arXiv:1311.0387v2 [math.NT] 17 Oct 2015 The shape of Z/ℓZ-number fields. Guillermo Mantilla-Soler, Marina Monsurr` o Abstract Let ℓ be a prime and let L/Q be a Galois number field with Galois group isomorphic to Z/ℓZ. We show that the shape of L, see definition 1.2, is either 1 2 A ℓ-1 or a fixed sub lattice depending only on ℓ; such a dichotomy in the value of the shape only depends on the type of ramification of L. This work is motivated by a result of Bhargava and Shnid- man, and a previous work of the first named author, on the shape of Z/3Z number fields. 1 Introduction Let L be a number field and let O L be its maximal order. Let O 0 L be the trace zero module of O L i.e., the set {x ∈ O L : tr L/Q (x)=0}. Let us now consider the symmetric Z-bilinear form obtained by restricting the trace pairing to O 0 L O 0 L × O 0 L → Z (x, y) → tr L/Q (xy); we will denote by q L the integral trace zero form i.e., the associated integral quadratic form. In [Bha-Sha], the authors use a sub-lattice of the binary quadratic form 〈O 0 L ,q L 〉 to count cubic fields. From their work, if L is a Galois cubic field, one can deduce that after scaling the form 〈O 0 L ,q L 〉 in such a way that the form is primitive, one obtains an integral binary quadratic form that is independent on the field. A straightforward calculation shows that the scaling factor is 2 · rad(d L ), where d L is the discriminant of L and rad(·) denotes the usual radical of an integer. Throughout the paper we will use the notation rad L := rad(d L ). The following result and the explicit calculation of the scaling factor can be found in [Man, Theorem 3.1]. Theorem 1.1. Let L be a Galois cubic field. Then, the rational binary quadratic form 1 2·rad L q L is integral, primitive, and does not depend on the field L. In particular, any two cubic fields of the same discriminant have isometric integral trace zero forms. Furthermore, ≠ O 0 L , 1 rad L q L ∑ ∼ =2x 2 − 2xy +2y 2 . 1