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

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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

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Plan

• 1. Number Field and its class group
• 2. 2-rank of 2-class group of the field Q( 4
• 4pq2)
• 3. 4-rank of 2-class group of classes of Q( 4
• 4pq2)
• 4. Applications

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Plan

• 1. Number Field and its class group
• 2. 2-rank of 2-class group of the field Q(

4

• 4pq2)
• 3. 4-rank of 2-class group of classes of Q(

4

• 4pq2)
• 4. Applications

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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 : (r1, r2) the signature, the discriminant DK, OK 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.

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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 : (r1, r2) the signature, the discriminant DK, OK 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.

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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]

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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 · 172 · 53, then

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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 · 172 · 53, then OK = Z + Zα − 1 2 + Zβ + Zγ − β 2 . such that α = √ 53, β =

4

√ 4 · 172 · 53 and γ = αβ. It is the same result given by T. Funakura in [5]

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Number Field in PariGp

Let OK be the ring of integers of K. We say that I fractional ideal

• f OK (or just of K), if I is an OK-submodule of K such that

there exists a non-zero d ∈ OK with dI ⊂ OK. 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 OK

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Number Field in PariGp

Let OK be the ring of integers of K. We say that I fractional ideal

• f OK (or just of K), if I is an OK-submodule of K such that

there exists a non-zero d ∈ OK with dI ⊂ OK. 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

• f K : U(K) 2 units and class group Cl(K).
• 2. The invertible elements of OK

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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

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A question

In the previous example, we have K = Q( 4

• 4pq2) 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).

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A question

In the previous example, we have K = Q( 4

• 4pq2) 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).

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A question

In the previous example, we have K = Q( 4

• 4pq2) 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) ?

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A question

In the previous example, we have K = Q( 4

• 4pq2) 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) ?

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A question

In the previous example, we have K = Q( 4

• 4pq2) 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) ?

p q

• q

p

• 4
• p

q

• 4

r2(K) r4(K) Cl(K) 13 17

• 1

1 2 1 [4, 2] 53 17

• 1
• 1

2 [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 [2, 2] 37 41

• 1

1 2 1 [4, 2] 61 41

• 1
• 1

2 [6, 2] 173 41

• 1
• 1

2 [10, 2]

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The main objective

Now we assume that Q( 4

• 4pq2) be a real pure quartic number

field, where p and q be two different odd prime numbers such that p ≡ 5 (mod 8), q ≡ 1 (mod 4) and p q

• = 1.

Our goal is to show that the 2-class group of this field, is of type (2, 2) if only if p q

• 4

= −1 and give its generators.

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Plan

• 1. Number Field and its class group
• 2. 2-rank of 2-class group of the field Q(

4

• 4pq2)
• 3. 4-rank of 2-class group of classes of Q(

4

• 4pq2)
• 4. Applications

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Definition of 2-rank and 4-rank

Let K be a number field and let Cl2(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 : r2(K) = dimF2(Cl2(K)/ Cl2(K)2) and r4(K) = dimF2(Cl2(K)2/ Cl2(K)4) where F2 is the finite field with 2 elements.

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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 aσ−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.

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The Ambiguous Class Number Formula [4]

Let be L/K a cyclic extension of prime degree ℓ, then # Am(L/K) = h(K) · ℓt−1 [EK : EK ∩ NL×], where EK is the unit group of K, and EK ∩ NL× the subgroup of units that are norms of elements of L. If h(K) is odd and ℓ = 2, then r2(L) = t − e − 1, with t is the number of ramified primes (including those at infinity) in L/K, 2e = [EK : EK ∩ NL×].

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Application of the ambiguous class number formula

We take K = Q(√p) and α = 2q√p such that p and q be two different odd prime numbers such that p ≡ 5 (mod 8), q ≡ 1 (mod 4) and p q

• = 1.

Then L = K(√α) = Q( 4

• 4pq2) and the relative discriminant of

L/K is (23q√p)OK = 23

I π1π2

√pOK. So the finite primes ramified in K/k are √pOK, π1, π2, 2I. 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 r2(L) = 4 − e. From the numerical examples obtained using PariGp e = 2.

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Hilbert Symbol

Let P be a prime ideal of K and let KP be the completion of K at

• P. For any α, β ∈ KP, we define the local Hilbert Symbol by

(α, β)P =

• 1,

if β is a norm in KP(√α); −1, if not.

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Hilbert Symbol

Let P be a prime ideal of K and let KP be the completion of K at

• P. For any α, β ∈ KP, we define the local Hilbert Symbol by

(α, β)P =

• 1,

if β is a norm in KP(√α); −1, if not. Let iP the natural injection between K and KP, then for any α, β ∈ K, we define the global Hilbert Symbol by α, β P

• = (iP(α), iP(β))P.

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Hilbert Symbol

Theorem ([6])

The Hilbert symbol satisfies the following conditions : i. α, β P

• =

β, α P

• .
• ii. If P∞ is a real infinite prime, then

α, β P∞

• = −1 if and only if

iP(α) < 0 and iP(β) < 0.

• iii. β is a norm in K(√α) if and only if

α, β P

• = 1 for all P.
• iv. The Hilbert symbol verifies
• P

α, β P

• = 1.

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the end of the proof of "r2(L) = 2"

It is known that EK = −1, εp, where εp is the fundamental unit

• f quadratic field K. Then To show that e = 2, it suffices to prove

that −1, εp, −εp / ∈ NL. This is true ?

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the end of the proof of "r2(L) = 2"

It is known that EK = −1, εp, where εp is the fundamental unit

• f quadratic field K. Then To show that e = 2, it suffices to prove

that −1, εp, −εp / ∈ NL. This is true ?

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the end of the proof of "r2(L) = 2"

It is known that EK = −1, εp, where εp is the fundamental unit

• f quadratic field K. Then To show that e = 2, it suffices to prove

that −1, εp, −εp / ∈ NL. This is true ? ? nf=nfinit(x^2-53); ? p1=idealprimedec(nf,53)[1]; ? p=idealprimedec(nf,2)[1]; ? nfhilbert(nf,-1, 2*17*x,p) %4 = -1 ? nfhilbert(nf,-1/2*x + 7/2, 2*17*x,p1) %5 = -1

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the end of the proof of "r2(L) = 2"

It is known that EK = −1, εp, where εp is the fundamental unit

• f quadratic field K. Then To show that e = 2, it suffices to prove

that −1, εp, −εp / ∈ NL. This is true ? ? nf=nfinit(x^2-53); ? p1=idealprimedec(nf,53)[1]; ? p=idealprimedec(nf,2)[1]; ? nfhilbert(nf,-1, 2*17*x,p) %4 = -1 ? nfhilbert(nf,-1/2*x + 7/2, 2*17*x,p1) %5 = -1

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the end of the proof of "r2(L) = 2"

It is known that EK = −1, εp, where εp is the fundamental unit

• f quadratic field K. Then To show that e = 2, it suffices to prove

that −1, εp, −εp / ∈ NL. This is true ? ? nf=nfinit(x^2-53); ? p1=idealprimedec(nf,53)[1]; ? p=idealprimedec(nf,2)[1]; ? nfhilbert(nf,-1, 2*17*x,p) %4 = -1 ? nfhilbert(nf,-1/2*x + 7/2, 2*17*x,p1) %5 = -1 This is true, because εp, 2q√p P∞

• =

−1, 2q√p 2I

• =

−εp, 2q√p (√p)

• = −1

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Plan

• 1. Number Field and its class group
• 2. 2-rank of 2-class group of the field Q(

4

• 4pq2)
• 3. 4-rank of 2-class group of classes of Q(

4

• 4pq2)
• 4. Applications

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Some work on 2-class group

Using the methods of Genus Theory, many mathematicians have calculated, r2(K) and r4(K) whenever K is a quartic number field having subfield with an odd class number. For instances we mention, the following works

• For biquadratic number field we indicate the works of A. Azizi,
• M. Taous, A. Zekhnini, Q.Yue in [2],[13],[14]. . . .
• For cyclic quartic fields we indicate the works of A. Azizi and
• M. Talbi in [3]. . . .
• For pure quartic number fields we indicate the works of C.J.

Parry in [10], [11]. . . . In this last, C. Parry has determined the exact power of 2 dividing the class number of pure quartic number field in many cases.

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4-rank formula

While in our case we assume that p ≡ 5 (mod 8), and q are two different odd primes such that q ≡ 1 (mod 8) and

• p

q

• = 1.

after we have shown that the 2-class group of L = Q( 4

• 4pq2) is 2,

we compute the 4-rank of the class group of fields L. Y. Qin provided a formula to calculate the 4-rank of a quadratic extension M( √ δ) of number fields M with odd class : r4(M( √ δ)) = t − 1 − rank(RM(

√ δ)/M),

where with t is the number of ramified primes M( √ δ)/M and RM(

√ δ)/M is the following matrix :

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The generalized Rédei-matrix

RM(

√ δ)/M =

      a1; δ P1

• . . .

an; δ P1

• . . .

an+r; δ P1

• . . .

. . . a1; δ Pt

• . . .

an; δ Pt

• . . .

an+r; δ Pt

• 

     . It is a matrix of type t × (n + r) with coefficients in F2, called the generalized Rédei-matrix, where (Pi)1≤i≤t are primes (finite and infinite) of M which ramify in M( √ δ), (aj)1≤j≤n+r is a family of elements of M determined by Y. Qin in [14, §2 Lemma 2.4, p.27] and −; δ Pi

• is the Hilbert symbol on M. Note that this matrix is

given by replacing the 1’s by 0’s and the -1’s by 1’s. For more details concerning the generalized Rédei-matrix, we refer to [14],

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Main Theorem

Keep the previous hypotheses and notations, then the 2-class group

• f L = Q( 4
• pq2), is of type (2, 2) if only if

p q

• 4

= −1. In this case Cl2(L) = Hh

1, H3

Where H1, H3 are the prime ideals of K above q, 2 respectively and h is the class number of Q(√p).

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Proof of Theorem

By assumption, q is a prime number and

• p

q

• = 1 .Then all finite

primes of k ramifying in K are : π1,π2,2I and (√p) where π1π2 = qOk, 2I = 2Ok and (√p)2 = pOk. So there exist a primes ideals.H1, H2, H3, H4 of K, such that H2

1 = π1OK, H2 2 = ˜

π1OK, H2

3 = 2IOK, H2 4 = √pOK.

Now, we have

• Hh

1σ(Hh 1) = Hh 1Hh 1OK = πh 1OK = ((x + y√p)/2)OK.

• Hh

2σ(Hh 2) = Hh 2Hh 2OK = ˜

πh

1OK = ((x − y√p)/2)OK.

• H3σ(H3) = 2OK.
• H4σ(H4) = √pOK.

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Proof of Theorem

We show first that −1; εp; −εp / ∈ EK ∩ N(L). which implies that the 5 × (4 + 2) generalized Rédei’s matrix of Hilbert symbols can be written as RL/K =       

2(x+√p);δ

π1

• 2(x−√p);δ

π1

• 2;δ

π1

• −√p;δ

π1

• −1;δ

π1

• εp;δ

π1

• 2(x+√p);δ

π2

• 2(x−√p);δ

π2

• 2;δ

π2

• −√p;δ

π2

• −1;δ

π2

• εp;δ

π2

• 2(x+√p);δ

2I

• 2(x−√p);δ

2I

• 2;δ

2I

• −√p;δ

2I

• −1;δ

2I

• εp;δ

2I

• x+√p;δ

√p

• 2(x−√p);δ

√p

• 2;δ

√p

• −√p;δ

√p

• −1;δ

√p

• εp;δ

√p

• 2(x+√p);δ

p∞

• 2(x−√p);δ

p∞

• 2;δ

p∞

• −√p;δ

p∞

• −1;δ

p∞

• εp;δ

p∞

• 

      Then RL/K =       

1

• p

q

• 4

1

• p

q

• 4

1

• p

q

• 4
• q

p

• 4
• p

q

• 4

1 1

• p

q

• 4

1

• p

q

• 4
• q

p

• 4
• p

q

• 4
• q

p

• 4
• p

q

• 4
• q

p

• 4

−1 1 −1 1

• q

p

• 4
• q

p

• 4

−1 −1 1 −1 1 1 1 −1 −1 −1

       ,

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Proof of Theorem (continued)

then by mapping 1 to 0 and −1 to 1, this matrix is with coefficients in F2 is

• If
• p

q

• 4 =
• q

p

• 4 = −1 then
• If
• p

q

• 4 = −
• q

p

• 4 = −1 then

RL/K =      

1 1 1 1 1 1 1 1 1 1 1 1 1 1

      , RL/K =      

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

      . so in the both cases, rank(RL/K) = 4 then use the Y.Qin formula ([14]) we find r4(L) = 5 − 1 − rank(RL/K) = 5 − 1 − 4 = 1.

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Discover the generators I

prang(v,m)=sum(i=1,#v,v[i]%m==0); ppart(v,m)=vector(prang(v,m),i,m^valuation(v[i],m)); bnfpclgp(P,m)=ppart(bnfinit(P).cyc,m); {forprime(p=3,400, forprime(q=3,400, if(Mod(q,8)==1 && Mod(p,8)==5, if(kronecker(p,q)==1, if(Mod(p^((q-1)/4),q)==Mod(1,q),a=1,a=-1);\\ (p/q)_4 if(Mod(q^((p-1)/4),p)==Mod(1,p),b=1,b=-1);\\ (q/p)_4 P=polredabs(x^4-4*p*q^2); T=polredabs(x^2-q); k0=bnfinit(x^2-p); h=k0.clgp.no; L=bnfinit(P); H1=idealpow(L,idealprimedec(L, q)[1],h); H3=idealprimedec(L, 2)[1];

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Discover the generators II

H4=idealprimedec(L, p)[1]; A=idealmul(L, H1, H3); B=idealmul(L, H1, H4); C=idealmul(L, H3, H4); PH1=bnfisprincipal(L,H1,0); PH3=bnfisprincipal(L,H3,0); PH4=bnfisprincipal(L,H4,0); PA=bnfisprincipal(L,A,0); PB=bnfisprincipal(L,B,0); PC=bnfisprincipal(L,C,0); if(a==-1, print(p":"q":"b":"PH1":"PH2 ":"PH3":"PA":"PB":"PC));););););} 5:41:1:[3, 0]~:[4, 2]~:[3, 1]~:[0, 1]~:[3, 0]~:[3, 1]~ 5:89:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 0]~:[1, 1]~ 5:241:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~ 5:281:1:[1, 1]~:[4, 2]~:[0, 1]~:[1, 0]~:[1, 1]~:[0, 1]~ 13:113:1:[1, 1]~:[4, 2]~:[1, 0]~:[0, 1]~:[1, 1]~:[1, 0]~

SLIDE 41

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Discover the generators III

13:233:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 0]~:[1, 1]~ 13:257:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 13:313:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~ 29:241:-1:[3, 1]~:[4, 2]~:[0, 0]~:[3, 1]~:[3, 0]~:[0, 1]~ 29:257:1:[0, 1]~:[4, 2]~:[3, 1]~:[3, 0]~:[0, 1]~:[3, 1]~ 29:313:1:[9, 0]~:[4, 2]~:[9, 1]~:[0, 1]~:[9, 0]~:[9, 1]~ 37:137:1:[1, 1]~:[4, 2]~:[0, 1]~:[1, 0]~:[1, 1]~:[0, 1]~ 53:17:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 53:89:1:[11, 0]~:[4, 2]~:[0, 1]~:[11, 1]~:[11, 0]~:[0, 1]~ 53:97:1:[1, 1]~:[4, 2]~:[1, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 53:241:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 53:281:1:[1, 1]~:[4, 2]~:[1, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 61:41:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 61:73:1:[1, 0]~:[4, 2]~:[0, 1]~:[1, 1]~:[1, 0]~:[0, 1]~ 61:113:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 61:137:1:[0, 1]~:[4, 2]~:[3, 1]~:[3, 0]~:[0, 1]~:[3, 1]~ 61:241:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~

SLIDE 42

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Discover the generators IV

61:257:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~ 101:137:1:[1, 1]~:[4, 2]~:[0, 1]~:[1, 0]~:[1, 1]~:[0, 1]~ 101:193:1:[0, 1]~:[4, 2]~:[1, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 101:233:1:[1, 1]~:[4, 2]~:[1, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 109:89:1:[0, 1]~:[4, 2]~:[1, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 109:97:1:[1, 1]~:[4, 2]~:[0, 1]~:[1, 0]~:[1, 1]~:[0, 1]~ 109:137:-1:[19, 1]~:[4, 2]~:[0, 0]~:[19, 1]~:[19, 0]~:[0, 1]~ 109:233:1:[3, 1]~:[4, 2]~:[3, 0]~:[0, 1]~:[3, 1]~:[3, 0]~ 149:73:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~ 149:113:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[7, 0]~:[7, 1]~ 149:281:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 149:337:1:[9, 0]~:[4, 2]~:[0, 1]~:[9, 1]~:[9, 0]~:[0, 1]~ 157:89:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~ 157:113:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~ 157:257:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~ 157:281:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~ 157:313:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 0]~:[1, 1]~

SLIDE 43

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Discover the generators V

157:353:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~ 173:41:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 173:73:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 173:89:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 173:137:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[3, 0]~:[3, 1]~ 173:233:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~ 173:257:1:[1, 0]~:[4, 2]~:[1, 1]~:[0, 1]~:[1, 0]~:[1, 1]~ 173:313:1:[1, 0]~:[4, 2]~:[0, 1]~:[1, 1]~:[1, 0]~:[0, 1]~ 181:73:1:[7, 0]~:[4, 2]~:[7, 1]~:[0, 1]~:[7, 0]~:[7, 1]~ 181:137:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[0, 1]~:[1, 1]~ 181:233:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[0, 1]~:[1, 1]~ 181:337:-1:[7, 1]~:[4, 2]~:[0, 0]~:[7, 1]~:[7, 0]~:[0, 1]~ 181:353:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[1, 0]~:[0, 1]~ 197:41:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 197:97:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[3, 0]~:[3, 1]~ 197:233:1:[9, 1]~:[4, 2]~:[9, 0]~:[0, 1]~:[9, 1]~:[9, 0]~ 197:313:-1:[5, 1]~:[4, 2]~:[0, 0]~:[5, 1]~:[0, 1]~:[5, 0]~

SLIDE 44

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Discover the generators VI

197:353:1:[1, 1]~:[4, 2]~:[1, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 229:17:1:[3, 0]~:[4, 2]~:[3, 1]~:[0, 1]~:[3, 0]~:[3, 1]~ 229:241:-1:[3, 0]~:[4, 2]~:[0, 0]~:[3, 0]~:[3, 1]~:[0, 1]~ 229:337:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[3, 0]~:[3, 1]~ 269:73:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[3, 1]~:[3, 0]~ 269:233:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 0]~:[1, 1]~ 277:89:-1:[3, 0]~:[4, 2]~:[0, 0]~:[3, 0]~:[3, 1]~:[0, 1]~ 277:113:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 0]~:[1, 1]~ 277:353:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 293:97:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[3, 1]~:[3, 0]~ 293:137:1:[1, 1]~:[4, 2]~:[0, 1]~:[1, 0]~:[1, 1]~:[0, 1]~ 293:193:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[0, 1]~:[1, 1]~ 293:233:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 293:257:-1:[5, 0]~:[4, 2]~:[0, 0]~:[5, 0]~:[0, 1]~:[5, 1]~ 293:353:1:[0, 1]~:[4, 2]~:[1, 0]~:[1, 1]~:[0, 1]~:[1, 0]~ 317:73:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[1, 1]~:[0, 1]~ 317:113:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~

SLIDE 45

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Discover the generators VII

317:193:1:[3, 1]~:[4, 2]~:[3, 0]~:[0, 1]~:[3, 1]~:[3, 0]~ 317:281:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 0]~:[1, 1]~ 317:353:1:[15, 1]~:[4, 2]~:[15, 0]~:[0, 1]~:[15, 1]~:[15, 0]~ 349:17:-1:[0, 1]~:[4, 2]~:[0, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 349:41:1:[1, 1]~:[4, 2]~:[0, 1]~:[1, 0]~:[1, 1]~:[0, 1]~ 349:241:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[1, 0]~:[0, 1]~ 349:281:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[0, 1]~:[1, 1]~ 349:337:-1:[1, 0]~:[4, 2]~:[0, 0]~:[1, 0]~:[0, 1]~:[1, 1]~ 373:89:-1:[3, 0]~:[4, 2]~:[0, 0]~:[3, 0]~:[0, 1]~:[3, 1]~ 373:257:1:[1, 1]~:[4, 2]~:[1, 0]~:[0, 1]~:[1, 1]~:[1, 0]~ 389:17:1:[3, 1]~:[4, 2]~:[0, 1]~:[3, 0]~:[3, 1]~:[0, 1]~ 389:41:-1:[1, 1]~:[4, 2]~:[0, 0]~:[1, 1]~:[1, 0]~:[0, 1]~ 389:73:1:[0, 1]~:[4, 2]~:[1, 1]~:[1, 0]~:[0, 1]~:[1, 1]~ 389:113:1:[1, 0]~:[4, 2]~:[0, 1]~:[1, 1]~:[1, 0]~:[0, 1]~ 389:353:-1:[3, 0]~:[4, 2]~:[0, 0]~:[3, 0]~:[0, 1]~:[3, 1]~

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Plan

• 1. Number Field and its class group
• 2. 2-rank of 2-class group of the field Q(

4

• 4pq2)
• 3. 4-rank of 2-class group of classes of Q(

4

• 4pq2)
• 4. Applications

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Cl2(L) is of type (2, 2)

As an application of our study, we take for example the case where q ≡ 1 (mod 8), p q

• = 1 and

p q

• 4

= −1, then r2(L) = 2 and r4(L) = 0 so the 2-classes group Cl2(L) of K is isomorphic to Z/2Z × Z/2Z (We say that Cl2(L) is of type (2, 2). By Class field theory we have this scheme :

SLIDE 48

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Cl2(L) is of type (2, 2)

As an application of our study, we take for example the case where q ≡ 1 (mod 8), p q

• = 1 and

p q

• 4

= −1, then r2(L) = 2 and r4(L) = 0 so the 2-classes group Cl2(L) of K is isomorphic to Z/2Z × Z/2Z (We say that Cl2(L) is of type (2, 2). By Class field theory we have this scheme : L(2)

2

L(1)

2

L2 L1 L3 L Q

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Cl2(L) is of type (2, 2)

We denote by L(1)

2

the Hilbert 2-class field of L, which is the maximal abelian unramified extension of L such that [L(1)

2

: K] is a power of 2. Let L(2)

2

be the Hilbert 2-class field of L(1)

2

and put G = Gal(L(2)

2 /L). It’s easy to see that

G/G ′ ≃ Cl2(L) ≃ Z/2Z × Z/2Z (Klein four-group) L(2)

2

L(1)

2

L2 L1 L3 L Q

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Cl2(L) is of type (2, 2) : what can we do ?

• 1. Constructed the extensions Li and L(1)

2 .

• 2. Find the 2-classes of Cl2(L) that capitulate in the extension Li.
• 3. As G/G ′ is of type (2, 2), then G is abelian of type (2, 2),

quaternion, dihedral or semi-dihedral, so can we find exactly the structure of G ? L(2)

2

L(1)

2

L2 L1 L3 L Q

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Cl2(L) is of type (2, 2) : what can we do ?

In [9, Theorem 2]], Kisilvsky gave a classification method of G, based on the capitulation problem

SLIDE 52

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Cl2(L) is of type (2, 2) : what can we do ?

In [9, Theorem 2]], Kisilvsky gave a classification method of G, based on the capitulation problem

Définition

Let M be a cyclic unramified extension of a number field k and j denote the basic homomorphism : jM/k : Cl(k) − → Cl(M), induced by extension of ideals from k to M. Thus, we say

• 1. M/k satisfies condition A iff |ker(jM/k) ∩ NL/k(Cl(M))| > 1.
• 2. M/k satisfies condition B iff |ker(jM/k) ∩ NL/k(Cl(M))| = 1.

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Cl2(L) is of type (2, 2) : what can we do ?

|ker j1| (A/B) |ker j2| (A/B) |ker j3| (A/B) G 4 4 4 (2, 2) 2A 2A 2A Q3 4 2B 2B Dm, m ≥ 3 2A 2B 2B Qm, m ≥ 4 2B 2B 2B Sm, m ≥ 4 With ker ji = ker jLi/L : Cl(L) − → Cl(Li),

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Cl2(K) is of type (2, 2) : Structure of G

suppose that q ≡ 1 (mod 8), p q

• = 1 and

p q

• 4

= −1, then Cl2(L) is of type (2, 2), where L = Q( 4

• 4pq2). To find the

structure of G = Gal(L(2)

2 /L), I only use the number fields

Li = L(√q) = Q( 4 √4pq, √q) and I show the following results :

• 1. Li/L is an quadratic unramified extension.
• 2. Class number of Q( 4

√p) is odd.

• 3. The fundamental system of units of Q( 4

√p) is S = {ε0; εp}, such that εp is the fundamental unit of Q(√p) and ε0 is the smallest unit of K satisfying ε0ε

′

0 = 1.whith ε

′

0 is the

conjugate relative of ε0 to Q(√p)

SLIDE 55

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Cl2(L) is of type (2, 2) : Structure of G

suppose that q ≡ 1 (mod 8), p q

• = 1 and

p q

• 4

= −1, then Cl2(L) is of type (2, 2), where L = Q( 4

• 4pq2). To find the

structure of G = Gal(L(2)

2 /L), I only use the number fields

Li = L(√q) = Q( 4 √4pq, √q) and I show the following results :

SLIDE 56

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Cl2(L) is of type (2, 2) : Structure of G

suppose that q ≡ 1 (mod 8), p q

• = 1 and

p q

• 4

= −1, then Cl2(L) is of type (2, 2), where L = Q( 4

• 4pq2). To find the

structure of G = Gal(L(2)

2 /L), I only use the number fields

Li = L(√q) = Q( 4 √4pq, √q) and I show the following results :

• 4. Cl2(Li) ≃ Z/2Z or Z/4Z.
• 5. Cl2(Li) ≃ Z/2Z iff

q p

• 4

= 1.

• 7. Only Hh

1H4 and trivial class capitulate in Li.

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Cl2(K) is of type (2, 2) : Structure of G

Theorem

suppose that q ≡ 1 (mod 8), p q

• = 1 and

p q

• 4

= −1, then Cl2(K) is of type (2, 2), where K = Q( 4

• pq2). In addition we

have :

• 1. G is of type (2, 2) iff

q p

• 4

= 1.

• 2. G ≃ Q3 iff

q p

• 4

= 1.

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Références I

[1]

• A. Azizi and M. Taous, Détermination des corps k = Q(

√ d, i) dont les 2-groupes de classes sont de type (2, 4) ou (2, 2, 2) Rend. Istit.

• Mat. Univ. Trieste. 40 (2008), 93-116.

[2]

• A. Azizi, M. Taous and A. Zekhnini, On the unit index of some real

biquadratic number fields, Turk. J. Math. (2018), no. 42, 703-715. [3]

• A. Azizi et M. Talbi, Capitulation des 2-classes d’idéaux de certains

corps biquadratiques cycliques, Acta Arithmetica 127 (2007),231-248. [4]

• C. Chevalley, Sur la théorie du corps de classes dans les corps finis

et les corps locaux, J. Fac.Sc. Tokyo, Sect. 1, t.2, (1933), 365-476. [5]

• T. Funakura, On integral bases of pure quartie fields, Math. J.

Okayama Univ., 26 (1984), 27 41. [6]

• G. Gras, Class field theory, from theory to practice, Springer Verlag

2003.

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Références II

[7]

• M. Haynou, M. Taous, The 4-Rank of the Class Group of Some

Real Pure Quartic Number Fields, Associative and Non-Associative Algebras and Applications, Springer, 2018. [8]

• J. A. Hymo and C. J. Parry, On relative integral bases for pure

quartic fields, Indian J. Pure Appl. Math., 23, 1992, 359-376. [9]

• H. Kisilevsky. Number fields with class number congruent to 4

(mod 8) and Hilbert’s theorem 94, J. Number Theory, vol. 8, no 3, (1976), 271-279. [10] C. J. Parry, Pure quartic number fields whose class numbers are even, J. Reine Angew. Math. 264 (1975), 102-112. [11] C. J. Parry, A genus theory for quartic fields, J. Reine Angew.

• Math. 314 (1980), 40-71.

[12] The PARI Group, PARI/GP version 2.10.0, Univ. Bordeaux, 2018, http ://pari.math.u-bordeaux.fr/.

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Références III

[13] M. Taous, Capitulation des 2-classes d’idéaux de certains corps Q( √ d, i) de type (2, 4), thèse, Université. Mohammed Premier Facultéé des Science, Oujda. 2008. [14] Y. Qin The generalized Rédei-matrix, Math. Z. 261 (2009), 23-37. [15] F. Lemmermeyer, Reciprocity Laws. From Euler to Eisenstein, Springer Monographs in Math. (2000) . [16] F. Lemmemeyer, Ideal class groups of cyclotomic number fields I, Acta Arith. 72.4 (1995), 347-359. [17] H. C Williams, The quadratic character of a certain quadratic surd, Utilitas Math.5 (1974), 49-55.

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