12. Quadratic Reciprocity in Number Fields

Transcription

1 1. Quadratic Reciprocity in Number Fields In [RL1, p. 73/74] we briefly mentioned how Eisenstein arrived at a hypothetical reciprocity law by assuming what later turned out to be an immediate consequence of the product formula for the Hilbert symbol, which in turn follows directly from Artin s reciprocity law. In this and the next chapter we will explain his beautiful idea in detail. 1.1 Quadratic Reciprocity in Z Eisenstein [] realized that the reciprocity laws for quadratic and cubic residues can be formulated in a way that allows the immediate generalization to arbitrary powers and number fields: his observation was that the inversion factor i(α, β ( α β ( β α 1 for p-th power residue symbols in Q(ζ p and odd primes p only depends on α, β mod (1 ζ m for some integer m. In this section we show that Eisenstein s version for p over Q is equivalent to the quadratic reciprocity law; then we will prove its generalization for quadratic number fields. In the next chapter we will then show how to extract the reciprocity law for p-th power residues from the above conjecture. For coprime odd integers a, b Z, define a symbol [a, b] with values in {0, 1} by ( a b ( b a ( 1[a,b]. Let us make the following assumption: Hypothesis (H. There exists an integer σ N such that [a, b] [a, b ] for all natural numbers a, b, a, b N with (a, b (a, b 1, and a a, b b mod σ. We could also allow integers a, b, a, b Z, but then we would have to add the condition that aa > 0 and bb > 0. In that case, (H may also be assumed to hold for b 1; note that ( a 1 ( a 1 1, since by definition of the Jacobi symbol ( a 1 p 1 ( a p, and this product is empty. In order to better understand what Hypothesis (H is all about we first derive the following result (which will not be used in the sequel: Proposition 1.1. If Hypothesis (H holds, then ( a ( a b b + σ a for all coprime numbers a, b N.

2 1. Quadratic Reciprocity in Number Fields In other words: Hypothesis (H states that the Legendre symbol ( a, as a function of its denominator, has conductor dividing σ a. This is a weak form of the reciprocity law as given by Euler: compare [RL1,Thm..8.iv]. Proof. Observe that ( a ( b ( a ( b + b a σ a ( a b b + σ a a b + a( σ a by Hypothesis (H and reduction modulo a. Let us now sketch Eisenstein s idea for deriving the quadratic reciprocity law from (H. The basic ingredient is a special case of Hensel s Lemma, i.e., the observation that integers a 1 mod 8 are squares modulo σ : Lemma 1.. If a Z is an integer with a 1 mod 8, then for every σ N there is an integer µ σ such that a µ σ mod σ. Proof. Assume that we know a µ e mod e for some e 3; then a µ e+ e b for some integer b. But then µ e+1 µ e + e 1 b will do it, since we have a (µ e + e 1 b mod e+1 in light of e 0 mod e+1 for e 3. This simple observation enables us to prove Proposition 1.3. If (H holds for some integer σ, then it holds for σ 3. Proof. Given an odd integer a, there is an integer r {1, 3, 5, 7} such that a r mod 8; this implies a/r 1 mod 8, so by Lemma 1., there is an integer m such that a rm mod σ. Similarly, we have b s mod 8 for some s {1, 3, 5, 7}, and thus b sn mod σ for some integer n. We have to show that if (a, b (r, s 1, then [a, b] [r, s]. Since we may choose m and n in such a way that (rm, sn 1, hypothesis (H then implies [a, b] [rm, sn ] [r, s]. Thus the value of [a, b] only depends on a, b mod 8. Corollary 1.4. We have [a, b] a 1 b 1 mod. Proof. All we have to do is compute [a, b] for all possible odd values of a, b mod 8. For example, if a b 3 mod 8, then [a, b] [3, 11] ( 3 11 ( , hence [a, b] 1 in this case. Checking every case then gives the following table: [a, b] Thus [a, b] a 1 b 1 mod as claimed. The preceding corollary immediately implies the following improvement of Prop. 1.3:

3 1.1 Quadratic Reciprocity in Z 3 Corollary 1.5. If (H holds for some σ, then it holds for σ. Deriving the quadratic reciprocity law (even for Jacobi symbols from (H was quite easy; the supplementary laws can be treated similarly. Adapting an idea of Eisenstein to the case of quadratic reciprocity, we now show that the supplementary laws follow from the general reciprocity law through a simple computation: Corollary 1.6. We have ( 1 a ( 1(a 1/ for a N. For any a, m N, the residue symbol ( a m only depends on the residue class of m mod 4a; in particular, ( a 1/8 for a Z \ {0}. ( 1(a Proof. For the first claim we consider the equation a b 4; reducing it modulo a and b gives (a/b ( b/a 1, hence ( 1/a (a/b(b/a ( 1 (a 1(b 1/4 ( 1 (a 1(a 5/4 ( 1 (a 1/ by the quadratic reciprocity law and the fact that a 1 a 5 mod. Now we are going to prove that (a/m only depends on m mod 4a (assuming that the denominator m remains positive; by induction, it suffices to show that (a/m (a/n for m, n N with m n 4a. Reducing this equation modulo m and n gives (m/n (a/n and ( n/m (a/m; thus (a/m(a/n (m/n( n/m. But but for positive integers m n mod 4, the reciprocity law can be written in the form (m/n ( n/m; thus the right hand side equals 1, and this implies the claim. In particular, (/a only depends on a mod 8, and then the formula ( a ( 1 (a 1/8 has only to be checked for a 1, 3, 5, 7. We have shown that Hypothesis (H implies the reciprocity law for positive odd integers a, b: ( a b ( b a ( 1 a 1 b 1. (1.1 Now we show that this formula is still valid if one of the numbers, say b, is negative. In that case, (1.1 shows ( a b ( b ( 1 a 1 b 1. a But ( a b ( a b b and ( a ( 1(a 1/ ( b a, hence we find ( a b ( b a ( 1 a 1 ( 1 a 1 b+1 ( 1 a 1 b 1. A similar calculation shows that if a and b are both negative, then ( a b ( b a 1 a ( 1 b 1. We can combine these formulas into a single formula valid for all odd integers as follows: Theorem 1.7. Let a and b be odd coprime integers. Then we have

4 4 1. Quadratic Reciprocity in Number Fields 1. The quadratic reciprocity law: ( a b. The supplementary laws: ( 1 a ( b a ( 1 a 1 b 1 ( 1 a 1 sgn (a 1 +, sgn (a 1 sgn (b 1 +. ( a ( 1 a 1. In Z, the correction factors involving the sgn -function are not really essential since for every nonzero a Z there is a unit ε ±1 such that εa > 0. In general number fields, this is not true in general, and the reciprocity law for totally positive elements is only a special case of the general law. 1. Eisenstein s Version of Quadratic Reciprocity In [RL1, Chapter 4] we have seen how Dirichlet reduced the quadratic reciprocity law in Z[i] to the one in Z. The same method works for every quadratic number field K Q( m, as was shown by Dörrie [18] for quadratic number fields with class number 1, and by Welmin [136] in the general case. Whereas Welmin stated the quadratic reciprocity law in quadratic number fields as a product formula for Hilbert symbols, our goal here is to provide explicit formulas that are as close to the classical formulation of the quadratic reciprocity law as possible. Let us now explain the notions of primary and primitive elements. An element α O K coprime to is called primary if α ξ mod 4 for some ξ O K coprime to. From classical results on ramification in Kummer extensions (see [RL1, Thm. 4.1] it is clear that α is primary if and only if K( α /K is unramified at the primes above. We call α O K primitive if it is not divisible by any rational prime number. Note that the set of primary elements is closed with respect to multiplication, whereas that of primitive elements is not. Below we will also need the signature of elements of quadratic number fields K. Let sgn denote the usual sign function; then for α K we let α denote its conjugate, put { (sgn α, sgn α if K is real, sign (α (+1, +1 if K is complex, and call it the signature of α K. In particular, we have sign (α (+1, +1 if and only if α is totally positive. We say that α β mod for α, β K if αβ is totally positive, that is, if α and β have the same signature. Before we state the quadratic reciprocity law in quadratic number fields, let us recall the relevant definition: for a prime ideal p of odd norm and

5 1. Eisenstein s Version of Quadratic Reciprocity 5 some element α O K coprime to p, define the quadratic Legendre symbol [α/p] {±1} by the congruence [ α p ] α(np 1/ mod p. We will now present a version of Eisenstein s quadratic reciprocity law for quadratic number fields: Theorem 1.8. Eisenstein s quadratic reciprocity law holds for quadratic number fields K Q( m : if α, β, γ, δ O K have odd norm, and if they satisfy (α, β (γ, δ (1 and α γ, β δ mod 4, then As a corollary we obtain [ α β ][ β α] [ γ δ ][ δ γ ]. Corollary 1.9. If α is primary and totally positive, then [ α β ] [ β α ]. In fact, in this case we have α ξ mod 4 for some ξ coprime to β, and we may simply choose γ ξ in Theorem 1.8. Our proof of Theorem 1.8 will go like this: we will introduce symbols [α] and {α} with values in { 1, +1}and then show 1. {α} only depends on the signature of α (Lemma 1.10;. [α] only depends on the residue class of α mod 4 (Lemma 1.1; 3. [ α β ][ β α ] [α][β ][αβ ]{α}{β }{αβ } (Lemma Using this result, Theorem 1.8 follows easily: If α γ mod 4 and β δ mod 4, then {α} {γ}, {β } {δ } and {αβ } {γδ } by 1, and similarly [α] [γ], [β ] [δ ] and [αβ ] [γδ ] by. Theorem 1.8 then follows from 3. Before we start, let us recall the following congruence, which we will use repeatedly: for odd integers r, s we have r 1 + s 1 rs 1 Signatures and Residue Classes modulo 4 mod. (1. In the following, let α a + bω denote an element in O K with odd norm A Nα. We will define two symbols [α] and {α} with values { 1, +1} that only depend on the residue class of α mod 4 and on the signature sign (α, respectively. We start with sgn A 1 sgn b 1 {( 1 if b 0, {α} 1 if b 0. Note that for complex quadratic fields K, we have {α} 1 for all α K. As an immediate consequence of the definition we obtain the useful observation

6 6 1. Quadratic Reciprocity in Number Fields sgn A 1 { α} ( 1 {α}. (1.3 Since {fα} {α} for integers f > 0, this actually shows that for all nonzero integers f. Now we can prove: {fα} ( 1 sgn A 1 sgn f 1 {α} (1.4 Lemma The value {α} only depends on the signature sign (α. Proof. This is trivial if b 0, i.e., if α Z. Assume therefore that b 0. Then the following table shows how {α} depends on the signature of α: sign (α sgn (A sgn (b {α} (+1, +1 +1? +1 (+1, ( 1, ( 1, 1 +1? +1 (1.5 If, for example, sign (α ( 1, +1, then a + b m < 0 and a b m > 0 imply that A Nα αα < 0 and b < 0, hence {α} 1. Next we define a symbol [ ] for nonzero elements in O K with odd norm. Observe that any such element can be written in the form fα for some positive odd integer f and a primitive α a + bω O K with odd norm A Nα. Then we put [fα] {( 1 A 1 f 1 ( b A {α} if b 0, 1 if b 0. (1.6 Since [α] ( b A {α} for primitive α O K \ Z (just put f 1 in (1.6, the definition shows that the analog of (1.4 is [fα] ( 1 A 1 f 1 [α]. (1.7 Comparing the factors [fα]/[α] and {fα}/{α} with Theorem 1.7 we see that their product is the inversion factor of ( f A ( A f. It seems to me that there is more to (1.4 and (1.7 than meets the eye; in particular it would be nice if we had a more conceptual definition of these factors. The following result will allow us to evaluate the symbols [α]: Lemma Let α a + bω be an element with odd norm A, and assume that b r b for some odd integer b. Then ( r [α] ( 1 A 1 b 1. (1.8 A

7 1. Eisenstein s Version of Quadratic Reciprocity 7 Proof. Write α fγ for some positive integer f and some primitive γ c+dω with Nγ C. Then, using the general quadratic reciprocity law in Z, we find [α] ( 1 C 1 ( 1 C 1 f 1 ( d C {γ} ( r ( d C f 1 ( 1 C 1 ( f 1 + sgn d 1 ( 1 A 1 b 1 ( A sgn C 1 sgn d ( 1 1 C ( r C r, where we have used (1. and A f C C mod 8. Now we claim (Skolem [1, Satz 1] proved this in the special case where α, β O K \ Z are primitive: Lemma 1.1. The symbol [α] only depends on the residue class of α mod 4: if α β mod 4 for α, β O K with odd norm, then [α] [β]. Proof. 1. If α, β Z, then [α] 1 [β].. Assume that b 0 and d 0. Since α ±1 mod 4 in this case, we must have r in the notation of Lemma This implies A 1 mod 8, hence [α] 1. Since we also have [β] 1 by definition, the claim follows. 3. Finally assume that bd 0. Write α a + bω and β c + dω with odd norms A Nα and B Nβ, and b r b and d s d for odd integers b, d. From α β mod 4 we immediately deduce that r > 1 s > 1, and that r s if r 1. Thus we have to consider the following cases: 1. r, s. Then A B 1 mod 8, hence [α] [β] 1.. r s 0. In this case α β mod 4 implies b d mod 4 and A B mod 4, hence [α] ( 1 A 1 b 1 ( 1 B 1 d 1 [β]. 3. r s 1. Here we distinguish some more cases: a m mod 4. Then A B 1 mod 8, hence [α] [β] 1. b m 3 mod 4. Now A B 5 mod 8, hence [α] [β] 1. c m 1 mod 4. Write m 4n + 1; then ω + ω 1 and ωω n. Moreover, A a + ab nb 1 + ab + 4n mod 8 and B 1 + cd + 4n mod 8. In particular, A B 3 mod 4, hence [α] ( 1/ A and [β] ( 1/ ( 1(b B. ( 1(d Now A B b d mod 8 shows that ( A ( B ( 1(b d/ ( 1 b d. The claim follows. This completes the proof.

8 8 1. Quadratic Reciprocity in Number Fields Reduction to Z In the following, let m be a squarefree integer and K Q( m a quadratic number field. Let {1, ω} denote the standard integral basis, that is, let ω m if m, 3 mod 4 and ω 1+ m if m 1 mod 4. We now want to compute the inversion factor [α/β][β/α] for coprime primitive elements α, β O K with odd norm. We start with a few preliminaries that should be familiar from [RL1, Chap. 5]. First we have the congruence dα ad + bdω ad bc mod β. (1.9 Next, αβ (a + bω(c + dω ac ad bdn (ad bcω, where n 0 if m, 3 mod 4 and n m 1 4 if m 1 mod 4. Thus if we write αβ e+fω, then (1.9 shows that dα f mod β. We will also use the formulas [ α a ] ( Nα a and [ a α ] ( a Nα for coprime elements α O K and a Z with odd norm; here ( denotes the Legendre symbol in Z. The central result of this subsection is Lemma Assume that α, β O K are coprime elements with odd norms. Then [ ][ ] α β [α][β ][αβ ]{α}{β }{αβ }. (1.10 β α Proof. Both sides are trivial if α, β Z. If α a and β c + dω with d 0, then the left hand side of (1.10 is [ ][ ] ( ( a β a B, (1.11 β a B a and, using (1.4 and (1.7, the right hand side can be transformed into [a][β ][aβ ]{a}{β }{aβ } ( 1 a 1 B 1 sgn a 1 sgn B 1 ( 1. (1.1 Now the quadratic reciprocity law 1.7 shows that the right hand sides of (1.11 and (1.1 agree, and the claim follows. For the rest of the proof we therefore may assume that bd 0. Assume first that β is primitive, i.e., that gcd(d, β gcd(c, d 1; with A Nα,B Nβ and αβ e + fω we find [ ] α β [ ][ ] d dα β β ( d B A similar calculation shows that [ ] β α ( ad bc B ( bf A ( df. B

9 1. Eisenstein s Version of Quadratic Reciprocity 9 for primitive α, hence [ ][ ] α β β α ( df B ( ( bf b A A [α][β ][αβ ]{α}{β }{αβ }. ( ( d f B AB This proves the claim for primitive α, β O K \ Z. Finally assume that α fγ and β gδ for odd positive integers f, g and primitive γ, δ O K \ Z. Using the fact that [ f g ] 1 for odd integers f, g, and putting C Nγ and D Nδ we get [ ][ ] [ ][ ][ ][ ][ ][ ] α β f β g α γ δ β α β f α g δ γ ( ( ( ( [ ][ ] f B g A γ δ, B f A g δ γ and since the claim holds for primitive elements, we deduce [ ][ ] ( ( ( ( α β f D g C [γ][δ ][γδ ]{γ}{δ }{γδ }. (1.13 β α D f C g On the other hand we know that [α][β ][αβ ] [fγ][gδ ][fgγδ ] [γ][δ ][γδ ]( 1 f 1 C 1 ( 1 g 1 D 1 ( 1 fg 1 Using (1. as well as the fact that f and g are positive we find CD 1. hence ( 1 f 1 C 1 ( 1 g 1 D 1 ( 1 fg 1 ( 1 f 1 C 1 ( 1 f 1 D 1 ( f D [α][β ][αβ ] ( D f ( 1 g 1 + g 1 ( f D ( g C CD 1 C 1 ( C g D 1 ( 1 ( f 1, ( ( D g f C + g 1 ( C 1 + D 1 ( C [γ][δ ][γδ ]. (1.14 g Since f, g > 0 we conclude that {α} {γ}, {β } {δ } and {αβ } {γδ }. Combining this with (1.13 and (1.14 we find [ ][ ] α β [α][β ][αβ ]{α}{β }{αβ }. β α This completes the proof.

10 10 1. Quadratic Reciprocity in Number Fields 1.3 The Weak Quadratic Reciprocity Law We will now derive various explicit versions of quadratic reciprocity in quadratic number fields. The following formulation gives a weak form of the reciprocity law that has several advantages: it is very similar to the quadratic reciprocity law in Z, and it is easy to guess the correct generalization to arbitrary number fields. A special case was obtained by Hecke [44, 45], but the general form is due to Hasse [41]: Theorem Let α, β 1 mod be coprime elements in the maximal order O K of a quadratic number field K Q( m. Then we have 1. The quadratic reciprocity law where S [ α β ][ β α] sgn α 1 sgn β 1 + sgn α 1. The first supplementary law [ 1 ] α sgn α 1 where U + sgn α The second supplementary law ( 1 S+T, (1.15 sgn β 1 and T Tr ( α 1 β 1. α 1 U+Tr ( 1, (1.16 [ α] ( 1 Tr α 1 8. (1.17 Note that this is not the most general quadratic reciprocity law in quadratic number fields: first, in the cases m, 3 mod 4 the condition α, β 1 mod is too restrictive since we also would like to know how to invert symbols involving elements congruent to 1+ m mod (if m mod 4 or to m mod (if m 3 mod 4. Second, the supplementary laws should allow us to compute [ε/α] for the fundamental units of real quadratic fields, as well as [α/β] for elements α with even norm. Corollary If m, 3 mod 4, then [ α β ][ β α ] 1 for all α β 1 mod such that α or β is totally positive. In fact, the condition α β 1 mod guarantees that T 0, and if one of these numbers is totally positive then S 0. Note that this corollary contains e.g. the quadratic reciprocity law in Z[i] (see [RL1, Sect. 5.1] as a special case. For proving the actual reciprocity law in Theorem 1.14 we need to compute the product {α}{β }{αβ }:

11 1.3 The Weak Quadratic Reciprocity Law 11 Lemma For nonzero α, β K we have where S sgn α 1 sgn β 1 Proof. There are three cases to consider: {α}{β }{αβ } ( 1 S, ( sgn α 1 1. For α, β Z, both sides of (1.18 are equal to 1. sgn β 1.. For α a + bω with b 0 and β c Z, (1.3 shows that On the other hand we have {α}{β }{αβ } ( 1 sgn c 1 sgn A 1. S sgn α 1 sgn c 1 + sgn α 1 sgn c 1 sgn c 1 ( sgn α 1 + sgn α 1 sgn c 1 sgn A 1. Thus in this case (1.18 also holds. 3. Finally assume that α a + bω and β c + dω with bd 0. Using table (1.5 we compute both {α}{β }{αβ } and ( 1 S as functions of the signatures of α (left column and β (top row, and find that they agree: This proves the claim. Now we can give the (+1, +1 (+1, 1 ( 1, +1 ( 1, 1 (+1, (+1, ( 1, ( 1, Proof of Theorem We know that [ α β ][ β α ] [α][β ][αβ ]{α}{β }{αβ }, and that {α}{β }{αβ } ( 1 S. Thus it remains to show that [α][β ][αβ ] ( 1 T whenever α β 1 mod. Since both sides are functions of the residue classes of α, β mod 4, this is a simple (and, for once, quite short calculation. In fact, if m, 3 mod 4, then Tr α 0 mod for any α O K, and if m 1 mod 4, than Tr α 0 mod whenever α 1 mod. Thus we only have to show that [α][β ][αβ ] 1 for all α β 1 mod.

12 1 1. Quadratic Reciprocity in Number Fields This is trivial if m mod 4, since [α] 1 whenever b according to Lemma 1.1. If m is odd and α ±1 mod 4, then [α] 1 and [αβ ] [β ], and the claim follows. If α, β ±1 + m mod 4, then [α] ( A, [β ] ( B and [αβ ] 1 since αβ ±1 mod 4. The claim now follows. It remains to prove the two supplementary laws. Proof of the First Supplementary Law. We have [ 1 α ] [ 1 α ] [ 1][α][ α]{ 1}{α}{ α}. Clearly [ 1] { 1} 1, and then (1.3 and (1.7 show that [ 1 A 1 sgn A 1 α ] ( 1 ( 1. But Lemma 1. shows that A 1 sgn A 1 α 1 sgn α 1 + α 1 and from these relations the claim follows. mod, + sgn α 1 mod, Proof of the Second Supplementary Law. The second supplementary law will be proved using the fact that [ α ] ( A ; the first supplementary law also can be proved along these lines. Note first that [ α ] ( Nα. Writing α a + b m we get Tr α 1 8 a +mb 1 4. Now we distinguish several cases. 1. m mod 4. If b is even, then Nα a + mb 1 mod 8 and therefore [ α ] 1; if b is odd, then Nα ±3 mod 8 and [ α ] 1. This shows that [ α ] ( 1b/ for α a + b m. On the other hand a simple calculation shows Tr α 1 8 a +mb 1 4 b mod. Thus the formula (1.17 holds as soon as Nα is odd.. m 3 mod 4. Since α a + b m 1 mod we have b, hence Nα 1+b mod 8 and therefore [ α ] ( 1b/. Again we find Tr α 1 a +mb 1 4 b 8 mod. Thus (1.17 holds whenever α 1 mod. 3. m 1 mod 4. Write α a + b m 1 mod. Then a or b is even. If b is even, then as above we find [ α ] ( 1b/ and Tr α 1 8 b mod. If b is odd, then [ α ] ( m 1 a +m ( a, and on the other hand Tr α 1 8 m a mod. The proof of (1.17 is now complete. 1.4 The Strong Quadratic Reciprocity Law The complete quadratic reciprocity law (except for the supplementary laws is given by the following

13 1.4 The Strong Quadratic Reciprocity Law 13 Theorem For any α O K with odd norm define elements t α, t α Z/Z by { m t α (1 + m t α ξ mod 4 if m 1 mod, α (1 + m tα ( 1 t α ξ mod 4 if m 0 mod for some ξ O K. Then the quadratic reciprocity law for coprime elements of odd norm is given by [ α ][ β ( 1 β α] S+T, (1.19 where S is as in Theorem 1.14 and { t α t β T + t αt β + t α t β mod if m 1, mod 4, t α t β + t αt β mod if m 3 mod 4. The first supplementary law has the form [ 1 ] α where U is as in Theorem { ( 1 U+tα if m 1, mod 4, ( 1 U if m 3 mod 4, This result (which is modeled after the quadratic reciprocity law in Z[i] given by Hilbert [51] clearly implies Theorem 1.8: the condition α γ mod 4 shows that t α t γ and t α t γ, and the fact that α γ mod shows that S does not change when α is replaced by γ. Since we know that {α}{β }{αβ } ( 1 S, we only have to show that [α][β ][αβ ] ( 1 T. For slightly simplifying the calculations we introduce the group M (O K /4O K. We will see below that the symbol [α] only depends on the coset αm. Note that M/M (Z/Z for every quadratic number field (see Exercise 1.4. The proof of the reciprocity law in Q( m is split up into three cases according to the residue class of m mod 4. In every case, we compute the symbols [α], t α and t α in terms of the cosets αm and then verify (1.19. m 3 mod 4. Write α a + b m and A a mb. We have to show that [α][β ][αβ ] ( 1 T. To this end, we observe that squares of elements with odd norm are congruent to ±1 mod 4 in this case; thus the symbols t α, t α are defined by α ± m tα (1 + m t α mod 4. (1.0 We now show that t α, t α and [α], are functions on M/M by computing their values on the residue classes modulo 4:

14 14 1. Quadratic Reciprocity in Number Fields αm α mod 4 t α t α [α] 1 ± m ±1 + m m ± m 1 0 ( 1 m + m ± m 1 1 ( 1 m In fact, observe that we always have A 1 mod 4, hence [α] ( A r, where b r b. Thus ( 1 T, where T t α t β +t αt β mod, is a function of the cosets of α modulo M. A little calculation shows that [α][β ][αβ ] ( 1 T, the values being given by the following table: 1 1 m + m m m The first supplementary law [ 1 α ] 1 is a trivial consequence of these calculations. Remark. The first supplementary law may be sharpened. The fact that [ 1 α ] 1 allows us to define the quartic symbol [ 1/α] 4 ( 1 ( Nα 1/4. For totally positive α, the following result is easily verified: Proposition For totally positve α Z[ m ] with m 3 mod 4 we have { [ 1 ] α t α + t 4 ( 1T α if m 3 mod 8, for T t α if m 7 mod 8. m mod 4. Write α a + b m, put A a mb and b r b for some odd integer b. Then [α] ( 1 A 1 b 1 ( A r. The values of [α], as well as the symbols t α and t α defined by α (1 + m tα ( 1 t α ξ mod 4 (1.1 again only depend on the cosets αm, as the following table shows: αm α mod 4 t α t α [α] 1 1, 3 + m , 1 + m m ±1 + m 1 0 ( 1 m 1 m ±1 m 1 1 ( 1 m

15 1.4 The Strong Quadratic Reciprocity Law 15 Using these tablse we compute the products [α][β ][αβ ] as well as ( 1 T for T t α t β + t αt β + t α t β. We find that both expressions agree; their values are given in the following table: m 1 m m m Remark. If α 1 mod and β 1 mod (, m, then T Tr ( α 1 m β 1 mod. I have not been able to find a similar (and simple formula covering all cases α, β 1 mod (, m. m 1 mod 4. Write m 4n + 1; the cosets αm with odd A Nα contain the following residue classes modulo 4: αm n n 1 1 1, n + ω 1 1 1, n ω m 1 + ω 1 + ω, n ω m 1 + ω 1 + ω, n + ω The difference is explained by the fact that there is only one coprime residue class modulo if splits (if m 1 mod 8, that is, if n, but three of them if is inert. Again it is easily checked that t α, t α and [α] only depend on the cosets αm : αm 1 1 m m t α t α [α] 1 1 ( m ( m Computing the values of [α][β ][αβ ] and ( 1 T gives the following table: 1 1 m m m m This proves the reciprocity law for m 1 mod 4. Plugging in β 1 gives the first supplementary law [ 1 α ] ( 1tα.

16 16 1. Quadratic Reciprocity in Number Fields 1.5 Applications Let us now see how to apply the explicit formulas in concrete situations. Scholz s Reciprocity Law Let p, q 1 mod 4 be primes; consider the fields K Q( p and K Q( q, and let ε p, ε q and h, h denote their fundamental units and their class numbers, respectively. If we assume that (p/q +1, then (q qq splits in K and (p pp splits in K. We know that q h is principal, and that q 3h is generated by an element α Z[ p ]. Similarly, p 3h (β for some β Z[ q ]. We want to evaluate the quadratic symbol (ε p /q [ε p /q]. Since raising this symbol to the power 3h does not change its value, we have (ε p /q [ε p /β]. Finally, replacing ε p by its cube if necessary allows us to write ε p t + u p for integers t, u, where clearly t is even. Write β c + d p. Then d is even, and and we can choose β totally positive. Then since [β/ε p ] 1 the quadratic reciprocity law tells us that [ εp β ] ( 1 T, where T Tr ε p 1 A simple calculation gives T (t 1(c 1+dup c 1 Now consider the equation [ εp ] β ( 1 c. β 1. mod, hence we get c pd q 3h. (1. Reducing this equation modulo c and applying the quadratic reciprocity law gives ( 1 c ( pq c ( c pq. Reducing (1. modulo p gives ( c p ( q p 4, and reducing (1. modulo q shows that ( c q ( 1 q 4( p q 4( d q. Writing d j e for some odd integer e we get ( d q ( q j ( q e ( q j. Now j 1 if and only if q 5 mod 8, and this implies ( q j ( 1 q 4. Collecting everything we find ( εp ( p q q q 4( p, 4 and this immediately implies Scholz s Reciprocity Law. Singular Numbers Let K be a number field. An element ω K is called singular if (ω c is the square of some (fractional ideal. The set of singular numbers in K modulo squares form a group called the Selmer group: Sel(K {ω K : (ω c }/F.

17 1.5 Applications 17 Lemma Let m be an ideal in O K. Then every class ωk Sel(K can be represented by some ω coprime to m. Proof. Write (ω c and find an ideal b coprime to m in the ideal class of c. Then γc b for some γ K, and now β ωγ represents the same element in Sel(K as ω, and we have (β b. As a corollary we observe Corollary 1.0. For singular numbers ω, the quadratic residue symbol [ ω α ] is defined for all α K coprime to. In fact, by Lemma 1.19 there is some γ K such that ωγ β with β coprime to α. Now put [ ω α ] [ β α ]. It is easily checked that this is well defined. Let us now introduce the following subgroups of the Selmer group: Sel + (K {α K : (α a, α 0}/K, Sel 4 (K {α K : (α a, α ξ mod 4}/K, Sel + 4 (K {α K : (α a, α ξ mod 4, α 0}/K. These subgroups will play a central role in later chapters; here we will only deal with the last group. Note that αk Sel + 4 (K if and only if the quadratic extension K( α /K is unramified everywhere. We now derive another surprising characterization of elements in Sel + 4 (F, bringing out a connection with quadratic reciprocity: Theorem 1.1. Let K be a quadratic number field. Then the following assertions are equivalent: 1. ωk Sel + 4 (K;. [ ω α ] 1 for all α coprime to ω Proof. (1 (: Since ω ξ mod 4 is totally positive, we have [ ω α ] [ α ω ] by the quadratic reciprocity law. Since ω is singular, we find [ α ω ] [ α c ] +1. ( (1: We first claim that (ω is an ideal square. Assume not; then we will show that [ ω β ] 1 for infinitely many ideals (β. In fact, write (ω p a1 1 par r and assume that a r is odd. Pick a set of prime ideals q 1,..., q s different from the p i and with odd norm, and let ν denote a quadratic nonresidue modulo p r. Then use the generalized Chinese Remainder Theorem (i.e., the approximation theorem to find a totally positive element β O K with β 1 mod q 1 q s, β 1 mod 4, β 1 mod p 1 p r 1, β ν mod p r.

18 18 1. Quadratic Reciprocity in Number Fields Then [ β ω ] [ β p i ] ai [ β p r ] 1; on the other hand, the reciprocity law for primary elements implies that [ ω β ] [ β ω ] 1. Now use Cor. 1.0 to make ω coprime to. Then we claim that ω ξ mod 4. If not, then an inspection of the tables giving the inversion factors shows that there is some residue class τ mod 4 such that for all totally positive β O K coprime to α and with β τ mod 4 we have [ ω β ] [ β ω ]. Now choose such a β with β 1 mod ω. Then [ ω β ] [ β ω ] 1. Finally we show that ω is totally positive. If not, then there is some signature σ such that for every β 1 mod 4 and sign (β σ we have [ ω β ] [ β ω ]. Choosing such a β with the additional property β 1 mod ω then leads to a contradiction as above. Notes The first mathematician who came up with a quadratic reciprocity law in a number field different from Q was Gauss: he discovered and proved the quadratic reciprocity law in Z[i] (see [RL1, Section 5.1]; a simpler proof was later given by Dirichlet. Next came Eisenstein s quadratic (actually sextic reciprocity law in Z[ζ 3 ] ([RL1, Section 7.3]. As we have seen in [RL1, Sections 6.5, 9.4], Goldscheider [37] proved the quadratic (as well as quartic and octic reciprocity law in Z[ζ 8 ]. In 189, Bonaventura [13] derived the quadratic reciprocity law in Z[ ]. In his 1898 dissertation, Dörrie [18] used Dirichlet s technique to derive the quadratic reciprocity law in quadratic fields with class number 1, and in 190 Rückle [111] and K.S. Hilbert [57] studied the product formula for quadratic Hilbert symbols in algebraic number fields, in particular in Z[ζ n]. Welmin [136] finally used Dirichlet s method to derive the quadratic reciprocity law in arbitrary quadratic number fields from the one in rational integers. The possibility of such a proof was realized independently by Hecke [43, footnote, p. 8], who derived the quadratic reciprocity law in real quadratic number fields using generalized Gauss sums. Hecke also proved the following version of the second supplementary law: Theorem 1.. Let k be a real quadratic number field, let l 1 and l be prime ideals above, and let λ O k be an algebraic integer with (λ l a1 1 la (put a 0 if there is only one prime ideal above in k. Assume that 4 if a 1 a 0 mod, λ ξ 4l 1 if a 1 1, a 0 mod, mod 4l if a 1 0, a 1 mod, 4l 1 l if a 1 a 1 mod. Then ( λ α ( 1 S, S sgn α 1 sgn λ 1 + sgn α 1 sgn λ 1.

19 Exercises 19 Our discussion of quadratic reciprocity in quadratic number fields is loosely based on Skolem s approach in [1]; Skolem rediscovered and then simplified Welmin s results, and was able to see how to generalize Dirichlet s technique to arbitrary number fields; in practice, however, this method will be far too technical for deriving explicit quadratic reciprocity laws in number fields of degree >. Eisenstein s way of formulating the quadratic reciprocity law (see Hypothesis (H on p. 1, and Theorem 1.8 goes back to [3], where the case of odd primes l is discussed. Eisenstein does not deal with quadratic reciprocity there, but he was aware of its exceptional status; [1, p. 619], he writes: dieses bildet übrigens hier einen Ausnahmefall, da der Unterschied zwischen pos. und neg. Zahlen nicht durch Congruenzen dargestellt werden kann. 1 This problem was later overcome by Hilbert and Furtwängler, who invented infinite primes in order to simplify statements in class field theory; Hasse then introduced congruences modulo infinite primes and was thus able to express the difference between positive and negative numbers by such congruences. E.F. Stueben also has studied quadratic reciprocity in quadratic fields, but his publication [16] was not available to me. The explicit quadratic reciprocity laws given in this chapter do certainly not exhaust the subject, and there are lots of problems remaining to be solved. Consider e.g. a quadratic number field K, let denote a prime ideal above, and let a and b denote ideals of odd norm in the ideal class [] 1. Then a (α and b (β for α, β O K, and we may ask when [ α q ][ β p ] is independent of the choices of α and β, and what its value is. Exercises 1.1 Assume that (/A only depends on A mod σ for some integer σ, and show that this implies that that we can take σ Let K/k be a quadratic extension. Show that the set of primitive elements does not form a monoid. Prove that if α and β are primitive and coprime, then so are αβ and α β. 1.3 Show that {α}{α } sgn A. 1.4 Let K be an algebraic number field of degree n, let O denote the ring of integers in K, and put M (O/4O. Show that squaring induces an exact sequence 1 M[] M M 1, where M[] α M : α 1. Next show that β +O 1+β +4O defines an isomorphism O/O M[]. Finally, use an integral basis to prove that O/O (Z/Z n, and deduce that this implies M/M (Z/Z n. 1 this forms an exception here, since the difference between pos[itive] and neg[ative] numbers cannot be represented by congruences.

20 0 1. Quadratic Reciprocity in Number Fields 1.5 Let k Q( m be a quadratic number field, and assume that α O k satisfies α 1 mod. Show that Nα 1 Tr α 1 mod. 1.6 (continued Deduce from the preceding exercises that, for totally positive α 1 mod, we have [ 1 ] α ( 1T, where T Tr α 1 1. Deduce that [ ] only α depends on the residue class α mod 4 (as long as α is totally positive. 1.7 Consider the quadratic number field k Q( m with m, 3 mod 4, and let l denote the prime ideal above (thus l (, m if m mod 4, and l (, 1 + m if m 3 mod 4. Show that T Tr ( α 1 β 1 Z whenever α β 1 mod l, and that the formula (α/β(β/α ( 1 T for the quadratic reciprocity law that we have proved for totally positive α β 1 mod does not hold in general under the weaker assumption α β 1 mod l. 1.8 For α a + b Z[ ] with α 1 mod, show that [ /α] ( /a. Hints: Use (1.15 to show that [ ] [ ] [ α+ α ] [ α α α α+ ] α+ ]. Now observe that α a mod. 1.9 Prove the following version of the second supplementary law in Z[ m ] for m 3 mod 4: Let λ Z[ m ] be an element of norm n ; for α Z[ m ] with odd positive norm and α 1 mod 4 we have h λ αi ( +1 if α ξ mod 4l, 1 if α ξ mod 4l, where l (, 1 + m is the prime ideal of norm. Hints: write λ r + s m and α a + b m with r s a 1 mod and 4 b. 1. Show that ( b Nα (. Show that [ λ α ] ( Nα. Nα rb sa (. Nα 3. Use quadratic reciprocity in Z to show that ( rb sa Nα ( rb sa Derive the quartic reciprocity law in Z[i] from Eisenstein s hypothesis (E for prime powers Derive the second supplementary law for quartic residues from the quartic reciprocity law in Z[i]. 1.1 Let k be a number field, let denote the product of all real places, and let O k l e 1 1 ler r be the prime ideal factorization of. Show that if an element α O k coprime to is a square modulo l e i+1 i, then it is a square modulo l r i for all r Let k be a number field, let denote the product of all real places, and let O k l e 1 1 ler r be the prime ideal factorization of. If Eisenstein s quadratic reciprocity law holds modulo m in k, where m is a product of powers of prime ideals dividing (, then it holds modulo 4l 1 l r Let k be a number field with class number 1; show that if Eisenstein s quadratic reciprocity law holds in k, then it holds in any quadratic extension K of k Consider the ring O K Z[ω] of algebraic integers in the cubic number field K Q(ω, where ω 3 ; its fundamental unit is ε ω Compute [ ε ] for the primes π 1 + ω and π 1 + π ω ;. Assuming that Eisenstein s quadratic reciprocity law holds in K, show that [ ε ] only depends on the residue class π mod 4OK if Nπ > 0; π

21 Exercises 1 3. Show that any element in O K with odd norm is associated to an element of the form a + bω + cω, where a 1 b 0 mod 4 and c 0 mod. 4. Prove that [ ε ] π ( 1c/, where π a+bω+cω and a 1 b 0 mod 4, c 0 mod. This result is due to Aigner [3] Assume that in the cubic number field K Q(ω with ω 3 we have the reciprocity law h α β ih β αi ( 1 Tr α 1 for all (totally positive α, β Z[ω]. Show that this implies the result of the preceding exercise Prove the following analog of Theorem 1.1: Let K be a quadratic number field. Then the following assertions are equivalent: 1. ωk Sel 4(K;. [ ω ] 1 for all totally positive α coprime to ω α 1.18 (continued Find an analog of Theorem 1.1 involving Sel + (K and prove it. β 1