1 Aendices We collect some results that might be covered in a first course in algebraic number theory. A. uadratic Recirocity Via Gauss Sums A1. Introduction In this aendix, is an odd rime unless otherwise secified. A quadratic equation modulo looks like ax + bx + c 0inF. Multilying by 4a, we have ( ax + b) b 4ac mod Thus in studying quadratic equations mod, it suffices to consider equations of the form x a mod. If a we have the uninteresting equation x 0, hence x 0, does not divide a. mod. Thus assume that A. Definition The Legendre symbol is given by χ(a) χ(a) ( ) a { 1 if a ( 1)/ 1 mod 1 if a ( 1)/ 1 mod. If b a ( 1)/ then b a 1 1 mod, sob ±1 mod and χ is well-defined. Thus A3. Theorem χ(a) a ( 1)/ mod. The Legendre symbol ( a ) is 1 if and only if a is a quadratic residue (from now on abbreviated R) mod. Proof. If a x mod then a ( 1)/ x 1 1 mod. (Note that if divides x then divides a, a contradiction.) Conversely, suose a ( 1)/ 1 mod. Ifg is a rimitive root mod, then a g r mod for some r. Therefore a ( 1)/ g r( 1)/ 1 mod, so 1 divides r( 1)/, hence r/ is an integer. But then (g r/ ) g r a mod, and a isarmod.
A4. Theorem The maing a χ(a) is a homomorhism from F to {±1}. Proof. We comute ( ) ab (ab) ( 1)/ a ( 1)/ b ( 1)/ so χ(ab) χ(a)χ(b). A5. Theorem ( a )( ) b If g is a rimitive root mod, then χ(g) 1, so g is a quadratic nonresidue (from now on abbreviated NR) mod. Consequently, exactly half of the integers 1,,..., 1 are R s and half are NR s. Proof. If h g mod then h is a rimitive root with twice the eriod of g, which is imossible. Thus by (A4), χ(ag) χ(a), so a ag gives a bijection between R s and NR s. A6. The First Sulementary Law From the definition (A) and the fact that ( 1)/ isevenif 1 mod 4 and odd if 3 mod 4, we have ( ) { 1 ( 1) ( 1)/ 1 if 1 mod 4 1 if 3 mod 4. A7. Definition Let K be a field of characteristic such that K contains the -th roots of unity. Let ζ K be a rimitive -th root of unity. Define the Gauss sum by A. Theorem 1 ( ) a τ ζ a. a1 τ ( 1) ( 1)/. Proof. From the definition of Gauss sum and (A4) we have τ 1 a,b1 ( ) ab ζ a+b.
3 For each a, we can sum over all c such that b ac mod. (As c ranges over 1,..., 1, ac also takes all values 1,..., 1.) Thus ( ) a Since 1, this simlifies to 1 1 τ 1 τ a1 c1 c1 ( 1 a1 ( a ) c ζ a+ac. )( ) c ζ a(1+c). If 1 + c 0 mod then 1,ζ 1+c,ζ (1+c),...,ζ ( 1)(1+c) runs through all the roots (zeros) of X 1 (note that ζ 1). But the coefficient of X 1 is 0, so the sum of the roots is 0. Therefore the sum of ζ 1+c,ζ (1+c),...,ζ ( 1)(1+c) is -1. If 1 + c 0 mod, then we are summing 1 ones. Consequently, ( ) c τ +( 1) c1 ( 1 (Note that 1 + c 0 mod c 1.) We can sum from 1 to 1ifweadd ( ) 1, hence 1 ( ) ( ) c 1 τ +. c1 The sum is 0 by (A5), and the result follows from (A6). A9. The Law of uadratic Recirocity Let and q be odd rimes, with q. Then ( )( ) q ( 1) ( 1)(q 1)/4. q Thus if either or q is congruent to 1 mod 4, then is a R mod q if and only if q is a R mod ; and if both and q are congruent to 3 mod 4, then isarmodq if and only if q is a NR mod. Proof. Let K have characteristic q and contain the -th roots of unity. For examle, take K to be the slitting field of X 1 over F q. Then τ q (τ ) (q 1)/ τ ). which by (A) is [ ( 1) ( 1)/ ] (q 1)/ τ.
4 Thus τ q ( 1) ( 1)(q 1)/4 (q 1)/ τ. But by the binomial exansion alied to the definition of τ in (A7), 1 ( ) a τ q ζ aq a1 (Recall that K has characteristic q, soa q a in K.) Let c aq mod and note that ( ) ( ) q 1 q because the roduct of the two terms is ( 1 ) 1. Thus 1 τ q c1 ( c )( ) q ζ c ( ) q τ. We now have two exressions (not involving summations) for τ q,so ( ) q ( 1) ( 1)(q 1)/4 (q 1)/. Since (q 1)/ ( q ) by (A), the above equation holds not only in K but in F q, hence can be written as a congruence mod q. Finally, multily both sides by ( q ) to comlete the roof. A10. The Second Sulementary Law so ( ) ( 1) ( 1)/ ( ) { 1 if ±1mod 1 if ±3mod. Thus if 1 or 7 mod, then is a R mod, and if 3or5mod,thenisa NR mod. Proof. Let K be a field of characteristic containing the th roots of unity, and let ζ be a rimitive th root of unity. Define τ ζ + ζ 1. Then τ ζ + ζ +. Now ζ and ζ 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. Modulo we have ( ) ( ) τ (τ ) ( 1)/ τ ( 1)/ τ τ (ζ + ζ 1 ).
5 But by definition of τ, τ (ζ + ζ 1 ) ζ + ζ. Now (again as in C) ζ + ζ and ζ + ζ 1 will coincide if ±1 mod, and will be negatives of each other if ±3mod. In other words, ζ + ζ ( 1) ( 1)/ (ζ + ζ 1 ). Equating the two exressions for τ, we get the desired result. We can justify the aeal to the comlex lane by requiring that K satisfy the constraints ζ + ζ 0 and ζ + ζ ( 1) ( 1)/ (ζ + ζ 1 ). A11. Examle We determine whether 113 is a R mod 17: ( 113 17 )(17 113 ) because 113 1 mod 4; ( 17 113 )(14 113 ) because ( a ) deends only on the residue class of a mod ; ( 14 113 )( 113 )( 7 113 )( 7 113 ) by (A4) and the fact that 113 1 mod ; ( 7 113 )(113 7 ) because 113 1 mod 4; ( 113 7 )(1 7 ) because 113 1 mod 7 and since ( 1 7 ) 1 by insection, 113 is a R mod 17. Exlicitly, 113 +13(17) 1764 4. A1. The Jacobi Symbol Let be an odd ositive integer with rime factorization q 1 q s. The Jacobi symbol is defined by ( ) a s ( a i1 q i ). It follows directly from the definition that ( )( ) ( ) a a a, ( )( ) ( ) a a aa. Also, a a mod ( ) a ( ) a because if a a mod then a a mod q i for all i 1,...,s. uadratic recirocity and the two sulementary laws can be extended to the Jacobi symbol, if we are careful.
6 A13. Theorem If is an odd ositive integer then ( ) 1 ( 1) ( 1)/. Proof. We comute ( ) 1 s ( ) 1 j1 q j Now if a and b are odd then ( ab 1 a 1 hence a 1 s j1 ( 1) (qj 1)/ ( 1) s j1 (qj 1)/. + b 1 ) (a 1)(b 1) 0 mod, + b 1 ab 1 mod. We can aly this result reeatedly (starting with a q 1,b q ) to get the desired formula. A14. Theorem If is an odd ositive integer, then ( ) ( 1) ( 1)/. Proof. As in (A13), ( ) s ( ) j1 q j s ( 1) (q j 1)/. j1 But if a and b are odd then a b ( 1 a 1 In fact a 1 Thus (a 1)(a+1) ) + b 1 (a 1)(b 1) 0 mod. is an integer and b 1 0 mod. (Just lug in b 1, 3, 5, 7.) a 1 + b 1 a b 1 mod and we can aly this reeatedly to get the desired result.
7 A15. Theorem If P and are odd, relatively rime ositive integers, then ( )( ) P ( 1) (P 1)( 1)/4. P Proof. Let the rime factorizations of P and be P r i1 i and s j1 q j. Then ( )( ) P ( )( ) i qj ( 1) i,j (i 1)(qj 1)/4. P q i,j j i But as in (A13), r ( i 1)/ (P 1)/ mod, i1 s (q j 1)/ ( 1)/ mod. j1 Therefore ( )( ) P [(P 1)/][( 1)/] ( 1) P as desired. A16. Remarks Not every roerty of the Legendre symbol extends to the Jacobi symbol. for examle, ( 15 )( 3 )( 5 )( 1)( 1) 1, but is a NR mod 15.
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 comact (hence comlete) in the toology induced by, then there is exactly one extension of to an absolute value on L, namely a N L/K (a) 1/n. We will need to do some reliminary work. B. Lemma Suose we are trying to rove 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>0such that for all a L, a 1 1+a C. Then satisfies the triangle inequality ( a + b a + b ). Proof. If a 1 a, then a a /a 1 satisfies a 1. We can take C without loss of generality (because we can relace C by C c ). Thus a 1 + a a 1 max{ a 1, a } so by induction, a 1 + + a r r max a j. If n is any ositive integer, choose r so that r 1 n r. Then a 1 + + a n r max a j n max a j. (Note that r 1 n r n. Also, we can essentially regard n as r by introducing zeros.) Now ( ) { ( ) } a + b n n n a j b n j (n +1) max a j b n j. j j j j0 But m 1+1+ +1 m for m 1, so a + b n 4(n +1) max j {( } n ) a j b n j. j The exression in braces is a single term in a binomial exansion, 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 aroaches ( a + b ) as n (take logarithms), and the result follows.
9 B3. Uniqueness Since L is a finite-dimensional vector sace over K, any two extensions to L are equivalent as norms, and therefore induce the same toology. Thus (see Section 9.1, Problem 3) for some c>0wehave 1 ( ) c. But a 1 a for every a K, soc must be 1. B4. Proof of Theorem B1 By (B), it suffices to find C>0 such that a 1 1+a C. Let b 1,...,b n be a basis for L over K. Ifa n i1 c ib i, then the max norm on L is defined by a 0 max 1 i n c i. The toology induced by 0 is the roduct toology determined by n coies of K. With resect to this toology, N L/K is continuous (it is a olynomial). Thus a a is a comosition of continuous functions, hence continuous. Consequently, is a nonzero continuous function on the comact set S {a L : a 0 1}. So there exist δ, > 0 such that for all a S we have o<δ a. If 0 a L we can find c K such that a 0 c. (We have a n i1 c ib i, and if c i is the maximum of the c j, 1 j n, take c c i.) Then a/c 0 1,soa/c S and Now because a/c 0 a 0 / c. Therefore 0 <δ a/c a/c a/c 0. a/c a/c 0 a a 0 0 <δ a a 0. ( Now suose a 1, so a 0 a /δ δ 1.Thus where ste (1) follows because 0 is a norm. 1+a 1+a 0 [ 1 0 + a 0 ] (1) [ 1 0 + δ 1 ]C
10 C. The Different C1. Definition Let O K be the ring of algebraic integers in the number field K. Let ω 1,...,ω n be an integral basis for O K, so that the field discriminant d K is det T (ω i ω j )), where T stands for trace. Define C. Theorem D 1 {x K : T (xo K ) Z}. D 1 is a fractional ideal with Z-basis ω1,...,ωn, the dual basis of ω 1,...,ω n referred to the vector sace K over. [The dual basis is determined by T (ω i ωj )δ ij, see (..9).] Proof. In view of (3..5), if we can show that D 1 is an O K -module, it will follow that D 1 is a fractional ideal. We have x O K,y D 1 T (xyo K ) T (yo K ) Z so xy D 1. By (..), the trace of ωi ω j is an integer for all j. ThusZω1 + + Zωn D 1.We must rove the reverse inclusion. Let x D 1,sox n i1 a iωi,a i. Then T (xω j )T ( a i ωi ω j )a j. i1 But x D 1 imlies that T (xω j ) Z, soa j Z and C3. Remarks D 1 Zωi. i1 Since K is the fraction field of O K, for each i there exists a i O K such that a i ωi O K. By (..), we can take each a i to be an integer. If m n i1 a i, then md 1 O K, which gives another roof 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 is called the co-different. In fact, D is an integral ideal of O K. We have 1 D 1 by definition of D 1 and (..). Thus D D1 DD 1 O K. The different can be defined in the general AKLB setu if A is integrally closed, so (..) alies.
11 C5. Theorem The norm of D is N(D) d K. Proof. Let m be a ositive integer such that md 1 O K [(see (C3)]. We have mωi ω i a ij ω j j1 b ij ωj j1 so there is a matrix equation (b ij )(a ij /m) 1.Now so T (ω i ω j ) b ik T (ωkω j ) b ik δ kj b ij k1 k1 det(b ij )d K. By (C), md 1 is an ideal of O K with Z-basis mωi,i1...,n, so by (4..5), d K/ (mω 1,...,mω n)n(md 1 ) d K and by (.3.), the left side of this equation is (det a ij ) d K.Thus det(a ij ) N(mD 1 )m n N(D 1 ) where the last ste follows because B/I B/mI / I/mI. Now DD 1 O K imlies that N(D 1 )N(D) 1,so d K det(b ij ) det(a ij /m) 1 [N(D 1 )] 1 N(D). C6. Some Comutations We calculate the different of K ( ). By (C5), N(D K ) d K 4. Now O K Z[ ] is a rincial ideal domain (in fact a Euclidean domain) so D (a+b ) for some a, b Z. Taking norms, we have a +b, and the only integer solution is a 0,b ±. Thus D ( ) ( ). We calculate the different of K ( 3). By (C5), N(D K ) d K 3. Now O K Z[ω], ω 1 + 1 3 and since O K is a PID, D (a + bω) for some a, b Z. Taking norms, we get ( 3 a ) b ( ) b +3, (a b) +3b 1.
1 There are 6 integer solutions: +ω, 1+ω, ω, 1 ω, 1+ω, 1 ω but all of these elements are associates, so they generate the same rincial ideal. Thus D (+ω). It can be shown that a rime ramifies in the number field K if and only if divides the different of K.