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