Diophantine equations and semistable elliptic curves over totally real number fields
Diophantine equations and semistable
elliptic curves over totally real
number fields
Samuele Anni
joint work with Samir Siksek
University of Warwick
Algebra, Geometry and Number Theory Seminar,
Universiteit Leiden, 18th April 2016
Diophantine equations and semistable elliptic curves over totally real number fields
Generalized Fermat Equation
1
Generalized Fermat Equation
2
x 2` + y 2m = z p
3
The proof
Diophantine equations and semistable elliptic curves over totally real number fields
Generalized Fermat Equation
Generalized Fermat Equation
Let (p, q, r ) ∈ Z 3≥2 . The equation
xp + yq = zr
is a Generalized Fermat Equation of signature (p, q, r ).
A solution (x, y , z) ∈ Z 3 is called
non-trivial if xyz 6= 0,
primitive if gcd(x, y , z) = 1.
Diophantine equations and semistable elliptic curves over totally real number fields
Generalized Fermat Equation
Conjecture (Darmon & Granville, Tijdeman, Zagier, Beal)
Suppose
1 1
1
+ + < 1.
p
q
r
The only non-trivial primitive solutions to x p + y q = z r are
1 + 23 = 32 ,
73 + 132 = 29 ,
35 + 114 = 1222 ,
14143 + 22134592 = 657 ,
438 + 962223 = 300429072 ,
25 + 72 = 34 ,
27 + 173 = 712 ,
177 + 762713 = 210639282 ,
92623 + 153122832 = 1137 ,
338 + 15490342 = 156133 .
Poonen–Schaefer–Stoll: (2, 3, 7).
Bruin: (2, 3, 8), (2, 8, 3), (2, 3, 9), (2, 4, 5), (2, 5, 4).
Many others . . .
Diophantine equations and semistable elliptic curves over totally real number fields
Generalized Fermat Equation
Infinite Families of Exponents:
Wiles: (p, p, p).
Darmon and Merel: (p, p, 2), (p, p, 3).
Many other infinite families by many people . . .
The study of infinite families uses Frey curves, modularity and
level-lowering over Q (or Q-curves).
Let us look at x p + y p = z ` for p and ` primes.
Diophantine equations and semistable elliptic curves over totally real number fields
Generalized Fermat Equation
Solve x p + y p = z `
Naı̈ve idea
To solve x p + y p = z ` factor over Q (ζ), where ζ is a p-th root of unity.
(x + y )(x + ζy ) . . . (x + ζ p−1 y ) = z ` .
x + ζ j y = αj ξj` ,
αj ∈ finite set.
∃j ∈ Q (ζ) such that 0 (x + y ) + 1 (x + ζy ) + 2 (x + ζ 2 y ) = 0.
γ0 ξ0` + γ1 ξ1` + γ2 ξ2` = 0
(γ0 , γ1 , γ2 ) ∈ finite set.
It looks like x ` + y ` + z ` = 0 solved by Wiles.
Problems
Problem 1: trivial solutions (±1, 0, ±1), (0, ±1, ±1) become non-trivial.
Problem 2: modularity theorems over non-totally real fields.
Diophantine equations and semistable elliptic curves over totally real number fields
Generalized Fermat Equation
Improvement: Freitas
factor LHS over K := Q (ζ + ζ −1 ).
xp + yp = z`
(p−1)/2
(x + y )
Y
x 2 + y 2 + θj xy = z `
θj := ζ j + ζ −j ∈ K .
j=1
x + y = α0 ξ0` ,
α0 ∈ finite set.
(x 2 + y 2 ) + θj xy = αj ξj` ,
|
{z
}
αj ∈ finite set.
fj (x,y )
γ0 ξ02` + γ1 ξ1` + γ2 ξ2` = 0
(γ0 , γ1 , γ1 ) ∈ finite set.
Theorem (Freitas)
Let ` > (1 + 318 )2 C . Then the only primitive solutions to x 7 + y 7 = 3z `
are (±1, ∓1, 0).
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
1
Generalized Fermat Equation
2
x 2` + y 2m = z p
3
The proof
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
Theorem (Anni-Siksek)
Let p = 3, 5, 7, 11 or 13. Let `, m ≥ 5 be primes, and if p = 13 suppose
moreover that `, m 6= 7. Then the only primitive solutions to
x 2` + y 2m = z p ,
are the trivial ones (x, y , z) = (±1, 0, 1) and (0, ±1, 1).
Remark: this is a bi-infinite family of equations.
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
Let `, m, p ≥ 5 be primes, ` 6= p, m 6= p.
x 2` + y 2m = z p ,
gcd(x, y , z) = 1.
Modulo 8 we get 2 - z so WLOG 2 | x. Only expected solution (0, ±1, 1).
(
x ` + y m i = (a + bi)p
a, b ∈ Z gcd(a, b) = 1.
x ` − y m i = (a − bi)p
p−1
x` =
Y
1
((a + bi)p + (a − bi)p ) = a ·
(a + bi) + (a − bi)ζ j
2
j=1
(p−1)/2
=a·
Y
j=1
(θj + 2)a2 + (θj − 2)b 2
θj = ζ j + ζ −j ∈ Q (ζ + ζ −1 ).
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
Let K := Q (ζ + ζ −1 ) then
(p−1)/2
x` = a ·
Y
j=1
(θj + 2)a2 + (θj − 2)b 2
|
{z
}
θj = ζ j + ζ −j ∈ K .
fj (a,b)
p - x =⇒ a = α` ,
fj (a, b) · OK = b `j ,
p | x =⇒ a = p `−1 α` ,
fj (a, b) · OK = p b `j ,
p = (θj − 2) | p.
(θ2 − 2)f1 (a, b) + (2 − θ1 )f2 (a, b) + 4(θ1 − θ2 )a2 = 0 .
|
{z
} |
{z
} |
{z
}
u
v
w
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
Frey curve
(∗)
E : Y 2 = X (X − u)(X + v ),
∆ = 16u 2 v 2 w 2 .
Problems
Problem 1: trivial solutions (0, ±1, 1) become non-trivial.
Trivial solution x = 0 implies a = 0, so w = 0 =⇒ ∆ = 0.
Problem 2: modularity theorems over non-totally real fields.
K := Q (ζ + ζ −1 )
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
Lemma
Suppose p - x. Let E be the Frey curve (∗). The curve E is semistable ,
with multiplicative reduction at all primes above 2 and good
reduction at p . It has minimal discriminant and conductor
2`
DE /K = 24`n−4 α4` b 2`
j bk ,
NE /K = 2 · Rad(αb j b k ).
Lemma
Suppose p | x. Let E be the Frey curve (∗). The curve E is semistable ,
with multiplicative reduction at p and at all primes above 2. It has
minimal discriminant and conductor
2`
DE /K = 24`n−4 p 2δ α4` b 2`
j bk ,
NE /K = 2p · Rad(αb j b k ).
Diophantine equations and semistable elliptic curves over totally real number fields
x 2` + y 2m = z p
Lemma
Suppose p - x. Let E be the Frey curve (∗). The curve E is semistable ,
with multiplicative reduction at all primes above 2 and good
reduction at p . It has minimal discriminant and conductor
2`
DE /K = 24`n−4 α4` b 2`
j bk ,
NE /K = 2 · Rad(αb j b k ).
Lemma
Suppose p | x. Let E be the Frey curve (∗). The curve E is semistable ,
with multiplicative reduction at p and at all primes above 2. It has
minimal discriminant and conductor
2`
DE /K = 24`n−4 p 2δ α4` b 2`
j bk ,
NE /K = 2p · Rad(αb j b k ).
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
1
Generalized Fermat Equation
2
x 2` + y 2m = z p
3
The proof
Residual irreducibility
Modularity
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Galois representations and Elliptic curves
Let ` be a prime, and E elliptic curve over totally real field K . The
mod ` Galois Representation attached to E is given by
ρE ,` : GK → Aut(E [`]) ∼
= GL2 (F` )
GK = Gal(K /K ).
The `-adic Galois Representation attached to E is given by
ρE ,` : GK → Aut(T` (E )) ∼
= GL2 (Z ` ),
where T` (E ) = lim E [`n ] is the `-adic Tate module.
←
−
Definition
E is modular if there exists a cuspidal Hilbert modular eigenform f such
that ρE ,` ∼ ρf,` .
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Hilbert modular forms
Let K be a totally real number field of degree d over Q . Let σ1 , . . . , σd
be the real embeddings of K , therefore we have a map
GL2 (K ) → GL2 (R )d .
Let OK be the ring of integers of K .
d
The group GL+
2 (OK ) acts on H component-wise via fractional
transformations: for every z = (z1 , . . . , zd ) ∈ Hd , for ever γ ∈ GL+
2 (OK )
we have γ · z = (σ1 (γ)z1 , . . . , σd (γ)zd ).
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Hilbert modular forms
For g =
a
c
b
d
∈ GL2 (R ) and z ∈ H define
j(g , z) = det(g )−1/2 (cz + d)
Definition
A Hilbert modular form f over K of weight (k1 , . . . , kd ) is an analytic
function on Hd such that for every γ ∈ GL+
2 (OK )
f (γz) =
d
Y
j(σi (γ), zi )ki f (z).
i=1
No extra condition is needed for the cusps (Koecher’s principle).
Hilbert modulars forms do have associated Fourier expansions
(q-expansions) and Galois representations (Taylor).
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Proof of Fermat’s Last Theorem
uses three big theorems:
1
Mazur: irreducibility of mod
` representations of elliptic
curves over Q for ` > 163
(i.e. absence of `-isogenies).
2
Wiles (and others):
modularity of elliptic curves
over Q .
3
Ribet: level lowering for mod
` representations—this
requires irreducibility and
modularity.
Over totally real fields we have
1
Merel’s uniform boundedness
theorem for torsion. No
corresponding result for
isogenies.
2
Partial modularity results, no
clean statements.
3
Level lowering for mod `
representations works exactly
as for Q : theorems of
Fujiwara, Jarvis and Rajaei.
Requires irreducibility and
modularity.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Reducible representations
Goal:
Want to bound ` such that ρE ,` is reducible.
Hypotheses:
1
E /K semistable elliptic curve, over Galois totally real field K .
2
` rational prime unramified in K .
ψ1 ∗
ρE ,` is reducible: ρE ,` ∼
,
0 ψ2
3
ψi : GK → F×
` .
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Fact:
υ - `,
υ finite
=⇒
ρE ,` |Iυ ∼
∗
.
1
1
0
Hence ψ1 , ψ2 are unramified at all finite υ - ` (i.e. ψi |Iυ = 1).
χ ∗
1 ∗
ρE ,` |Iυ ∼
Serre:
υ|`
=⇒
or
0 1
0 χ
χ(σ)
σ
χ : GK → F×
` is the mod ` cyclotomic character: ζ` = ζ`
.
Hence, for υ | `,
either ψ1 |Iυ = χ|Iυ and ψ2 |Iυ = 1;
or ψ1 |Iυ = 1 and ψ2 |Iυ = χ|Iυ .
Let
S` = {υ : υ | `},
S = { υ ∈ S` : ψ1 |Iυ = χ|Iυ }.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Lemma
Suppose
hK+ = 1;
S = ∅.
Then E (K )[`] 6= 0.
Proof.
S = ∅ =⇒ ψ1 : GK → F×
` is unramified at all finite places
=⇒ ψ1 = 1.
ρE ,` : GK → Aut(E [`]) ∼
= GL2 (F` ),
In this case, can bound ` by Merel.
ρE ,` ∼
1
0
∗
.
ψ2
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Lemma
Suppose
hK+ = 1;
S = S` .
Then E 0 (K )[`] 6= 0, where E 0 is `-isogenous to K .
Proof.
ρE ,` ∼
ψ1
0
∗
,
ψ2
ρE 0 ,` ∼
ψ2
0
∗
.
ψ1
S = S` =⇒ ψ2 : GK → F×
` is unramified at all finite places . . .
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Class Field Theory (Momose?, Kraus?, David?)
∃ non-empty proper subset S ⊂ S` , and ψ : GK → F×
` such that
ψ|Iυ = 1 for (finite) υ ∈
/ S, and
ψ|Iυ = χ|Iυ for υ ∈ S.
Let L = K (ψ). View ψ : Gal(L/K ) → F×
` .
Local Artin map Θυ : Kυ× → Gal(L/K ).
Let u ∈ OK be a totally positive unit.
Will compute ψ(Θυ (u)) as υ ranges over the places of K .
Suppose υ | ∞. Then u > 0 in Kυ . So Θυ (u) = 1. So
ψ(Θυ (u)) = 1.
Suppose υ - ∞. By local reciprocity Θυ (u) ∈ Iυ .
If υ ∈
/ S then ψ(Θυ (u)) = 1.
If υ ∈ S then ψ(Θυ (u)) = χ(Θυ (u)) = NormFυ /F` (u)−1 .
Global reciprocity =⇒
Y
Θυ (u) = 1
=⇒
Y
NormFυ /F` (u) = 1
υ∈S
(1 ∈ F` ).
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Question: Is there a non-empty proper subset S ⊂ S` and a character
ψ : GK → F×
/ S, and ψ|Iυ = χ|Iυ for
` such that ψ|Iυ = 1 for (finite) υ ∈
υ ∈ S?
If answer is no then can bound ` by Merel.
Suppose answer is YES. Let u be a totally positive unit. Then
Y
(1 ∈ F` ).
NormFυ /F` (u) = 1
υ∈S
Therefore, there is a non-empty proper subset T ⊂ Gal(K /Q ) such that
!
!
Y
σ
u
−1 .
` | BT (u)
BT (u) := Norm
σ∈T
Lemma (Freitas–Siksek)
For each non-empty proper subset T ⊂ Gal(K /Q ), there exists totally
positive unit u such that BT (u) 6= 0.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Residual irreducibility
Reducible representations
Let E be a Frey curve as in (∗).
Lemma
Suppose ρE ,` is reducible. Then either E /K has non-trivial `-torsion, or
is `-isogenous to an an elliptic curve over K that has non-trivial `-torsion.
Lemma
For p = 5, 7, 11, 13, and ` ≥ 5, with ` 6= p, the mod ` representation
ρE ,` is irreducible.
Sketch of the proof: use hK+ = 1 for all these p, class field theory and
Classification of `-torsion over fields of degree 2 (Kamienny), degree
3 (Parent), degrees 4, 5, 6 (Derickx, Kamienny, Stein, and Stoll).
“A criterion to rule out torsion groups for elliptic curves over
number fields”, Bruin and Najman.
Computations of K -points on modular curves.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Modularity
Three kinds of modularity theorems:
Kisin, Gee, Breuil, . . . : if ` = 3, 5 or 7 and ρE ,` (GK ) is ‘big’ then E is
modular.
√
Thorne: if ` = 5, and 5 ∈
/ K and PρE ,` (GK ) is dihedral then E is
modular.
Skinner & Wiles: if ρE ,` (GK ) is reducible (and other conditions) then E
is modular.
Fix ` = 5 and suppose
√
5∈
/ K . Remaining case ρE ,` (GK ) reducible.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Skinner & Wiles
K totally real field,
E /K semistable elliptic curve,
5 unramified in K ,
ρE ,5 is reducible:
ρE ,5 ∼
ψ1
0
∗
,
ψ2
ψi : GK → F×
5 .
Theorem (Skinner & Wiles)
Suppose K (ψ1 /ψ2 ) is an abelian extension of Q . Then E is modular.
Plan: Start with K abelian over Q . Find sufficient conditions so that
K (ψ1 /ψ2 ) ⊆ K (ζ5 ). Then (assuming these conditions) E is modular.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Reducible Representations
K real abelian field.
ρE ,5 ∼
ψ1
0
∗
,
ψ2
ψi : GK → F×
5 .
Fact: ψ1 ψ2 = χ
χ(σ)
σ
χ : GK → F×
5 satisfies ζ5 = ζ5
where
.
ψ1
ψ2
χ
= 2 = 1.
ψ2
χ
ψ2
K (ψ1 /ψ2 ) ⊆ K (ζ5 )K (ψ22 ),
K (ψ1 /ψ2 ) ⊆ K (ζ5 )K (ψ12 ).
Plan: If K (ψ12 ) = K or K (ψ22 ) = K then K (ψ1 /ψ2 ) ⊆ K (ζ5 ). Then E is
modular.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Modularity
Theorem (Anni-Siksek)
Let K be a real abelian number field. Write S5 = {q | 5}. Suppose
(a) 5 is unramified in K ;
(b) the class number of K is odd;
(c) for each non-empty proper subset S of S5 , there is some totally
positive unit u of OK such that
Y
NormFq /F5 (u mod q) 6= 1 .
q∈S
Then every semistable elliptic curve E over K is modular.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Proof.
By Kisin, . . . and Thorne, can suppose that ρE ,5 is reducible.
By (c), ψ1 or ψ2 is unramified at all finite places.
So ψ12 or ψ22 is unramified at all places.
By (b), K (ψ12 ) = K or K (ψ22 ) = K .
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Corollary
For p = 5, 7, 11, 13, the Frey curve E is modular.
Proof.
√
For p = 7, 11, 13 apply the above. For p = 5 we have K = Q ( 5).
Modularity of elliptic curves over quadratic fields was proved by Freitas,
Le Hung & Siksek.
Corollary
Let K be a real abelian field of conductor n < 100 with 5 - n and n 6= 29,
87, 89. Let E be a semistable elliptic curve over K . Then E is modular.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Let E /K be the Frey curve (∗), then ρE ,` is modular and irreducible.
Then ρE ,` ∼ ρf,λ for some Hilbert cuspidal eigenform f over K of parallel
weight 2 that is new at level N` , where
(
2OK if p - x
N` =
2p
if p | x .
Here λ | ` is a prime of Q f , the field generated over Q by the eigenvalues
of f.
For p = 3 the modular forms to consider are classical newform of weight
2 and level 6: there is no such newform and so we conclude.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
If p ≡ 1 (mod 4), the field K = Q (ζ + ζ −1 ) has a unique subfield K 0 of
degree (p − 1)/4.
In the case p - x the Frey curve E is defined over K 0 .
This not true in the case p | x, but we can take a twist of the Frey curve
some that it is defined over K 0 .
This way we can work with Hilbert modular cuspforms over a totally real
field of lower degree. The conductor of this Frey curve can be computed
similarly to the previous case.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
p
5
7
11
Case
5-x
5|x
7-x
7|x
11 - x
11 | x
Field K
K
K
K
K
K
K
Frey curve E
E
E
E
E
E
E
Level N
2K
2p
2K
2p
2K
2p
13 - x
K
E
2K
13 | x
K0
E0
2B
13
Eigenforms f
–
–
–
f1
f2
f3 , f4
f5 , f6
f7
f8
f9 , f10
f11 , f12
[Q f : Q ]
–
–
–
1
2
5
1
2
3
1
3
Table : Frey curve and Hilbert eigenform information. Here p is the unique
prime of K above p, and B is the unique prime of K 0 above p.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
In almost each case we deduce a contradiction using the q-expansions of
the Hilbert modular forms in the table and the study of the Frey curve
described before.
The only case left is the case p = 13 and ` = 7: we strongly suspect that
reducibility of ρf11 ,λ (where λ is the unique prime above 7 of Q f11 ) but
we are unable to prove it.
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Final remarks
In order to solve x 2` + y 2m = z p for p ≥ 17 we need to be able to
compute Hilbert modular cuspforms over totally real field of high degree.
Anyway the following theorem hold:
Theorem (Anni-Siksek)
Let p be an odd prime and let K = Q (ζ + ζ −1 ). Let OK be the ring of
integers of K and p be the unique prime ideal above p. Suppose that
there are no elliptic curves E /K with full 2-torsion and conductors 2OK ,
2p . Then there is an ineffective constant Cp (depending only on p) such
that for all primes `, m ≥ Cp , the only primitive solutions to
x 2` + y 2m = z p are (x, y , z) = (±1, 0, 1) and (0, ±1, 1).
Diophantine equations and semistable elliptic curves over totally real number fields
The proof
Modularity
Diophantine equations and semistable
elliptic curves over totally real
number fields
Samuele Anni
joint work with Samir Siksek
University of Warwick
Algebra, Geometry and Number Theory Seminar,
Universiteit Leiden, 18th April 2016
Thanks!