Jacobsthal identity for Q( √−2)

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Jacobsthal identity for Q( √

−2)

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

National Chiao Tung University, Taiwan

5 November 2010, NCKU Colloquium

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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.

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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.

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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.

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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.

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Properties of Legendre symbols

Definition

If f (x ) ∈ Z[x], then we call

J_f(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.

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Properties of Legendre symbols

Definition

J_f(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.

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Properties of Legendre symbols

Definition

J_f(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.

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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) = . . . .

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Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x³− nx p

.

S(n).

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Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x³− nx p

.

S(n).

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Gauss’ proof of the Jacobsthal identity

• Set S(n) =

p−1

X

x =0

x³− nx p

.

S(n).

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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

δ_x2,y² =2(p − 1)p.

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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 δ_x2,y² =2(p − 1)p.

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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(E_n/Q, s)⁻¹=Y

p



1 +

p−1

X

x =0

x³− nx p

p^−s+p^1−2s



.

Since E₁and E_n 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.

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Arithmetic-geometric approach

Idea.

#En(Fp) =1 +

p−1

X

x =0

1 + x³− nx p

=p + 1 +

p−1

X

x =0

x³− nx p

.

Thus,

L(E_n/Q, s)⁻¹=Y

p



1 +

p−1

X

x =0

x³− nx p

p^−s+p^1−2s



. Since E₁and E_n 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

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Arithmetic-geometric approach

Idea.

#En(Fp) =1 +

p−1

X

x =0

1 + x³− nx p

=p + 1 +

p−1

X

x =0

x³− nx p

.

Thus,

L(E_n/Q, s)⁻¹=Y

p



1 +

p−1

X

x =0

x³− nx p

p^−s+p^1−2s



.

Since E₁and E_n 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.

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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 G_K = Gal(Q/K ) acts on T`(E ), yielding a Galois representation

ρE ,`:G_K → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

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Tate modules and Galois representations

T_`(E ) = lim

←−E [`ⁿ].

ρE ,`:G_K → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

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Tate modules and Galois representations

T_`(E ) = lim

←−E [`ⁿ].

ρE ,`:G_K → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

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Tate modules and Galois representations

T_`(E ) = lim

←−E [`ⁿ].

ρE ,`:G_K → GL(2, Q`).

Then L(ρ_{E ,`},s) = L(E /K , s).

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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.

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A lemma

Lemma (Clifford)

Hand ρ₂

Then

ρ₁' ρ₂⊗ χ

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A lemma

Lemma (Clifford)

Hand ρ₂

Then

ρ₁' ρ₂⊗ χ

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Arithmetic-geometric approach

Let E_n: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 G_K = Gal(Q/K ) and GL= Gal(Q/L).

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

GK = πn⊕ π_n, where π_nare representations of G_K of degree 1.

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Arithmetic-geometric approach

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Arithmetic-geometric approach

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Arithmetic-geometric approach

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

G_L ' π_n G_L. By the lemma above,

π_n= π₁⊗ χ

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

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Arithmetic-geometric approach

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

G_L ' π_n G_L. By the lemma above,

π_n= π₁⊗ χ

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

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Arithmetic-geometric approach

A character on G_K with G_L⊂ 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 Frob_p is the element σ^j ∈ Gal(L/K ) such that

σ^j(√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(Frob_p) =i^jk

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Arithmetic-geometric approach

σ :√⁴

n 7−→ i√⁴ n.

σ^j(√⁴

n) ≡ (√⁴

Then there exists k ∈ {1, 3} such that χ satisfies χ(Frob_p) =i^jk for all p.

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Arithmetic-geometric approach

σ :√⁴

n 7−→ i√⁴ n.

σ^j(√⁴

n) ≡ (√⁴

Then there exists k ∈ {1, 3} such that χ satisfies χ(Frob_p) =i^jk

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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(Frob_p) = ±i.

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Proof of Jacobsthal’s identity

(√⁴

n)^Np≡ ±i√⁴

n mod p.

That is,

χ(Frob_p) = ±i.

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Proof of Jacobsthal’s identity

(√⁴

n)^Np≡ ±i√⁴

n mod p.

That is,

χ(Frob_p) = ±i.

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Proof of Jacobsthal’s identity

It is well-known that L(E₁/Q, s) = Y

p≡1 mod 4

1

1 − 2_pa_pp^−s+p^1−2s

Y

p≡3 mod 4

1 1 + p^1−2s, where for p ≡ 1 mod 4, apand bpare positive integers with ap odd and b_p even such that p = a²_p+b²_p, and

p= −1 ap

(−1)^b^p^/2.

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

Then

π_n(p) = π₁(p)χ(p) = ±b_p± api

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Proof of Jacobsthal’s identity

p≡1 mod 4

1

1 − 2_pa_pp^−s+p^1−2s

Y

p≡3 mod 4

p= −1 ap

(−1)^b^p^/2.

Then

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Proof of Jacobsthal’s identity

p≡1 mod 4

1

1 − 2_pa_pp^−s+p^1−2s

Y

p≡3 mod 4

p= −1 ap

(−1)^b^p^/2.

Then

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Proof of Jacobsthal’s identity

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

(1 ± 2b_pp^−s+p^1−2s)⁻¹. That is,

p−1

X

x =0

x³− nx p

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

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Proof of Jacobsthal’s identity

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

(1 ± 2b_pp^−s+p^1−2s)⁻¹. That is,

p−1

X

x =0

x³− nx p

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

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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.

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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.

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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?

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Question

Let

f (x , y ) =

−d p

. Question. When

−d p

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Question

Let

f (x , y ) =

−d p

. Question. When

−d p

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Jacobsthal identity for Q( √

−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.

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Jacobsthal identity for Q( √

−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.

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The elliptic curve y

²

= x

³

+ 4x

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 − 2papp^−s+p^1−2s

Y

p≡5,7 mod 8

1 1 + p^1−2s, where a_pand b_pare positive integers such that p = a²_p+2b²_pand

_p =







2(−1)^b^p^/2

−2 ap

, if p ≡ 1 mod 8,

−2

−2 ap

, if p ≡ 3 mod 8.

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The hyperelliptic curve y

²

= x

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).

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The hyperelliptic curve y

²

= x

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).

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L-function of y

²

= x

⁵

+ x

Corollary

For C : y²=x⁵+x , let

1

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

• If p ≡ 1 mod 8, then α_p,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 α_p,j = ±a ± b√

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

√ √

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L-function of y

²

= x

⁵

+ x

Corollary

For C : y²=x⁵+x , let

1

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

• If p ≡ 1 mod 8, then α_p,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 α_p,j = ±a ± b√

−2, where a and b are

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The curve y

²

= x

⁶

+ 4x

⁵

+ 10x

⁴

− 20x

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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The curve y

²

= x

⁶

+ 4x

⁵

+ 10x

⁴

− 20x

²

− 16x − 8

Lemma

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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The curve y

²

= x

⁶

+ 4x

⁵

+ 10x

⁴

− 20x

²

− 16x − 8

Lemma

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?

−2)(a − b√

−2).

(1 ± (a + b√

−2)p^−s)(1 ± (a − b√

−2)p^−s), then

p−1

X

x =0

f (x) p

= ±2a.

−2].

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How do we find the curves?

−2)(a − b√

−2).

(1 ± (a + b√

−2)p^−s)(1 ± (a − b√

−2)p^−s), then

p−1

X

x =0

f (x) p

= ±2a.

−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?

−2)(ζ8+ ζ₈³) + (a − b√

−2)(ζ₈⁵+ ζ₈⁷) = −4b.

1 (1 ± ζ₈(a + b√

−2)p^−s) . . . (1 ± ζ₈⁷(a − b√

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How do we find the curves?

−2)(ζ8+ ζ₈³) + (a − b√

−2)(ζ₈⁵+ ζ₈⁷) = −4b.

1 (1 ± ζ₈(a + b√

−2)p^−s) . . . (1 ± ζ₈⁷(a − b√

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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 = (

(x_v) ∈Y

v

K_v^∗ :x_v ∈ O^∗_v 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

IK = (

(x_v) ∈Y

v

Definition

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Hecke characters

Let

IK = (

(x_v) ∈Y

v

Definition

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Hecke L-functions and their functional equations

Definition

Let χ be a Hecke character. Write χ =Q

vχ_v. TheHecke L-functionis defined by

L(s, χ) = Y

v finite, χv(O^∗v)=1

1

1 − χv(πv)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 −1}for all x ∈ IK/K^∗. Setting

Λ(s, χ) = 2π pd_Kd_χ

!−s

Γ(s)L(s, χ),

we have

Λ(s, χ) = Λ(k − s, χ)

for some root of unity , where d_K is the discriminant of K and d_χis 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 −1}for all x ∈ IK/K^∗. Setting

Λ(s, χ) = 2π pd_Kd_χ

!−s

Γ(s)L(s, χ),

we have

Λ(s, χ) = Λ(k − s, χ)

for some root of unity , where d_K is the discriminant of K and d_χis 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 ζ₈^j(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).

(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).

(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.