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# Jacobsthal identity for Q( √−2)

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(1)

## −2)

Yifan Yang (with Ki-Ichiro Hashimoto and Ling Long)

National Chiao Tung University, Taiwan

5 November 2010, NCKU Colloquium

(2)

## Jacobsthal’s identity

Theorem (Fermat)

An odd prime p is a sum of two integer squares if and only if p ≡ 1 mod 4.

Theorem (Jacobsthal)

Let p be a prime congruent to 1 modulo 4 and n be a quadratic nonresidue modulo p. Set

A = 1 2

p−1

X

x =0

 x3− x p



, B = 1

2

p−1

X

x =0

 x3− nx p

 .

Then A, B ∈ Z and satisfies p = A2+B2.

(3)

## Jacobsthal’s identity

Theorem (Fermat)

An odd prime p is a sum of two integer squares if and only if p ≡ 1 mod 4.

Theorem (Jacobsthal)

Let p be a prime congruent to 1 modulo 4 and n be a quadratic nonresidue modulo p. Set

A = 1 2

p−1

X

x =0

 x3− x p



, B = 1

2

p−1

X

x =0

 x3− nx p

 .

Then A, B ∈ Z and satisfies p = A2+B2.

(4)

## Legendre symbols

Definition

Let p be an odd prime. An integer a relatively prime to p is aquadratic residue(resp.quadratic nonresidue) modulo p if the congruence equation

x2≡ a mod p is solvable (resp. unsolvable) in integers.

Definition

Let p be an odd prime. Then theLegendre symbol

· p



is defined by

 a p



=





0, if p|a,

1, if a is a quadratic residue modulo p,

−1, if a is a quadratic nonresidue modulo p.

(5)

## Legendre symbols

Definition

Let p be an odd prime. An integer a relatively prime to p is aquadratic residue(resp.quadratic nonresidue) modulo p if the congruence equation

x2≡ a mod p is solvable (resp. unsolvable) in integers.

Definition

Let p be an odd prime. Then theLegendre symbol

· p



is defined by

 a p



=





0, if p|a,

1, if a is a quadratic residue modulo p,

−1, if a is a quadratic nonresidue modulo p.

(6)

## Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

Jf(p) :=

p−1

X

x =0

 f (x) p



aJacobsthal sum.

Proposition We have

•  ab p



= a p

  b p

 ,

•  a

≡ a(p−1)/2 mod p.

(7)

## Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

Jf(p) :=

p−1

X

x =0

 f (x) p



aJacobsthal sum.

Proposition We have

•  ab p



= a p

  b p

 ,

•  a p



≡ a(p−1)/2 mod p.

(8)

## Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

Jf(p) :=

p−1

X

x =0

 f (x) p



aJacobsthal sum.

Proposition We have

•  ab p



= a p

  b p

 ,

•  a

≡ a(p−1)/2 mod p.

(9)

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

 x3− nx p

 .

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r2n) = r p

 S(n).

• Let g be a primitive root modulo p. The above shows S(g) = −S(g3) =S(g5) = −S(g7) = . . . , S(g2) = −S(g4) =S(g6) = −S(g8) = . . . .

(10)

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

 x3− nx p

 .

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r2n) = r p

 S(n).

• Let g be a primitive root modulo p. The above shows S(g) = −S(g3) =S(g5) = −S(g7) = . . . , S(g2) = −S(g4) =S(g6) = −S(g8) = . . . .

(11)

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

 x3− nx p

 .

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r2n) = r p

 S(n).

• Let g be a primitive root modulo p. The above shows S(g) = −S(g3) =S(g5) = −S(g7) = . . . , S(g2) = −S(g4) =S(g6) = −S(g8) = . . . .

(12)

## Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

 x3− nx p

 .

• Pairing the term x = a with the term x = p − a, we find S(n) is always even.

• Replacing x by rx , we find S(r2n) = r p

 S(n).

• Let g be a primitive root modulo p. The above shows S(g) = −S(g3) =S(g5) = −S(g7) = . . . , S(g2) = −S(g4) =S(g6) = −S(g8) = . . . .

(13)

## Gauss’ proof of the Jacobsthal identity, continued

• Let S(g) = 2A and S(g2) =2B. Then 2(p − 1)(A2+B2) =

p−1

X

n,x ,y =0

 x3− nx p

  y3− ny p



=

p−1

X

x ,y =0

 xy p

p−1 X

n=0

 (x2− n)(y2− n) p

 .

• Using

p−1

X

z=0

 z(z + r ) p



=

(p − 1, if r ≡ 0 mod p,

−1, if r 6≡ 0 mod p, we find

2(p − 1)(A2+B2) =p

p−1

X

x ,y =0

δx2,y2 =2(p − 1)p.

(14)

## Gauss’ proof of the Jacobsthal identity, continued

• Let S(g) = 2A and S(g2) =2B. Then 2(p − 1)(A2+B2) =

p−1

X

n,x ,y =0

 x3− nx p

  y3− ny p



=

p−1

X

x ,y =0

 xy p

p−1 X

n=0

 (x2− n)(y2− n) p

 .

• Using

p−1

X

z=0

 z(z + r ) p



=

(p − 1, if r ≡ 0 mod p,

−1, if r 6≡ 0 mod p, we find

2(p − 1)(A2+B2) =p

p−1

X δx2,y2 =2(p − 1)p.

(15)

## Arithmetic-geometric approach

Idea.

Consider the elliptic curve En:y2=x3− nx. We have

#En(Fp) =1 +

p−1

X

x =0



1 + x3− nx p



=p + 1 +

p−1

X

x =0

 x3− nx p

 .

Thus,

L(En/Q, s)−1=Y

p

1 +

p−1

X

x =0

 x3− nx p



p−s+p1−2s

.

Since E1and En are isomorphic over Q(√4

n), the two L-functions L(E1/Q, s) and L(En/Q, s) must be related in some way, which give information about the Jacobsthal sums.

(16)

## Arithmetic-geometric approach

Idea.

Consider the elliptic curve En:y2=x3− nx. We have

#En(Fp) =1 +

p−1

X

x =0



1 + x3− nx p



=p + 1 +

p−1

X

x =0

 x3− nx p

 .

Thus,

L(En/Q, s)−1=Y

p

1 +

p−1

X

x =0

 x3− nx p



p−s+p1−2s

. Since E1and En are isomorphic over Q(√4

n), the two L-functions L(E1/Q, s) and L(En/Q, s) must be related in some way, which give

(17)

## Arithmetic-geometric approach

Idea.

Consider the elliptic curve En:y2=x3− nx. We have

#En(Fp) =1 +

p−1

X

x =0



1 + x3− nx p



=p + 1 +

p−1

X

x =0

 x3− nx p

 .

Thus,

L(En/Q, s)−1=Y

p

1 +

p−1

X

x =0

 x3− nx p



p−s+p1−2s

.

Since E1and En are isomorphic over Q(√4

n), the two L-functions L(E1/Q, s) and L(En/Q, s) must be related in some way, which give information about the Jacobsthal sums.

(18)

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and E [`n]be the subgroup of `n-torsion points.

Consider the Tate module

T`(E ) = lim

←−E [`n].

The absolute Galois group GK = Gal(Q/K ) acts on T`(E ), yielding a Galois representation

ρE ,`:GK → GL(2, Q`).

Then L(ρE ,`,s) = L(E /K , s).

(19)

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and E [`n]be the subgroup of `n-torsion points.

Consider the Tate module

T`(E ) = lim

←−E [`n].

The absolute Galois group GK = Gal(Q/K ) acts on T`(E ), yielding a Galois representation

ρE ,`:GK → GL(2, Q`).

Then L(ρE ,`,s) = L(E /K , s).

(20)

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and E [`n]be the subgroup of `n-torsion points.

Consider the Tate module

T`(E ) = lim

←−E [`n].

The absolute Galois group GK = Gal(Q/K ) acts on T`(E ), yielding a Galois representation

ρE ,`:GK → GL(2, Q`).

Then L(ρE ,`,s) = L(E /K , s).

(21)

## Tate modules and Galois representations

Let ` be a prime. Let E be an elliptic curve over a number field K and E [`n]be the subgroup of `n-torsion points.

Consider the Tate module

T`(E ) = lim

←−E [`n].

The absolute Galois group GK = Gal(Q/K ) acts on T`(E ), yielding a Galois representation

ρE ,`:GK → GL(2, Q`).

Then L(ρE ,`,s) = L(E /K , s).

(22)

## A lemma

Lemma (Clifford)

(Under suitable conditions on G and ρ) Assume that H G and G/H is cyclic of finite order.

Assume that ρ1:G → GL(V1)and ρ2:G → GL(V2)are irreducible representations over an algebraically closed of characteristic not dividing |G/H| such that ρ1

Hand ρ2

Hhave a common isomorphic irreducible subrepresentations of H.

Then

ρ1' ρ2⊗ χ

for some representation of G of degree 1 that is lifted from that of G/H.

(23)

## A lemma

Lemma (Clifford)

(Under suitable conditions on G and ρ) Assume that H G and G/H is cyclic of finite order.

Assume that ρ1:G → GL(V1)and ρ2:G → GL(V2)are irreducible representations over an algebraically closed of characteristic not dividing |G/H| such that ρ1

Hand ρ2

Hhave a common isomorphic irreducible subrepresentations of H.

Then

ρ1' ρ2⊗ χ

for some representation of G of degree 1 that is lifted from that of G/H.

(24)

## A lemma

Lemma (Clifford)

(Under suitable conditions on G and ρ) Assume that H G and G/H is cyclic of finite order.

Assume that ρ1:G → GL(V1)and ρ2:G → GL(V2)are irreducible representations over an algebraically closed of characteristic not dividing |G/H| such that ρ1

Hand ρ2

Hhave a common isomorphic irreducible subrepresentations of H.

Then

ρ1' ρ2⊗ χ

for some representation of G of degree 1 that is lifted from that of G/H.

(25)

## Arithmetic-geometric approach

Let En:y2=x3− nx. It is isomorphic to E1over Q(√4

n), which is not abelian over Q.

Extend the base field to K = Q(i). Then L = Q(√4

n, i) is cyclic over Q.

Let GK = Gal(Q/K ) and GL= Gal(Q/L).

The elliptic curves Enhave CM by Z[i], so ρEn,`

GK = πn⊕ πn, where πnare representations of GK of degree 1.

(26)

## Arithmetic-geometric approach

Let En:y2=x3− nx. It is isomorphic to E1over Q(√4

n), which is not abelian over Q.

Extend the base field to K = Q(i). Then L = Q(√4

n, i) is cyclic over Q.

Let GK = Gal(Q/K ) and GL= Gal(Q/L).

The elliptic curves Enhave CM by Z[i], so ρEn,`

GK = πn⊕ πn, where πnare representations of GK of degree 1.

(27)

## Arithmetic-geometric approach

Let En:y2=x3− nx. It is isomorphic to E1over Q(√4

n), which is not abelian over Q.

Extend the base field to K = Q(i). Then L = Q(√4

n, i) is cyclic over Q.

Let GK = Gal(Q/K ) and GL= Gal(Q/L).

The elliptic curves Enhave CM by Z[i], so ρEn,`

GK = πn⊕ πn, where πnare representations of GK of degree 1.

(28)

## Arithmetic-geometric approach

E1/K and En/K are isomorphic over L, so π1

GL ' πn GL. By the lemma above,

πn= π1⊗ χ

for some linear character χ on GK with GL⊂ kerχ, i.e., a character on GK/GL' Gal(L/K ).

(29)

## Arithmetic-geometric approach

E1/K and En/K are isomorphic over L, so π1

GL ' πn GL. By the lemma above,

πn= π1⊗ χ

for some linear character χ on GK with GL⊂ kerχ, i.e., a character on GK/GL' Gal(L/K ).

(30)

## Arithmetic-geometric approach

A character on GK with GL⊂ kerχ has the following description. The Galois group Gal(L/K ) is generated by

σ :√4

n 7−→ i√4 n.

For each prime p of K not dividing 2n, the Frobenius Frobp is the element σj ∈ Gal(L/K ) such that

σj(√4

n) ≡ (√4

n)Np mod p, where Np denotes the norm of p.

Then there exists k ∈ {1, 3} such that χ satisfies χ(Frobp) =ijk

(31)

## Arithmetic-geometric approach

A character on GK with GL⊂ kerχ has the following description. The Galois group Gal(L/K ) is generated by

σ :√4

n 7−→ i√4 n.

For each prime p of K not dividing 2n, the Frobenius Frobp is the element σj ∈ Gal(L/K ) such that

σj(√4

n) ≡ (√4

n)Np mod p, where Np denotes the norm of p.

Then there exists k ∈ {1, 3} such that χ satisfies χ(Frobp) =ijk for all p.

(32)

## Arithmetic-geometric approach

A character on GK with GL⊂ kerχ has the following description. The Galois group Gal(L/K ) is generated by

σ :√4

n 7−→ i√4 n.

For each prime p of K not dividing 2n, the Frobenius Frobp is the element σj ∈ Gal(L/K ) such that

σj(√4

n) ≡ (√4

n)Np mod p, where Np denotes the norm of p.

Then there exists k ∈ {1, 3} such that χ satisfies χ(Frobp) =ijk

(33)

## Proof of Jacobsthal’s identity

Now for a prime p ≡ 1 mod 4, a prime of K lying over p has norm p.

If n is a quadratic nonresidue modulo p, then n(p−1)/2 ≡ −1 mod p, which implies that

(√4

n)Np≡ ±i√4

n mod p.

That is,

χ(Frobp) = ±i.

(34)

## Proof of Jacobsthal’s identity

Now for a prime p ≡ 1 mod 4, a prime of K lying over p has norm p.

If n is a quadratic nonresidue modulo p, then n(p−1)/2 ≡ −1 mod p, which implies that

(√4

n)Np≡ ±i√4

n mod p.

That is,

χ(Frobp) = ±i.

(35)

## Proof of Jacobsthal’s identity

Now for a prime p ≡ 1 mod 4, a prime of K lying over p has norm p.

If n is a quadratic nonresidue modulo p, then n(p−1)/2 ≡ −1 mod p, which implies that

(√4

n)Np≡ ±i√4

n mod p.

That is,

χ(Frobp) = ±i.

(36)

## Proof of Jacobsthal’s identity

It is well-known that L(E1/Q, s) = Y

p≡1 mod 4

1

1 − 2papp−s+p1−2s

Y

p≡3 mod 4

1 1 + p1−2s, where for p ≡ 1 mod 4, apand bpare positive integers with ap odd and bp even such that p = a2p+b2p, and

p= −1 ap



(−1)bp/2.

Thus, for a prime p of K = Q(i) lying over p ≡ 1 mod 4, π1(p) = ±ap± bpi

Then

πn(p) = π1(p)χ(p) = ±bp± api

(37)

## Proof of Jacobsthal’s identity

It is well-known that L(E1/Q, s) = Y

p≡1 mod 4

1

1 − 2papp−s+p1−2s

Y

p≡3 mod 4

1 1 + p1−2s, where for p ≡ 1 mod 4, apand bpare positive integers with ap odd and bp even such that p = a2p+b2p, and

p= −1 ap



(−1)bp/2.

Thus, for a prime p of K = Q(i) lying over p ≡ 1 mod 4, π1(p) = ±ap± bpi

Then

πn(p) = π1(p)χ(p) = ±bp± api

(38)

## Proof of Jacobsthal’s identity

It is well-known that L(E1/Q, s) = Y

p≡1 mod 4

1

1 − 2papp−s+p1−2s

Y

p≡3 mod 4

1 1 + p1−2s, where for p ≡ 1 mod 4, apand bpare positive integers with ap odd and bp even such that p = a2p+b2p, and

p= −1 ap



(−1)bp/2.

Thus, for a prime p of K = Q(i) lying over p ≡ 1 mod 4, π1(p) = ±ap± bpi

Then

πn(p) = π1(p)χ(p) = ±bp± api

(39)

## Proof of Jacobsthal’s identity

Therefore, the p-factor of L(En,s) is

(1 ± 2bpp−s+p1−2s)−1. That is,

p−1

X

x =0

 x3− nx p



= ±2bp, which gives us the Jacobsthal identity.

(40)

## Proof of Jacobsthal’s identity

Therefore, the p-factor of L(En,s) is

(1 ± 2bpp−s+p1−2s)−1. That is,

p−1

X

x =0

 x3− nx p



= ±2bp, which gives us the Jacobsthal identity.

(41)

## Cubic analogue of the Jacobsthal identity

Theorem (Chan-Long-Y)

Let p ≡ 1 mod 6. Assume that n is an integer such that x3≡ n mod p is not solvable in integers. Set

A =

p−1

X

x =0

 x3− 1 p



, B =

p−1

X

x =0

 x3− n p

 .

Then

A2+AB + B2=3p.

(42)

## Cubic analogue of the Jacobsthal identity

Theorem (Chan-Long-Y)

Let p ≡ 1 mod 6. Assume that n is an integer such that x3≡ n mod p is not solvable in integers. Set

A =

p−1

X

x =0

 x3− 1 p



, B =

p−1

X

x =0

 x3− n p

 .

Then

A2+AB + B2=3p.

(43)

## Question

Let −d be the discriminant of an imaginary quadratic number field such that Q(√

−d ) has class number 1.

Let

f (x , y ) =

(x2+ (d /4)y2, if d ≡ 0 mod 4, x2+xy + ((1 + d )/4)y2, if d ≡ 3 mod 4.

Then whether p = f (x , y ) is solvable depends only on

−d p

 . Question. When

−d p



=1, can we express the integers A and B in p = f (A, B) in terms of Jacobsthal sums in a uniform way?

(44)

## Question

Let −d be the discriminant of an imaginary quadratic number field such that Q(√

−d ) has class number 1.

Let

f (x , y ) =

(x2+ (d /4)y2, if d ≡ 0 mod 4, x2+xy + ((1 + d )/4)y2, if d ≡ 3 mod 4.

Then whether p = f (x , y ) is solvable depends only on

−d p

 . Question. When

−d p



=1, can we express the integers A and B in p = f (A, B) in terms of Jacobsthal sums in a uniform way?

(45)

## Question

Let −d be the discriminant of an imaginary quadratic number field such that Q(√

−d ) has class number 1.

Let

f (x , y ) =

(x2+ (d /4)y2, if d ≡ 0 mod 4, x2+xy + ((1 + d )/4)y2, if d ≡ 3 mod 4.

Then whether p = f (x , y ) is solvable depends only on

−d p

 . Question. When

−d p



=1, can we express the integers A and B in p = f (A, B) in terms of Jacobsthal sums in a uniform way?

(46)

## −2), Part I

Theorem (Hashimoto-Long-Y)

Assume that p ≡ 1 mod 8 and n is a quadratic nonresidue modulo p.

Set

A = 1 2

p−1

X

x =0

 x3+4x2+2x p



, B = 1

4

p−1

X

x =0

 x5+nx p

 .

Then A and B are integers satisfying p = A2+2B2.

(47)

## −d ), Part II

Theorem (Hashimoto-Long-Y) Assume that p ≡ 3 mod 8. Set

A = 1 2

p−1

X

x =0

 x3+4x2+2x p

 ,

B = 1 4

1 +

p−1

X

x =0

 x6+4x5+10x4− 20x2− 16x − 8 p



. Then A and B are integers satisfying p = A2+2B2.

(48)

2

3

2

## + 2x

Lemma

The elliptic curve y2=x3+4x2+2x has CM by Z[√

−2] and its L-function is

Y

p≡1,3 mod 8

1

1 − 2papp−s+p1−2s

Y

p≡5,7 mod 8

1 1 + p1−2s, where apand bpare positive integers such that p = a2p+2b2pand

p =

2(−1)bp/2

−2 ap



, if p ≡ 1 mod 8,

−2

−2 ap



, if p ≡ 3 mod 8.

(49)

2

5

## + x

Lemma

For C : y2=x5+x , we have

L(C/Q, s) = L(E1/Q, s)L(E2/Q, s), (1) where E1:y2=x3+4x2+2x , E2:y2=x3− 4x2+2x .

Proof.

There are 2-to-1 coverings

(x , y ) 7−→ (X , Y ) = (x ± 1)2

x ,y (x ± 1) x2



from C to E1and E2. Considering the associated Galois representations, we get (1).

(50)

2

5

## + x

Lemma

For C : y2=x5+x , we have

L(C/Q, s) = L(E1/Q, s)L(E2/Q, s), (1) where E1:y2=x3+4x2+2x , E2:y2=x3− 4x2+2x .

Proof.

There are 2-to-1 coverings

(x , y ) 7−→ (X , Y ) = (x ± 1)2

x ,y (x ± 1) x2



from C to E1and E2. Considering the associated Galois representations, we get (1).

(51)

2

5

## + x

Corollary

For C : y2=x5+x , let

1

(1 − αp,1p−s) . . . (1 − αp,4p−s) be the p-factor of L(C/Q, s).

• If p ≡ 1 mod 8, then αp,j = −2

a



(−1)b/2(a ± b√

−2),

each with multiplicity 2, where a and b are the positive integers such that p = a2+2b2.

• If p ≡ 3 mod 8, then αp,j = ±a ± b√

−2, where a and b are integers such that p = a2+2b2.

(52)

2

5

## + x

Corollary

For C : y2=x5+x , let

1

(1 − αp,1p−s) . . . (1 − αp,4p−s) be the p-factor of L(C/Q, s).

• If p ≡ 1 mod 8, then αp,j = −2

a



(−1)b/2(a ± b√

−2),

each with multiplicity 2, where a and b are the positive integers such that p = a2+2b2.

• If p ≡ 3 mod 8, then αp,j = ±a ± b√

−2, where a and b are

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2

6

5

4

2

## − 16x − 8

Lemma

The hyperelliptic curve X1:y2=x6+4x5+10x4− 20x2− 16x − 8 is isomorphic to X2:y2=x5+x over a field of degree 16 over Q, which is cyclic of degree 4 over Q(ζ8).

Proof.

Setting

x =

2(x1+1)

x1− 1 , y = y1 (x1− 1)3, we get y12=128(2 +√

2)x1(x14+3 − 2√ 2).

The proof of the theorem follows the argument in the case of the classical Jacobsthal identity (although more complicated).

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2

6

5

4

2

## − 16x − 8

Lemma

The hyperelliptic curve X1:y2=x6+4x5+10x4− 20x2− 16x − 8 is isomorphic to X2:y2=x5+x over a field of degree 16 over Q, which is cyclic of degree 4 over Q(ζ8).

Proof.

Setting

x =

2(x1+1)

x1− 1 , y = y1 (x1− 1)3, we get y12=128(2 +√

2)x1(x14+3 − 2√ 2).

The proof of the theorem follows the argument in the case of the

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2

6

5

4

2

## − 16x − 8

Lemma

The hyperelliptic curve X1:y2=x6+4x5+10x4− 20x2− 16x − 8 is isomorphic to X2:y2=x5+x over a field of degree 16 over Q, which is cyclic of degree 4 over Q(ζ8).

Proof.

Setting

x =

2(x1+1)

x1− 1 , y = y1 (x1− 1)3, we get y12=128(2 +√

2)x1(x14+3 − 2√ 2).

The proof of the theorem follows the argument in the case of the classical Jacobsthal identity (although more complicated).

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## How do we find the curves?

Assume p ≡ 1, 3 mod 8 and p = a2+2b2= (a + b√

−2)(a − b√

−2).

If C : y2=f (x ) is a curve such that the p-factor of L(C/Q, s) is 1

(1 ± (a + b√

−2)p−s)(1 ± (a − b√

−2)p−s), then

p−1

X

x =0

 f (x) p



= ±2a.

Thus, we are looking at elliptic curves with CM by Z[√

−2].

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## How do we find the curves?

Assume p ≡ 1, 3 mod 8 and p = a2+2b2= (a + b√

−2)(a − b√

−2).

If C : y2=f (x ) is a curve such that the p-factor of L(C/Q, s) is 1

(1 ± (a + b√

−2)p−s)(1 ± (a − b√

−2)p−s), then

p−1

X

x =0

 f (x) p



= ±2a.

Thus, we are looking at elliptic curves with CM by Z[√

−2].

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## How do we find the curves?

Assume p ≡ 1, 3 mod 8 and p = a2+2b2= (a + b√

−2)(a − b√

−2).

If C : y2=f (x ) is a curve such that the p-factor of L(C/Q, s) is 1

(1 ± (a + b√

−2)p−s)(1 ± (a − b√

−2)p−s), then

p−1

X

x =0

 f (x) p



= ±2a.

Thus, we are looking at elliptic curves with CM by Z[√

−2].

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## How do we find the curves?

To get b, we observe that (a + b√

−2)(ζ8+ ζ83) + (a − b√

−2)(ζ85+ ζ87) = −4b.

Thus, we are looking for curves y2=f (x ) whose L-function has p-factor

1 (1 ± ζ8(a + b√

−2)p−s) . . . (1 ± ζ87(a − b√

−2)p−s), i.e., a hyperelliptic curve of genus 2.

To find such a curve, we shall find “its L-function” first.

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## How do we find the curves?

To get b, we observe that (a + b√

−2)(ζ8+ ζ83) + (a − b√

−2)(ζ85+ ζ87) = −4b.

Thus, we are looking for curves y2=f (x ) whose L-function has p-factor

1 (1 ± ζ8(a + b√

−2)p−s) . . . (1 ± ζ87(a − b√

−2)p−s), i.e., a hyperelliptic curve of genus 2.

To find such a curve, we shall find “its L-function” first.

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## How do we find the curves?

To get b, we observe that (a + b√

−2)(ζ8+ ζ83) + (a − b√

−2)(ζ85+ ζ87) = −4b.

Thus, we are looking for curves y2=f (x ) whose L-function has p-factor

1 (1 ± ζ8(a + b√

−2)p−s) . . . (1 ± ζ87(a − b√

−2)p−s), i.e., a hyperelliptic curve of genus 2.

To find such a curve, we shall find “its L-function” first.

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## Hecke characters

Let K be a number field. For each place v , let Kv be the completion of K with respect to | · |v and Ov be the valuation ring of Kv when v is a finite place.

Let

IK = (

(xv) ∈Y

v

Kv :xv ∈ Ov for all but finitely many v )

be the idele group of K , equipped with the product topology.

Definition

AHecke character (Grössencharakter)χis a continuous group homomorphism from the idele class group IK/K to C.

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## Hecke characters

Let K be a number field. For each place v , let Kv be the completion of K with respect to | · |v and Ov be the valuation ring of Kv when v is a finite place.

Let

IK = (

(xv) ∈Y

v

Kv :xv ∈ Ov for all but finitely many v )

be the idele group of K , equipped with the product topology.

Definition

AHecke character (Grössencharakter)χis a continuous group homomorphism from the idele class group IK/K to C.

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## Hecke characters

Let K be a number field. For each place v , let Kv be the completion of K with respect to | · |v and Ov be the valuation ring of Kv when v is a finite place.

Let

IK = (

(xv) ∈Y

v

Kv :xv ∈ Ov for all but finitely many v )

be the idele group of K , equipped with the product topology.

Definition

AHecke character (Grössencharakter)χis a continuous group homomorphism from the idele class group IK/K to C.

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## Hecke L-functions and their functional equations

Definition

Let χ be a Hecke character. Write χ =Q

vχv. TheHecke L-functionis defined by

L(s, χ) = Y

v finite, χv(Ov)=1

1

1 − χvv)Nv−s,

where πv is any uniformizer of Kv and Nv is the norm of the prime ideal corresponding to v .

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## Hecke L-functions and their functional equations

Proposition

Let K be an imaginary quadratic number field. Suppose that k is the positive integer such that |χ(x )| = |x |k −1for all x ∈ IK/K. Setting

Λ(s, χ) = 2π pdKdχ

!−s

Γ(s)L(s, χ),

we have

Λ(s, χ) = Λ(k − s, χ)

for some root of unity , where dK is the discriminant of K and dχis the norm of the modulus of χ.

Remark

We getCM modular formsfrom Hecke characters on imaginary

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## Hecke L-functions and their functional equations

Proposition

Let K be an imaginary quadratic number field. Suppose that k is the positive integer such that |χ(x )| = |x |k −1for all x ∈ IK/K. Setting

Λ(s, χ) = 2π pdKdχ

!−s

Γ(s)L(s, χ),

we have

Λ(s, χ) = Λ(k − s, χ)

for some root of unity , where dK is the discriminant of K and dχis the norm of the modulus of χ.

Remark

We getCM modular formsfrom Hecke characters on imaginary quadratic number field.

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## Finding curves

• Let K = Q(√

−2). We construct Hecke characters χ1and χ2of modulus 8 so that χ takes value ζ8j(a + b√

−2).

• We then look for a hyperelliptic curve whose L-function coincide with L(s, χ1)L(s, χ2). Specifically, we look for such a curve among hyperelliptic curves with an automorphism defined over Q(√

−2).

• In practice, we consider curves

y2=x6+mx5+nx4− 2nx2− 4mx − 8, which has an automorphism

(x , y ) 7−→ 2 x,

√−8y x3

 ,

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## Finding curves

• Let K = Q(√

−2). We construct Hecke characters χ1and χ2of modulus 8 so that χ takes value ζ8j(a + b√

−2).

• We then look for a hyperelliptic curve whose L-function coincide with L(s, χ1)L(s, χ2). Specifically, we look for such a curve among hyperelliptic curves with an automorphism defined over Q(√

−2).

• In practice, we consider curves

y2=x6+mx5+nx4− 2nx2− 4mx − 8, which has an automorphism

(x , y ) 7−→ 2 x,

√−8y x3

 ,

and search for m and n such that the L-function is L(s, χ1)L(s, χ2).

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## Finding curves

• Let K = Q(√

−2). We construct Hecke characters χ1and χ2of modulus 8 so that χ takes value ζ8j(a + b√

−2).

• We then look for a hyperelliptic curve whose L-function coincide with L(s, χ1)L(s, χ2). Specifically, we look for such a curve among hyperelliptic curves with an automorphism defined over Q(√

−2).

• In practice, we consider curves

y2=x6+mx5+nx4− 2nx2− 4mx − 8, which has an automorphism

(x , y ) 7−→ 2 x,

√−8y x3

 ,

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## Problem

Problem. For each imaginary quadratic number field K with class number 1, find an analogous identity.

We now describe the architecture of the Q-function of the bridge-bidding problem. We initialize l separate Q-functions, where l is the total number of bids. For the first Q-

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