1 is called the co-different. In fact, D is an integral ideal of OK . We have 1 ∈D by definition of D−1 and (2.2.2). Thus

−1 D = D1 ⊆DD = OK .

The different can be defined in the general AKLB setup if A is integrally closed, so (2.2.2) applies. 11

C5. Theorem

The norm of D is N(D)=|dK |. −1 Proof. Let m be a positive integer such that mD ⊆OK [(see (C3)]. We have n ∗ mωi = aijωj j=1 n ∗ ωi = bijωj j=1

−1 so there is a matrix equation (bij)=(aij/m) .Now n n ∗ T (ωiωj)= bikT (ωkωj)= bikδkj = bij k=1 k=1 so

det(bij)=dK . D−1 O Z ∗ By (C2), m is an ideal of K with -basis mωi ,i=1... ,n, so by (4.2.5), ∗ ∗ D−1 2 dK/Q(mω1 ,... ,mωn)=N(m ) dK

2 and by (2.3.2), the left side of this equation is (det aij) dK .Thus

−1 n −1 | det(aij)| = N(mD )=m N(D )

−1 where the last step follows because |B/I| = |B/mI|/|I/mI|.NowDD = OK implies that N(D−1)=N(D)−1,so

−1 −1 −1 |dK | = | det(bij)| = | det(aij/m)| =[N(D )] = N(D). ♣

C6. Some Computations √ Q − D | | × We calculate√ the different of K = ( 2). By (C5), N( K )= dK =4 2=8.Now√ OK = Z[ −2] is a principal ideal domain (in fact a Euclidean domain) so D =(a+b −2) for some a, b ∈ Z. Taking norms, we have 8 = a2 +2b2, and the only integer solution is a =0,b= ±2. Thus √ √ D =(2 −2) = (−2 −2). √ We calculate the different of K = Q( −3). By (C5), N(DK )=|dK | =3.Now 1 1√ O = Z[ω],ω= − + −3 K 2 2 and since OK is a PID, D =(a + bω) for some a, b ∈ Z. Taking norms, we get b 2 b 2 3= a − +3 , (2a − b)2 +3b2 =12. 2 2 12

There are 6 integer solutions:

2+ω, −1+ω, −2 − ω, 1 − ω, 1+2ω, −1 − 2ω but all of these elements are associates, so they generate the same principal ideal. Thus D =(2+ω). It can be shown that a prime p ramifies in the number field K if and only if p divides the different of K.