The generators of $3$-class group of some fields of degree $6$ over $\mathbb{Q}$

Let $\mathrm{k}=\mathbb{Q}\left(\sqrt[3]{p},\zeta_3\right)$, where $p$ is a prime number such that $p \equiv 1 \pmod 9$, and let $C_{\mathrm{k},3}$ be the $3$-component of the class group of $\mathrm{k}$. In \cite{GERTH3}, Frank Gerth III proves a conjecture made by Calegari and Emerton \cite{Cal-Emer} which gives necessary and sufficient conditions for $C_{\mathrm{k},3}$ to be of $\operatorname{rank}\,$ two. The purpose of the present work is to determine generators of $C_{\mathrm{k},3}$, whenever it is isomorphic to $\mathbb{Z}/9\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.


Introduction
Let Γ = Q( 3 √ p) be a pure cubic field, where p is a prime number such that p ≡ 1 (mod 9). We denote by ζ 3 = −1/2 + i √ 3/2 the normalized primitive third roots of unity, k = Q( 3 √ p, ζ 3 ) the normal closure of Γ and C k,3 the 3-component of the class group of k.
Assuming 9 divides exactly the 3-class number of Γ. Then C k,3 Z/9Z × Z/3Z if and only if u = 1, where u is an index of units that will be defined in the notations below. In this paper, we will determine the generators of C k,3 when C k,3 is of type (9, 3) and 3 is not a cubic residue modulo p. We spot that Calegari and Emerton ([2, Lemma 5.11]) proved that the rank of the 3-class group of Q( 3 √ p, ζ 3 ), with p ≡ 1 (mod 9), is equal to two if 9 divides the 3-class number of Q( 3 √ p). Moreover, in his work [6, Theorem 1, p.471], Frank Gerth III proves that the converse to Calegari-Emerton's result is also true. The present work can be viewed as a continuation of the works [2] and [6] . After reviewing some basic properties of the norm residue symbols and prime factorization in the normal closure of a pure cubic field that will be needed later, we will establish in section 3 some preliminary results of the 3-class group C k, 3 . Using this, we arrive to determine the generators of 3-class groups C k,3 of type (9,3). All the study cases are illustrated by numerical examples and summarized in tables in section 4. The usual notations on which the work is based is as follows: : a pure cubic field, where d is a cube-free natural number; • k 0 = Q(ζ 3 ) : the third cyclotomic field ; • k = Q( 3 √ d, ζ 3 ) : the normal closure of the pure cubic field Γ; • u = [E k : E 0 ] : the index of the sub-group E 0 generated by the units of intermediate fields of the extension k/Q in E k the group of units of k; • For an algebraic number field L: and p is a prime number congruent to 1(mod 3).
2 Norm residue symbol and ideal factorization theory

The norm residue symbol
Let L/K an abelian extension of number fields with conductor f . For each finite or infinite prime ideal P of K, we note by f P the largest power of P that divides f . Let a ∈ K * , we determine an auxiliary number a 0 by the two conditions a 0 ≡ a (mod f P ) and a 0 ≡ 1 (mod f f P ). Let Q an ideal co-prime with P such that (a 0 ) = P e Q (b = 0 if P is infinite). We note by a, L P = L/K Q the Artin map in L/K applied to Q.
Definition 2.1. Let K be a number field containing the l th -roots of units, where l ∈ N, then for each a, b ∈ K * and prime ideal P of K, we define the norm residue symbol by: Therefore, if the prime ideal P is unramified in the field K( l √ b), then we write Remark 2.1. Notice that a, b P l and b P l are two l th -roots of units.
Following [9], the principal properties of the norm residue symbol are given as follows: Properties 1. The product formula: σa, σb σP l = σ a, b P l , for each automorphism σ of K; 5. If P is not divisible by the conductor f b of K( l √ b) and appears in (a) with the exponent e, then: • P is infinite (e = 0) ⇒ a, b P l = 1; 6. The classical reciprocity law: let a, b ∈ K * , and the conductors f a and f b of respectively K( l √ a) and K( l √ b) are co-prime, then: where the product is taken on the finite and infinite prime ideals; 8. Let L is a finite extension of K, a ∈ L and b ∈ K * , then: for each prime ideal P of K.
For more basic properties of the norm residue symbol in the number fields, we refer the reader to the papers [3], [8] and [9]. Notice that in section 3, we will use the norm cubic residue symbols (l = 3). As the ring of integer O k 0 is principal, h k 0 = 1, we will write the norm cubic residue symbol as follows: where a, b ∈ k * 0 and π is a prime integer of O k 0 .

Prime factorization in a pure cubic field and in its normal closure
Let be Γ = Q( 3 √ d) a pure cubic field, and O Γ the ring of integers of Γ. We write the natural integer d in form d = ab 2 , where a and b are cube-free and co-prime positive integers. In his paper [3], Dedekind has defined two different types of pure cubic fields as follows: Definition 2.2. Using the same notations as above:

We say that
Now, let p be a prime number. In the following Proposition, we give the decomposition of the prime p in the pure cubic field Γ = Q( 3 √ ab 2 ). We denote by P, P i prime ideals of Γ, and by N the absolute norm N Γ/Q .
Let p a prime number such that p = 3, then: 1. If p divides ab and p = 3, then pO Γ = P 3 , N (P) = p.
The ramification of the prime 3 need a particular treatment, it is the purpose of the following Proposition: The decomposition into prime factors of 3 is: The ideal factorization rules for the 3rd cyclotomic field k 0 (see [11]) is as follows: Next, let k be the normal closure of Γ. We note by O k the ring of integers of k, P and P s are prime ideals of k, N = N k/Q the norm of k on Q. Combining the ideal factorization rules for Γ with those of the field k 0 . The decomposition of the prime 3 in k is the purpose of the following Theorem: The prime 3 decomposes in k as follows:
2) Conversely, suppose that Γ is of second kind, then 3O Γ = P 2 P 1 . It follows that However, we have the following Proposition in which we characterize the decomposition of prime ideals of p = 3 in k.

Proposition 2.4.
Let p a prime number such that p = 3, then: 2. If p does not divides D Γ and p ≡ 1 (mod 3), then: (a) p decomposes completely in k if and only if D Γ is a cubic residue modulo p.
(b) pO k = P 1 P 2 , with N (P 1 ) = N (P 2 ) = p 3 , if and only if D Γ is not a cubic residue modulo p.
Proof. 1. We use Proposition 2.1 and the decomposition of prime ideals in the quadratic fields k 0 = Q(ζ 3 ).
2. Suppose that p does not divide D Γ and p ≡ 1 (mod 3), then −3 is a quadratic residue modulo p, then the multiplication formula gives On the one hand, by the Euler's Theorem we have On the other hand, the quadratic reciprocity law gives Thus, p decomposes completely in k 0 .
(a) If D Γ is a cubic residue modulo p, then by Proposition 2.1 we have p split completely in Γ. Hence p split completely in k.
(b) If D Γ is not a cubic residue modulo p, we have p remains prime in Γ. Hence pO k = P 1 P 2 .
3. We have pO Γ = PP 1 , and p remains inert in k 0 , hence the result.

The generators of C k,3
First, we let C (σ) k,3 = {A ∈ C k,3 | A σ = A} be the group of ambiguous ideal classes of k/k 0 , where σ is a generator of Gal (k/k 0 ), and put q * = 0 or 1 according to ζ 3 is not norm or norm of an element of k\{0}. Let t be the number of primes ramifies in k/k 0 . Then according to [4], we have |C By the exact sequence : The fact that C (σ) k,3 and C k,3 /C 1−σ k,3 are elementary abelian 3-groups imply that: and let s be the positive integer such that C The following Proposition gives the structure of the 3-class group C k,3 when 27 divides exactly the class number of k: Proposition 3.1. Let be Γ a pure cubic field, k its normal closure and u the index of units defined as above, then:

2)
We have the same proof as above. Proof. Since p ≡ 1 (mod 3), then according to section 2.2, we have p = π 1 π 2 , where π 1 and π 2 are two primes of k 0 such that π 2 = π τ 1 and π 1 ≡ π 2 ≡ 1 ( mod 3O k 0 ). We study all cases depending on the congruence class of p modulo 9, then: • If p ≡ 4 or 7 (mod 9), then according to section 2.2, the prime 3 is ramified in the field L, so the prime ideal (1 − ζ 3 ) is ramified in k/k 0 . Also π 1 and π 2 are totally ramified in k. So t = 3. In addition, the fact that p ≡ 4 or 7 (mod 9) imply that where the symbol ( , ) 3 is the cubic Hilbert symbol. We deduce that ζ 3 is not a norm in the extension k/k 0 , so q * = 0. Hence rank C • If p ≡ 1 (mod 9), the prime ideals which ramified in k/k 0 are π 1 and π 2 , so t = 2.
, then according to [4], the cubic Hilbert symbol: We conclude that ζ 3 is a norm in the extension k/k 0 , then q * = 1, so rank C The basic result for determining the generators of the 3-class group of k = Q( 3 √ p, ζ 3 ) when the 3-class number of k is divisible by 27 exactly, where p is a prime number such that p ≡ 1 (mod 9), is summarized in the following Theorem: where p is a prime number such that p ≡ 1 (mod 9), k = Q( 3 √ p, ζ 3 ) its normal closure and C k,3 the 3-class group of k. Assuming 9 divides the 3-class number of Γ exactly, then: The 3-class group C k,3 is isomorphic to Z/9Z × Z/3Z if and only if u = 1.
where p is a prime number such that p ≡ 1 (mod 9), k = Q( 3 √ p, ζ 3 ) it's normal closure, and C k,3 be the 3-class group of k. Assume that 9 divides the 3-class number of Γ exactly and u = 1. Put A = C + k,3 , where A ∈ C k,3 such that A 9 = 1 and A 3 = 1. Let C (σ) k,3 be the 3-group of ambiguous ideal classes of k/k 0 and C 1−σ k,3 = {A 1−σ | A ∈ C k,3 } be the principal genus of C k,3 . Then: where C + k,3 and C − k,3 are defined in Lemma 2.1 of [5].
Proof. 1. Since 9 divides the 3-class number of Γ exactly and u = 1, then according to Theorem 3.1, C k,3 is of type (9,3), this implies by [6] that the integer s defined above is equal 3, and according to Case 4 of [6], we conclude that| C (σ) k,3 k,3 is a subgroup of C + k,3 . Therefore, A 3 is the unique subgroup of order 3 of C + k,3 and C (σ) k,3 is cyclic of order 3, then C . This is impossible because B σ = B. As B 3 = 1 and B 1+σ+σ 2 = 1, then B σ 2 = B 2+2σ . This equality makes it possible to show that B 1−σ is an ambiguous class. We conclude that C 2. We reason as in the assertion 1. Since A 2 ∈ C − k,3 , we deduce that C − k,3 = (A 2 ) 1−σ . then C − k,3 and C (σ) k,3 are contained in C 1−σ k,3 which is of order 9, because |C k,3 | = 27 and |C (σ) k,3 | = 3. Consequently, Our principal result can be stated as follows: where p is a prime number such that p ≡ 1 (mod 9).
The prime 3 decomposes in k as 3O k = P 2 Q 2 R 2 , where P, Q and R are prime ideals of k . Put h = h k 27 , where h k is the class number of k. Assume that 9 divides exactly the 3-class number of Q( 3 √ p) and u = 1. If 3 is not a cubic residue modulo p, then: In Appendix of this paper, we illustrated this results by the numerical examples with the aid of Pari programming [12] and summarized in some tables in section 4.

Proof.
We start our proof by showing that [R h ] is of order 9: Since the field Γ = Q( 3 √ p) with p ≡ 1 (mod 9) is of second kind, then by Proposition 2.2 we have 3O Γ = H 2 S, where H and S are prime of Γ, since HO k = PQ and SO k = R 2 , then 3O k = P 2 Q 2 R 2 , where P, Q and R are prime ideals of k. Moreover, the prime ideal R is invariant by τ , then [R] ∈ {χ ∈ C k,3 |χ τ = χ}. If 9 divides the 3-class number of Q( 3 √ p) exactly and u = 1, then by Theorem 3.1 we have C k,3 is of type (9,3). According to Proposition 3.1, we have C + k,3 is cyclic of order 9, thus [R h ] 9 = 1. Hence the class [R h ] is of order 9 if and only if R h and R 3h are not principal. We argue by the absurd: assume that R h is principal, we have , where π 1 and π 2 are two primes of k 0 such that p = π 1 π 2 . Hence, by property (5) we have: for all ideal P of k 0 . In particular, we calculate this symbol for P = π 1 O k 0 or P = π 2 O k 0 . For P = π 1 O k 0 , using the property (1) of the norm residue symbol, we have: the properties (2) and (5) imply that: and from the properties (1) and (6) we have Since the two primes π 1 and π 2 play symmetric roles, then we obtain a similar relation when P = π 2 : The equation ( * ) imply that λ π 1 The fact that 3 is not a cubic residue modulo p imply that λ π 1 π 2 3 = 1 then λ π 1 3 = 1 or λ π 2 3 = 1.
Since 3 does not divide h, then which is a contradiction. Consequently, the ideal R h is not principal.
Since the class [R h ] is invariant by τ , we deduce that the ideal R 3h is principal if and only if [R h ] = C (σ) k,3 . Since 9 divides exactly the 3-class number of Q( 3 √ p) and u = 1, then by we get |C k,3 | = 27, so the positive integer s defined above is equal 3, then C (1−σ) 3 k,3 = 1, this implies that is of order 9. This completes the proof of the first statement.
The fact that the ideals P h , Q h and R h are not principals, we prove the assertions (1), (2), (3) and (4) by applying the decomposition of 3 in the normal closure k. For the assertion (5), since the ideals P h , Q h and R h are not principal, we obtain the result by the same reasoning above. The assertions (6) and (7) follows by using Proposition 3.2.