L-functions and Elliptic Curves - Técnico, Lisboagmp/slides... · Analytic L-functions De nition A...

L-functions and Elliptic Curves

Nuno Freitas

Universitat Bayreuth

January 2014

Motivation

Let m(P) denote the logarithmic Mahler measure of a polynomialP ∈ C[x±1, y±1].

I In 1981, Smyth proved the following formula:

m(1 + x + y) = L′(χ−3,−1),

where χ−3 is the Dirichlet character associated to thequadratic field Q(

√−3).

I In 1997, Deninger conjectured the following formula

m(x +1

x+ y +

y+ 1) = L′(E , 0),

where E is the elliptic curve that is the projective closure ofthe polynomial in the left hand side.

Our goal: Sketch the basic ideas that allow to make sense of theright hand side of these formulas.

Motivation

m(1 + x + y) = L′(χ−3,−1),

√−3).

m(x +1

x+ y +

y+ 1) = L′(E , 0),

Motivation

m(1 + x + y) = L′(χ−3,−1),

√−3).

m(x +1

x+ y +

y+ 1) = L′(E , 0),

Motivation

m(1 + x + y) = L′(χ−3,−1),

√−3).

m(x +1

x+ y +

y+ 1) = L′(E , 0),

The Riemann Zeta function

The L-functions are constructed on the model of the Riemann Zetafunction ζ(s), so let us recall properties of this function.

The Riemann Zeta function ζ(s) is defined on C, for Re(s) > 1,by the formula

ζ(s) =∑n≥1

Euler showed that

ζ(s) =∏p

1− p−s.

In particular, Euler’s equality provides an alternative proof of theexistence of infinitely many prime numbers.

ζ(s) =∑n≥1

Euler showed that

ζ(s) =∏p

1− p−s.

ζ(s) =∑n≥1

Euler showed that

ζ(s) =∏p

1− p−s.

ζ(s) =∑n≥1

Euler showed that

ζ(s) =∏p

1− p−s.

Theorem (Riemann)

The Riemann Zeta function ζ(s) can be analytical continued to ameromorphic function of the complex plane. Its only pole is ats = 1, and its residue is 1.Moreover, the function Λ defined by

Λ(s) := π−s/2Γ(s/2)ζ(s)

satisfies the functional equation

Λ(s) = Λ(1− s).

The Gamma function

The function Γ in the previous theorem is defined by

Γ(s) :=

∫ ∞0

e−tts−1dt.

It admits a meromorphic continuation to all C and satisfies thefunctional equation

Γ(s + 1) = sΓ(s).

The function Γ(s/2) has simple poles at the negative even integers.To compensate these poles we have ζ(−2n) = 0. These are calledthe trivial zeros of ζ(s).

Conjecture (Riemann Hypothesis)

All the non-trivial zeros of ζ(s) satisfy Re(s) = 1/2.

The Gamma function

Γ(s) :=

∫ ∞0

e−tts−1dt.

Γ(s + 1) = sΓ(s).

The Gamma function

Γ(s) :=

∫ ∞0

e−tts−1dt.

Γ(s + 1) = sΓ(s).

The function Γ(s/2) has simple poles at the negative even integers.

To compensate these poles we have ζ(−2n) = 0. These are calledthe trivial zeros of ζ(s).

The Gamma function

Γ(s) :=

∫ ∞0

e−tts−1dt.

Γ(s + 1) = sΓ(s).

The Gamma function

Γ(s) :=

∫ ∞0

e−tts−1dt.

Γ(s + 1) = sΓ(s).

Analytic L-functions

DefinitionA Dirichlet series is a formal series of the form

F (s) =∞∑n=1

anns, where an ∈ C.

We call an Euler product to a product of the form

F (s) =∏p

Lp(s).

The factors Lp(s) are called the local Euler factors.

An analytic L-function is a Dirichlet series that has an Eulerproduct and satisfies a certain type of functional equation.

F (s) =∞∑n=1

F (s) =∏p

Lp(s).

F (s) =∞∑n=1

F (s) =∏p

Lp(s).

Dirichlet characters

A function χ : Z→ C is called a Dirichlet character modulo N ifthere is a group homomorphism χ : (Z/NZ)∗ → C∗ such that

χ(x) = χ( x (mod N)) if (x ,N) = 1

andχ(x) = 0 if (x ,N) 6= 1.

Moreover, we say that χ is primitive if there is no strict divisorM | N and a character χ0 : (Z/MZ)∗ → C∗ such that

χ(x) = χ0( x (mod M)) if (x ,M) = 1.

In particular, if N = p is a prime every non-trivial character moduloN is primitive. Moreover, any Dirichlet character is induced from aunique primitive character χ0 as above. We call M its conductor.

χ(x) = χ( x (mod N)) if (x ,N) = 1

andχ(x) = 0 if (x ,N) 6= 1.

χ(x) = χ0( x (mod M)) if (x ,M) = 1.

χ(x) = χ( x (mod N)) if (x ,N) = 1

andχ(x) = 0 if (x ,N) 6= 1.

χ(x) = χ0( x (mod M)) if (x ,M) = 1.

χ(x) = χ( x (mod N)) if (x ,N) = 1

andχ(x) = 0 if (x ,N) 6= 1.

χ(x) = χ0( x (mod M)) if (x ,M) = 1.

In particular, if N = p is a prime every non-trivial character moduloN is primitive.

Moreover, any Dirichlet character is induced from aunique primitive character χ0 as above. We call M its conductor.

χ(x) = χ( x (mod N)) if (x ,N) = 1

andχ(x) = 0 if (x ,N) 6= 1.

χ(x) = χ0( x (mod M)) if (x ,M) = 1.

Dirichlet L-functions

DefinitionWe associate to a Dirichlet character χ an L-function given by

L(χ, s) =∑n≥1

1− χ(p)p−s

For example,

L(χ−3, s) =∞∑n=1

ns= 1− 1

4s− 1

5s+ ...,

where the sign is given by the symbol

1 if n is a square mod 3−1 if n is not a square mod 30 if 3 | n

L(χ, s) =∑n≥1

ns=∏p

1− χ(p)p−s

For example,

L(χ−3, s) =∞∑n=1

ns= 1− 1

4s− 1

5s+ ...,

L(χ, s) =∑n≥1

ns=∏p

1− χ(p)p−s

For example,

L(χ−3, s) =∞∑n=1

ns= 1− 1

4s− 1

5s+ ...,

Let χ be a Dirichlet character. We say that χ is even ifχ(−1) = 1; we say that χ is odd if χ(−1) = −1.

Define also, if χ is even,

Λ(χ, s) := π−s/2Γ(s/2)L(χ, s)

or, if χ is odd,

Λ(χ, s) := π−(s+1)/2Γ((s + 1)/2)L(χ, s)

Let χ be a Dirichlet character. We say that χ is even ifχ(−1) = 1; we say that χ is odd if χ(−1) = −1.

Define also, if χ is even,

Λ(χ, s) := π−s/2Γ(s/2)L(χ, s)

or, if χ is odd,

Λ(χ, s) := π−(s+1)/2Γ((s + 1)/2)L(χ, s)

TheoremLet χ be a primitive Dirichlet character of conductor N 6= 1. Then,L(χ, s) has an extension to C as an entire function and satisfiesthe functional equation

Λ(χ, s) = ε(χ)N1/2−sΛ(χ, 1− s),

ε(χ) =

{τ(χ)√

Nif χ is even

−i τ(χ)√N

if χ is odd

andτ(χ) =

∑x (mod N)

χ(x)e2iπx/N

Elliptic Curves

DefinitionAn elliptic curve over a field k is a non-singular projective planecurve given by an affine model of the form

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6,

where all ai ∈ k . Write O = (0 : 1 : 0) for the point at infinity.

The change of variables fixing O are of the form

x = u2x ′ + r y = u3y ′ + u2sx ′ + t,

where u, r , s, t ∈ k , u 6= 0. If char(k) 6= 2, 3, after a change ofvariables, E can be writen as

y2 = x3 + Ax + B, A,B ∈ k , ∆(E ) = 4A3 + 27B2.

If ∆(E ) 6= 0 then E is nonsingular.

Elliptic Curves

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6,

x = u2x ′ + r y = u3y ′ + u2sx ′ + t,

where u, r , s, t ∈ k , u 6= 0.

If char(k) 6= 2, 3, after a change ofvariables, E can be writen as

y2 = x3 + Ax + B, A,B ∈ k , ∆(E ) = 4A3 + 27B2.

Elliptic Curves

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6,

x = u2x ′ + r y = u3y ′ + u2sx ′ + t,

where u, r , s, t ∈ k , u 6= 0. If char(k) 6= 2, 3, after a change ofvariables, E can be writen as

y2 = x3 + Ax + B, A,B ∈ k , ∆(E ) = 4A3 + 27B2.

Example

Consider the curve

E : y2 = x3 − 2x + 1,

having attached quantities

∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

Another example

Consider the set defined by

x+ y +

y+ 1 = 0

Multiplication by xy followed by homogenization gives

x2y + yz2 + y2x + xz2 + xyz = 0.

Applying the isomorphism (x , y , z) 7→ (y , x − y , z − x) yelds

x3 − 2x2z + xyz − y2z + xz2 = 0.

After setting z = 1 and rearranging we get the elliptic curve withconductor 15 given by

y2 − xy = x3 − 2x2 + x .

Another example

x+ y +

y+ 1 = 0

x2y + yz2 + y2x + xz2 + xyz = 0.

x3 − 2x2z + xyz − y2z + xz2 = 0.

y2 − xy = x3 − 2x2 + x .

Another example

x+ y +

y+ 1 = 0

x2y + yz2 + y2x + xz2 + xyz = 0.

x3 − 2x2z + xyz − y2z + xz2 = 0.

y2 − xy = x3 − 2x2 + x .

Another example

x+ y +

y+ 1 = 0

x2y + yz2 + y2x + xz2 + xyz = 0.

x3 − 2x2z + xyz − y2z + xz2 = 0.

y2 − xy = x3 − 2x2 + x .

TheoremLet E/k be an elliptic curve. There is an abelian group structureon the set of points E (k).

Theorem (Mordell–Weil)

Let E/k be an elliptic curve over a number field k . The groupE (k) is finitely generated.

TheoremLet E/k be an elliptic curve. There is an abelian group structureon the set of points E (k).

Let E/k be an elliptic curve over a number field k . The groupE (k) is finitely generated.

Example

Consider the curve

E : y2 = x3 − 2x + 1,

∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

Its rational torsion points are

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)},

and they form a cyclic group of order 4.

Example

Consider the curve

E : y2 = x3 − 2x + 1,

∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)},

and they form a cyclic group of order 4.

Reduction modulo p

Let E/Q be an elliptic curve. There exists a model E/Z such that|∆(E )| is minimal. For such a model and a prime p, we set ai = ai(mod p) and consider the reduced curve over Fp

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

It can be seen that E has at most one singular point.

Definition (type of reduction)

Let p be a prime. We say that E

I has good reduction at p if E is an elliptic curve.

I has bad multiplicative reduction at p if E admits a doublepoint with two distinct tangents. We say it is split ornon-split if the tangents are defined over Fp or Fp2 ,respectively.

I has bad additive reduction at p if E admits a double pointwith only one tangent.

Reduction modulo p

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

Reduction modulo p

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

Reduction modulo p

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

I has bad multiplicative reduction at p if E admits a doublepoint with two distinct tangents.

We say it is split ornon-split if the tangents are defined over Fp or Fp2 ,respectively.

Reduction modulo p

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

Reduction modulo p

E : y2 + a1xy + a3y = x3 + a2x2 + a4x + a6.

The Conductor of an elliptic curve.

DefinitionThe conductor NE of an elliptic curve E/Q is an integer. It iscomputed by Tate’s algorithm, and is of the form

NE =∏p

where the exponents fp satisfy

0 if E has good reduction at p,1 if E has bad multiplicative reduction at p,2 if E has bad additive reduction at p ≥ 5,2 + δp, 0 ≤ δp ≤ 6 if E has bad additive reduction at p = 2, 3.

In particular, NE | ∆(E ) for the discriminant associated with anymodel of E .

Example

Consider the curve

E : y2 = x3 − 2x + 1, which is a minimal model

∆ = 24 · 5, j = 211 · 33 · 5−1.

The reduction type at p = 5 is bad split multiplicative reductionand at p = 2 is bad additive reduction. Furthermore,

NE = 23 · 5 = 40

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Example

Consider the curve

∆ = 24 · 5, j = 211 · 33 · 5−1.

NE = 23 · 5 = 40

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Example

Consider the curve

∆ = 24 · 5, j = 211 · 33 · 5−1.

NE = 23 · 5 = 40

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Artin Zeta FunctionLet E/Fp be an elliptic curve given by

y2 + a1xy + a3y − x3 + a2x2 + a4x + a6 = 0.

Consider the associated Dedekind domain

A = Fp[X ,Y ]/(E )

For a non-zero ideal I of A we define its norm

N(I) = #(A/I).

The Zeta function associated to A is

ζA(s) =∑I6=0

N(I)s=∏P

1− N(P)−s

DefinitionFor s ∈ C such that Re(s) > 1, we set

ζE (s) =1

1− p−sζA(s)

y2 + a1xy + a3y − x3 + a2x2 + a4x + a6 = 0.

A = Fp[X ,Y ]/(E )

N(I) = #(A/I).

ζA(s) =∑I6=0

N(I)s=∏P

1− N(P)−s

ζE (s) =1

1− p−sζA(s)

y2 + a1xy + a3y − x3 + a2x2 + a4x + a6 = 0.

A = Fp[X ,Y ]/(E )

N(I) = #(A/I).

ζA(s) =∑I6=0

N(I)s=∏P

1− N(P)−s

ζE (s) =1

1− p−sζA(s)

y2 + a1xy + a3y − x3 + a2x2 + a4x + a6 = 0.

A = Fp[X ,Y ]/(E )

N(I) = #(A/I).

ζA(s) =∑I6=0

1− N(P)−s

ζE (s) =1

1− p−sζA(s)

y2 + a1xy + a3y − x3 + a2x2 + a4x + a6 = 0.

A = Fp[X ,Y ]/(E )

N(I) = #(A/I).

ζA(s) =∑I6=0

N(I)s=∏P

1− N(P)−s

ζE (s) =1

1− p−sζA(s)

y2 + a1xy + a3y − x3 + a2x2 + a4x + a6 = 0.

A = Fp[X ,Y ]/(E )

N(I) = #(A/I).

ζA(s) =∑I6=0

N(I)s=∏P

1− N(P)−s

ζE (s) =1

1− p−sζA(s)

Artin Zeta Function

Theorem (Artin)

Let E/Fp be an elliptic curve and set

aE := p + 1−#E (Fp).

ζE (s) =1− aE · p−s + p · p−2s

(1− p−s)(1− p · p−s)

andζE (s) = ζE (1− s).

Artin Zeta Function

Theorem (Artin)

Let E/Fp be an elliptic curve and set

aE := p + 1−#E (Fp).

ζE (s) =1− aE · p−s + p · p−2s

(1− p−s)(1− p · p−s)

andζE (s) = ζE (1− s).

The Hasse-Weil L-function of E/QLet E/Q be an elliptic curve. For a prime p of good reduction, letE be the reduction of E mod p, and set

Lp(s) = (1− aE · p−s + p · p−2s)−1.

Define also Euler factors for primes p of bad reduction by

Lp(s) =

(1− p−s)−1 if E has bad split multiplicative reduction at p,(1 + p−s)−1 if E has bad non-split mult. reduction at p,1 if E has bad additive reduction at p.

DefinitionThe L-function of E is defined by

L(E , s) =∏p

Lp(s) = (1− aE · p−s + p · p−2s)−1.

Lp(s) =

L(E , s) =∏p

Lp(s) = (1− aE · p−s + p · p−2s)−1.

Lp(s) =

L(E , s) =∏p

A really brief incursion into modular cuspforms

I A modular form is a function on the upper-half plane thatsatisfies certain transformation and holomorphy conditions.

I Let N ≥ 1 be an integer. Define

Γ0(N) =

{[a bc d

]∈ SL2(Z) :

[a bc d

]≡[∗ ∗0 ∗

](mod N)

}I In particular, a cuspform f for Γ0(N) (of weight 2) admits a

Fourier expansion

f (τ) =∞∑n=1

an(f )qn/N , an(f ) ∈ C, q = e2πiτ .

I There is a family of Hecke operators {Tn}n≥1 acting on theC-vector space of cuspforms for Γ0(N) of weight 2.

I To a cuspform that is an eigenvector of all Tn we call aneigenform. Furthermore, we assume they are normalizedsuch that a1(f ) = 1.

Γ0(N) =

{[a bc d

]∈ SL2(Z) :

[a bc d

]≡[∗ ∗0 ∗

](mod N)

Fourier expansion

f (τ) =∞∑n=1

Γ0(N) =

{[a bc d

]∈ SL2(Z) :

[a bc d

]≡[∗ ∗0 ∗

](mod N)

Fourier expansion

f (τ) =∞∑n=1

The L-function of an eigenform

DefinitionThe L-function attached to an eigenform for Γ0(N) is defined by

L(f , s) =∞∑n≥1

an(f )

TheoremLet f be an eigenform for Γ0(N) of weight 2. The function L(f , s)has an entire continuation to C. Moreover, the function

Λf (s) := (

2π)−sΓ(s)L(f , s)

satisfies the functional equation

Λf (s) = wΛf (2− s),

where w = ±1.

Modularity and the L-function of E/Q

Theorem (Wiles, Breuil–Conrad–Diamond–Taylor)

Let E/Q be an elliptic curve of conductor NE . There is aneigenform f for Γ0(NE ) (of weight 2) such that

L(E , s) = L(f , s).

Corollary

Let E/Q be an elliptic curve of conductor NE . Define the function

ΛE (s) := (

2π)−sΓ(s)L(E , s).

The function L(E , s) has an entire continuation to C and ΛE (s)satisfies

ΛE (s) = wΛE (2− s),

where w = ±1.

Modularity and the L-function of E/Q

Theorem (Wiles, Breuil–Conrad–Diamond–Taylor)

Let E/Q be an elliptic curve of conductor NE . There is aneigenform f for Γ0(NE ) (of weight 2) such that

L(E , s) = L(f , s).

Corollary

Let E/Q be an elliptic curve of conductor NE . Define the function

ΛE (s) := (

2π)−sΓ(s)L(E , s).

The function L(E , s) has an entire continuation to C and ΛE (s)satisfies

ΛE (s) = wΛE (2− s),

where w = ±1.

Example

Consider the curve

E : y2 = x3 − 2x + 1, ∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

It has conductor NE = 23 · 5 = 40.

The cuspform of weight 2 forΓ0(40) corresponding to E by modularity is

f := q + q5 − 4q7 − 3q9 + O(q10).

The rational torsion points are

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Example

Consider the curve

E : y2 = x3 − 2x + 1, ∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

It has conductor NE = 23 · 5 = 40.The cuspform of weight 2 forΓ0(40) corresponding to E by modularity is

f := q + q5 − 4q7 − 3q9 + O(q10).

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Example

Consider the curve

E : y2 = x3 − 2x + 1, ∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

It has conductor NE = 23 · 5 = 40.The cuspform of weight 2 forΓ0(40) corresponding to E by modularity is

f := q + q5 − 4q7 − 3q9 + O(q10).

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

The BSD conjecture

Let E/Q be an elliptic curve. Then the group E (Q) is finitelygenerated. More precisely,

E (Q) ∼= E (Q)Tor ⊕ ZrE

Conjecture (Birch–Swinnerton-Dyer)

The rank rE of the Mordell-Weil group of an elliptic E/Q is equalto the order of the zero of L(E , s) at s = 1.

The BSD conjecture

Let E/Q be an elliptic curve. Then the group E (Q) is finitelygenerated. More precisely,

E (Q) ∼= E (Q)Tor ⊕ ZrE

Conjecture (Birch–Swinnerton-Dyer)

The rank rE of the Mordell-Weil group of an elliptic E/Q is equalto the order of the zero of L(E , s) at s = 1.

Example

Consider the curve

E : y2 = x3 − 2x + 1, ∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

It has conductor NE = 23 · 5 = 40. The cuspform of weight 2 forΓ0(40) corresponding to E by modularity is

f := q + q5 − 4q7 − 3q9 + O(q10).

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Moreover, the rank rE = 0 since the function L(E , s) satisfies

L(E , 1) = 0.742206236711.

Thus E (Q) ∼= (Z/4Z).

Example

Consider the curve

E : y2 = x3 − 2x + 1, ∆ = 24 · 5 6= 0, j = 211 · 33 · 5−1.

It has conductor NE = 23 · 5 = 40. The cuspform of weight 2 forΓ0(40) corresponding to E by modularity is

f := q + q5 − 4q7 − 3q9 + O(q10).

E (Q)Tor = {O, (0 : −1 : 1), (0 : 1 : 1), (1 : 0 : 1)} ∼= (Z/4Z)

Moreover, the rank rE = 0 since the function L(E , s) satisfies

L(E , 1) = 0.742206236711.

Thus E (Q) ∼= (Z/4Z).

Counting Points on Varieties

Let V /Fq be a projective variety, given by the set of zeros

f1(x0, . . . , xN) = · · · = fm(x0, . . . , xN) = 0

of a collection of homogeneous polynomials. The number of pointsin V (Fqn) is encoded in the zeta function

DefinitionThe Zeta function of V /Fq is the power series

Z (V /Fq;T ) := exp(∑n≥1

#V (Fqn)T n

The Zeta function of the Projective space

Let N ≥ 1 and V = PN . A point in V (Fqn) is given byhomogeneous coordinates (x0 : .. : xN) with xi not all zero. Twochoices of coordinates give the same point if they differ bymultiplication of a non-zero element in Fqn . Hence,

#V (Fqn) =qn(N+1) − 1

qn − 1=

N∑i=0

qni so

logZ (V /Fq;T ) =∞∑n=0

(N∑i=0

qni )T n

N∑i=0

− log(1− qiT ).

Z (PN/Fq;T ) =1

(1− T )(1− qT ) . . . (1− qNT )

The Zeta function of E/Fp

TheoremLet E/Fp be an elliptic curve and define

aE = p + 1−#E (Fp).

Z (E/Fp;T ) =1− aET + pT 2

(1− T )(1− pT )

Moreover,

1− aET + pT 2 = (1− α)(1− β) with |α| = |β| =√p

Note that by setting T = p−s we obtain the equality

Z (E/Fp; p−s) = ζE (s)

The Zeta function of E/Fp

TheoremLet E/Fp be an elliptic curve and define

aE = p + 1−#E (Fp).

Z (E/Fp;T ) =1− aET + pT 2

(1− T )(1− pT )

Moreover,

1− aET + pT 2 = (1− α)(1− β) with |α| = |β| =√p

Note that by setting T = p−s we obtain the equality

Z (E/Fp; p−s) = ζE (s)

Example

Consider the curve E : y2 = x3 − 2x + 1 which has bad additivereduction at 2.Let p = 2. Its mod p reduction is given by

E2 : (y − 1)2 = x3

and satisfies #E2(F2n) = 2n + 1. Hence,

logZ (E2/F2n ;T ) =∞∑n=1

2n + 1

= log(1

1− 2T) + log(

1− T)

Z (E2/F2n ;T ) =1

(1− 2T )(1− T )

Bibliography

I C.J. Smyth, On measures of polynomials in several variables,Bull. Austral. Math. Soc. 23 (1981), 49–63;

I C. Deninger, Deligne periods of mixed motives, K -theory andthe entropy of certain Zn-actions, J. Amer. Math. Soc. 10:2(1997), 259–281;

I J.H. Silverman, The Arithmetic of Elliptic Curves, GTM 106,Springer, 1986;

I F. Diamond and J. Shurman, A First Course on ModularForms, GTM 228, Springer, 2005;

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