12. Quadratic Reciprocity in Number Fieldsfranz/rl2/rlb12.pdf · Deriving the quadratic reciprocity...

12. Quadratic Reciprocity in Number Fields

In [RL1, p. 273/274] we briefly mentioned how Eisenstein arrived at a hypo-thetical reciprocity law by assuming what later turned out to be an immediateconsequence of the product formula for the Hilbert symbol, which in turn fol-lows directly from Artin’s reciprocity law. In this and the next chapter wewill explain his beautiful idea in detail.

12.1 Quadratic Reciprocity in Z

Eisenstein [22] realized that the reciprocity laws for quadratic and cubicresidues can be formulated in a way that allows the immediate generalizationto arbitrary powers and number fields: his observation was that the inver-sion factor i(α, β) = (αβ )(βα )−1 for p-th power residue symbols in Q(ζp) andodd primes p only depends on α, β mod (1− ζ)m for some integer m. In thissection we show that Eisenstein’s version for p = 2 over Q is equivalent tothe quadratic reciprocity law; then we will prove its generalization for quad-ratic number fields. In the next chapter we will then show how to extract thereciprocity 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 (ab )(

ba ) = (−1)[a,b]. Let us make the following assumption:

Hypothesis (H). There exists an integer σ ∈ N such that [a, b] = [a′, b′] forall natural numbers a, b, a′, b′ ∈ N with (a, b) = (a′, b′) = 1, and a ≡ a′, b ≡b′ mod 2σ.

We could also allow integers a, b, a′, b′ ∈ Z, but then we would have toadd the condition that aa′ > 0 and bb′ > 0. In that case, (H) may also beassumed to hold for b = −1; note that ( a

−1 ) = (a1 ) = 1, since by definition ofthe Jacobi symbol (a1 ) =

∏p|1(

ap ), and this product is empty.

In order to better understand what Hypothesis (H) is all about we firstderive the following result (which will not be used in the sequel):

Proposition 12.1. If Hypothesis (H) holds, then(ab

)=

( a

b+ 2σa

)for all coprime numbers a, b ∈ N.

2 12. Quadratic Reciprocity in Number Fields

In other words: Hypothesis (H) states that the Legendre symbol (a· ), as afunction of its denominator, has conductor dividing 2σa. This is a weak formof the reciprocity law as given by Euler: compare [RL1,Thm. 2.28.iv)].

Proof. Observe that(ab

)( ba

)=

( a

b+ 2σa

)(b+ 2σaa

)=

( a

b+ 2σa

)( ba

)by Hypothesis (H) and reduction modulo a.

Let us now sketch Eisenstein’s idea for deriving the quadratic reciprocitylaw 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 2σ:

Lemma 12.2. If a ∈ Z is an integer with a ≡ 1 mod 8, then for every σ ∈ Nthere is an integer µσ such that a ≡ µ2

σ mod 2σ.

Proof. Assume that we know a ≡ µ2e mod 2e for some e ≥ 3; then a = µ2

e+2ebfor some integer b. But then µe+1 = µe + 2e−1b will do it, since we havea ≡ (µe + 2e−1b)2 mod 2e+1 in light of 22e−2 ≡ 0 mod 2e+1 for e ≥ 3.

This simple observation enables us to prove

Proposition 12.3. 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 thata ≡ r mod 8; this implies a/r ≡ 1 mod 8, so by Lemma 12.2, there is aninteger m such that a ≡ rm2 mod 2σ. Similarly, we have b ≡ s mod 8 forsome s ∈ {1, 3, 5, 7}, and thus b ≡ sn2 mod 2σ for some integer n. We haveto show that if (a, b) = (r, s) = 1, then [a, b] = [r, s]. Since we may choose mand n in such a way that (rm, sn) = 1, hypothesis (H) then implies [a, b] =[rm2, sn2] = [r, s]. Thus the value of [a, b] only depends on a, b mod 8.

Corollary 12.4. We have [a, b] ≡ a−12

b−12 mod 2.

Proof. All we have to do is compute [a, b] for all possible odd values ofa, b mod 8. For example, if a ≡ b ≡ 3 mod 8, then [a, b] = [3, 11] = ( 3

11 )( 113 ) =

−1, hence [a, b] = 1 in this case. Checking every case then gives the followingtable:

[a, b] 1 3 5 71 0 0 0 03 0 1 0 15 0 0 0 07 0 1 0 1

Thus [a, b] ≡ a−12

b−12 mod 2 as claimed.

The preceding corollary immediately implies the following improvementof Prop. 12.3:

12.1 Quadratic Reciprocity in Z 3

Corollary 12.5. If (H) holds for some σ, then it holds for σ = 2.

Deriving the quadratic reciprocity law (even for Jacobi symbols) from (H)was quite easy; the supplementary laws can be treated similarly. Adapting anidea of Eisenstein to the case of quadratic reciprocity, we now show that thesupplementary laws follow from the general reciprocity law through a simplecomputation:

Corollary 12.6. We have (−1a ) = (−1)(a−1)/2 for a ∈ N. For any a,m ∈ N,

the residue symbol ( am ) only depends on the residue class of m mod 4a; inparticular, ( 2

a ) = (−1)(a2−1)/8 for a ∈ Z \ {0}.

Proof. For the first claim we consider the equation a − b = 4; reducing itmodulo 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)/2 by the quadratic reci-procity law and the fact that a−1

2 ≡ a−52 mod 2.

Now we are going to prove that (a/m) only depends on m mod 4a (as-suming that the denominator m remains positive); by induction, it sufficesto show that (a/m) = (a/n) for m,n ∈ N with m − n = 4a. Reducing thisequation 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 theright hand side equals 1, and this implies the claim.

In particular, (2/a) only depends on a mod 8, and then the formula ( 2a ) =

(−1)(a2−1)/8 has only to be checked for a = 1, 3, 5, 7.

We have shown that Hypothesis (H) implies the reciprocity law for posi-tive odd integers a, b: (a

b

)( ba

)= (−1)

a−12 · b−1

2 . (12.1)

Now we show that this formula is still valid if one of the numbers, say b, isnegative. In that case, (12.1) shows( a

−b

)(−ba

)= (−1)

a−12 ·−b−1

2 .

But ( a−b ) = (ab ) and (−ba ) = (−1)(a−1)/2( ba ), hence we find(a

b

)( ba

)= (−1)

a−12 (−1)

a−12 · b+1

2 = (−1)a−12 · b−1

2 .

A similar calculation shows that if a and b are both negative, then(ab )(

ba ) = −(−1)

a−12 · b−1

2 . We can combine these formulas into a single for-mula valid for all odd integers as follows:

Theorem 12.7. Let a and b be odd coprime integers. Then we have


1. The quadratic reciprocity law:(ab

)( ba

)= (−1)

a−12 · b−1

2 +sgn (a)−1

2 · sgn (b)−12 .

2. The supplementary laws:(−1a

)= (−1)

a−12 +

sgn (a)−12 ,

(2a

)= (−1)

a2−12 .

In Z, the correction factors involving the sgn -function are not really es-sential 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 lawfor totally positive elements is only a special case of the general law.

12.2 Eisenstein’s Version of Quadratic Reciprocity

In [RL1, Chapter 4] we have seen how Dirichlet reduced the quadratic reci-procity law in Z[i] to the one in Z. The same method works for every quadraticnumber field K = Q(

√m ), as was shown by Dorrie [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 provideexplicit formulas that are as close to the classical formulation of the quadraticreciprocity law as possible.

Let us now explain the notions of primary and primitive elements. Anelement α ∈ OK coprime to 2 is called primary if α ≡ ξ2 mod 4 for someξ ∈ OK coprime to 2. From classical results on ramification in Kummerextensions (see [RL1, Thm. 4.12]) it is clear that α is primary if and only ifK(√α )/K is unramified at the primes above 2. We call α ∈ OK primitive if

it is not divisible by any rational prime number. Note that the set of primaryelements is closed with respect to multiplication, whereas that of primitiveelements is not.

Below we will also need the signature of elements of quadratic numberfields K. Let sgn denote the usual sign function; then for α ∈ K× we let α′

denote its conjugate, put

sign (α) =

{(sgn α, sgn α′) if K is real,(+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

12.2 Eisenstein’s Version of Quadratic Reciprocity 5

some element α ∈ OK coprime to p, define the quadratic Legendre symbol[α/p] ∈ {±1} by the congruence [αp ] ≡ α(Np−1)/2 mod p.

We will now present a version of Eisenstein’s quadratic reciprocity law forquadratic number fields:

Theorem 12.8. Eisenstein’s quadratic reciprocity law holds for quadraticnumber fields K = Q(

√m ): if α, β, γ, δ ∈ OK have odd norm, and if they

satisfy (α, β) = (γ, δ) = (1) and α ≡ γ, β ≡ δ mod 4∞, then[αβ

][βα

]=

[γδ

][ δγ

].

As a corollary we obtain

Corollary 12.9. If α is primary and totally positive, then [αβ ] = [βα ].

In fact, in this case we have α ≡ ξ2 mod 4 for some ξ coprime to β, andwe may simply choose γ = ξ2 in Theorem 12.8.

Our proof of Theorem 12.8 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 12.10);2. [α] only depends on the residue class of α mod 4 (Lemma 12.12);3. [αβ ][βα ] = [α][β′][αβ′]{α}{β′}{αβ′} (Lemma 12.13).

Using this result, Theorem 12.8 follows easily: If α ≡ γ mod 4∞ andβ ≡ δ mod 4∞, then {α} = {γ}, {β′} = {δ′} and {αβ′} = {γδ′} by 1), andsimilarly [α] = [γ], [β′] = [δ′] and [αβ′] = [γδ′] by 2). Theorem 12.8 thenfollows from 3).

Before we start, let us recall the following congruence, which we will userepeatedly: for odd integers r, s we have

r − 12

+s− 1

2≡ rs− 1

2mod 2. (12.2)

Signatures and Residue Classes modulo 4

In the following, let α = a + bω denote an element in OK with odd normA = Nα. We will define two symbols [α] and {α} with values {−1,+1} thatonly depend on the residue class of α mod 4 and on the signature sign (α),respectively. We start with

{α} =

{(−1)

sgn A−12 · sgn b−1

2 if b 6= 0,1 if b = 0.

Note that for complex quadratic fields K, we have {α} = 1 for all α ∈ K×. Asan immediate consequence of the definition we obtain the useful observation


{−α} = (−1)sgn A−1

2 {α}. (12.3)

Since {fα} = {α} for integers f > 0, this actually shows that

{fα} = (−1)sgn A−1

2sgn f−1

2 {α} (12.4)

for all nonzero integers f .Now we can prove:

Lemma 12.10. 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 6= 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) −1 −1 −1(−1,−1) +1 ? +1

(12.5)

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 OK with odd norm.Observe that any such element can be written in the form fα for some positiveodd integer f and a primitive α = a + bω ∈ OK with odd norm A = Nα.Then we put

[fα] =

{(−1)

A−12

f−12

(bA

){α} if b 6= 0,

1 if b = 0.(12.6)

Since [α] = ( bA ){α} for primitive α ∈ OK \ Z (just put f = 1 in (12.6)),the definition shows that the analog of (12.4) is

[fα] = (−1)A−1

2f−12 [α]. (12.7)

Comparing the factors [fα]/[α] and {fα}/{α} with Theorem 12.7 we see thattheir product is the inversion factor of ( fA )(Af ). It seems to me that there ismore to (12.4) and (12.7) than meets the eye; in particular it would be niceif we had a more conceptual definition of these factors.

The following result will allow us to evaluate the symbols [α]:

Lemma 12.11. Let α = a+ bω be an element with odd norm A, and assumethat b = 2rb′ for some odd integer b′. Then

[α] = (−1)A−1

2b′−1

2

(2A

)r. (12.8)


Proof. Write α = fγ for some positive integer f and some primitive γ = c+dωwith Nγ = C. Then, using the general quadratic reciprocity law in Z, we find

[α] = (−1)C−1

2f−12

(dC

){γ}

= (−1)C−1

2f−12

(2C

)r(d′C

)(−1)

sgn C−12

sgn d′−12

= (−1)C−1

2 ( f−12 + sgn d′−1

2 )(

2C

)r= (−1)

A−12

b′−12

(2A

)r,

where we have used (12.2) and A = f2C ≡ C mod 8.

Now we claim (Skolem [122, Satz 1] proved this in the special case whereα, β ∈ OK \ Z are primitive):

Lemma 12.12. The symbol [α] only depends on the residue class of α mod 4:if α ≡ β mod 4 for α, β ∈ OK with odd norm, then [α] = [β].

Proof. 1. If α, β ∈ Z, then [α] = 1 = [β].

2. Assume that b 6= 0 and d = 0. Since α ≡ ±1 mod 4 in this case, we musthave r ≥ 2 in the notation of Lemma 12.11. 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 6= 0. Write α = a + bω and β = c + dω with oddnorms A = Nα and B = Nβ, and b = 2rb′ and d = 2sd′ for odd integersb′, d′. From α ≡ β mod 4 we immediately deduce that r > 1 ⇐⇒ s > 1, andthat r = s if r ≤ 1. Thus we have to consider the following cases:

1. r, s ≥ 2. Then A ≡ B ≡ 1 mod 8, hence [α] = [β] = 1.2. r = s = 0. In this case α ≡ β mod 4 implies b ≡ d mod 4 and A ≡B mod 4, hence [α] = (−1)

A−12

b−12 (−1)

B−12

d−12 = [β].

3. r = s = 1. Here we distinguish some more cases:a) m ≡ 2 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 = a2 + ab − nb2 ≡ 1 + ab + 4n mod 8 and B ≡1 + cd + 4n mod 8. In particular, A ≡ B ≡ 3 mod 4, hence [α] =( 2A )(−1)(b

′−1)/2 and [β] = ( 2B )(−1)(d

′−1)/2.Now A − B ≡ b − d mod 8 shows that ( 2

A )( 2B ) = (−1)(b−d)/2 =

(−1)b′−d′ . The claim follows.

This completes the proof.


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 ≡ 2, 3 mod 4 and ω = 1+

√m

2 if m ≡ 1 mod 4.We now want to compute the inversion factor [α/β][β/α] for coprime

primitive elements α, β ∈ OK 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. (12.9)

Next, αβ′ = (a+ bω)(c+ dω′) = ac− ad− bdn− (ad− bc)ω, where n = 0 ifm ≡ 2, 3 mod 4 and n = m−1

4 if m ≡ 1 mod 4. Thus if we write αβ′ = e+fω,then (12.9) shows that dα ≡ −f mod β.

We will also use the formulas [αa ] =(Nαa

)and [ aα ] =

(aNα

)for coprime

elements α ∈ OK and a ∈ Z with odd norm; here ( ·· ) denotes the Legendresymbol in Z.

The central result of this subsection is

Lemma 12.13. Assume that α, β ∈ OK are coprime elements with oddnorms. Then [

α

β

][β

α

]= [α][β′][αβ′]{α}{β′}{αβ′}. (12.10)

Proof. Both sides are trivial if α, β ∈ Z. If α = a and β = c+ dω with d 6= 0,then the left hand side of (12.10) is[

a

β

][β

a

]=

(a

B

)(B

a

), (12.11)

and, using (12.4) and (12.7), the right hand side can be transformed into

[a][β′][aβ′]{a}{β′}{aβ′} = (−1)a−12

B−12 (−1)

sgn a−12

sgn B−12 . (12.12)

Now the quadratic reciprocity law 12.7 shows that the right hand sides of(12.11) and (12.12) agree, and the claim follows.

For the rest of the proof we therefore may assume that bd 6= 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

β

][dα

β

]=

(d

B

)(ad− bc

B

)=

(−dfB

).

A similar calculation shows that[β

α

]=

(bf

A

)


for primitive α, hence[α

β

][β

α

]=

(−dfB

)(bf

A

)=

(b

A

)(−dB

)(f

AB

)= [α][β′][αβ′]{α}{β′}{αβ′}.

This proves the claim for primitive α, β ∈ OK \ Z.Finally assume that α = fγ and β = gδ for odd positive integers f, g and

primitive γ, δ ∈ OK \Z. Using the fact that [ fg ] = 1 for odd integers f, g, andputting C = Nγ and D = Nδ we get[

α

β

][β

α

]=

[f

β

][β

f

][g

α

][α

g

][γ

δ

][δ

γ

]=

(f

B

)(B

f

)(g

A

)(A

g

)[γ

δ

][δ

γ

],

and since the claim holds for primitive elements, we deduce[α

β

][β

α

]=

(f

D

)(D

f

)(g

C

)(C

g

)[γ][δ′][γδ′]{γ}{δ′}{γδ′}. (12.13)

On the other hand we know that

[α][β′][αβ′] = [fγ][gδ′][fgγδ′]

= [γ][δ′][γδ′](−1)f−12

C−12 (−1)

g−12

D−12 (−1)

fg−12

CD−12 .

Using (12.2) as well as the fact that f and g are positive we find

(−1)f−12

C−12 (−1)

g−12

D−12 (−1)

fg−12

CD−12

= (−1)f−12

C−12 (−1)

g−12

D−12 (−1)(

f−12 + g−1

2 )( C−12 + D−1

2 )

= (−1)f−12

D−12 + g−1

2C−1

2

=(f

D

)(D

f

)(g

C

)(C

g

),

hence

[α][β′][αβ′] =(f

D

)(D

f

)(g

C

)(C

g

)[γ][δ′][γδ′]. (12.14)

Since f, g > 0 we conclude that {α} = {γ}, {β′} = {δ′} and {αβ′} = {γδ′}.Combining this with (12.13) and (12.14) we find[

α

β

][β

α

]= [α][β′][αβ′]{α}{β′}{αβ′}.

This completes the proof.


12.3 The Weak Quadratic Reciprocity Law

We will now derive various explicit versions of quadratic reciprocity in quad-ratic number fields. The following formulation gives a weak form of the reci-procity law that has several advantages: it is very similar to the quadraticreciprocity law in Z, and it is easy to guess the correct generalization to ar-bitrary number fields. A special case was obtained by Hecke [44, 45], but thegeneral form is due to Hasse [41]:

Theorem 12.14. Let α, β ≡ 1 mod 2 be coprime elements in the maximalorder OK of a quadratic number field K = Q(

√m ). Then we have

1. The quadratic reciprocity law[αβ

][βα

]= (−1)S+T , (12.15)

where S = sgn α−12

sgn β−12 + sgn α′−1

2sgn β′−1

2 and T = Tr(α−1

2β−1

2

).

2. The first supplementary law[−1α

]= (−1)U+Tr α−1

2 , (12.16)

where U = sgn α−12 + sgn α′−1

2 .3. The second supplementary law[ 2

α

]= (−1)Tr α2−1

8 . (12.17)

Note that this is not the most general quadratic reciprocity law inquadratic number fields: first, in the cases m ≡ 2, 3 mod 4 the conditionα, β ≡ 1 mod 2 is too restrictive since we also would like to know how to in-vert symbols involving elements congruent to 1+

√m mod 2 (if m ≡ 2 mod 4)

or to√m mod 2 (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 12.15. If m ≡ 2, 3 mod 4, then [αβ ][βα ] = 1 for all α ≡ β ≡1 mod 2 such that α or β is totally positive.

In fact, the condition α ≡ β ≡ 1 mod 2 guarantees that T = 0, and ifone of these numbers is totally positive then S = 0. Note that this corollarycontains e.g. the quadratic reciprocity law in Z[i] (see [RL1, Sect. 5.1]) as aspecial case.

For proving the actual reciprocity law in Theorem 12.14 we need to com-pute the product {α}{β′}{αβ′}:

12.3 The Weak Quadratic Reciprocity Law 11

Lemma 12.16. For nonzero α, β ∈ K× we have

{α}{β′}{αβ′} = (−1)S , (12.18)

where

S =sgn α− 1

2sgn β − 1

2+

sgn α′ − 12

sgn β′ − 12

.

Proof. There are three cases to consider:

1. For α, β ∈ Z, both sides of (12.18) are equal to 1.

2. For α = a+ bω with b 6= 0 and β = c ∈ Z, (12.3) shows that

{α}{β′}{αβ′} = (−1)sgn c−1

2sgn A−1

2 .

On the other hand we have

S =sgn α− 1

2sgn c− 1

2+

sgn α′ − 12

sgn c− 12

=sgn c− 1

2

( sgn α− 12

+sgn α′ − 1

2

)=

sgn c− 12

sgn A− 12

.

Thus in this case (12.18) also holds.

3. Finally assume that α = a + bω and β = c + dω with bd 6= 0. Usingtable (12.5) we compute both {α}{β′}{αβ′} and (−1)S as functions of thesignatures of α (left column) and β (top row), and find that they agree:

(+1,+1) (+1,−1) (−1,+1) (−1,−1)(+1,+1) +1 +1 +1 +1(+1,−1) +1 −1 +1 −1(−1,+1) +1 +1 −1 −1(−1,−1) +1 −1 −1 +1

This proves the claim.

Now we can give the

Proof of Theorem 12.14. We know that [αβ ][βα ] = [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 2. Since both sides are functions of theresidue classes of α, β mod 4, this is a simple (and, for once, quite short)calculation.

In fact, if m ≡ 2, 3 mod 4, then Trα ≡ 0 mod 2 for any α ∈ OK , and ifm ≡ 1 mod 4, than Trα ≡ 0 mod 2 whenever α ≡ 1 mod 2. Thus we onlyhave to show that [α][β′][αβ′] = 1 for all α ≡ β ≡ 1 mod 2.


This is trivial if m ≡ 2 mod 4, since [α] = 1 whenever 2 | b according toLemma 12.12.

If m is odd and α ≡ ±1 mod 4, then [α] = 1 and [αβ′] = [β′], and theclaim follows. If α, β ≡ ±1 + 2

√m mod 4, then [α] = ( 2

A ), [β′] = ( 2B ) 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 [−1α ] = [−1

α′ ] =[−1][α][−α]{−1}{α}{−α}. Clearly [−1] = {−1} = 1, and then (12.3) and(12.7) show that [−1

α ] = (−1)A−1

2 (−1)sgn A−1

2 . But Lemma 12.2 shows that

A− 12

≡ α− 12

+α′ − 1

2mod 2,

sgn A− 12

≡ sgn α− 12

+sgn α′ − 1

2mod 2,

and from these relations the claim follows.

Proof of the Second Supplementary Law. The second supplementarylaw will be proved using the fact that [ 2

α ] = ( 2A ); the first supplementary law

also can be proved along these lines.Note first that [ 2

α ] = ( 2Nα ). Writing α = a + b

√m we get Tr α

2−18 =

a2+mb2−14 . Now we distinguish several cases.

1. m ≡ 2 mod 4. If b is even, then Nα = a2 +mb2 ≡ 1 mod 8 and therefore[ 2α ] = 1; if b is odd, then Nα ≡ ±3 mod 8 and [ 2

α ] = −1. This shows that[ 2α ] = (−1)b/2 for α = a+ b

√m.

On the other hand a simple calculation shows Tr α2−18 = a2+mb2−1

4 ≡b2 mod 2. Thus the formula (12.17) holds as soon as Nα is odd.

2. m ≡ 3 mod 4. Since α = a + b√m ≡ 1 mod 2 we have 2 | b, hence

Nα ≡ 1+b2 mod 8 and therefore [ 2α ] = (−1)b/2. Again we find Tr α

2−18 =

a2+mb2−14 ≡ b

2 mod 2. Thus (12.17) holds whenever α ≡ 1 mod 2.3. m ≡ 1 mod 4. Write α = a+ b

√m ≡ 1 mod 2. Then a or b is even.

If b is even, then as above we find [ 2α ] = (−1)b/2 and Tr α

2−18 ≡ b

2 mod 2.If b is odd, then [ 2

α ] = ( 2a2+m ) = (−1)

m−14 + a

2 , and on the other hand

Tr α2−18 ≡ m−1

4 + a2 mod 2.

The proof of (12.17) is now complete.

12.4 The Strong Quadratic Reciprocity Law

The complete quadratic reciprocity law (except for the supplementary laws)is given by the following

12.4 The Strong Quadratic Reciprocity Law 13

Theorem 12.17. For any α ∈ OK with odd norm define elements tα, t′α ∈Z/2Z by

α ≡

{√mtα(1 + 2

√m )t

′αξ2 mod 4 if m ≡ 1 mod 2,

(1 +√m )tα(−1)t

′αξ2 mod 4 if m ≡ 0 mod 2

for some ξ ∈ OK . Then the quadratic reciprocity law for coprime elements ofodd norm is given by [α

β

][βα

]= (−1)S+T , (12.19)

where S is as in Theorem 12.14 and

T ≡

{tαt

′β + t′αtβ + tαtβ mod 2 if m ≡ 1, 2 mod 4,

tαt′β + t′αtβ mod 2 if m ≡ 3 mod 4.

The first supplementary law has the form

[−1α

]=

{(−1)U+tα if m ≡ 1, 2 mod 4,(−1)U if m ≡ 3 mod 4,

where U is as in Theorem 12.14.

This result (which is modeled after the quadratic reciprocity law in Z[i]given by Hilbert [51]) clearly implies Theorem 12.8: the condition α ≡ γ mod4 shows that tα = tγ and t′α = t′γ , and the fact that α ≡ γ mod ∞ showsthat 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 introducethe group M = (OK/4OK)×. We will see below that the symbol [α] onlydepends on the coset αM2. Note that M/M2 ' (Z/2Z)2 for every quadraticnumber field (see Exercise 12.4).

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 thesymbols [α], tα and t′α in terms of the cosets αM2 and then verify (12.19).

m ≡ 3 mod 4. Write α = a + b√m and A = a2 −mb2. We have to show

that [α][β′][αβ′] = (−1)T . To this end, we observe that squares of elementswith odd norm are congruent to ±1 mod 4 in this case; thus the symbolstα, t

′α are defined by

α ≡ ±√mtα(1 + 2

√m )t

′α mod 4. (12.20)

We now show that tα, t′α and [α], are functions on M/M2 by computing theirvalues on the residue classes modulo 4:


αM2 α mod 4 tα t′α [α]

1 ±1 0 0 +1

1 + 2√m ±1 + 2

√m 0 1 +1

√m ±

√m 1 0 ( 2

1−m )

2 +√m 2±

√m 1 1 −( 2

1−m )

In fact, observe that we always have A ≡ 1 mod 4, hence [α] = ( 2A )r, where

b = 2rb′.Thus (−1)T , where T ≡ tαt

′β+ t′αtβ mod 2, is a function of the cosets of α

modulo M2. A little calculation shows that [α][β′][αβ′] = (−1)T , the valuesbeing given by the following table:

1 −1√m 2 +

√m

1 +1 +1 +1 +1−1 +1 +1 −1 −1√m +1 −1 +1 −1

2 +√m +1 −1 −1 +1

The first supplementary law [−1α ] = 1 is a trivial consequence of these

calculations.

Remark. The first supplementary law may be sharpened. The fact that[−1α ] = 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 12.18. For totally positve α ∈ Z[√m ] with m ≡ 3 mod 4 we

have [−1α

]4

= (−1)T for T =

{tα + t′α if m ≡ 3 mod 8,t′α if m ≡ 7 mod 8.

m ≡ 2 mod 4. Write α = a + b√m, put A = a2 − mb2 and b = 2rb′ for

some odd integer b′. Then [α] = (−1)A−1

2b′−1

2 ( 2A )r. The values of [α], as well

as the symbols tα and t′α defined by

α ≡ (1 +√m )tα(−1)t

′αξ2 mod 4 (12.21)

again only depend on the cosets αM2, as the following table shows:

αM2 α mod 4 tα t′α [α]

1 1, 3 + 2√m 0 0 +1

− 1 −1, 1 + 2√m 0 1 +1

1 +√m ±1 +

√m 1 0 ( 2

1−m )

1−√m ±1−

√m 1 1 −( 2

1−m )

12.4 The Strong Quadratic Reciprocity Law 15

Using these tablse we compute the products [α][β′][αβ′] as well as (−1)T

for T = tαt′β + t′αtβ + tαtβ . We find that both expressions agree; their values

are given in the following table:

1 −1 1 +√m 1−

√m

1 +1 +1 +1 +1−1 +1 +1 −1 −1

1 +√m +1 −1 −1 +1

1−√m +1 −1 +1 −1

Remark. If α ≡ 1 mod 2 and β ≡ 1 mod (2,√m ), then

T ≡ Tr(α− 1

2√m

β − 12

)mod 2.

I have not been able to find a similar (and simple) formula covering all casesα, β ≡ 1 mod (2,

√m ).

m ≡ 1 mod 4. Write m = 4n + 1; the cosets αM2 with odd A = Nαcontain the following residue classes modulo 4:

αM2 2 | n 2 - n1 1 1, n+ ω

−1 −1 −1,−n− ω√m −1 + 2ω −1 + 2ω, n− ω

−√m 1 + 2ω 1 + 2ω,−n+ ω

The difference is explained by the fact that there is only one coprime residueclass modulo 2 if 2 splits (if m ≡ 1 mod 8, that is, if 2 | n), but three of themif 2 is inert.

Again it is easily checked that tα, t′α and [α] only depend on the cosetsαM2:

αM2 1 −1√m −

√m

tα 0 0 1 1t′α 0 1 0 1[α] 1 1 ( 2

m ) −( 2m )

Computing the values of [α][β′][αβ′] and (−1)T gives the following table:

1 −1√m −

√m

1 +1 +1 +1 +1−1 +1 +1 −1 −1√m +1 −1 −1 +1

−√m +1 −1 +1 −1

This proves the reciprocity law for m ≡ 1 mod 4. Plugging in β = −1 givesthe first supplementary law [−1

α ] = (−1)tα .


12.5 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 εp, εq 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 qh is principal, and that q3h is generated by anelement α ∈ Z[

√p ]. Similarly, p3h′ = (β) for some β ∈ Z[

√q ].

We want to evaluate the quadratic symbol (εp/q) = [εp/q]. Since raisingthis symbol to the power 3h does not change its value, we have (εp/q) =[εp/β]. Finally, replacing εp by its cube if necessary allows us to write εp =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/εp] = 1 the quadratic reciprocity law tells us that[εpβ

]= (−1)T , where T = Tr

εp − 12

β − 12

.

A simple calculation gives T = (t−1)(c−1)+dup2 ≡ c−1

2 mod 2, hence we get[εpβ

]=

(−1c

).

Now consider the equation

c2 − pd2 = q3h. (12.22)

Reducing this equation modulo c and applying the quadratic reciprocity lawgives (−1

c ) = (pqc ) = ( cpq ). Reducing (12.22) modulo p gives ( cp ) = ( qp )4, andreducing (12.22) modulo q shows that ( cq ) = (−1

q )4(pq )4(dq ). Writing d = 2je

for some odd integer e we get (dq ) = ( 2q )j( qe ) = ( 2

q )j . Now j = 1 if and only

if q ≡ 5 mod 8, and this implies ( 2q )j = (−1

q )4. Collecting everything we find(εpq

)=

(pq

)4

(qp

)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) = c2

is the square of some (fractional) ideal. The set of singular numbers in Kmodulo squares form a group called the Selmer group:

Sel(K) = {ω ∈ K× : (ω) = c2}/F× 2.

12.5 Applications 17

Lemma 12.19. Let m be an ideal in OK . Then every class ωK× 2 ∈ Sel(K)can be represented by some ω coprime to m.

Proof. Write (ω) = c2 and find an ideal b coprime to m in the ideal class ofc. Then γc = b for some γ ∈ K×, and now β = ωγ2 represents the sameelement in Sel(K) as ω, and we have (β) = b2.

As a corollary we observe

Corollary 12.20. For singular numbers ω, the quadratic residue symbol [ωα ]is defined for all α ∈ K× coprime to 2.

In fact, by Lemma 12.19 there is some γ ∈ K× such that ωγ2 = β withβ coprime to 2α. Now put [ωα ] = [βα ]. It is easily checked that this is welldefined.

Let us now introduce the following subgroups of the Selmer group:

Sel+(K) = {α ∈ K× : (α) = a2, α� 0}/K× 2,

Sel4(K) = {α ∈ K× : (α) = a2, α ≡ ξ2 mod 4}/K× 2,

Sel+4 (K) = {α ∈ K× : (α) = a2, α ≡ ξ2 mod 4, α� 0}/K× 2.

These subgroups will play a central role in later chapters; here we willonly deal with the last group. Note that αK× 2 ∈ Sel+4 (K) if and only ifthe quadratic extension K(

√α )/K is unramified everywhere. We now derive

another surprising characterization of elements in Sel+4 (F ), bringing out aconnection with quadratic reciprocity:

Theorem 12.21. Let K be a quadratic number field. Then the followingassertions are equivalent:

1. ωK× 2 ∈ Sel+4 (K);2. [ωα ] = 1 for all α coprime to 2ω

Proof. (1) =⇒ (2): Since ω ≡ ξ2 mod 4 is totally positive, we have [ωα ] = [αω ]by the quadratic reciprocity law. Since ω is singular, we find [αω ] = [αc ]2 = +1.

(2) =⇒ (1): We first claim that (ω) is an ideal square. Assume not; thenwe will show that [ωβ ] = −1 for infinitely many ideals (β).

In fact, write (ω) = pa11 · · · par

r and assume that ar is odd. Pick a setof prime ideals q1, . . . , qs different from the pi and with odd norm, and letν denote a quadratic nonresidue modulo pr. Then use the generalized Chi-nese Remainder Theorem (i.e., the approximation theorem) to find a totallypositive element β ∈ OK with

β ≡ 1 mod q1 · · · qs,β ≡ 1 mod 4,β ≡ 1 mod p1 · · · pr−1,

β ≡ ν mod pr.


Then [ βω ] =∏

[ βpi]ai = [ βpr

] = −1; on the other hand, the reciprocity law forprimary elements implies that [ωβ ] = [ βω ] = −1.

Now use Cor. 12.20 to make ω coprime to 2. Then we claim that ω ≡ξ2 mod 4. If not, then an inspection of the tables giving the inversion factorsshows that there is some residue class τ mod 4 such that for all totally positiveβ ∈ OK coprime to 2α and with β ≡ τ mod 4 we have [ωβ ] = −[ βω ]. Nowchoose such a β with β ≡ 1 mod ω. Then [ωβ ] = −[ βω ] = −1.

Finally we show that ω is totally positive. If not, then there is somesignature σ such that for every β ≡ 1 mod 4 and sign (β) = σ we have[ωβ ] = −[ βω ]. Choosing such a β with the additional property β ≡ 1 mod ωthen leads to a contradiction as above.

Notes

The first mathematician who came up with a quadratic reciprocity law in anumber field different from Q was Gauss: he discovered and proved the quad-ratic reciprocity law in Z[i] (see [RL1, Section 5.1]); a simpler proof was latergiven by Dirichlet. Next came Eisenstein’s quadratic (actually sextic) reci-procity law in Z[ζ3] ([RL1, Section 7.3]). As we have seen in [RL1, Sections6.5, 9.4], Goldscheider [37] proved the quadratic (as well as quartic and oc-tic) reciprocity law in Z[ζ8]. In 1892, Bonaventura [13] derived the quadraticreciprocity law in Z[

√−2 ]. In his 1898 dissertation, Dorrie [18] used Dirich-

let’s technique to derive the quadratic reciprocity law in quadratic fields withclass number 1, and in 1902 Ruckle [111] and K.S. Hilbert [57] studied theproduct formula for quadratic Hilbert symbols in algebraic number fields, inparticular in Z[ζ2n ]. Welmin [136] finally used Dirichlet’s method to derivethe quadratic reciprocity law in arbitrary quadratic number fields from theone in rational integers. The possibility of such a proof was realized indepen-dently by Hecke [43, footnote 2, p. 28], who derived the quadratic reciprocitylaw in real quadratic number fields using generalized Gauss sums. Hecke alsoproved the following version of the second supplementary law:

Theorem 12.22. Let k be a real quadratic number field, let l1 and l2 beprime ideals above 2, and let λ ∈ Ok be an algebraic integer with (λ) = la1

1 la22

(put a2 = 0 if there is only one prime ideal above 2 in k). Assume that

λ ≡ ξ2 mod

4 if a1 ≡ a2 ≡ 0 mod 2,4l1 if a1 ≡ 1, a2 ≡ 0 mod 2,4l2 if a1 ≡ 0, a2 ≡ 1 mod 2,4l1l2 if a1 ≡ a2 ≡ 1 mod 2.

Then(λα

)= (−1)S , S =

sgn α− 12

sgn λ− 12

+sgn α′ − 1

2sgn λ′ − 1

2.

Exercises 19

Our discussion of quadratic reciprocity in quadratic number fields isloosely based on Skolem’s approach in [122]; Skolem rediscovered and thensimplified Welmin’s results, and was able to see how to generalize Dirichlet’stechnique to arbitrary number fields; in practice, however, this method willbe far too technical for deriving explicit quadratic reciprocity laws in numberfields of degree > 2.

Eisenstein’s way of formulating the quadratic reciprocity law (see Hy-pothesis (H) on p. 1, and Theorem 12.8) goes back to [23], where the case ofodd primes ` is discussed. Eisenstein does not deal with quadratic reciprocitythere, but he was aware of its exceptional status; [21, p. 619], he writes:

dieses bildet ubrigens hier einen Ausnahmefall, da der Unterschiedzwischen pos. und neg. Zahlen nicht durch Congruenzen dargestelltwerden kann.1

This problem was later overcome by Hilbert and Furtwangler, who inventedinfinite primes in order to simplify statements in class field theory; Hasse thenintroduced congruences modulo infinite primes and was thus able to expressthe “difference between positive and negative numbers” by such congruences.

E.F. Stueben also has studied quadratic reciprocity in quadratic fields,but his publication [126] was not available to me.

The explicit quadratic reciprocity laws given in this chapter do certainlynot exhaust the subject, and there are lots of problems remaining to besolved. Consider e.g. a quadratic number field K, let 2 denote a prime idealabove 2, and let a and b denote ideals of odd norm in the ideal class [2]−1.Then 2a = (α) and 2b = (β) for α, β ∈ OK , and we may ask when [αq ][βp ] isindependent of the choices of α and β, and what its value is.

Exercises

12.1 Assume that (2/A) only depends on A mod 2σ for some integer σ, and showthat this implies that that we can take σ = 3.

12.2 Let K/k be a quadratic extension. Show that the set of primitive elementsdoes not form a monoid. Prove that if α and β are primitive and coprime,then so are αβ′ and α′β.

12.3 Show that {α}{α′} = sgn A.

12.4 Let K be an algebraic number field of degree n, let O denote the ring ofintegers in K, and put M = (O/4O)×. Show that squaring induces an exactsequence

1 −−−−−→ M [2] −−−−−→ M −−−−−→ M2 −−−−−→ 1,

where M [2] = α ∈M : α2 = 1. Next show that β+2O 7−→ 1+2β+4O definesan isomorphism O/2O −→ M [2]. Finally, use an integral basis to prove thatO/2O ' (Z/2Z)n, and deduce that this implies M/M2 ' (Z/2Z)n.

1 this forms an exception here, since the difference between pos[itive] and neg[ative]numbers cannot be represented by congruences.


12.5 Let k = Q(√m ) be a quadratic number field, and assume that α ∈ Ok satisfies

α ≡ 1 mod 2. Show that Nα−12

≡ Tr α−12

mod 2.

12.6 (continued) Deduce from the preceding exercises that, for totally positive α ≡1 mod 2, we have [−1

α] = (−1)T , where T = Tr α−1

2. Deduce that [−1

α] only

depends on the residue class α mod 4 (as long as α is totally positive).

12.7 Consider the quadratic number field k = Q(√m ) with m ≡ 2, 3 mod 4, and

let l denote the prime ideal above 2 (thus l = (2,√m ) if m ≡ 2 mod 4, and

l = (2, 1 +√m ) if m ≡ 3 mod 4). Show that T = Tr (α−1

2β−1

2) ∈ Z whenever

α ≡ β ≡ 1 mod l, and that the formula (α/β)(β/α) = (−1)T for the quadraticreciprocity law that we have proved for totally positive α ≡ β ≡ 1 mod 2 doesnot hold in general under the weaker assumption α ≡ β ≡ 1 mod l.

12.8 For α = a+ b√

2 ∈ Z[√

2 ] with α ≡ 1 mod 2, show that [√

2/α] = (−2/a).

Hints: Use (12.15) to show that [√

2α

] = [ 2√

2α

] = [α+2√

2α

] = [ α

α+2√

2] =

√2

α+2√

2].

Now observe that α ≡ a mod 2√

2.

12.9 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 2n; for α ∈ Z[

√m ] with

odd positive norm and α ≡ 1 mod 4 we have

hλα

i=

(+1 if α ≡ ξ2 mod 4l,

−1 if α 6≡ ξ2 mod 4l,

where l = (2, 1 +√m ) is the prime ideal of norm 2.

Hints: write λ = r + s√m and α = a + b

√m with r ≡ s ≡ a ≡ 1 mod 2 and

4 | b.1. Show that ( b

Nα) = ( 2

Nα).

2. Show that [ λα] = ( 2

Nα)( rb−sa

Nα).

3. Use quadratic reciprocity in Z to show that ( rb−saNα

) = ( 2rb−sa

).

12.10 Derive the quartic reciprocity law in Z[i] from Eisenstein’s hypothesis (E) forprime powers.

12.11 Derive the second supplementary law for quartic residues from the quarticreciprocity law in Z[i].

12.12 Let k be a number field, let ∞ denote the product of all real places, and let2Ok = le1

1 · · · lerr be the prime ideal factorization of 2. Show that if an element

α ∈ Ok coprime to 2 is a square modulo l2ei+1i , then it is a square modulo lri

for all r ≥ 1.

12.13 Let k be a number field, let ∞ denote the product of all real places, and let2Ok = le1

1 · · · lerr be the prime ideal factorization of 2. If Eisenstein’s quadratic

reciprocity law holds modulo m∞ in k, where m is a product of powers of primeideals dividing (2), then it holds modulo 4l1 · · · lr∞.

12.14 Let k be a number field with class number 1; show that if Eisenstein’s quad-ratic reciprocity law holds in k, then it holds in any quadratic extension K ofk.

12.15 Consider the ring OK = Z[ω] of algebraic integers in the cubic number fieldK = Q(ω), where ω3 = 2; its fundamental unit is ε = ω − 1.1. Compute [ ε

π] for the primes π = 1 + ω and π = 1 + ω2;

2. Assuming that Eisenstein’s quadratic reciprocity law holds in K, showthat [ ε

π] only depends on the residue class π mod 4OK if Nπ > 0;

Exercises 21

3. Show that any element in OK with odd norm is associated to an elementof the form a+ bω + cω2, where a− 1 ≡ b ≡ 0 mod 4 and c ≡ 0 mod 2.

4. Prove that [ επ] = (−1)c/2, where π = a+bω+cω2 and a−1 ≡ b ≡ 0 mod 4,

c ≡ 0 mod 2.This result is due to Aigner [3].

12.16 Assume that in the cubic number field K = Q(ω) with ω3 = 2 we have thereciprocity law hα

β

ihβα

i= (−1)Tr α−1

2β−1

2

for all (totally) positive α, β ∈ Z[ω]. Show that this implies the result of thepreceding exercise.

12.17 Prove the following analog of Theorem 12.21:Let K be a quadratic number field. Then the following assertions are equiva-lent:1. ωK× 2 ∈ Sel4(K);2. [ω

α] = 1 for all totally positive α coprime to 2ω

12.18 (continued) Find an analog of Theorem 12.21 involving Sel+(K) and prove it.

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