number theory seminar, feb.16, 2011 noam d. elkiespeople.math.harvard.edu/~elkies/j3.pdf ·...
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Refined 3-adic approximations to j(nζ3)
Number Theory Seminar, Feb.16, 2011
Noam D. Elkies
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
Refined 3-adic approximations to j(nζ3)
Number Theory Seminar, Feb.16, 2011
Noam D. Elkies
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
Recall: in char. 0, an elliptic curve E/k has (k-)endomorphism
ring End(E) either Z (ordinary case) or in imag. quad. order
(CM or “singular” case). If k ⊆ C, write E = C/Λ for some
lattice Λ = (Z + Zτ)ω1; then CM ⇐⇒ [Q(τ) : Q] = 2. This
depends only on the j-invariant j(E). If
E : y2 = x3 + a4x + a6
(possible except in characteristic 2 or sometimes 3) then
j(E) = 123 4a34
4a34 + 27a2
6
.
If k ⊆ C then j = j(τ) = q−1 + 744 + O(q) where q = e±2πiτ .
Examples of CM values of j, in terms of ∆ = disc(End(E)):
∆ −3 −4 −7 −8 −11 −12 −27 −48
j 0 123 −153 203 −323 2 · 303 −3 · 1603 4(450± 255√
3)3
1
Recall: in char. 0, an elliptic curve E/k has (k-)endomorphism
ring End(E) either Z (ordinary case) or in imag. quad. order
(CM or “singular” case). If k ⊆ C, write E = C/Λ for some
lattice Λ = (Z + Zτ)ω1; then CM ⇐⇒ [Q(τ) : Q] = 2. This
depends only on the j-invariant j(E). If
E : y2 = x3 + a4x + a6
(possible except in characteristic 2 or sometimes 3) then
j(E) = 123 4a34
4a34 + 27a2
6
.
If k ⊆ C then j = j(τ) = q−1 + 744 + O(q) where q = e±2πiτ .
Examples of CM values of j, in terms of ∆ = disc(End(E)):
∆ −3 −4 −7 −8 −11 −12 −27 −48
j 0 123 −153 203 −323 2 · 303 −3 · 1603 4(450± 255√
3)3
1
Recall: in char. 0, an elliptic curve E/k has (k-)endomorphism
ring End(E) either Z (ordinary case) or in imag. quad. order
(CM or “singular” case). If k ⊆ C, write E = C/Λ for some
lattice Λ = (Z + Zτ)ω1; then CM ⇐⇒ [Q(τ) : Q] = 2. This
depends only on the j-invariant j(E). If
E : y2 = x3 + a4x + a6
(possible except in characteristic 2 or sometimes 3) then
j(E) = 123 4a34
4a34 + 27a2
6
.
If k ⊆ C then j = j(τ) = q−1 + 744 + O(q) where q = e±2πiτ .
Examples of CM values of j, in terms of ∆ = disc(End(E)):
∆ −3 −4 −7 −8 −11 −12 −27 −48
j 0 123 −153 203 −323 2 · 303 −3 · 1603 4(450± 255√
3)3
1
Recall: in char. 0, an elliptic curve E/k has (k-)endomorphism
ring End(E) either Z (ordinary case) or in imag. quad. order
(CM or “singular” case). If k ⊆ C, write E = C/Λ for some
lattice Λ = (Z + Zτ)ω1; then CM ⇐⇒ [Q(τ) : Q] = 2. This
depends only on the j-invariant j(E). If
E : y2 = x3 + a4x + a6
(possible except in characteristic 2 or sometimes 3) then
j(E) = 123 4a34
4a34 + 27a2
6
.
If k ⊆ C then j = j(τ) = q−1 + 744 + O(q) where q = e±2πiτ .
Examples of CM values of j, in terms of ∆ = disc(End(E)):
∆ −3 −4 −7 −8 −11 −12 −27 −48
j 0 123 −153 203 −323 2 · 303 −3 · 1603 4(450± 255√
3)3
1
Recall: in char. 0, an elliptic curve E/k has (k-)endomorphism
ring End(E) either Z (ordinary case) or in imag. quad. order
(CM or “singular” case). If k ⊆ C, write E = C/Λ for some
lattice Λ = (Z + Zτ)ω1; then CM ⇐⇒ [Q(τ) : Q] = 2. This
depends only on the j-invariant j(E). If
E : y2 = x3 + a4x + a6
(possible except in characteristic 2 or sometimes 3) then
j(E) = 123 4a34
4a34 + 27a2
6
.
If k ⊆ C then j = j(τ) = q−1 + 744 + O(q) where q = e±2πiτ .
Examples of CM values of j, in terms of ∆ = disc(End(E)):
∆ −3 −4 −7 −8 −11 −12 −27 −48
j 0 123 −153 203 −323 2 · 303 −3 · 1603 4(450± 255√
3)3
1
Recall: in char. 0, an elliptic curve E/k has (k-)endomorphism
ring End(E) either Z (ordinary case) or in imag. quad. order
(CM or “singular” case). If k ⊆ C, write E = C/Λ for some
lattice Λ = (Z + Zτ)ω1; then CM ⇐⇒ [Q(τ) : Q] = 2. This
depends only on the j-invariant j(E). If
E : y2 = x3 + a4x + a6
(possible except in characteristic 2 or sometimes 3) then
j(E) = 123 4a34
4a34 + 27a2
6
.
If k ⊆ C then j = j(τ) = q−1 + 744 + O(q) where q = e±2πiτ .
Examples of CM values of j, in terms of ∆ = disc(End(E)):
∆ −3 −4 −7 −8 −11 −12 −27 −48
j 0 123 −153 203 −323 2 · 303 −3 · 1603 4(450± 255√
3)3
1
• Galois properties: CM values of j, and of other modular
functions, generate abelian extensions of the CM field =⇒Jugendtraum (explicit class field theory) for quad. imag. fields
• Diophantine properties from maps between mod. curves, e.g.
j
1 +√−163
2
= −6403203 = 123 − 163 · 401330162
“because” j1/3 and (j − 123)1/2 are Hauptmoduln for curves
intermediate between X(1) and X(p) for p = 3 and p = 2
• Congruences: j1 6= j2 congruent mod p iff corresponding CM
curves E1, E2 reduce mod p to the same curve E; Ei both CM
=⇒ E supersingular with End(E) containing both End(Ei)’s,
=⇒ p is small, etc. [Gross and Zagier, “On Singular Moduli”].
E.g. 640320 = 26 3 5 23 29, 40133016 = 23 33 7 11 19 127.
But it turns out there’s more. . .2
• Galois properties: CM values of j, and of other modular
functions, generate abelian extensions of the CM field =⇒Jugendtraum (explicit class field theory) for quad. imag. fields
• Diophantine properties from maps between mod. curves, e.g.
j
1 +√−163
2
= −6403203 = 123 − 163 · 401330162
“because” j1/3 and (j − 123)1/2 are Hauptmoduln for curves
intermediate between X(1) and X(p) for p = 3 and p = 2
• Congruences: j1 6= j2 congruent mod p iff corresponding CM
curves E1, E2 reduce mod p to the same curve E; Ei both CM
=⇒ E supersingular with End(E) containing both End(Ei)’s,
=⇒ p is small, etc. [Gross and Zagier, “On Singular Moduli”].
E.g. 640320 = 26 3 5 23 29, 40133016 = 23 33 7 11 19 127.
But it turns out there’s more. . .2
• Galois properties: CM values of j, and of other modular
functions, generate abelian extensions of the CM field =⇒Jugendtraum (explicit class field theory) for quad. imag. fields
• Diophantine properties from maps between mod. curves, e.g.
j
1 +√−163
2
= −6403203 = 123 − 163 · 401330162
“because” j1/3 and (j − 123)1/2 are Hauptmoduln for curves
intermediate between X(1) and X(p) for p = 3 and p = 2
• Congruences: j1 6= j2 congruent mod p iff corresponding CM
curves E1, E2 reduce mod p to the same curve E; Ei both CM
=⇒ E supersingular with End(E) containing both End(Ei)’s,
=⇒ p is small, etc. [Gross and Zagier, “On Singular Moduli”].
E.g. 640320 = 26 3 5 23 29, 40133016 = 23 33 7 11 19 127.
But it turns out there’s more. . .2
• Galois properties: CM values of j, and of other modular
functions, generate abelian extensions of the CM field =⇒Jugendtraum (explicit class field theory) for quad. imag. fields
• Diophantine properties from maps between mod. curves, e.g.
j
1 +√−163
2
= −6403203 = 123 − 163 · 401330162
“because” j1/3 and (j − 123)1/2 are Hauptmoduln for curves
intermediate between X(1) and X(p) for p = 3 and p = 2
• Congruences: j1 6= j2 congruent mod p iff corresponding CM
curves E1, E2 reduce mod p to the same curve E; Ei both CM
=⇒ E supersingular with End(E) containing both End(Ei)’s,
=⇒ p is small, etc. [Gross and Zagier, “On Singular Moduli”].
E.g. 640320 = 26 3 5 23 29, 40133016 = 23 33 7 11 19 127.
But it turns out there’s more. . .2
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
From: Brian Conrad Fri Jan 21 11:26
To: NDE
[...] Samit [Dasgupta] & John Voight need [...] the following:
Claim: For n > 0 not divisible by 3, j(n zeta_3) mod 81 Zbar
is 0 or -27 depending respectively on whether n = 1 mod 3
or n = -1 mod 3.
I gave [...] a proof of this claim by using deformation theory
of elliptic curves and Serre-Tate to switch to deformation
theory of 3-divisible groups, for which a calculation with
Grothendieck-Messing theory (to "see" the CM property in terms
of a deformation parameter) does the job.
From: NDE
To: self
!?3
From: Brian Conrad Fri Jan 21 11:26
To: NDE
[...] Samit [Dasgupta] & John Voight need [...] the following:
Claim: For n > 0 not divisible by 3, j(n zeta_3) mod 81 Zbar
is 0 or -27 depending respectively on whether n = 1 mod 3
or n = -1 mod 3.
I gave [...] a proof of this claim by using deformation theory
of elliptic curves and Serre-Tate to switch to deformation
theory of 3-divisible groups, for which a calculation with
Grothendieck-Messing theory (to "see" the CM property in terms
of a deformation parameter) does the job.
From: NDE
To: self
!?3
Brian continued:
I couldn’t see a way to prove it using less esoteric methods
(whereas the weaker claim that j(n zeta_3) = 0 mod 27
for such n does follow from arguments in the spirit of the
Gross-Zagier "Singular Moduli" paper).
In view of the concrete nature of the assertion in the Claim,
it does seem natural to wonder if there may be a more classical
approach for proving it.
From: NDE
To: self
In view of the concrete nature of the assertion in the Claim,
it does seem natural to wonder if there may be a more classical
approach for proving it.
4
Brian continued:
I couldn’t see a way to prove it using less esoteric methods
(whereas the weaker claim that j(n zeta_3) = 0 mod 27
for such n does follow from arguments in the spirit of the
Gross-Zagier "Singular Moduli" paper).
In view of the concrete nature of the assertion in the Claim,
it does seem natural to wonder if there may be a more classical
approach for proving it.
From: NDE
To: self
In view of the concrete nature of the assertion in the Claim,
it does seem natural to wonder if there may be a more classical
approach for proving it.
4
Brian’s Jan.21 e-mail concludes:
Do you think there should be such a method, or see how to
probably carry it out?
Brian
From: NDE
To: self
Hm...
5
Brian’s Jan.21 e-mail concludes:
Do you think there should be such a method, or see how to
probably carry it out?
Brian
From: NDE
To: self
Hm...
5
NDE to Brian, Mon Jan 24 22:27 : Yes, and indeed we get
a congruence mod 39/2Z, possibly even mod 35 in the case
n ≡ −1 mod 3 (in which case j ≡ 54).
Brian to NDE et al., Mon Jan 24 23:51 :
Good news: the method with Grothendieck-Messing proves
the stronger result you mentioned (congruences modulo 39/2,
not just 34) with no extra work. [. . . ] Your suggested stronger
result modulo 35 (for n ≡ −1 mod 3) doesn’t fit as easily [. . . ]
How extensive is the numerical evidence in favor of this mod-35
congruence?
6
NDE to Brian, Mon Jan 24 22:27 : Yes, and indeed we get
a congruence mod 39/2Z, possibly even mod 35 in the case
n ≡ −1 mod 3 (in which case j ≡ 54).
Brian to NDE et al., Mon Jan 24 23:51 :
Good news: the method with Grothendieck-Messing proves
the stronger result you mentioned (congruences modulo 39/2,
not just 34) with no extra work. [. . . ] Your suggested stronger
result modulo 35 (for n ≡ −1 mod 3) doesn’t fit as easily [. . . ]
How extensive is the numerical evidence in favor of this mod-35
congruence?
6
Brian to NDE et al., Wed Jan 26 11:09 :
With a bit more care in the deformation theory, I can prove
your conjectured mod-35 congruence using the Grothendieck-
Messing method. Do you see a way to a proof of it by using
the more elementary methods [. . . ]?
NDE to Brian et al., Wed Jan 26 11:24 :
I had only a few examples when I first wrote, but now I’ve
checked it for all prime n < 200, and all n = 2p with n < 150,
subject to the congruence n ≡ 2 mod 3. Better yet, I have a
proof for all n. . .
7
Brian to NDE et al., Wed Jan 26 11:09 :
With a bit more care in the deformation theory, I can prove
your conjectured mod-35 congruence using the Grothendieck-
Messing method. Do you see a way to a proof of it by using
the more elementary methods [. . . ]?
NDE to Brian et al., Wed Jan 26 11:24 :
I had only a few examples when I first wrote, but now I’ve
checked it for all prime n < 200, and all n = 2p with n < 150,
subject to the congruence n ≡ 2 mod 3. Better yet, I have a
proof for all n. . .
7
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
Let E : y2 = x3 + a4x + a6 be a curve with j(E) = j(nζ3).
Then E is n-isogenous with a curve E0 of j-invariant zero.
That is, E0 contains a cyclic group G with #(G) = n such
that E0/G ∼= E. Plan: get at E, and thus at its a4, a6, and
j = 123 · 4a34/(4a3
4 + 27a26), in terms of G, which is to say in
terms of the coordinates on E0 of the nonzero elements of G.
We’ll use formulas of Velu (1971) that we’ll motivate by com-
parison with Weierstraß’s construction of the functions ℘, ℘′
on C/Λ.
8
We want an even meromorphic function on C that’s Λ-periodic
and has only a double pole at the origin of C/Λ. Start with the
function 1/z2 on C and try to form its trace w.r.t. translations
by Λ. It doesn’t quite converge, so renormalize:
℘(z,Λ) :=1
z2+
∑ω∈Λ
′(
1
(z − ω)2−
1
ω2
).
Now suppose [Λ : Λ0] = n and we already know ℘(z,Λ0). Then
℘(z,Λ) = ℘(z,Λ0) +∑
ω∈Λ\Λ0
(1
(z − ω)2−
1
ω2
)
= ℘(z,Λ0) +∑′
g∈Λ/Λ0
∑[ω]=g
(· · ·)
= ℘(z,Λ0) +∑′
g∈Λ/Λ0
(℘(z + g,Λ0)− ℘(g,Λ0))
which is a rational function in ℘(z,Λ0), ℘′(z,Λ0), and the co-
ordinates ℘(g,Λ), ℘′(g,Λ).
9
We want an even meromorphic function on C that’s Λ-periodic
and has only a double pole at the origin of C/Λ. Start with the
function 1/z2 on C and try to form its trace w.r.t. translations
by Λ. It doesn’t quite converge, so renormalize:
℘(z,Λ) :=1
z2+
∑ω∈Λ
′(
1
(z − ω)2−
1
ω2
).
Now suppose [Λ : Λ0] = n and we already know ℘(z,Λ0). Then
℘(z,Λ) = ℘(z,Λ0) +∑
ω∈Λ\Λ0
(1
(z − ω)2−
1
ω2
)
= ℘(z,Λ0) +∑′
g∈Λ/Λ0
∑[ω]=g
(· · ·)
= ℘(z,Λ0) +∑′
g∈Λ/Λ0
(℘(z + g,Λ0)− ℘(g,Λ0))
which is a rational function in ℘(z,Λ0), ℘′(z,Λ0), and the co-
ordinates ℘(g,Λ), ℘′(g,Λ).
9
By either analogy or generalization from
℘(z,Λ) =1
z2+
∑ω∈Λ
′(
1
(z − ω)2−
1
ω2
)
and
℘(z,Λ) = ℘(z,Λ0) +∑′
g∈Λ/Λ0
(℘(z + g,Λ0)− ℘(g,Λ0)) ,
we thus form an even function X on E0/G:
X(p + G) =∑g∈G
X0(p + g) −∑g∈G
′X0(g)
for any elliptic curve E0/k with a finite subgroup G of E0(k).
This is almost the trace of X0 w.r.t. translations by G: we sub-
tracted∑′
g∈G X0(g) to kill the constant coeff. of the Laurent
expansion of X about z = 0.
10
Now to get at a Weierstraß equation Y 2 = X3 + a4X + a6 for
E/Λ, we need not just X = ℘ but also an odd function Y :
Y = −1
2
℘(z,Λ)
dz=
∑ω∈Λ
1
(z − ω)3.
Then the Laurent expansions of ℘, ℘′ about z = 0 give the
coefficients
a4 = −15∑ω∈Λ
′ω−4, a6 = −35
∑ω∈Λ
′ω−6.
If [Λ : Λ0] = n as before then
−1
2℘′(z,Λ) =
∑g∈Λ/Λ0
∑[ω]=g
1
(z − ω)3,
which is exactly the trace of Y0 w.r.t. translations by G.
11
Now to get at a Weierstraß equation Y 2 = X3 + a4X + a6 for
E/Λ, we need not just X = ℘ but also an odd function Y :
Y = −1
2
℘(z,Λ)
dz=
∑ω∈Λ
1
(z − ω)3.
Then the Laurent expansions of ℘, ℘′ about z = 0 give the
coefficients
a4 = −15∑ω∈Λ
′ω−4, a6 = −35
∑ω∈Λ
′ω−6.
If [Λ : Λ0] = n as before then
−1
2℘′(z,Λ) =
∑g∈Λ/Λ0
∑[ω]=g
1
(z − ω)3,
which is exactly the trace of Y0 w.r.t. translations by G.
11
So again for an elliptic curve E0 : Y 20 = X3
0 + (a4)0X0 + (a6)0over any field k, and a finite subgroup G of E0(k), we take
X(p + G) =∑g∈G
X0(p + g) −∑g∈G
′X0(g)
as above and
Y (p + G) =∑g∈G
Y0(p + g),
and these generate the function field of E = E0/G.
Comparing Y 2 and powers of X near the origin X0 = ∞, Velu
found the Weierstraß equation Y 2 = X3 + a4X + a6 for E,
where
a4 = (a4)0 − 5∑
g∈G
′ (3X0(g)
2 + (a4)0)
,
a6 = (a6)0 − 7∑
g∈G
′ (5X0(g)
3 + 3(a4)0X0(g) + 2(a6)0)
.
12
So again for an elliptic curve E0 : Y 20 = X3
0 + (a4)0X0 + (a6)0over any field k, and a finite subgroup G of E0(k), we take
X(p + G) =∑g∈G
X0(p + g) −∑g∈G
′X0(g)
as above and
Y (p + G) =∑g∈G
Y0(p + g),
and these generate the function field of E = E0/G.
Comparing Y 2 and powers of X near the origin X0 = ∞, Velu
found the Weierstraß equation Y 2 = X3 + a4X + a6 for E,
where
a4 = (a4)0 − 5∑
g∈G
′ (3X0(g)
2 + (a4)0)
,
a6 = (a6)0 − 7∑
g∈G
′ (5X0(g)
3 + 3(a4)0X0(g) + 2(a6)0)
.
12
Exercise: Recall that if E0 is an ordinary elliptic curve in char-
acteristic 3, or any elliptic curve in characteristic 2, then E0 has
no “narrow Weierstraß equation” Y 20 = X3
0 +(a4)0X0 +(a6)0,
though it always has an “extended Weierstraß equation” that
incorporates also multiples of X0Y0 , X20, Y0 . For such an
equation, and a finite subgroup G of E0(k), generalize Velu’s
construction to find coordinates and an extended Weierstraß
equation for the quotient curve E0/G.
13
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
In our setting, E0 has j-invariant zero, i.e. (a4)0 = 0. We
can choose any convenient nonzero α for (a6)0, so E0 has
Weierstraß equation Y 20 = X3
0 − α, and then Velu’s formulas
simplify to
a4 = −15∑g∈G
′X0(g)
2, a6 = α− 7∑g∈G
′ (5X0(g)
3 + 2α)
.
How to approximate 3-adically these a4 and a6?
Lemma. Let α = 1. If g is a nontrivial torsion point of order
prime to 3 then X0(g) ≡ 1 mod ρ.
(Here ρ = 2ζ3 + 1 so ρ2 = −3.)
This will let us distinguish n ≡ 1 mod 3 from n ≡ −1 mod 3:
the number of terms in each∑′ is n− 1.
14
In our setting, E0 has j-invariant zero, i.e. (a4)0 = 0. We
can choose any convenient nonzero α for (a6)0, so E0 has
Weierstraß equation Y 20 = X3
0 − α, and then Velu’s formulas
simplify to
a4 = −15∑g∈G
′X0(g)
2, a6 = α− 7∑g∈G
′ (5X0(g)
3 + 2α)
.
How to approximate 3-adically these a4 and a6?
Lemma. Let α = 1. If g is a nontrivial torsion point of order
prime to 3 then X0(g) ≡ 1 mod ρ.
(Here ρ = 2ζ3 + 1 so ρ2 = −3.)
This will let us distinguish n ≡ 1 mod 3 from n ≡ −1 mod 3:
the number of terms in each∑′ is n− 1.
14
[repeat] Lemma. Let α = 1. If g is a nontrivial torsion point
of order prime to 3 then X0(g) ≡ 1 mod ρ.
(Here ρ = 2ζ3 + 1 so ρ2 = −3.)
Proof : Since j = 0 ∈ Z3 the curve E0 has potential good
reduction at 3. But the model Y 20 = X3
0 − 1 is singular mod 3
at (X0, Y0) = (1,0). Good reduction is attained over Q[ρ1/2]:
(X0, Y0) = (1 + ρX1, ρ3/2Y1) gives Y 21 = X3
1 − ρ2X21 −X1 ,
which reduces mod ρ to the smooth curve Y 21 = X3
1 − X1.
Hence X1(g) ∈ Z3, whence X0(g) ≡ 1 mod ρ, Q.E.D.
Remark: The need for a fourth root of an element of 3Z∗3could be predicted from ∆(E0) = −432 = −2433, since the
valuation of ∆ is constant mod 12.
15
[repeat] Lemma. Let α = 1. If g is a nontrivial torsion point
of order prime to 3 then X0(g) ≡ 1 mod ρ.
(Here ρ = 2ζ3 + 1 so ρ2 = −3.)
Proof : Since j = 0 ∈ Z3 the curve E0 has potential good
reduction at 3. But the model Y 20 = X3
0 − 1 is singular mod 3
at (X0, Y0) = (1,0). Good reduction is attained over Q[ρ1/2]:
(X0, Y0) = (1 + ρX1, ρ3/2Y1) gives Y 21 = X3
1 − ρ2X21 −X1 ,
which reduces mod ρ to the smooth curve Y 21 = X3
1 − X1.
Hence X1(g) ∈ Z3, whence X0(g) ≡ 1 mod ρ, Q.E.D.
Remark: The need for a fourth root of an element of 3Z∗3could be predicted from ∆(E0) = −432 = −2433, since the
valuation of ∆ is constant mod 12.
15
We can now prove:
Theorem. i) If n ≡ 1 mod 3 then j(nζ3) ≡ 0 mod ρ9.
ii) If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod ρ9.
Proof: We saw that j(nζ3) = 123 · 4a34/(4a3
4 + 27a26) where
a4 = −15∑g∈G
′X0(g)
2, a6 = α− 7∑g∈G
′ (5X0(g)
3 − 2)
and the sums extend over n− 1 nonzero points g on the curve
E0 : Y 20 = X3
0 − 1, each satisfying n · g = 0. By our Lemma
each of these satisfies X0(g) ≡ 1 mod ρ. Hence
(a4, a6) ≡ (3(n− 1), −1− 3(n− 1)) mod ρ3,
from which
(4a34,27a2
6) ≡ (108(n− 1)3, 27− 81(n− 1)) mod ρ9
[NB the residues of X0 and a4/3 mod ρ determine their cubes
mod ρ3]. The claimed congruences soon follow, Q.E.D.
16
We can now prove:
Theorem. i) If n ≡ 1 mod 3 then j(nζ3) ≡ 0 mod ρ9.
ii) If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod ρ9.
Proof: We saw that j(nζ3) = 123 · 4a34/(4a3
4 + 27a26) where
a4 = −15∑g∈G
′X0(g)
2, a6 = α− 7∑g∈G
′ (5X0(g)
3 − 2)
and the sums extend over n− 1 nonzero points g on the curve
E0 : Y 20 = X3
0 − 1, each satisfying n · g = 0. By our Lemma
each of these satisfies X0(g) ≡ 1 mod ρ. Hence
(a4, a6) ≡ (3(n− 1), −1− 3(n− 1)) mod ρ3,
from which
(4a34,27a2
6) ≡ (108(n− 1)3, 27− 81(n− 1)) mod ρ9
[NB the residues of X0 and a4/3 mod ρ determine their cubes
mod ρ3]. The claimed congruences soon follow, Q.E.D.
16
We can now prove:
Theorem. i) If n ≡ 1 mod 3 then j(nζ3) ≡ 0 mod ρ9.
ii) If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod ρ9.
Proof: We saw that j(nζ3) = 123 · 4a34/(4a3
4 + 27a26) where
a4 = −15∑g∈G
′X0(g)
2, a6 = α− 7∑g∈G
′ (5X0(g)
3 − 2)
and the sums extend over n− 1 nonzero points g on the curve
E0 : Y 20 = X3
0 − 1, each satisfying n · g = 0. By our Lemma
each of these satisfies X0(g) ≡ 1 mod ρ. Hence
(a4, a6) ≡ (3(n− 1), −1− 3(n− 1)) mod ρ3,
from which
(4a34,27a2
6) ≡ (108(n− 1)3, 27− 81(n− 1)) mod ρ9
[NB the residues of X0 and a4/3 mod ρ determine their cubes
mod ρ3]. The claimed congruences soon follow, Q.E.D.
16
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)and mod 36 (for n ≡ 1 mod 3)
Further remarks
Can we give congruences modulo powers of ρ past ρ9?
For n ≡ +1 mod 3, not in this form: The congruence would
have to hold for all conjugates of j(nζ3), but already for n = 4
and n = 7, when j(nζ3) is only a quadratic irrationality, the dis-
criminant of the quadratic has 3-valuation 9 (as is predictable
from a formula in Gross and Zagier’s “On Singular Moduli”),
so a 3-adic ball that contains them both must have radius at
least 3−9/2. Explicitly,
j(4ζ3) = 1417905000 + 500 · 17 · 29 · 41 · 39/2,
j(7ζ3) = −17424252776448000− 2165311223 · 29 · 71 · 34√
21.
But for n ≡ −1 mod 3 there’s hope, e.g.
j(5ζ3) = −327201914880− 21517 · 23 · 47 · 35√
5.
17
Can we give congruences modulo powers of ρ past ρ9?
For n ≡ +1 mod 3, not in this form: The congruence would
have to hold for all conjugates of j(nζ3), but already for n = 4
and n = 7, when j(nζ3) is only a quadratic irrationality, the dis-
criminant of the quadratic has 3-valuation 9 (as is predictable
from a formula in Gross and Zagier’s “On Singular Moduli”),
so a 3-adic ball that contains them both must have radius at
least 3−9/2. Explicitly,
j(4ζ3) = 1417905000 + 500 · 17 · 29 · 41 · 39/2,
j(7ζ3) = −17424252776448000− 2165311223 · 29 · 71 · 34√
21.
But for n ≡ −1 mod 3 there’s hope, e.g.
j(5ζ3) = −327201914880− 21517 · 23 · 47 · 35√
5.
17
Can we give congruences modulo powers of ρ past ρ9?
For n ≡ +1 mod 3, not in this form: The congruence would
have to hold for all conjugates of j(nζ3), but already for n = 4
and n = 7, when j(nζ3) is only a quadratic irrationality, the dis-
criminant of the quadratic has 3-valuation 9 (as is predictable
from a formula in Gross and Zagier’s “On Singular Moduli”),
so a 3-adic ball that contains them both must have radius at
least 3−9/2. Explicitly,
j(4ζ3) = 1417905000 + 500 · 17 · 29 · 41 · 39/2,
j(7ζ3) = −17424252776448000− 2165311223 · 29 · 71 · 34√
21.
But for n ≡ −1 mod 3 there’s hope, e.g.
j(5ζ3) = −327201914880− 21517 · 23 · 47 · 35√
5.
17
Can we give congruences modulo powers of ρ past ρ9?
For n ≡ +1 mod 3, not in this form: The congruence would
have to hold for all conjugates of j(nζ3), but already for n = 4
and n = 7, when j(nζ3) is only a quadratic irrationality, the dis-
criminant of the quadratic has 3-valuation 9 (as is predictable
from a formula in Gross and Zagier’s “On Singular Moduli”),
so a 3-adic ball that contains them both must have radius at
least 3−9/2. Explicitly,
j(4ζ3) = 1417905000 + 500 · 17 · 29 · 41 · 39/2,
j(7ζ3) = −17424252776448000− 2165311223 · 29 · 71 · 34√
21.
But for n ≡ −1 mod 3 there’s hope, e.g.
j(5ζ3) = −327201914880− 21517 · 23 · 47 · 35√
5.
17
Theorem. If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod 35.
Proof: Since Q(ρ, j(nζ3)) is unramified over Q3(ρ), we need
only prove ρ−9(j(nζ3)− 54) ≡ 0 modulo each prime above 3.
Velu’s formulas let us compute the residue of ρ−9(j(nζ3)−54)
in terms of the residues of the X1(g) for nonzero g ∈ G, which
are the x1-coordinates of the images of these g on the curve
E0 : y21 = x3
1 − x1 in characteristic 3.
We find:
18
Theorem. If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod 35.
Proof: Since Q(ρ, j(nζ3)) is unramified over Q3(ρ), we need
only prove ρ−9(j(nζ3)− 54) ≡ 0 modulo each prime above 3.
Velu’s formulas let us compute the residue of ρ−9(j(nζ3)−54)
in terms of the residues of the X1(g) for nonzero g ∈ G, which
are the x1-coordinates of the images of these g on the curve
E0 : y21 = x3
1 − x1 in characteristic 3.
We find:
18
Theorem. If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod 35.
Proof: Since Q(ρ, j(nζ3)) is unramified over Q3(ρ), we need
only prove ρ−9(j(nζ3)− 54) ≡ 0 modulo each prime above 3.
Velu’s formulas let us compute the residue of ρ−9(j(nζ3)−54)
in terms of the residues of the X1(g) for nonzero g ∈ G, which
are the x1-coordinates of the images of these g on the curve
E0 : y21 = x3
1 − x1 in characteristic 3.
We find:
18
Theorem. If n ≡ −1 mod 3 then j(nζ3) ≡ 54 mod 35.
Proof: Since Q(ρ, j(nζ3)) is unramified over Q3(ρ), we need
only prove ρ−9(j(nζ3)− 54) ≡ 0 modulo each prime above 3.
Velu’s formulas let us compute the residue of ρ−9(j(nζ3)−54)
in terms of the residues of the X1(g) for nonzero g ∈ G, which
are the x1-coordinates of the images of these g on the curve
E0 : y21 = x3
1 − x1 in characteristic 3.
We find:
18
a4 ≡ 3− 3ρ2∑′g x1(g) mod 9, whence
4a34 ≡ 108 + 81ρ2∑
g
′(x1(g)
3 − x1) mod 35;
and a6 ≡ −4− 3ρ2∑′g(x1(g)
3 − x1) mod 9, whence
27a26 ≡ −54 + 81ρ2∑
g
′(x1(g)
3 − x1) mod 35.
The bad news is that the effect of the new term∑′
g(x1(g)3−x1)
does not cancel out in the computation of j(E).
The good news is:
Proposition. Let G be a subgroup of order n ≡ −1 mod 3 of
the elliptic curve E0 : y21 = x3
1 − x1 in characteristic 3. Then∑g∈G
′(x1(g)
3 − x1(g)) = 0.
19
a4 ≡ 3− 3ρ2∑′g x1(g) mod 9, whence
4a34 ≡ 108 + 81ρ2∑
g
′(x1(g)
3 − x1) mod 35;
and a6 ≡ −4− 3ρ2∑′g(x1(g)
3 − x1) mod 9, whence
27a26 ≡ −54 + 81ρ2∑
g
′(x1(g)
3 − x1) mod 35.
The bad news is that the effect of the new term∑′
g(x1(g)3−x1)
does not cancel out in the computation of j(E).
The good news is:
Proposition. Let G be a subgroup of order n ≡ −1 mod 3 of
the elliptic curve E0 : y21 = x3
1 − x1 in characteristic 3. Then∑g∈G
′(x1(g)
3 − x1(g)) = 0.
19
a4 ≡ 3− 3ρ2∑′g x1(g) mod 9, whence
4a34 ≡ 108 + 81ρ2∑
g
′(x1(g)
3 − x1) mod 35;
and a6 ≡ −4− 3ρ2∑′g(x1(g)
3 − x1) mod 9, whence
27a26 ≡ −54 + 81ρ2∑
g
′(x1(g)
3 − x1) mod 35.
The bad news is that the effect of the new term∑′
g(x1(g)3−x1)
does not cancel out in the computation of j(E).
The good news is:
Proposition. Let G be a subgroup of order n ≡ −1 mod 3 of
the elliptic curve E0 : y21 = x3
1 − x1 in characteristic 3. Then∑g∈G
′(x1(g)
3 − x1(g)) = 0.
19
In characteristic 3 we have∑
x3 = (∑
x)3, so∑g∈G
′(x1(g)
3−x1(g)) = 0 ⇐⇒∑g∈G
′x1(g) ∈ F3 ⇐⇒
∑g∈G
′(x1(g))
3 ∈ F3.
Now E0 is supersingular with Frob9 = −3, so G is stable under
Frob9 and we automatically have∑′
g x1(g) ∈ F9 = F3[i].
Also j(E0) = 0 is the unique supersingular j-invariant in char-
acteristic 3, and all of End(E0) is defined over F9. Hence E0/G
is F9-isogenous with E0, and therefore F9-isomorphic with E0.
We can thus apply Velu’s formulas to this isogeny, here with
((a4)0, (a6)0)) = (−1,0). We find (a4, a6) = (+1,∑′
g x1(g)3).
But y2 = x3 + x + a6 is F9-isomorphic with E0 iff ia6 = u3 − u
for some u ∈ F9, which is exactly the condition a6 ∈ F3, Q.E.D.
20
In characteristic 3 we have∑
x3 = (∑
x)3, so∑g∈G
′(x1(g)
3−x1(g)) = 0 ⇐⇒∑g∈G
′x1(g) ∈ F3 ⇐⇒
∑g∈G
′(x1(g))
3 ∈ F3.
Now E0 is supersingular with Frob9 = −3, so G is stable under
Frob9 and we automatically have∑′
g x1(g) ∈ F9 = F3[i].
Also j(E0) = 0 is the unique supersingular j-invariant in char-
acteristic 3, and all of End(E0) is defined over F9. Hence E0/G
is F9-isogenous with E0, and therefore F9-isomorphic with E0.
We can thus apply Velu’s formulas to this isogeny, here with
((a4)0, (a6)0)) = (−1,0). We find (a4, a6) = (+1,∑′
g x1(g)3).
But y2 = x3 + x + a6 is F9-isomorphic with E0 iff ia6 = u3 − u
for some u ∈ F9, which is exactly the condition a6 ∈ F3, Q.E.D.
20
In characteristic 3 we have∑
x3 = (∑
x)3, so∑g∈G
′(x1(g)
3−x1(g)) = 0 ⇐⇒∑g∈G
′x1(g) ∈ F3 ⇐⇒
∑g∈G
′(x1(g))
3 ∈ F3.
Now E0 is supersingular with Frob9 = −3, so G is stable under
Frob9 and we automatically have∑′
g x1(g) ∈ F9 = F3[i].
Also j(E0) = 0 is the unique supersingular j-invariant in char-
acteristic 3, and all of End(E0) is defined over F9. Hence E0/G
is F9-isogenous with E0, and therefore F9-isomorphic with E0.
We can thus apply Velu’s formulas to this isogeny, here with
((a4)0, (a6)0)) = (−1,0). We find (a4, a6) = (+1,∑′
g x1(g)3).
But y2 = x3 + x + a6 is F9-isomorphic with E0 iff ia6 = u3 − u
for some u ∈ F9, which is exactly the condition a6 ∈ F3, Q.E.D.
20
In characteristic 3 we have∑
x3 = (∑
x)3, so∑g∈G
′(x1(g)
3−x1(g)) = 0 ⇐⇒∑g∈G
′x1(g) ∈ F3 ⇐⇒
∑g∈G
′(x1(g))
3 ∈ F3.
Now E0 is supersingular with Frob9 = −3, so G is stable under
Frob9 and we automatically have∑′
g x1(g) ∈ F9 = F3[i].
Also j(E0) = 0 is the unique supersingular j-invariant in char-
acteristic 3, and all of End(E0) is defined over F9. Hence E0/G
is F9-isogenous with E0, and therefore F9-isomorphic with E0.
We can thus apply Velu’s formulas to this isogeny, here with
((a4)0, (a6)0)) = (−1,0). We find (a4, a6) = (+1,∑′
g x1(g)3).
But y2 = x3 + x + a6 is F9-isomorphic with E0 iff ia6 = u3 − u
for some u ∈ F9, which is exactly the condition a6 ∈ F3, Q.E.D.
20
So what happens for n ≡ +1 mod 3?
Same kind of analysis shows the residue of j(nζ3) mod 35, and
indeed mod 36, includes a term∑′
g x1(g)3 (not x1(g)
3−x1(g)).
This time Velu gives an isogeny from E0 to a curve E0/G:
y2 = x3 − x + a6 with a6 =∑′
g x1(g)3. The fact that this is
F9-isomorphic with E0 then yields a6 ∈ iF3.
We’ve thus obtained:
Theorem. If n ≡ +1 mod 3 then
j(nζ3) ≡ 324ρ∑g∈G
′(x1(g))
3 ≡ 0 or ± 324√
3 mod 36.
21
So what happens for n ≡ +1 mod 3?
Same kind of analysis shows the residue of j(nζ3) mod 35, and
indeed mod 36, includes a term∑′
g x1(g)3 (not x1(g)
3−x1(g)).
This time Velu gives an isogeny from E0 to a curve E0/G:
y2 = x3 − x + a6 with a6 =∑′
g x1(g)3. The fact that this is
F9-isomorphic with E0 then yields a6 ∈ iF3.
We’ve thus obtained:
Theorem. If n ≡ +1 mod 3 then
j(nζ3) ≡ 324ρ∑g∈G
′(x1(g))
3 ≡ 0 or ± 324√
3 mod 36.
21
So what happens for n ≡ +1 mod 3?
Same kind of analysis shows the residue of j(nζ3) mod 35, and
indeed mod 36, includes a term∑′
g x1(g)3 (not x1(g)
3−x1(g)).
This time Velu gives an isogeny from E0 to a curve E0/G:
y2 = x3 − x + a6 with a6 =∑′
g x1(g)3. The fact that this is
F9-isomorphic with E0 then yields a6 ∈ iF3.
We’ve thus obtained:
Theorem. If n ≡ +1 mod 3 then
j(nζ3) ≡ 324ρ∑g∈G
′(x1(g))
3 ≡ 0 or ± 324√
3 mod 36.
21
CM review
Statement of the problem
Velu’s isogeny formulas
Evaluation of j(nζ3) mod 34 (indeed mod ρ9 = 39/2)
Congruences for j(nζ3) mod 35 (for n ≡ −1 mod 3)
and mod 36 (for n ≡ 1 mod 3)
Further remarks
Some further questions:
How to tell whether j(nζ3) is 0 or ±324√
3 mod 36 in the case
n ≡ 1 mod 3?
Sorry, this doesn’t make sense: there’s more than one conju-
gate of j(nζ3) [or: more than one (conjugate of the) prime
above 3], and different ones may give different residues.
Indeed that’s what’s nice about the congruences mod ρ9 and
mod ρ10: they hold for all choices of conjugate.
OK, so how many of the conjugates of j(nζ3) are 0 mod 36?
That’s a better question. . .
22
Some further questions:
How to tell whether j(nζ3) is 0 or ±324√
3 mod 36 in the case
n ≡ 1 mod 3?
Sorry, this doesn’t make sense: there’s more than one conju-
gate of j(nζ3) [or: more than one (conjugate of the) prime
above 3], and different ones may give different residues.
Indeed that’s what’s nice about the congruences mod ρ9 and
mod ρ10: they hold for all choices of conjugate.
OK, so how many of the conjugates of j(nζ3) are 0 mod 36?
That’s a better question. . .
22
Some further questions:
How to tell whether j(nζ3) is 0 or ±324√
3 mod 36 in the case
n ≡ 1 mod 3?
Sorry, this doesn’t make sense: there’s more than one conju-
gate of j(nζ3) [or: more than one (conjugate of the) prime
above 3], and different ones may give different residues.
Indeed that’s what’s nice about the congruences mod ρ9 and
mod ρ10: they hold for all choices of conjugate.
OK, so how many of the conjugates of j(nζ3) are 0 mod 36?
That’s a better question. . .
22
Some further questions:
How to tell whether j(nζ3) is 0 or ±324√
3 mod 36 in the case
n ≡ 1 mod 3?
Sorry, this doesn’t make sense: there’s more than one conju-
gate of j(nζ3) [or: more than one (conjugate of the) prime
above 3], and different ones may give different residues.
Indeed that’s what’s nice about the congruences mod ρ9 and
mod ρ10: they hold for all choices of conjugate.
OK, so how many of the conjugates of j(nζ3) are 0 mod 36?
That’s a better question. . .
22
. . . and the answer is: 1/18 times the number of solutions of
n = ‖µ‖+ ‖3ν‖ = ‖µ‖+ 9‖ν‖ with µ, ν ∈ Z[ρ] ∼= A2〈1/2〉 and
ν 6= 0.
This comes from calculation in End(E0) = Z[ρ]⊕ iZ[ρ], where
ρ : (x1, y1) 7→ (x1 + 1, y1), i : (x1, y1) 7→ (−x1, iy1)
so the norm form is Z[ρ]⊕ Z[ρ].
This is consistent with the “Singular Moduli” picture: an en-
domorphism (µ, ν) ∈ End(E0) = Z[ρ] ⊕ iZ[ρ] survives lift to
j ≡ 0 mod 34 iff ρ|ν, and to j ≡ 0 mod 36 iff 3|ν.
23
. . . and the answer is: 1/18 times the number of solutions of
n = ‖µ‖+ ‖3ν‖ = ‖µ‖+ 9‖ν‖ with µ, ν ∈ Z[ρ] ∼= A2〈1/2〉 and
ν 6= 0.
This comes from calculation in End(E0) = Z[ρ]⊕ iZ[ρ], where
ρ : (x1, y1) 7→ (x1 + 1, y1), i : (x1, y1) 7→ (−x1, iy1)
so the norm form is Z[ρ]⊕ Z[ρ].
This is consistent with the “Singular Moduli” picture: an en-
domorphism (µ, ν) ∈ End(E0) = Z[ρ] ⊕ iZ[ρ] survives lift to
j ≡ 0 mod 34 iff ρ|ν, and to j ≡ 0 mod 36 iff 3|ν.
23
. . . and the answer is: 1/18 times the number of solutions of
n = ‖µ‖+ ‖3ν‖ = ‖µ‖+ 9‖ν‖ with µ, ν ∈ Z[ρ] ∼= A2〈1/2〉 and
ν 6= 0.
This comes from calculation in End(E0) = Z[ρ]⊕ iZ[ρ], where
ρ : (x1, y1) 7→ (x1 + 1, y1), i : (x1, y1) 7→ (−x1, iy1)
so the norm form is Z[ρ]⊕ Z[ρ].
This is consistent with the “Singular Moduli” picture: an en-
domorphism (µ, ν) ∈ End(E0) = Z[ρ] ⊕ iZ[ρ] survives lift to
j ≡ 0 mod 34 iff ρ|ν, and to j ≡ 0 mod 36 iff 3|ν.
23
Is there a similar 2-adic result for j(ni) with n odd?
Indeed there is:
Theorem. If n ≡ +1 mod 4 then j(ni) ≡ 123 mod 210.
If n ≡ −1 mod 4 then j(ni) ≡ 123 + 29i mod 210.
Exercise: Prove this using the generalization of Velu’s formulas
to extended Weierstraß form. A choice of twist that yields
a curve over Q(i) with j = 123 and good reduction at 2 is
Y 2 = X3 − (1 + 2i)X, for which the change of variables
(X, Y ) = (−2iX1 − 1, (1− i)3Y1 + 2X1 − (1 + i))
yields the model with coefficients [i−1,−i,1,0,0] that’s smooth
at 2. Can you obtain congruences modulo 210(1+i) and higher
powers of 1 + i?
24
Is there a similar 2-adic result for j(ni) with n odd?
Indeed there is:
Theorem. If n ≡ +1 mod 4 then j(ni) ≡ 123 mod 210.
If n ≡ −1 mod 4 then j(ni) ≡ 123 + 29i mod 210.
Exercise: Prove this using the generalization of Velu’s formulas
to extended Weierstraß form. A choice of twist that yields
a curve over Q(i) with j = 123 and good reduction at 2 is
Y 2 = X3 − (1 + 2i)X, for which the change of variables
(X, Y ) = (−2iX1 − 1, (1− i)3Y1 + 2X1 − (1 + i))
yields the model with coefficients [i−1,−i,1,0,0] that’s smooth
at 2. Can you obtain congruences modulo 210(1+i) and higher
powers of 1 + i?
24
Is there a similar 2-adic result for j(ni) with n odd?
Indeed there is:
Theorem. If n ≡ +1 mod 4 then j(ni) ≡ 123 mod 210.
If n ≡ −1 mod 4 then j(ni) ≡ 123 + 29i mod 210.
Exercise: Prove this using the generalization of Velu’s formulas
to extended Weierstraß form. A choice of twist that yields
a curve over Q(i) with j = 123 and good reduction at 2 is
Y 2 = X3 − (1 + 2i)X, for which the change of variables
(X, Y ) = (−2iX1 − 1, (1− i)3Y1 + 2X1 − (1 + i))
yields the model with coefficients [i−1,−i,1,0,0] that’s smooth
at 2. Can you obtain congruences modulo 210(1+i) and higher
powers of 1 + i?
24
Anything else you want to mention?
In fact there is. . . This model Y 2 = X3 − (1 + 2i)X is special
not just for having good 2-adic reduction [for which 1 + 2i is
only the simplest of many inequivalent choices] but for having
a 5-torsion point over Q(i) [for which 1 + 2i is unique up to
multiplication by (Q[i]∗)4]. The torsion group is generated by
(X, Y ) = (−1,−1 − i), i.e. (X1, Y1) = (0,0). This is a (2 + i)
torsion point where i acts by (X, Y ) 7→ (−X, iY ).
So shouldn’t something like this happen also for j = 0?
Yes, but it’s somewhat trickier. . .
25
Anything else you want to mention?
In fact there is. . . This model Y 2 = X3 − (1 + 2i)X is special
not just for having good 2-adic reduction [for which 1 + 2i is
only the simplest of many inequivalent choices] but for having
a 5-torsion point over Q(i) [for which 1 + 2i is unique up to
multiplication by (Q[i]∗)4]. The torsion group is generated by
(X, Y ) = (−1,−1 − i), i.e. (X1, Y1) = (0,0). This is a (2 + i)
torsion point where i acts by (X, Y ) 7→ (−X, iY ).
So shouldn’t something like this happen also for j = 0?
Yes, but it’s somewhat trickier. . .
25
For j = 0, our choice of twist Y 2 = X3 − 1 gave the curve
full 2-torsion over Q(ρ), but this is not maximal. Choosing
Y 2 = X3+1 keeps all the 2-torsion and adds a ρ-torsion point
at X = 0. [Here 32 = 23 + 1 is 2ρ-torsion (thus 6-torsion).]
More exotic: [2,−ρ,−ρ,0,0] has j = 0 and 7-torsion! Again
annihilated by a norm-7 element of Z[ρ]; here the Y 2 = X3+α
model is somewhat less pleasant: α = 48(ρ + 5).
This isn’t hard to explain by CM theory, and to generalize
beyond elliptic curves, as to p-torsion on Jacobians of curves
y2 = x(p−1)/2+α. If this is known already, it’s not well-known;
but that’s a topic for a different talk, and I have only ε minutes
left. . .
. . . and probably ε < 0.
THE END
26
For j = 0, our choice of twist Y 2 = X3 − 1 gave the curve
full 2-torsion over Q(ρ), but this is not maximal. Choosing
Y 2 = X3+1 keeps all the 2-torsion and adds a ρ-torsion point
at X = 0. [Here 32 = 23 + 1 is 2ρ-torsion (thus 6-torsion).]
More exotic: [2,−ρ,−ρ,0,0] has j = 0 and 7-torsion! Again
annihilated by a norm-7 element of Z[ρ]; here the Y 2 = X3+α
model is somewhat less pleasant: α = 48(ρ + 5).
This isn’t hard to explain by CM theory, and to generalize
beyond elliptic curves, as to p-torsion on Jacobians of curves
y2 = x(p−1)/2+α. If this is known already, it’s not well-known;
but that’s a topic for a different talk, and I have only ε minutes
left. . .
. . . and probably ε < 0.
THE END
26
For j = 0, our choice of twist Y 2 = X3 − 1 gave the curve
full 2-torsion over Q(ρ), but this is not maximal. Choosing
Y 2 = X3+1 keeps all the 2-torsion and adds a ρ-torsion point
at X = 0. [Here 32 = 23 + 1 is 2ρ-torsion (thus 6-torsion).]
More exotic: [2,−ρ,−ρ,0,0] has j = 0 and 7-torsion! Again
annihilated by a norm-7 element of Z[ρ]; here the Y 2 = X3+α
model is somewhat less pleasant: α = 48(ρ + 5).
This isn’t hard to explain by CM theory, and to generalize
beyond elliptic curves, as to p-torsion on Jacobians of curves
y2 = x(p−1)/2+α. If this is known already, it’s not well-known;
but that’s a topic for a different talk, and I have only ε minutes
left. . .
. . . and probably ε < 0.
THE END
26
For j = 0, our choice of twist Y 2 = X3 − 1 gave the curve
full 2-torsion over Q(ρ), but this is not maximal. Choosing
Y 2 = X3+1 keeps all the 2-torsion and adds a ρ-torsion point
at X = 0. [Here 32 = 23 + 1 is 2ρ-torsion (thus 6-torsion).]
More exotic: [2,−ρ,−ρ,0,0] has j = 0 and 7-torsion! Again
annihilated by a norm-7 element of Z[ρ]; here the Y 2 = X3+α
model is somewhat less pleasant: α = 48(ρ + 5).
This isn’t hard to explain by CM theory, and to generalize
beyond elliptic curves, as to p-torsion on Jacobians of curves
y2 = x(p−1)/2+α. If this is known already, it’s not well-known;
but that’s a topic for a different talk, and I have only ε minutes
left. . .
. . . and probably ε < 0.
THE END
26