2.1 A Gram matrix representation of the trace-zero form

In this section, we use a trace-zero basis, coming from a normal integral basis as in the previous section, to find a canonical Gram matrix for the integral trace zero form.

Lemma 2.2 Let K be a number field of degree m, and let $G_K := {\mathbb {Z}} + {\mathfrak {o}}_K^{0}$ . We have

$$\begin{align*}\left|{\mathfrak{o}}_K / G_K \right| = \left| \mathrm{Tr}_{K / {\mathbb{Q}}} ({\mathfrak{o}}_K) / m{\mathbb{Z}} \right|.\end{align*}$$

Corollary If K is a tamely ramified number field of degree m, then

$$ \begin{align*}\left|{\mathfrak{o}}_K / G_K \right| = m.\end{align*} $$

Proof Since K is tamely ramified, by [Reference Narkiewicz12, Corollary 5 to Theorem 4.24],

$$\begin{align*}\mathrm{Tr}_{K/ {\mathbb{Q}}}({\mathfrak{o}}_K) = {\mathbb{Z}}.\end{align*}$$

Using this, the result follows from Lemma 2.2.

Theorem 2.4 Let K and L be tamely ramified cyclic number fields of degree m. Then,

$$\begin{align*}\left<{\mathfrak{o}}_K^{0} , \mathrm{Tr}_{K / {\mathbb{Q}}} ()|_{{\mathfrak{o}}_K}\right> \simeq \left<{\mathfrak{o}}_L^{0} , \mathrm{Tr}_{L / {\mathbb{Q}}} ()|_{{\mathfrak{o}}_L} \right> \ \mbox{ if and only if} \ \ \mathfrak{d}(L) = \mathfrak{d}(K). \end{align*}$$

Proof Suppose that $ \left <{\mathfrak {o}}_K^{0} , \mathrm {Tr}_{K / {\mathbb {Q}}} ()|_{{\mathfrak {o}}_K}\right> \simeq \left <{\mathfrak {o}}_L^{0} , \mathrm {Tr}_{L / {\mathbb {Q}}} ()|_{{\mathfrak {o}}_L} \right> $ . Since the decomposition of $G_{K}={\mathbb {Z}} + {\mathfrak {o}}_K^{0}$ (resp. $G_{L}={\mathbb {Z}} + {\mathfrak {o}}_L^{0}$ ) is an orthogonal decomposition with respect to the trace pairing, we have the following equalities between determinants of the trace in the respective modules:

$$\begin{align*}\mathfrak{d}(G_K) = \mathfrak{d}({\mathfrak{o}}_K^{0}) m = \mathfrak{d}({\mathfrak{o}}_L^{0}) m = \mathfrak{d}(G_L).\end{align*}$$

Since

$$\begin{align*}\mathfrak{d}(G_K) =\left| {\mathfrak{o}}_K / G_K \right|{}^2 \mathfrak{d}({\mathfrak{o}}_K) \ \mbox{and} \ \mathfrak{d}(G_L) =\left| {\mathfrak{o}}_L / G_L \right|{}^2 \mathfrak{d}({\mathfrak{o}}_L), \end{align*}$$

the result follows thanks to Corollary 2.3.

On the other hand, if $\mathfrak {d}(K) = \mathfrak {d}(L)$ , then, by [Reference Mantilla-Soler and Bolaños9, Theorems 4.2 and 4.5], there exist normal integral bases $B:=\{ {e}_{1},\ldots , {e}_{m}\}$ and $B':=\{ {e}^{\prime }_{1},\ldots , {e}^{\prime }_{m}\}$ of ${\mathfrak {o}}_K$ and ${\mathfrak {o}}_L$ , respectively, such that $\left <{\mathfrak {o}}_K , \mathrm {Tr}_{K / {\mathbb {Q}}} ()\right> $ and $ \left <{\mathfrak {o}}_L , \mathrm { Tr}_{L / {\mathbb {Q}}} () \right>$ are isometric via an isometry

$$\begin{align*}\gamma: {\mathfrak{o}}_K \to {\mathfrak{o}}_L\end{align*}$$

such that $\gamma ({e}_{i})={e}^{\prime }_{i}$ for all $1 \leq i \leq m$ . It follows from Lemma 2.1 that such an isometry $\gamma $ restricts to an isometry between the quadratic modules $\left <{\mathfrak {o}}_K^{0} , \mathrm {Tr}_{K / {\mathbb {Q}}} ()|_{{\mathfrak {o}}_K}\right>$ and $ \left <{\mathfrak {o}}_L^{0} , \mathrm {Tr}_{L / {\mathbb {Q}}} ()|_{{\mathfrak {o}}_L} \right>$ .

Definition 2.2 Let m be a positive integer. For every d a positive divisor of m, we let $A_d$ be an $m \times m $ matrix defined as

$$\begin{align*}(A_d)_{i,j} := \left\{ \begin{array}{ll} 1, & \mbox{ if } \frac{m}{d} \big| (i-j), \\ & \\ 0, & \mathrm{otherwise}. \end{array} \right.\end{align*}$$

Suppose that K is a tame cyclic number field of degree m. Theorems 4.2 and 4.5 of [Reference Mantilla-Soler and Bolaños9] state that there exists a normal integral basis B of ${\mathfrak {o}}_K$ such that the Gram matrix of $\left <{\mathfrak {o}}_K , \mathrm {Tr}_{K / {\mathbb {Q}}} ()\right> $ in such basis is equal to

$$\begin{align*}\sum_{d|m} a_d A_d,\end{align*}$$

where $a_d$ are integers given in Lemma 4.3 of [Reference Mantilla-Soler and Bolaños9] that only depend on the discriminant of K. For the reader’s convenience, see also Definition 2.5.

We devote the rest of this section to describe a similar canonical decomposition, i.e., a Gram matrix depending only on the discriminant of the field, for the trace-zero module $\left <{\mathfrak {o}}^{0}_K , \mathrm {Tr}_{K / {\mathbb {Q}}} ()\right> $ .

Let $n \ge 2$ be an integer. Then, we define $\mathbb {B}_n$ as the $n \times n$ matrix

$$\begin{align*}\begin{pmatrix} 1 & 0 & 0& \dots & -1 \\ -1 & 1 & 0 & \dots & 0 \\ 0 & -1 & 1 & \dots & 0 \\ \vdots & 0 & \ddots & \ddots & \vdots \\ 0& 0 & \cdots & -1 & 1 \\ \end{pmatrix}.\end{align*}$$

In other words,

$$\begin{align*}(\mathbb{B}_n)_{i,j}:= \left\{ \begin{array}{rl} 1, & \mbox{ if } i=j, \\ -1, & \mbox{ if } i=j+1 \ \mathrm{or} \ (i,j) =(1,n),\\ 0, & \mbox{ otherwise. } \end{array} \right. \end{align*}$$

Proof Let $i,j$ be integers such that $1\leq i,j < m$ . Then,

$$ \begin{align*} (\mathbb{B}_m^T M \mathbb{B}_m)_{i,j} &= \sum_{k=1}^{m}\sum_{l=1}^{m}\mathbb{B}_{k,i}M_{k,l}\mathbb{B}_{l,j} \\ &=M_{i,j}-M_{i,j+1}-M_{i+1,j}+ M_{i+1, j+1} \\ &= \mathrm{Tr}_{K / {\mathbb{Q}}} (e_{i}e_{j}) - \mathrm{Tr}_{K / {\mathbb{Q}}} (e_{i}e_{j+1})- \mathrm{Tr}_{K / {\mathbb{Q}}} (e_{i+1}e_{j})+ \mathrm{Tr}_{K / {\mathbb{Q}}} (e_{i+1}e_{j+1}) \\ &= \mathrm{Tr}_{K / {\mathbb{Q}}}\left( (e_{i}-e_{i+1})(e_{j}-e_{j+1})\right) \\ &= (M_{0})_{i,j}.\\[-35pt] \end{align*} $$

Corollary Let K be a tame cyclic number field of degree $m \neq 1$ . There exists a basis of ${\mathfrak {o}}_K^{0}$ such that the Gram matrix of $\left <{\mathfrak {o}}_K^{0} , \mathrm {Tr}_{K / {\mathbb {Q}}} ()|_{{\mathfrak {o}}_K^{0}} \right> $ with respect to such basis is equal to the $(m,m)$ minor of

$$\begin{align*}\sum _{d | m} a_d \mathbb{B}_m^T A_d \mathbb{B}_m,\end{align*}$$

where the coefficients $a_d$ are given in Lemmas 3.7, 3.10, and 4.3 of [Reference Mantilla-Soler and Bolaños9] and depend solely on the discriminant of K. See also Definition 2.5.

Proof Since

$$\begin{align*}M = \sum _{d | m} a_d A_d, \end{align*}$$

the corollary is an immediate consequence of Proposition 2.5 and [Reference Mantilla-Soler and Bolaños9, Theorems 4.2 and 4.5].

2.1.1 The matrix $A(\mathfrak {d})$

Given a prime p, we denote by $v_{p}$ the usual p-adic valuation on the rationals.

Definition 2.3 Suppose that $m>0$ and $\mathfrak {d}$ are integers, and let p be a prime. Suppose that $m \neq v_{p}(\mathfrak {d})$ . The p-ramification index of $\mathfrak {d}$ is the rational number defined as

$$\begin{align*}e_{p}(\mathfrak{d}):=\frac{m}{m-v_{p}(\mathfrak{d})}.\end{align*}$$

Remark 2.7 The above definition is motivated by the following fact: If K is a Galois number field of degree m and discriminant $\mathfrak {d}$ , then $e_{p}(\mathfrak {d})$ is the ramification index of p in K for any prime p that is not wildly ramified in K.

Definition 2.4 Let $m>1$ and $\mathfrak {d} \neq 0$ be integers. Let $\mathrm {div}(\mathfrak {d})$ be the set of prime divisors of $\mathfrak {d}$ . Let $1 = d_1 < d_2 < \dots < d_{\tau (m)} = m$ be the set of positive divisors of m. Let

$$\begin{align*}P(m) := \left\{ \left(d_2^{\varepsilon_2}, d_3^{\varepsilon_3}, \dots , d_{\tau(m)}^{\varepsilon_{\tau(m)}} \right) \in {\mathbb{Z}}^{\tau(m)-1} : \varepsilon_i \in \{0,1\} \ \mbox{for all} \ i \right\},\end{align*}$$

and for every $\vec {v} \in P(m)$ , let

$$\begin{align*}\operatorname{\mathrm{lcm}}(\vec{v}) := \operatorname{\mathrm{lcm}}\left[ d_2^{\varepsilon_2}, d_3^{\varepsilon_3}, \dots , d_{\tau(m)}^{\varepsilon_{\tau(m)}}\right],\end{align*}$$

$$\begin{align*}\gcd(\vec{v}) := \gcd\left(d_2^{\varepsilon_2}, d_3^{\varepsilon_3}, \dots , d_{\tau(m)}^{\varepsilon_{\tau(m)}}\right),\end{align*}$$

and for every divisor d of m, let

$$\begin{align*}P_d := \{ \vec{v} \in P(m) : \operatorname{\mathrm{lcm}}(\vec{v}) = d \}. \end{align*}$$

If $d>1$ is a divisor of m, let

$$\begin{align*}\mathbb{P}_d:= \{ p \in \mathrm{div}(\mathfrak{d}) : e_{p}(\mathfrak{d}) = d \}\end{align*}$$

and

$$\begin{align*}w_{d}:=\prod_{p \in \mathbb{P}_d}p \ \mbox{and} \ f_d := \frac{w_d - 1}{d}. \end{align*}$$

Definition 2.5 Let $m>0$ and $\mathfrak {d}$ be integers. For d, a divisor of m not equal to $1$ , we let

$$\begin{align*}a_d := \sum\limits_{\vec{v} \in P_d} \left(\gcd(\vec{v}) \prod\limits_{\varepsilon_i = 0} w_{d_i} \prod\limits_{\varepsilon_j = 1}(- f_{d_j}) \right) \end{align*}$$

and let $a_1 := \prod \limits _{p \in \mathrm {div}(\mathfrak {d})} p $ . Let $A(\mathfrak {d})$ be the $(m-1)\times (m-1)$ matrix defined as

$$\begin{align*}A(\mathfrak{d}):=\sum_{ \substack{d \mid m \\ d \neq m}}a_{d}\widehat{A}_{d},\end{align*}$$

where $\displaystyle \widehat {A}_{d}:= \left (\mathbb {B}_m^T A_d \mathbb {B}_m \right )_{(m,m)}.$

Example 2.8 Let $m=9$ and $\mathfrak {d}= 9,644,443,241,083,841,416,681= 7^{6}\cdot 13^{6}\cdot 19^{8}.$ Here, the set of positive divisors of m is $1<3<9$ and $\mathrm {div}(\mathfrak {d})= \{7,13, 19\}$ . Hence, $P(9)$ is the subset of ${\mathbb {Z}}^{2}$ given by

$$\begin{align*}P(9)=\{ (1,1), (1,9), (3,1) ,(3, 9)\},\end{align*}$$

and thus

$$\begin{align*}P_{3}= \{ (3,1)\} \ \mbox{and} \ P_{9}= \{ (1,9), (3,9)\}.\end{align*}$$

On the other hand, $e_{7}(\mathfrak {d})=3, e_{13}(\mathfrak {d})=3$ , and $e_{19}(\mathfrak {d})=9$ . Thus, $\mathbb {P}_3=\{7,13\}$ and $\mathbb {P}_9=\{19\}.$ Therefore, $w_{3}=91, f_{3}=30$ , and $w_{9}=19, f_{9}=2$ . From this, we calculate that

$$\begin{align*}a_{3}= 1\cdot 19 \cdot (-30) =-570\ \mbox{and} \ a_{9}=1\cdot 91 \cdot (-2) + 3\cdot 1 \cdot (60)=-2; \end{align*}$$

moreover, $a_{1}=7\cdot 13\cdot 19=1,729.$ Since $A_{1}$ is the identity matrix of dimension $9$ , we have that

$$\begin{align*}A(\mathfrak{d})= 1,729 (\mathbb{B}_{9}^{t}\mathbb{B}_{9})_{(9,9)}-570(\mathbb{B}_{9}^{t}A_{3}\mathbb{B}_{9})_{(9,9)}.\end{align*}$$

In other words,

$$\begin{align*}A(\mathfrak{d})= \begin{pmatrix} \ 2,318 & -1,159 & \ 570 & -1,140 & \ 570 & \ 570 & -1,140 & \ 570 \\ -1,159 & \ 2,318 &-1,159 & \ 570 &-1,140 & \ 570 & \ 570 &-1,140 \\ \ 570 & -1,159 & \ 2,318 &-1,159 & \ 570 &-1,140 & \ 570 & 570\\ -1,140 & \ 570 &-1,159 & \ 2,318 &-1,159 & \ 570 &-1,140 & \ 570 \\ \ 570 &-1,140 & \ 570 &-1,159 & \ 2,318 &-1,159 & \ 570 &-1,140 \\ \ 570 & \ 570 &-1,140 & \ 570 &-1,159 & \ 2,318 &-1,159 & \ 570 \\ -1,140 & \ 570 & \ 570 &-1,140 & \ 570 &-1,159 & \ 2,318 &-1,159 \\ \ 570 &-1,140 & \ 570 & \ 570 &-1,140 & \ 570 &-1,159 & \ 2,318 \end{pmatrix}. \end{align*}$$

Example 2.9 Suppose that $m=\ell $ is a prime number. In such case,

$$\begin{align*}A(\mathfrak{d})=a_{1}\widehat{A}_{1}=\mathrm{rad}(\mathfrak{d})\left(\mathbb{B}_{\ell}^T A_1 \mathbb{B}_{\ell} \right)_{(\ell,\ell)}=\mathrm{ rad}(\mathfrak{d})\left(\mathbb{B}_{\ell}^T \mathbb{B}_{\ell} \right)_{(\ell,\ell)}=\mathrm{rad}(\mathfrak{d})\mathcal{A}_{\ell -1}.\end{align*}$$

Now, we are in position to state and prove one of the main theorems in the paper.

Theorem 2.10 Let m be an integer not equal to $1$ , and let K be a tame $ \mathbb {Z}/m \mathbb {Z}$ -number field of discriminant $\mathfrak {d}(K)$ . There exists a ${\mathbb {Z}}$ -basis of ${\mathfrak {o}}_K^{0}$ such that the Gram matrix of

$$\begin{align*}\left<{\mathfrak{o}}_K^{0} , \mathrm{Tr}_{K / {\mathbb{Q}}} ()|_{{\mathfrak{o}}_K}\right>\end{align*}$$

with respect to such basis is equal to

$$\begin{align*}A(\mathfrak{d}(K)).\end{align*}$$

In particular, if L is a tame $ \mathbb {Z}/m \mathbb {Z}$ -number field, then

$$\begin{align*}\left<{\mathfrak{o}}_K^{0} , \mathrm{Tr}_{K / {\mathbb{Q}}} ()|_{{\mathfrak{o}}_K}\right> \simeq \left<{\mathfrak{o}}_L^{0} , \mathrm{Tr}_{L / {\mathbb{Q}}} ()|_{{\mathfrak{o}}_L}\right> \ \mbox{if and only if} \ \mathfrak{d}(K)= \mathfrak{d}(L). \end{align*}$$

Proof For each d divisor of m, let

$$\begin{align*}\widehat{A}_{d}:= \left(\mathbb{B}_m^T A_d \mathbb{B}_m \right)_{(m,m)} \end{align*}$$

be the $(m-1) \times (m-1)$ matrix obtained by erasing the last row and column of $\mathbb {B}_m^T A_d \mathbb {B}_m$ . By definition, all the entries in $A_m$ are equal to $1$ , and hence the columns of $\mathbb {B}_m$ belong to $ \text {Ker}(A_m)$ . In particular, $\mathbb {B}_m^T A_m \mathbb {B}_m =0$ , and thus the result follows from Corollary 2.6 and [Reference Mantilla-Soler and Bolaños9, Theorems 4.2 and 4.5].

Example 2.11 There are four ${\mathbb {Z}}/9{\mathbb {Z}}$ -number fields with discriminant $\mathfrak {d}=7^{6}\cdot 13^{6}\cdot 19^{8}$ (see John Jones’ database [Reference Jones and Roberts5]). Since 3 is unramified in any of those fields, neither of the fields have wild ramification. Such fields are defined, respectively, by the following polynomials:

• • $x^9 - x^8 - 578x^7 - 1,855x^6 + 87,155x^5 + 310,749x^4 - 4,599,958x^3 - 6,198,626x^2 + 102,071,235x - 169,800,379$ .

• • $x^9 - x^8 - 578x^7 - 1,855x^6 + 87,155x^5 + 518,229x^4 - 2,594,318x^3 - 22,409,730x^2 - 36,985,319x - 7,889,903.$

• • $x^9 - x^8 - 578x^7 + 1,603x^6 + 88,884x^5 - 430,992x^4 - 3,668,027x^3 + 27,283,459x^2 - 40,339,579x - 7,447,279.$

• • $x^9 - x^8 - 578x^7 + 1,603x^6 + 88,884x^5 - 119,772x^4 - 5,379,737x^3 - 3,169,418x^2 + 113,584,646x + 256,187,183.$

Let $A(\mathfrak {d})$ be the matrix calculated in Example 2.8. It follows from Theorem 2.10 that for each of those fields, there is a basis of the trace-zero integral module such that all the Gram matrices of the trace form in that basis are equal to the matrix $A(\mathfrak {d})$ . This can be verified computationally, for instance, in MAGMA [Reference Bosma, Cannon and Playoust2], using the code found in [Reference Mantilla-Soler8, Section 3.1].

2.2 Another explicit description

In this subsection, we show that if we apply the results obtained here to the case $m=\ell $ , a prime, we recover the formulas obtained in [Reference Mantilla and Monsurrò6]. We also see how some of the polynomial descriptions of the trace-zero form of [Reference Mantilla and Monsurrò6] are extended to the general case of this paper.

Let n be a positive integer bigger than $1$ . The extended n-dimensional $\mathcal {A'}_n$ lattice is the lattice associated with the matrix

$$\begin{align*}\mathcal{A'}_n := \mathbb{B}_n^T \mathbb{B}_n = \begin{pmatrix} \ 2 & - 1 & \ 0 & \ 0 & \dots & -1 \\ - 1 & \ 2 & -1 & \ 0 & \cdots & \ 0 \\ \ 0 & -1 & \ 2 & -1 & \dots & \ 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\ \ 0 & \ 0 & \ 0 & -1 & \ 2 & -1 \\ -1 & \ 0 & \ 0 & \ 0 & -1 & \ 2 \end{pmatrix}.\end{align*}$$

For any positive integer n, we denote by $1_{n}$ the $n\times n$ matrix with all its entries equal to $1$ and by $\mathrm {I}_{n}$ the identity matrix of dimension n

Lemma 2.12 Let m be an integer bigger than $1$ , and let $d\neq m$ be a positive divisor of m. Then,

$$\begin{align*}\mathbb{B}_m^T A_d \mathbb{B}_m = 1_{d}\otimes \mathcal{A'}_{\frac{m}{d}}.\end{align*}$$

Applying Lemma 2.12 to the situation of Theorem 2.10, we deduce that the Gram matrix $M_0$ of $\left <{\mathfrak {o}}_K^{0} , \mathrm {Tr}_{K / {\mathbb {Q}}} ()|_{{\mathfrak {o}}_K} \right> $ in the given basis has the form

(2.1) $$ \begin{align} M_0 = \left( \sum_{\substack{ d|m \\ d<m}} a_d 1_{d} \otimes \mathcal{A'}_{\frac{m}{d}}. \right)_{(m,m)}. \end{align} $$

If m is equal to a prime $\ell $ , then the above equation is simply

$$\begin{align*}M_0 = (a_1 \mathcal{A'}_{\ell} )_{(\ell, \ell)}.\end{align*}$$

Since $a_1 = \mathrm {rad}(\mathfrak {d}(K))$ and $(\mathcal {A'}_{\ell })_{(\ell ,\ell )}$ is the usual $(\ell -1)$ -dimensional root lattice $\mathcal {A}_{\ell -1}$ , the equation $\displaystyle M_0 = (a_1 \mathcal {A'}_{\ell } )_{(\ell ,\ell )}$ is precisely [Reference Mantilla and Monsurrò6, Theorem 2.9].

An explicit polynomial description of the integral trace-zero form is the following.

Corollary Let K be a tame totally real cyclic number field of degree $m>1$ . Then,

$$\begin{align*}\left<{\mathfrak{o}}_K^{0} , \mathrm{Tr}_{K / {\mathbb{Q}}} ()\Big|_{{\mathfrak{o}}_K} \right> \end{align*}$$

is the lattice associated with the form

$$\begin{align*}\sum_{1 \leq i\leq j \leq m-1} c_{i,j} x_ix_j,\end{align*}$$

where

$$\begin{align*}c_{i,j}:= \sum_{\substack{ d \mid m, d \neq m \\ m \big| (i-j)d }} 2a_d - \sum_ {\substack{ d \mid m, d \neq m \\ m \big| (|i-j| -1)d }} a_d. \end{align*}$$

Proof Note that, for every $d|m$ and $d \neq m$ , we have

$$\begin{align*}1_d \otimes \mathcal{A'}_{\frac{m}{d}} := \left\{ \begin{array}{rl} 2, & \mbox{ if } (i-j) \equiv 0\quad \mod (\frac{m}{d}), \\ -1, & \mbox{ if } |i -j| \equiv 1\quad \mod (\frac{m}{d}), \\ 0, & \mbox{ otherwise. } \end{array} \right. \end{align*}$$

Then, the result follows from formula 2.1.

2.3 Equivalences on the shape of cyclic number fields

Finally, we show that for real cyclic fields, that have no wild ramification, the shape of the field is characterized by the discriminant. More specifically, we show the following theorem.