CIRM - Luminy, 2 - 6 March 2020 Francophone Computer Algebra Days On - PowerPoint PPT Presentation

CIRM - Luminy, 2 - 6 March 2020 Francophone Computer Algebra Days On the generators of 2 -class group of some number fields illustrated by PARI/GP Mohammed TAOUS Moulay Ismaïl university of Meknes Faculty of Sciences and Technology Errachidia. Morocco 05/03/2020 1/45

Plan 1. Number Field and its class group � 2. 2-rank of 2-class group of the field Q ( 4 4 pq 2 ) � 3. 4-rank of 2-class group of classes of Q ( 4 4 pq 2 ) 4. Applications 2/45

Plan 1. Number Field and its class group � 2. 2-rank of 2-class group of the field Q ( 4 4 pq 2 ) � 3. 4-rank of 2-class group of classes of Q ( 4 4 pq 2 ) 4. Applications 3/45

Number Field in PariGp 1 A number field K of degree n ∈ N ∗ can be represented in equivalent ways. In GP they based on the following representation : K = Q [ X ] / ( T ) with T is a monic irreducible polynomial of Z [ X ] of degree n ∈ N ∗ . The function nfinit contains the basic arithmetic data attached to the number field : ( r 1 , r 2 ) the signature, the discriminant D K , O K the ring of integers of K . For example if T = X 4 − 61268, then 1. Initialement développé par H. Cohen et ses collaborateurs (U. Bordeaux I), PARI est maintenant maintenu par K. Belabas, avec l’aide de nombreux col- laborateurs bénévoles. 4/45

Number Field in PariGp 1 A number field K of degree n ∈ N ∗ can be represented in equivalent ways. In GP they based on the following representation : K = Q [ X ] / ( T ) with T is a monic irreducible polynomial of Z [ X ] of degree n ∈ N ∗ . The function nfinit contains the basic arithmetic data attached to the number field : ( r 1 , r 2 ) the signature, the discriminant D K , O K the ring of integers of K . For example if T = X 4 − 61268, then ? K = nfinit(X^4-61268); ? K.sign %2 = [2, 1] ? K.disc %3 = -2753628992 ? K.zk %4 = [1, 1/68*X^2 - 1/2, X, 1/68*X^3 - 1/2*X] 1. Initialement développé par H. Cohen et ses collaborateurs (U. Bordeaux I), PARI est maintenant maintenu par K. Belabas, avec l’aide de nombreux col- laborateurs bénévoles. 4/45

Number Field in PariGp ? K = nfinit(X^4-61268); ? K.sign %2 = [2, 1] ? K.disc %3 = -2753628992 ? K.zk %4 = [1, 1/68*X^2 - 1/2, X, 1/68*X^3 - 1/2*X] 5/45

Number Field in PariGp ? K = nfinit(X^4-61268); ? K.sign %2 = [2, 1] ? K.disc %3 = -2753628992 ? K.zk %4 = [1, 1/68*X^2 - 1/2, X, 1/68*X^3 - 1/2*X] As 68 = 4 · 17 and 61268 = 4 · 17 2 · 53, then 5/45

Number Field in PariGp ? K = nfinit(X^4-61268); ? K.sign %2 = [2, 1] ? K.disc %3 = -2753628992 ? K.zk %4 = [1, 1/68*X^2 - 1/2, X, 1/68*X^3 - 1/2*X] As 68 = 4 · 17 and 61268 = 4 · 17 2 · 53, then O K = Z + Z α − 1 + Z β + Z γ − β . 2 2 √ √ 4 · 17 2 · 53 and γ = αβ . It is the same 4 such that α = 53, β = result given by T. Funakura in [5] 5/45

Number Field in PariGp Let O K be the ring of integers of K . We say that I fractional ideal of O K (or just of K ), if I is an O K -submodule of K such that there exists a non-zero d ∈ O K with d I ⊂ O K . The class group de K is the quotient group Cl ( K ) = I ( K ) / P ( K ) . Where I ( K ) is the group of fractional ideals of K and P ( K ) is its subgroup of principal ideals. It is a finite group, its cardinal is called the class number of K and noted h ( K ) . 2. The invertible elements of O K 6/45

Number Field in PariGp Let O K be the ring of integers of K . We say that I fractional ideal of O K (or just of K ), if I is an O K -submodule of K such that there exists a non-zero d ∈ O K with d I ⊂ O K . The class group de K is the quotient group Cl ( K ) = I ( K ) / P ( K ) . Where I ( K ) is the group of fractional ideals of K and P ( K ) is its subgroup of principal ideals. It is a finite group, its cardinal is called the class number of K and noted h ( K ) . The function bnfinit contains nfinit and the deeper invariants of K : U ( K ) 2 units and class group Cl ( K ) . 2. The invertible elements of O K 6/45

Number Field in Gp ? K = nfinit(X^4-61268); ? K.clgp.cyc *** at top-level: K.clgp.cyc *** ^-------- *** _.clgp: incorrect type in clgp (t_VEC). *** Break loop: type ’break’ to go back to GP prompt break> break ? K = bnfinit(X^4-61268); ? K.clgp.cyc %3 = [2, 2] ? K.clgp.no %4 = 4 7/45

A question � In the previous example, we have K = Q ( 4 4 pq 2 ) such that p = 53 ≡ 5 ( mod 5 ) , q = 17 ≡ 1 ( mod 8 ) and 2-class group of the field K , is of type ( 2 , 2 ) . 8/45

A question � In the previous example, we have K = Q ( 4 4 pq 2 ) such that p = 53 ≡ 5 ( mod 5 ) , q = 17 ≡ 1 ( mod 8 ) and 2-class group of the field K , is of type ( 2 , 2 ) . 8/45

A question � In the previous example, we have K = Q ( 4 4 pq 2 ) such that p = 53 ≡ 5 ( mod 5 ) , q = 17 ≡ 1 ( mod 8 ) and 2-class group of the field K , is of type ( 2 , 2 ) . If We keep the same conditions on p and q , the 2-group of classes of K remains of type ( 2 , 2 ) ? 8/45

A question � In the previous example, we have K = Q ( 4 4 pq 2 ) such that p = 53 ≡ 5 ( mod 5 ) , q = 17 ≡ 1 ( mod 8 ) and 2-class group of the field K , is of type ( 2 , 2 ) . If We keep the same conditions on p and q , the 2-group of classes of K remains of type ( 2 , 2 ) ? 8/45

A question � In the previous example, we have K = Q ( 4 4 pq 2 ) such that p = 53 ≡ 5 ( mod 5 ) , q = 17 ≡ 1 ( mod 8 ) and 2-class group of the field K , is of type ( 2 , 2 ) . If We keep the same conditions on p and q , the 2-group of classes of K remains of type ( 2 , 2 ) ? � � � � q p p q r 2 ( K ) r 4 ( K ) Cl ( K ) p q 4 4 13 17 -1 1 2 1 [4, 2] 53 17 -1 -1 2 0 [2, 2] 101 17 -1 -1 2 1 [8, 2] 149 17 1 1 2 2 [28, 4] 157 17 1 1 2 2 [8, 4] 5 41 1 -1 2 0 [2, 2] 37 41 -1 1 2 1 [4, 2] 61 41 -1 -1 2 0 [6, 2] 173 41 -1 -1 2 0 [10, 2] 8/45

The main objective � Now we assume that Q ( 4 4 pq 2 ) be a real pure quartic number field, where p and q be two different odd prime numbers such that � p � p ≡ 5 ( mod 8 ) , q ≡ 1 ( mod 4 ) and = 1 . q Our goal is to show that the 2-class group of this field, is of type � p � ( 2 , 2 ) if only if = − 1 and give its generators. q 4 9/45

Plan 1. Number Field and its class group � 2. 2-rank of 2-class group of the field Q ( 4 4 pq 2 ) � 3. 4-rank of 2-class group of classes of Q ( 4 4 pq 2 ) 4. Applications 10/45

Definition of 2-rank and 4-rank Let K be a number field and let Cl 2 ( K ) denote its 2-class group that is the 2-Sylow subgroup of its class group Cl ( K ) . We define the 2-rank and the 4-rank of Cl ( K ) respectively as follows : r 2 ( K ) = dim F 2 ( Cl 2 ( K ) / Cl 2 ( K ) 2 ) and r 4 ( K ) = dim F 2 ( Cl 2 ( K ) 2 / Cl 2 ( K ) 4 ) where F 2 is the finite field with 2 elements. 11/45

The Ambiguous Class Number Formula Let K be a number field, L / K a cyclic extension of prime degree ℓ , and σ a generator of its Galois group G = Gal ( L / K ) . We say that an ideal class [ a ] ∈ Cl ( L ) is ambiguous if [ a ] σ = [ a ] , that is, if there exists an element α ∈ L × such that a σ − 1 = ( α ) . The group Am ( L / K ) of ambiguous ideal classes is the subgroup of the class group Cl ( L ) consisting of ambiguous ideal classes. 12/45

The Ambiguous Class Number Formula [4] Let be L / K a cyclic extension of prime degree ℓ , then ℓ t − 1 # Am ( L / K ) = h ( K ) · [ E K : E K ∩ NL × ] , where E K is the unit group of K , and E K ∩ NL × the subgroup of units that are norms of elements of L . If h ( K ) is odd and ℓ = 2, then r 2 ( L ) = t − e − 1 , with t is the number of ramified primes (including those at infinity) in L / K , 2 e = [ E K : E K ∩ NL × ] . 13/45

Application of the ambiguous class number formula We take K = Q ( √ p ) and α = 2 q √ p such that p and q be two different odd prime numbers such that � p � p ≡ 5 ( mod 8 ) , q ≡ 1 ( mod 4 ) and = 1 . q Then L = K ( √ α ) = Q ( 4 � 4 pq 2 ) and the relative discriminant of L / K is ( 2 3 q √ p ) O K = 2 3 √ p O K . I π 1 π 2 So the finite primes ramified in K / k are √ p O K , π 1 , π 2 , 2 I . As the signature of K is ( 2 , 0 ) and ( 2 , 1 ) for L , then there is unique infinite prime of K ramified in L , thus t = 5 and r 2 ( L ) = 4 − e . From the numerical examples obtained using PariGp e = 2. 14/45

Hilbert Symbol Let P be a prime ideal of K and let K P be the completion of K at P . For any α, β ∈ K P , we define the local Hilbert Symbol by if β is a norm in K P ( √ α ); � 1 , ( α, β ) P = − 1 , if not. 15/45

Hilbert Symbol Let P be a prime ideal of K and let K P be the completion of K at P . For any α, β ∈ K P , we define the local Hilbert Symbol by if β is a norm in K P ( √ α ); � 1 , ( α, β ) P = − 1 , if not. Let i P the natural injection between K and K P , then for any α, β ∈ K , we define the global Hilbert Symbol by � α, β � = ( i P ( α ) , i P ( β )) P . P 15/45

Hilbert Symbol Theorem ([6]) The Hilbert symbol satisfies the following conditions : � α, β � � β, α � i. = . P P � α, β � ii. If P ∞ is a real infinite prime, then = − 1 if and only if P ∞ i P ( α ) < 0 and i P ( β ) < 0 . iii. β is a norm in K ( √ α ) if and only if � α, β � = 1 for all P . P � α, β � � iv. The Hilbert symbol verifies = 1 . P P 16/45