Computing abelian extensions of quartic fields using complex - - PowerPoint PPT Presentation

Computing abelian extensions of quartic fields using complex multiplication

Jared Asuncion Supervision: Andreas Enge and Marco Streng JNCF 2020

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 1

degree [L : K] of a field extension L of K is dimK(L) (as a vector space). Q( √ 29) = Q + Q √ 29 [Q( √ 29) : Q] = dimQ(Q( √ 29)) = 2.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 2

degree [L : K] of a field extension L of K is dimK(L) (as a vector space). Q( √ 29) = Q + Q √ 29 [Q( √ 29) : Q] = dimQ(Q( √ 29)) = 2. number field: field extension K/Q such that [K : Q] < ∞ Examples: Q, Q(i), Q( √ 29) Non-examples: F2, Q[X], R, C = R(i)

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 3

degree [L : K] of a field extension L of K is dimK(L) (as a vector space). Q( √ 29) = Q + Q √ 29 [Q( √ 29) : Q] = dimQ(Q( √ 29)) = 2. number field: field extension K/Q such that [K : Q] < ∞ Examples: Q, Q(i), Q( √ 29) Non-examples: F2, Q[X], R, C = R(i) Aut(L/K) = {σ : L → L : σ(x) = x for each x ∈ K} Aut

• Q(

√ 29)/Q

• is generated by a + b

√ 29 → a − b √ 29.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 4

degree [L : K] of a field extension L of K is dimK(L) (as a vector space). Q( √ 29) = Q + Q √ 29 [Q( √ 29) : Q] = dimQ(Q( √ 29)) = 2. number field: field extension K/Q such that [K : Q] < ∞ Examples: Q, Q(i), Q( √ 29) Non-examples: F2, Q[X], R, C = R(i) Aut(L/K) = {σ : L → L : σ(x) = x for each x ∈ K} Aut

• Q(

√ 29)/Q

• is generated by a + b

√ 29 → a − b √ 29. abelian extension L of K: extension of number fields L/K such that ∗ | Aut(L/K)| = [L : K] ∗ means Galois extension Aut(L/K) is abelian

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 5

Let K be a number field. Class Field Theory There exists a set {HK(m) : m ∈ Z} such that: For any finite degree abelian extension L of K, there exists f ∈ Z>0 such that L ⊆ HK(f ). HK(m) is called the ray class field of K for the modulus m.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 6

Let K be a number field. Class Field Theory There exists a set {HK(m) : m ∈ Z} such that: For any finite degree abelian extension L of K, there exists f ∈ Z>0 such that L ⊆ HK(f ). HK(m) is called the ray class field of K for the modulus m. Kronecker-Weber Theorem (1896) Let m ∈ Z>0. Then HQ(m) = Q(ζm) = Q(exp(2πiτ) : τ ∈ 1

mZ).

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 7

Kronecker-Weber Theorem (1896) Let m ∈ Z>0. Then HQ(m) = Q(ζm) = Q(exp(2πiτ) : τ ∈ 1

mZ).

τ → exp(2πiτ) R/Z

6 6

S1(C)

6 1 6 2 6 3 6 4 6 5 6

analytic function: map τ → exp(2πiτ) from R/Z to the unit circle special arguments: torsion points of the group R/Z

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 8

Kronecker-Weber Theorem (1896) Let m ∈ Z>0. Then HQ(m) = Q(ζm) = Q(exp(2πiτ) : τ ∈ 1

mZ).

analytic function: map τ → exp(2πiτ) from R/Z to the unit circle special arguments: torsion points of the group R/Z Hilbert’s 12th Problem (1900) Let K be a number field. Let m ∈ Z>0. Then HK(m) = ???. analytic function: ??? special arguments: ???

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 9

K = Q( √ −D) with D ∈ Z>0

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 10

Definition (CM field) CM field K: number field which is a totally imaginary quadratic extension (no embeddings of K into R) of a totally real number field K0 (all embeddings of K0 in C land in R). K = Q( √ −D) with D ∈ Z>0

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 11

Definition (CM field) CM field K: number field which is a totally imaginary quadratic extension (no embeddings of K into R) of a totally real number field K0 (all embeddings of K0 in C land in R). Example K = Q( √ −D) with D ∈ Z>0 is a CM field of degree 2. K0 = Q

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 12

Definition (CM field) CM field K: number field which is a totally imaginary quadratic extension (no embeddings of K into R) of a totally real number field K0 (all embeddings of K0 in C land in R). Example K = Q( √ −D) with D ∈ Z>0 is a CM field of degree 2. K0 = Q An Oversimplification of CM Theory Let K be a CM field of degree 2g. Let m ∈ Z>0. CM theory gives an abelian extension CMK(m) ⊆ HK(m) of K.

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An Oversimplification of CM Theory Let K be a CM field of degree 2g. Let m ∈ Z>0. CM theory gives an abelian extension CMK(m) ⊆ HK(m) of K. Imaginary Quadratic Number Field Let K be a CM field of degree 2. Let m ∈ Z>0. Then

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 14

An Oversimplification of CM Theory Let K be a CM field of degree 2g. Let m ∈ Z>0. CM theory gives an abelian extension CMK(m) ⊆ HK(m) of K. Imaginary Quadratic Number Field Let K be a CM field of degree 2. Let m ∈ Z>0. Then CMK(1) = K(j(τ)) CMK(m) = K(j(τ), h(P) : P ∈ Eτ[m]) where τ depends on K and j(τ) is an invariant of C2/(Z + τZ) ∼ = Eτ.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 15

Imaginary Quadratic Number Field Let K be a CM field of degree 2. Let m ∈ Z>0. Then CMK(1) = K(j(τ)) CMK(m) = K(j(τ), h(P) : P ∈ Eτ[m]) where τ depends on K and j(τ) is an invariant of C2/(Z + τZ) ∼ = Eτ. analytic function: map P → h(P) from Eτ to C special arguments: torsion points of the group Eτ

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 16

Imaginary Quadratic Number Field Let K be a CM field of degree 2. Let m ∈ Z>0. Then CMK(1) = K(j(τ)) CMK(m) = K(j(τ), h(P) : P ∈ Eτ[m]) where τ depends on K and j(τ) is an invariant of C2/(Z + τZ) ∼ = Eτ. analytic function: map P → h(P) from Eτ to C special arguments: torsion points of the group Eτ A Miracle Let K be a CM field of degree 2. Let m ∈ Z>0. Then CMK(m) = HK(m).

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 17

Primitive Quartic CM Field Let K be a primitive CM field of degree 4. primitive: not bicyclic Galois Let m ∈ Z>0. Then

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 18

Primitive Quartic CM Field Let K be a primitive CM field of degree 4. primitive: not bicyclic Galois Let m ∈ Z>0. Then CMK(1) = K(i1(τ), i2(τ), i3(τ)) CMK(m) = K(i1(τ), i2(τ), i3(τ), h(P) : P ∈ Aτ[m]) where τ depends on K and ik(τ) are invariants of C2/(Z2 + τZ2) ∼ = Aτ.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 19

Primitive Quartic CM Field Let K be a primitive CM field of degree 4. primitive: not bicyclic Galois Let m ∈ Z>0. Then CMK(1) = K(i1(τ), i2(τ), i3(τ)) CMK(m) = K(i1(τ), i2(τ), i3(τ), h(P) : P ∈ Aτ[m]) where τ depends on K and ik(τ) are invariants of C2/(Z2 + τZ2) ∼ = Aτ. analytic function: map P → h(P) from Aτ to C special arguments: torsion points of the group Aτ

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 20

Primitive Quartic CM Field Let K be a primitive CM field of degree 4. primitive: not bicyclic Galois Let m ∈ Z>0. Then CMK(1) = K(i1(τ), i2(τ), i3(τ)) CMK(m) = K(i1(τ), i2(τ), i3(τ), h(P) : P ∈ Aτ[m]) where τ depends on K and ik(τ) are invariants of C2/(Z2 + τZ2) ∼ = Aτ. analytic function: map P → h(P) from Aτ to C special arguments: torsion points of the group Aτ No miracles here. Let K be a primitive CM field of degree 4. Let m ∈ Z>0. Then CMK(m) ⊆ HK(m).

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 21

Primitive Quartic CM Field Let K be a primitive CM field of degree 4. primitive: not bicyclic Galois Let m ∈ Z>0. Then CMK(1) = K(i1(τ), i2(τ), i3(τ)) CMK(m) = K(i1(τ), i2(τ), i3(τ), h(P) : P ∈ Aτ[m]) where τ depends on K and ik(τ) are invariants of C2/(Z2 + τZ2) ∼ = Aτ. analytic function: map P → h(P) from Aτ to C special arguments: torsion points of the group Aτ Streng (2010) Let K be a primitive CM field of degree 4. Let m ∈ Z>0. Then HK0(m) CMK(m) ⊆ HK(m) is of exponent at most 2.

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Streng (2010) Let K be a primitive CM field of degree 4. Then HK0(1) CMK(1) ⊆ HK(1) is of exponent at most 2.

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Streng (2010) Let K be a primitive CM field of degree 4. Then HK0(1) CMK(1) ⊆ HK(1) is of exponent at most 2. Optimist: Sometimes HK0(1) CMK(1) = HK(1)! experimentally, 80%

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 24

Streng (2010) Let K be a primitive CM field of degree 4. Then HK0(1) CMK(1) ⊆ HK(1) is of exponent at most 2. Optimist: Sometimes HK0(1) CMK(1) = HK(1)! experimentally, 80% Pessimist: Sometimes it’s not. experimentally, 20%

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Streng (2010) Let K be a primitive CM field of degree 4. Then HK0(1) CMK(1) ⊆ HK(1) is of exponent at most 2. Optimist: Sometimes HK0(1) CMK(1) = HK(1)! experimentally, 80% Pessimist: Sometimes it’s not. experimentally, 20% Shimura (1968) Let K be a primitive CM field of degree 4. Then there is m ∈ Z>0 such that HK(1) ⊆ HK0(m) CMK(m). (⋆)

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Shimura (1968) Let K be a primitive CM field of degree 4. Then there is m ∈ Z>0 such that HK(1) ⊆ HK0(m) CMK(m). (⋆)

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Shimura (1968) Let K be a primitive CM field of degree 4. Then there is m ∈ Z>0 such that HK(1) ⊆ HK0(m) CMK(m). (⋆) Let OK be the ring of integers of K. ClK(1) = { ideals of OK} { ideals of OK generated by a single element }

• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of primes of Z divisible by the elements of S. Take m = 4

p∈P p. Then, (⋆) is true.

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• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of rational primes below the prime ideals in S. Take m = 4

p∈P p. Then, HK(1) ⊆ HK0(m) CMK(m) is true.

Let K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 29

• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of rational primes below the prime ideals in S. Take m = 4

p∈P p. Then, HK(1) ⊆ HK0(m) CMK(m) is true.

Let K = Q(y) where y4 + 24y2 + 28 = 0. Let ClK(1) = [p5]8 where p5 = 5OK + (y − 2)OK.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 30

• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of rational primes below the prime ideals in S. Take m = 4

p∈P p. Then, HK(1) ⊆ HK0(m) CMK(m) is true.

Let K = Q(y) where y4 + 24y2 + 28 = 0. Let ClK(1) = [p5]8 where p5 = 5OK + (y − 2)OK. Let S = { p2 , p5 } where p2 is the only prime ideal dividing 2OK.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 31

• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of rational primes below the prime ideals in S. Take m = 4

p∈P p. Then, HK(1) ⊆ HK0(m) CMK(m) is true.

Let K = Q(y) where y4 + 24y2 + 28 = 0. Let ClK(1) = [p5]8 where p5 = 5OK + (y − 2)OK. Let S = { p2 , p5 } where p2 is the only prime ideal dividing 2OK. Then we take P = {2, 5}.

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• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of rational primes below the prime ideals in S. Take m = 4

p∈P p. Then, HK(1) ⊆ HK0(m) CMK(m) is true.

Let K = Q(y) where y4 + 24y2 + 28 = 0. Let ClK(1) = [p5]8 where p5 = 5OK + (y − 2)OK. Let S = { p2 , p5 } where p2 is the only prime ideal dividing 2OK. Then we take P = {2, 5}. Take m = 40.

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• A. (2019)

Let K be a primitive CM field of degree 4. Let S be a finite set of prime ideals of OK such that S contains all prime ideals which divide 2OK and | ClK(1)/S| is odd. Let P be the set of rational primes below the prime ideals in S. Take m = 4

p∈P p. Then, HK(1) ⊆ HK0(m) CMK(m) is true.

Let K = Q(y) where y4 + 24y2 + 28 = 0. Let ClK(1) = [p5]8 where p5 = 5OK + (y − 2)OK. Let S = { p2 , p5 } where p2 is the only prime ideal dividing 2OK. Then we take P = {2, 5}. Take m = 40. HK(1) ⊆ HK0(40) CMK(40). [HK0(40) CMK(40) : HK(1)] = 49 152.

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Given m ∈ Z>0. HK(1) ⊆ HK0(m) CMK(m) ?

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 35

Given m ∈ Z>0. HK(1) ⊆ HK0(m) CMK(m) ? By Galois theory, this is equivalent to asking... Given m ∈ Z>0. A ⊇ B ∩ C ? A := Aut(HK(m)/HK(1)) B := Aut(HK(m)/HK0(m)) C := Aut(HK(m)/ CMK(m))

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 36

Given m ∈ Z>0. HK(1) ⊆ HK0(m) CMK(m) ? By Galois theory, this is equivalent to asking... Given m ∈ Z>0. A ⊇ B ∩ C ? A := Aut(HK(m)/HK(1)) ∼ = ker( ClK(m) → ClK(1) ) B := Aut(HK(m)/HK0(m)) ∼ = ker( NK/K0 : ClK(m) → ClK0(m) ) C := Aut(HK(m)/ CMK(m)) ∼ = ker( N : ClK(m) → CK(m) )

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A := Aut(HK(m)/HK(1)) ∼ = ker( ClK(m) → ClK(1) ) B := Aut(HK(m)/HK0(m)) ∼ = ker( NK/K0 : ClK(m) → ClK0(m) ) C := Aut(HK(m)/ CMK(m)) ∼ = ker( N : ClK(m) → CK(m) ) Computing Aut(HK(m)/ CMK(m)). ClK(m), ClK(1), ClK(m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pari/GP NK/K0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pari/GP CK(m), m = 1 . . . . . . . . .Recip (Streng) / cmh (Enge / Thom´ e) m ≥ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (A. 2020)

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A := Aut(HK(m)/HK(1)) ∼ = ker( ClK(m) → ClK(1) ) B := Aut(HK(m)/HK0(m)) ∼ = ker( NK/K0 : ClK(m) → ClK0(m) ) C := Aut(HK(m)/ CMK(m)) ∼ = ker( N : ClK(m) → CK(m) ) Computing Aut(HK(m)/ CMK(m)). ClK(m), ClK(1), ClK(m) . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pari/GP NK/K0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pari/GP CK(m), m = 1 . . . . . . . . .Recip (Streng) / cmh (Enge / Thom´ e) m ≥ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (A. 2020) 0 → O+

K0/NK/K0(OK,1) → CK(m) → ker(ClK(m) → Cl+ K0(1)) → 0

• perations on group morphisms . . . . . . . . . . . . . . . . . . . . . . abgrp.gp

kernels, images, quotients, group extensions

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 39

Let K = Q(y) where y4 + 24y2 + 28 = 0. m = 1 A = Aut(HK(1)/HK(1)) ∼ = [(1)]1 B = Aut(HK(1)/HK0(1)) ∼ = [p5]8 C = Aut(HK(1)/ CMK(1)) ∼ = [p4

5]2

A ⊇ B ∩ C ?

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 40

Let K = Q(y) where y4 + 24y2 + 28 = 0. m = 1 A = Aut(HK(1)/HK(1)) ∼ = [(1)]1 B = Aut(HK(1)/HK0(1)) ∼ = [p5]8 C = Aut(HK(1)/ CMK(1)) ∼ = [p4

5]2

A ⊇ B ∩ C ? NO. ⇒ HK(1) ⊆ HK0(1) CMK(1)

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 41

Let K = Q(y) where y4 + 24y2 + 28 = 0. m = 1 A = Aut(HK(1)/HK(1)) ∼ = [(1)]1 B = Aut(HK(1)/HK0(1)) ∼ = [p5]8 C = Aut(HK(1)/ CMK(1)) ∼ = [p4

5]2

A ⊇ B ∩ C ? NO. ⇒ HK(1) ⊆ HK0(1) CMK(1) m = 2 A = Aut(HK(2)/HK(1)) ∼ = [(α1)]2 × [(α2)]2 B = Aut(HK(2)/HK0(2)) ∼ = [p5]8 × [(α1)]2 × [(α2)]2 C = Aut(HK(2)/ CMK(2)) ∼ = [(α1)]2 A ⊇ B ∩ C ? YES. ⇒ HK(1) ⊆ HK0(2) CMK(2)

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 42

Let K = Q(y) where y4 + 24y2 + 28 = 0. m = 1 A = Aut(HK(1)/HK(1)) ∼ = [(1)]1 B = Aut(HK(1)/HK0(1)) ∼ = [p5]8 C = Aut(HK(1)/ CMK(1)) ∼ = [p4

5]2

A ⊇ B ∩ C ? NO. ⇒ HK(1) ⊆ HK0(1) CMK(1) m = 2 A = Aut(HK(2)/HK(1)) ∼ = [(α1)]2 × [(α2)]2 B = Aut(HK(2)/HK0(2)) ∼ = [p5]8 × [(α1)]2 × [(α2)]2 C = Aut(HK(2)/ CMK(2)) ∼ = [(α1)]2 A ⊇ B ∩ C ? YES. ⇒ HK(1) ⊆ HK0(2) CMK(2) A ⊇ C! ⇒ HK(1) ⊆ CMK(2).

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 43

Let K be a primitive quartic CM field. HK(1) ⊆ HK0(m) CMK(m) (⋆) So Far... Find m such that ⋆ is true. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine if ⋆ is true for any m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computing CMK(m). m = 1 m = 2 m = 1, 2

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 44

Let K be a primitive quartic CM field. HK(1) ⊆ HK0(m) CMK(m) (⋆) So Far... Find m such that ⋆ is true. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine if ⋆ is true for any m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computing CMK(m). m = 1 . . . . . . . . . . . . . . . . . Recip (Streng) / cmh (Enge, Thom´ e)) m = 2 m = 1, 2

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 45

Let K be a primitive quartic CM field. HK(1) ⊆ HK0(m) CMK(m) (⋆) So Far... Find m such that ⋆ is true. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine if ⋆ is true for any m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computing CMK(m). m = 1 . . . . . . . . . . . . . . . . . Recip (Streng) / cmh (Enge, Thom´ e)) m = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . example after this slide m = 1, 2

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 46

Let K be a primitive quartic CM field. HK(1) ⊆ HK0(m) CMK(m) (⋆) So Far... Find m such that ⋆ is true. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine if ⋆ is true for any m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computing CMK(m). m = 1 . . . . . . . . . . . . . . . . . Recip (Streng) / cmh (Enge, Thom´ e)) m = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . example after this slide Assume ⋆ for m = 2 and [HK(1) : K] is large: can find HK(1) faster than Kummer theory m = 1, 2

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 47

Let K be a primitive quartic CM field. HK(1) ⊆ HK0(m) CMK(m) (⋆) So Far... Find m such that ⋆ is true. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine if ⋆ is true for any m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computing CMK(m). m = 1 . . . . . . . . . . . . . . . . . Recip (Streng) / cmh (Enge, Thom´ e)) m = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . example after this slide Assume ⋆ for m = 2 and [HK(1) : K] is large: can find HK(1) faster than Kummer theory m = 1, 2 . . . . . . . . . . . . . . . . map P → h(P) from Aτ to C (slide 5)

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 48

Let K be a primitive quartic CM field. HK(1) ⊆ HK0(m) CMK(m) (⋆) So Far... Find m such that ⋆ is true. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Determine if ⋆ is true for any m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computing CMK(m). m = 1 . . . . . . . . . .Recip (Streng) / cmh (Enge, Thom´ e)) ∼ 80% m = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . example after this slide ∼ 10% Assume ⋆ for m = 2 and [HK(1) : K] is large: can find HK(1) faster than Kummer theory m = 1, 2 . . . . . . . . . map P → h(P) from Aτ to C (slide 5) ∼ 10%

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 49

HK(1) ⊆ CMK(2) = K(λk(τ) : k ∈ {1, 2, 3}) where λk(τ) are in terms of theta constants (analytic functions). where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 50

HK(1) ⊆ CMK(2) = K(λk(τ) : k ∈ {1, 2, 3}) where λk(τ) are in terms of theta constants (analytic functions). Some costly theta computations give: approximations ˜ λk(τ)σ for each σ ∈ Aut(CMK(2)/K). where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 51

HK(1) ⊆ CMK(2) = K(λk(τ) : k ∈ {1, 2, 3}) where λk(τ) are in terms of theta constants (analytic functions). Some costly theta computations give: approximations ˜ λk(τ)σ for each σ ∈ Aut(CMK(2)/K). G = Aut(CMK(2)/K) = σ18 × σ22 H = Aut(CMK(2)/HK(1)) = σ22 G/H = Aut(HK(1)/K) = σ18 where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 52

Some costly theta computations give: approximations ˜ λk(τ)σ for each σ ∈ Aut(CMK(2)/K). G = Aut(CMK(2)/K) = σ18 × σ22 H = Aut(CMK(2)/HK(1)) = σ22 G/H = Aut(HK(1)/K) = σ18 HK(1) = CMK(2)H = K(λk(τ) + λk(τ)σ2 : k ∈ {1, 2, 3}). where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 53

G = Aut(CMK(2)/K) = σ18 × σ22 H = Aut(CMK(2)/HK(1)) = σ22 G/H = Aut(HK(1)/K) = σ18 HK(1) = CMK(2)H = K(λk(τ) + λk(τ)σ2 : k ∈ {1, 2, 3}). ˜ p1(X) =

• σ∈G/H
• X −
• ˜

λ1(τ) + ˜ λ1(τ)σ2 σ . ≈ X 8 − 36.591404X 7 + 530.70821X 6 − 3769.6792X 5 + · · · where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 54

HK(1) = CMK(2)H = K(λk(τ) + λk(τ)σ2 : k ∈ {1, 2, 3}). ˜ p1(X) =

• σ∈G/H
• X −
• ˜

λ1(τ) + ˜ λ1(τ)σ2 σ . ≈ X 8 − 36.591404X 7 + 530.70821X 6 − 3769.6792X 5 + · · · Use an algorithm involving LLL to find

dp1(X) = dX 8 + (94151707542711000000y 2 + 1869571596200694750000)X 7+ (38252084655413295380000y 2 + 874990832290501062080000)X 6 + . . .

where d = 58 74 134 234. where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 55

˜ p1(X) =

• σ∈G/H
• X −
• ˜

λ1(τ) + ˜ λ1(τ)σ2 σ . ≈ X 8 − 36.591404X 7 + 530.70821X 6 − 3769.6792X 5 + · · · Use an algorithm involving LLL to find

dp1(X) = dX 8 + (94151707542711000000y 2 + 1869571596200694750000)X 7+ (38252084655413295380000y 2 + 874990832290501062080000)X 6 + . . .

where d = 58 74 134 234. p1(X) is a degree 8 irreducible polynomial in K[X]. So HK(1) = K(α) where p1(α) = 0. where K = Q(y) where y4 + 24y2 + 28 = 0.

Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 56