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 ]
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 dim K ( L ) (as a vector space). √ √ √ √ Q ( 29) = Q + Q 29 [ Q ( 29) : Q ] = dim Q ( 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 dim K ( L ) (as a vector space). √ √ √ √ Q ( 29) = Q + Q 29 [ Q ( 29) : Q ] = dim Q ( Q ( 29)) = 2. number field: field extension K / Q such that [ K : Q ] < ∞ √ Examples: Q , Q ( i ) , Q ( 29) Non-examples: F 2 , 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 dim K ( L ) (as a vector space). √ √ √ √ Q ( 29) = Q + Q 29 [ Q ( 29) : Q ] = dim Q ( Q ( 29)) = 2. number field: field extension K / Q such that [ K : Q ] < ∞ √ Examples: Q , Q ( i ) , Q ( 29) Non-examples: F 2 , 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 dim K ( L ) (as a vector space). √ √ √ √ Q ( 29) = Q + Q 29 [ Q ( 29) : Q ] = dim Q ( Q ( 29)) = 2. number field: field extension K / Q such that [ K : Q ] < ∞ √ Examples: Q , Q ( i ) , Q ( 29) Non-examples: F 2 , 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 { H K ( m ) : m ∈ Z } such that: For any finite degree abelian extension L of K , there exists f ∈ Z > 0 such that L ⊆ H K ( f ). H K ( 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 { H K ( m ) : m ∈ Z } such that: For any finite degree abelian extension L of K , there exists f ∈ Z > 0 such that L ⊆ H K ( f ). H K ( m ) is called the ray class field of K for the modulus m . Kronecker-Weber Theorem (1896) Let m ∈ Z > 0 . Then H Q ( m ) = Q ( ζ m ) = Q (exp(2 π i τ ) : τ ∈ 1 m Z ). Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 7
Kronecker-Weber Theorem (1896) Let m ∈ Z > 0 . Then H Q ( m ) = Q ( ζ m ) = Q (exp(2 π i τ ) : τ ∈ 1 m Z ). R / Z τ �→ exp(2 π i τ ) S 1 ( C ) 0 1 2 3 4 5 6 6 6 6 6 6 6 6 analytic function: map τ �→ exp(2 π i τ ) 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 H Q ( m ) = Q ( ζ m ) = Q (exp(2 π i τ ) : τ ∈ 1 m Z ). analytic function: map τ �→ exp(2 π i τ ) 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 H K ( m ) = ???. analytic function: ??? special arguments: ??? Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 9
√ − D ) with D ∈ Z > 0 K = Q ( 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 K 0 (all embeddings of K 0 in C land in R ). √ − D ) with D ∈ Z > 0 K = Q ( 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 K 0 (all embeddings of K 0 in C land in R ). Example √ − D ) with D ∈ Z > 0 is a CM field of degree 2 . K = Q ( K 0 = 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 K 0 (all embeddings of K 0 in C land in R ). Example √ − D ) with D ∈ Z > 0 is a CM field of degree 2 . K = Q ( K 0 = Q An Oversimplification of CM Theory Let K be a CM field of degree 2 g . Let m ∈ Z > 0 . CM theory gives an abelian extension CM K ( m ) ⊆ H K ( m ) of K . Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 13
An Oversimplification of CM Theory Let K be a CM field of degree 2 g . Let m ∈ Z > 0 . CM theory gives an abelian extension CM K ( m ) ⊆ H K ( 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 2 g . Let m ∈ Z > 0 . CM theory gives an abelian extension CM K ( m ) ⊆ H K ( m ) of K . Imaginary Quadratic Number Field Let K be a CM field of degree 2. Let m ∈ Z > 0 . Then CM K (1) = K ( j ( τ )) CM K ( m ) = K ( j ( τ ) , h ( P ) : P ∈ E τ [ m ]) where τ depends on K and j ( τ ) is an invariant of C 2 / ( Z + τ Z ) ∼ = E τ . 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 CM K (1) = K ( j ( τ )) CM K ( m ) = K ( j ( τ ) , h ( P ) : P ∈ E τ [ m ]) where τ depends on K and j ( τ ) is an invariant of C 2 / ( Z + τ Z ) ∼ = E τ . analytic function: map P �→ h ( P ) from E τ to C special arguments: torsion points of the group E τ 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 CM K (1) = K ( j ( τ )) CM K ( m ) = K ( j ( τ ) , h ( P ) : P ∈ E τ [ m ]) where τ depends on K and j ( τ ) is an invariant of C 2 / ( Z + τ Z ) ∼ = E τ . analytic function: map P �→ h ( P ) from E τ to C special arguments: torsion points of the group E τ A Miracle Let K be a CM field of degree 2. Let m ∈ Z > 0 . Then CM K ( m ) = H K ( 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 CM K (1) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ )) CM K ( m ) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ ) , h ( P ) : P ∈ A τ [ m ]) where τ depends on K and i k ( τ ) are invariants of C 2 / ( Z 2 + τ Z 2 ) ∼ = A τ . 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 CM K (1) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ )) CM K ( m ) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ ) , h ( P ) : P ∈ A τ [ m ]) where τ depends on K and i k ( τ ) are invariants of C 2 / ( Z 2 + τ Z 2 ) ∼ = A τ . analytic function: map P �→ h ( P ) from A τ to C special arguments: torsion points of the group A τ 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 CM K (1) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ )) CM K ( m ) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ ) , h ( P ) : P ∈ A τ [ m ]) where τ depends on K and i k ( τ ) are invariants of C 2 / ( Z 2 + τ Z 2 ) ∼ = A τ . analytic function: map P �→ h ( P ) from A τ to C special arguments: torsion points of the group A τ No miracles here. Let K be a primitive CM field of degree 4. Let m ∈ Z > 0 . Then CM K ( m ) ⊆ H K ( 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 CM K (1) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ )) CM K ( m ) = K ( i 1 ( τ ) , i 2 ( τ ) , i 3 ( τ ) , h ( P ) : P ∈ A τ [ m ]) where τ depends on K and i k ( τ ) are invariants of C 2 / ( Z 2 + τ Z 2 ) ∼ = A τ . analytic function: map P �→ h ( P ) from A τ to C special arguments: torsion points of the group A τ Streng (2010) Let K be a primitive CM field of degree 4. Let m ∈ Z > 0 . Then H K 0 ( m ) CM K ( m ) ⊆ H K ( m ) is of exponent at most 2. Computing abelian extensions of quartic fields via CM Jared Asuncion JNCF 2020 22
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