Calculating Generators of Power Integral Bases in Sextic Fields with a Quadratic Subfield: The General Case

Abstract

In some previous works we gave algorithms for determining generators of power integral basis in sextic fields with a quadratic subfield, under certain restrictions. The purpose of the present paper is to extend those methods to the general case, when the relative integral basis of the sextic field over the quadratic subfield is of general form. This raises several technical difficulties, which we consider here.

1. Introduction

Monogenity and power integral bases is a classical topic in algebraic number theory, which is intensively studied even today. See [1] for classical results and [2,3] for more recent results.

A number field K of degree n with ring of integers ${\mathbb{Z}}_K$ is called monogenic (cf. [2]) if there exists $\xi \in {\mathbb{Z}}_K$ such that $(1,\xi ,\dots ,{\xi }^{n-1})$ is an integral basis, called power integral basis. We call $\xi$ the generator of this power integral basis.

An irreducible polynomial $f\left(x\right)\in \mathbb{Z}\left[x\right]$ is called monogenic if a root $\xi$ of $f\left(x\right)$ generates a power integral basis in $K=\mathbb{Q}\left(\xi \right)$. If $f\left(x\right)$ is monogenic, then K is also monogenic, but the converse is not true.

For $\alpha \in {\mathbb{Z}}_K$ (generating K over $\mathbb{Q}$) the module index

$$I\left(\alpha \right)=({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$$

is called the index of $\alpha$. The element $\alpha$ generates a power integral basis in K if and only if $I\left(\alpha \right)=1$. Searching for elements of ${\mathbb{Z}}_K$, generating power integral bases, leads to a Diophantine equation, called an index form equation (cf. [2]).

There are certain algorithms to determine “all solutions” of these equations, that is, all generators of power integral bases. This “complete resolution” very often requires excessive CPU time. On the other hand, there are some very fast methods for determining generators of power integral bases with “small” coefficients, for example, being <${10}^{100}$ in absolute value, with respect to an integral basis. All our experiences show that generators of power integral bases have very small coefficients in the integral basis; therefore, these “small” solutions cover all solutions with high probability, and certainly all generators that can be used in practice for further calculations. It is also usual to apply such algorithms if we need to solve a large number of equations (cf. [2]).

In this paper we consider sextic number fields. Monogenity in various types of sextic fields has been considered by several authors; as examples, we refer to [4,5,6,7]. Járási [8] gave an algorithm to determine generators of power integral bases in sextic fields with a cubic subfield.

In sextic fields with a quadratic subfield, we developed some efficient methods for calculating “small” solutions of the index form equation; see [9]. For simplicity, in this and some related results, we assumed that the basis of the sextic field is of special type

$$(1,\alpha ,{\alpha }^2,\omega ,\omega \alpha ,\omega {\alpha }^2),$$

where $(1,\omega )$ is an integral basis of the quadratic subfield. This implicitly yields that the sextic field a priori has a relative power integral basis over the quadratic subfield. In the present paper we extend this special case to the general case, when the relative integral basis is of arbitrary form.

This paper was initiated by the recent work of Harrington and Jones [10], where they consider sextic trinomials of the form $f\left(x\right)=x^6+a{x}^3+b$. Considering the sextic fields in [10], generated by a root of such a trinomial, we find that in most cases the root of the polynomial does not generate a power integral basis over the quadratic subfield. In the present paper we intend to give a fast algorithm to calculate “small” solutions of the index form equation in such sextic fields. We shall see that some crucial ingredients of the method are similar to the formerly considered simpler cases; however, several complications occur that make it worthwhile to provide a description in the general case. In other words, we describe how the previous algorithms can be extended to the general case. Also, note that the present method can be easily transformed to a process to calculate “all” solutions.

2. Sextic Fields with a Quadratic Subfield

Let M be a quadratic number field with integral basis $(1,\omega )$, and let $f\left(x\right)=x^3+C_2{x}^2+C_{1}x+C_0\in {\mathbb{Z}}_M\left[x\right]$ be the relative defining polynomial of $\alpha$ over M, with $K=M\left(\alpha \right)$. For sextic fields with a quadratic subfield a crucial step, the reduction, only works for complex quadratic subfields; therefore, we assume that M is complex.

We are going to determine generators of power integral bases of K.

To present our formulas explicitly, we write the relative integral basis of K over M in the form

$$\left(1,{\displaystyle \frac{A\alpha +B}{k}},{\displaystyle \frac{C{\alpha }^2+D\alpha +E}{\ell }}\right),$$

(1)

where $A,B,C,D,E\in {\mathbb{Z}}_M$, $0<k,\ell \in \mathbb{Z}$. Note that if K is (absolutely) monogenic, then it is also relatively monogenic over M, implying that K has a relative integer basis over M.

Using the relative integral basis (1) we can represent any $\gamma \in {\mathbb{Z}}_K$ in the form

$$\gamma =X_0+X_1{\displaystyle \frac{A\alpha +B}{k}}+X_2{\displaystyle \frac{C{\alpha }^2+D\alpha +E}{\ell }},$$

(2)

with unknown $X_i=x_{i1}+\omega x_{i2}\in {\mathbb{Z}}_M(i=0,1,2)$. Our purpose is to construct a fast algorithm to determine all tuples $(x_{02},x_{11},x_{12},x_{21},x_{22})\in {\mathbb{Z}}^5$ with

$$max\left(\right|x_{02}|,|x_{11}|,|x_{12}|,|x_{21}|,|x_{22}\left|\right)<C,$$

(3)

with, for example, $C={10}^{100}$, such that $\gamma$ generates a power integral basis in K (the index of $\gamma$ is independent from $x_{01}$).

We have

$$\gamma =Y_0+Y_1\alpha +Y_2{\alpha }^2,$$

where

$$Y_0=X_0+X_1{\displaystyle \frac{B}{k}}+X_2{\displaystyle \frac{E}{\ell }},Y_1=X_1{\displaystyle \frac{A}{k}}+X_2{\displaystyle \frac{D}{\ell }},Y_2=X_2{\displaystyle \frac{C}{\ell }},$$

(4)

are not necessarily integer elements in M.

Let ${\mu }^{\left(i\right)}$ be the conjugates of any $\mu \in M$, corresponding to ${\omega }^{\left(i\right)},(i=1,2)$. We denote by ${\alpha }^{(i,j)}$ the roots of $f^{\left(i\right)}\left(x\right)=x^3+C_2^{\left(i\right)}{x}^2+C_1^{\left(i\right)}x+C_0^{\left(i\right)},(i=1,2,j=1,2,3)$. The conjugates of any $\mu \in K$ corresponding to ${\alpha }^{(i,j)}$ will also be denoted by ${\mu }^{(i,j)}$.

For $i=1,2,1\le j_1,j_2\le 3,j_1\ne j_2$ we have

$${\displaystyle \frac{{\gamma }^{(i,j_1)}-{\gamma }^{(i,j_2)}}{{\alpha }^{(i,j_1)}-{\alpha }^{(i,j_2)}}}=Y_1+({\alpha }^{(i,j_1)}+{\alpha }^{(i,j_2)})Y_2=Y_1-{\delta }^{(i,j_3)}{Y}_2,$$

where $-{\delta }^{(i,j_3)}={\alpha }^{(i,j_1)}+{\alpha }^{(i,j_2)}=-C_2^{\left(i\right)}-{\alpha }^{(i,j_3)}$, $C_2\in {\mathbb{Z}}_M$ being the quadratic coefficient of the relative defining polynomial $f\left(x\right)$ of $\alpha$ over M, and $\left\{j_3\right\}=\{1,2,3\}\setminus \{j_1,j_2\}$.

By the representation (1) of the relative integral basis of K over M, for the relative discriminant $D_{K/M}$ we have

$$N_{M/\mathbb{Q}}\left(D_{K/M}\right)={\displaystyle \frac{N_{M/\mathbb{Q}}\left(d_{K/M}\left(\alpha \right)\right)}{{(k\ell )}^2}}={\displaystyle \frac{1}{{(k\ell )}^2}}\prod _{i=1}^2\prod _{1\le j_1<j_2\le 3}{\left({\alpha }^{(i,j_1)}-{\alpha }^{(i,j_2)}\right)}^2.$$

Therefore, we obtain

$$I_{K/M}\left(\gamma \right)={\displaystyle \frac{1}{\sqrt{|N_{M/\mathbb{Q}}\left(D_{K/M}\right)|}}}\prod _{i=1}^2\prod _{1\le j_1<j_2\le 3}\left|{\gamma }^{(i,j_1)}-{\gamma }^{(i,j_2)}\right|$$

$$=(k\ell )\prod _{i=1}^2\prod _{j=1}^3\left|Y_1^{\left(i\right)}-{\delta }^{(i,j)}{Y}_2^{\left(i\right)}\right|=(k\ell )\left|N_{M/\mathbb{Q}}\left(N_{K/M}(Y_1-\delta Y_2)\right)\right|.$$

(5)

As it is known (see [2], Chapter 1, Theorem 1.6), if $I\left(\gamma \right)=1$, then both

$$I_{K/M}\left(\gamma \right)=1,\mathrm{and}J\left(\gamma \right)=1,$$

where

$$J\left(\gamma \right)={\displaystyle \frac{1}{{\left(\sqrt{|D_M|}\right)}^3}}\prod _{j_1=1}^3\prod _{j_2=1}^3\left|{\gamma }^{(1,j_1)}-{\gamma }^{(2,j_2)}\right|.$$

(6)

3. Elementary Estimates

By $I_{K/M}\left(\gamma \right)=1$, (5) implies

$$N_{M/\mathbb{Q}}\left(N_{K/M}(Z_1-\delta Z_2)\right)=\pm {(k\ell )}^5,$$

(7)

where

$$Z_1=(k\ell )Y_1,Z_2=(k\ell )Y_2\in {\mathbb{Z}}_M.$$

(8)

Using an algebraic number theory package like Magma or Kash, we can determine a complete set of non-associated elements $\mu \in {\mathbb{Z}}_M$ of norm $\pm {(k\ell )}^5$. Let $\epsilon$ be one of the finitely many units in M. We confer

$$N_{K/M}(Z_1-\delta Z_2)=\epsilon \mu ,$$

(9)

with certain possible values of $\mu ,\epsilon$. In complex quadratic fields the conjugated elements have equal absolute values; therefore, (9) implies

$$\prod _{j=1}^3\left|Z_1^{\left(1\right)}-{\delta }^{(1,j)}{Z}_2^{\left(1\right)}\right|={|k\ell |}^{5/2}.$$

(10)

Denote by $j_0$ the conjugate with

$$\left|Z_1^{\left(1\right)}-{\delta }^{(1,j_0)}{Z}_2^{\left(1\right)}\right|=\underset{1\le j\le 3}{min}\left|Z_1^{\left(1\right)}-{\delta }^{(1,j)}{Z}_2^{\left(1\right)}\right|.$$

Then

$$\left|Z_1^{\left(1\right)}-{\delta }^{(1,j_0)}{Z}_2^{\left(1\right)}\right|\le c_1$$

(11)

with $c_1={|k\ell |}^{5/6}$, and for $j\ne j_0$ we have

$$\left|Z_1^{\left(1\right)}-{\delta }^{(1,j)}{Z}_2^{\left(1\right)}\right|\ge \left|{\delta }^{(1,j)}-{\delta }^{(1,j_0)}\right||Z_2^{\left(1\right)}|-c_1\ge c_2\left|Z_2^{\left(1\right)}\right|,$$

(12)

with $c_2=0.9·{min}_{j\ne j_0}|{\delta }^{(1,j)}-{\delta }^{(1,j_0)}|$, if $|Z_2^{\left(1\right)}|>10c_1/{min}_{j\ne j_0}|{\delta }^{(1,j)}-{\delta }^{(1,j_0)}|$. Small coordinates of $Z_2$, not satisfying this inequality, are tested separately.

We set $Z_1=z_{11}+\omega z_{12},Z_2=z_{21}+\omega z_{22}$ with $z_{11},z_{12},z_{21},z_{22}\in \mathbb{Z}$ and let

$$A=max\left(\right|z_{11}|,|z_{12}|,|z_{21}|,|z_{22}\left|\right).$$

Note that to find all suitable $(x_{02},x_{11},x_{12},x_{21},x_{22})\in {\mathbb{Z}}^5$ satisfying (3), in view of (4) and (8), we have to consider all $(z_{11},z_{12},z_{21},z_{22})$ with

$$A\le 6(k\ell )C(1+\overline{\left|\omega \right|})max\left(1,{\displaystyle \frac{B}{k}},{\displaystyle \frac{E}{\ell }},{\displaystyle \frac{A}{k}},{\displaystyle \frac{D}{\ell }},{\displaystyle \frac{C}{\ell }}\right).$$

(13)

(11) implies

$$max\left(\right|z_{11}|,|z_{12}\left|\right)\le 2|Z_1^{\left(1\right)}|\le 2(c_1+\overline{\left|\delta \right|}|Z_2^{\left(1\right)}\left|\right)\le 2(0.1+\overline{\left|\delta \right|})\left|Z_2^{\left(1\right)}\right|,$$

if $0.1|Z_2^{\left(1\right)}|>c_1$ (small coordinates of $Z_2$ are tested separately). Here $\overline{\left|\delta \right|}$ is the size $\delta$ (the maximum absolute values of its conjugates). Similarly,$max\left(\right|z_{21}|,|z_{22}|)\le 2|Z_2^{\left(1\right)}|$; therefore,

$$A\le 2(0.1+\overline{\left|\delta \right|})\left|Z_2^{\left(1\right)}\right|.$$

By (10) and (12), we obtain

$$\left|Z_1^{\left(1\right)}-{\delta }^{(1,j_0)}{Z}_2^{\left(1\right)}\right|\le {\displaystyle \frac{{(k\ell )}^{5/2}}{c_2^2}}{\left|Z_2^{\left(1\right)}\right|}^{-2}\le c_3{A}^{-2},$$

with

$$c_3={\displaystyle \frac{{(k\ell )}^{5/2}}{4c_2^2{(0.1+\overline{\left|\delta \right|})}^2}},$$

whence

$$\left|z_{11}+{\omega }^{\left(1\right)}{z}_{12}-{\delta }^{(1,j_0)}{z}_{21}-{\delta }^{(1,j_0)}{\omega }^{\left(1\right)}{z}_{22}\right|\le c_3{A}^{-2}.$$

(14)

4. Reduction

The reduction procedure is based on inequality (14). The bound in (13) is reduced in several consecutive steps. We start with $A_0=A_{max}$, $A_{max}$ being the bound in (13). We assign a suitable large constant H and perform the following reduction step, which produces a new bound for A. We set this new bound in place of $A_0$ and continue the reduction until the reduced bound is smaller than the original one.

Consider the lattice generated by the columns of the matrix

$$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ H& H\Re \left({\omega }^{\left(1\right)}\right)& \Re (-{\delta }^{(1,j_0)})& H\Re (-{\delta }^{(1,j_0)}{\omega }^{\left(1\right)})\\ 0& H\Im \left({\omega }^{\left(1\right)}\right)& H\Im (-{\delta }^{(1,j_0)})& H\Im (-{\delta }^{(1,j_0)}{\omega }^{\left(1\right)})\end{array}\right).$$

Denote by $b_1$ the first vector of the LLL reduced basis of this lattice. According to Lemma 5.3 of [2] (which statement is based on (14)), if $A\le A_0$ and H is large enough to have

$$|b_1|\ge \sqrt{40}·A_0,$$

(15)

then

$$A\le {\left({\displaystyle \frac{c_3·H}{A_0}}\right)}^{1/2}.$$

For a certain $A_0$ the suitable H is of magnitude $A_0^2$. A typical sequence of reduced bounds staring from $A_0={10}^{100}$ was the following:

A	${10}^{100}$	$1.5805\times {10}^{51}$	$6.2833\times {10}^{26}$	$1.2528\times {10}^{15}$
H	${10}^{202}$	$2.4979\times {10}^{104}$	$3.9481\times {10}^{55}$	$1.5695\times {10}^{32}$
$\mathrm{new}A$	$1.5805\times {10}^{51}$	$6.2833\times {10}^{26}$	$1.2528\times {10}^{15}$	$5.5942\times {10}^8$

A	$5.5942\times {10}^8$	$3.7382\times {10}^5$	8663	1553
H	$3.1295\times {10}^{19}$	$1.3974\times {10}^{13}$	$9.3381\times {10}^9$	$2.4139\times {10}^8$
$\mathrm{new}A$	$3.7382\times {10}^5$	8663	1553	622

A	622	394	313	280
H	$3.8810\times {10}^7$	$1.5562\times {10}^7$	$9.8543\times {10}^6$	$7.8416\times {10}^6$
$\mathrm{new}A$	394	313	280	264

If, in a certain step, H was not sufficiently large, we replaced it with $10H$.

The reduction procedure was executed with 250 digits accuracy and took only a few seconds. It has to be performed for all possible values of $j_0$, and the final reduced bound for A is the maximum of the reduced bounds obtained for $j_0=1,2,3$.

5. Enumeration, Test

The reduced bound obtained in the previous section gives an upper bound among others for $|z_{11}|,|z_{12}|$, and hence we can enumerate all possible $Z_1$. Further, for all possible $\epsilon ,\mu$, Equation (9) gives a cubic equation for $Z_2\in {\mathbb{Z}}_M$. Testing the roots of this cubic equation in $Z_2$ we can determine all $Z_2\in {\mathbb{Z}}_M$ corresponding to $Z_1$.

From (8) and (4) we can determine $Y_1,Y_2$ and then the coordinates $x_{11},x_{12}$ and $x_{21},x_{22}$ of $X_1,X_2$, corresponding to $Z_1,Z_2$. Finally, we use (6) to determine $x_{02}$ in the representation (2) of $\gamma$ (the index of $\gamma$ is independent of $x_{01}$). Substituting the possible tuples $x_{11},x_{12},x_{21},x_{22}$ into $J\left(\gamma \right)$, we obtain a polynomial $F\left(x\right)=a_9{x}^9+\dots +a_{1}x+a_0$ in $x_{02}$ of degree 9, such that

$$\left|F\right(x_{02}\left)\right|=1.$$

For the roots $x_{02}$ of absolute value >1 we have

$$|x_{02}|\le {\displaystyle \frac{|a_8|+\dots |a_1|+|a_0|+1}{|a_9|}}.$$

We test the possible integer values of $x_{02}$ and obtain the solutions. Note that $x_{11},x_{12}$ and $x_{21},x_{22}$ are usually small values, and therefore the bound for $|x_{02}|$ is also reasonably small.

6. Example

We developed and tested our method by taking the trinomial

$$f\left(x\right)=x^6+3x^3+9$$

with Galois group $C_3\times S_3$ from the paper [10] of Harrington and Jones. These trinomials have several interesting features, which may be the topic of a separate paper. This polynomial is not monogenic, but the number field K generated by a root $\alpha$ of it is monogenic.

The quadratic subfield of K is determined by the equation $x^2+3x+9=0$. Its root is $\beta =(-3+3i\sqrt{3})/2$, and therefore $M=\mathbb{Q}\left(i\sqrt{3}\right)$. We set $\omega =(1+i\sqrt{3})/2$, then $\beta =3\omega -3$ and $\alpha =\sqrt[3]{\beta }$. A relative integers basis of $K=\mathbb{Q}\left(\alpha \right)$ over $\mathbb{Q}$ is given by

$$\left(1,\alpha ,{\displaystyle \frac{{\alpha }^2(1+\omega )}{3}}\right).$$

We have

$$\gamma =X_0+X_1\alpha +X_2{\displaystyle \frac{{\alpha }^2(1+\omega )}{3}}=Y_0+Y_1\alpha +Y_2{\alpha }^2,$$

with

$$Y_0=X_0,Y_1=X_1,Y_2=X_2{\displaystyle \frac{1+\omega }{3}}.$$

Moreover, $\delta =\alpha ,k=1,\ell =3$,

$$3·N_{M/\mathbb{Q}}\left(N_{K/M}(Y_1-\delta Y_2)\right)=\pm 1,$$

and

$$N_{M/\mathbb{Q}}\left(N_{K/M}(Z_1-\delta Z_2)\right)=\pm 3^5$$

with $Z_1=3Y_1,Z_2=3Y_2$, whence

$$|Z_1^3-\beta Z_2^3|=\left|N_{K/M}(Z_1-\delta Z_2)\right|=3^{5/2}.$$

Taking $C={10}^{100}$ we have to reduce A from $24C$. The reduction procedure gives a bound 250 for the absolute values of the coordinates $z_{11},z_{12},z_{21},z_{22}$. In our case $Y_1$ is also integer, and hence $z_{11},z_{12}$ are divisible by 3, which considerably reduces the number of possible pairs $z_{11},z_{12}$.

We used Magma to calculate the elements in ${\mathbb{Z}}_M$ of norm $\pm 3^5$. It turned out that up to associates the only such element is $9-18\omega$. We set $\epsilon =\pm 1,{\displaystyle \frac{\pm 1\pm i\sqrt{3}}{2}}$ and used

$$Z_1^3-\beta Z_2^3=\epsilon \mu$$

to determine the possible values of $Z_1$, corresponding to $Z_2$. Finally, we calculated $Y_1,Y_2$, then $X_1,X_2$ and substituted the coordinates of $X_1,X_2$ into $J\left(\gamma \right)=1$ to determine the appropriate values of $x_{02}$. We obtained that up to sign the solutions are as follows:

$$\begin{array}{|ccccc|}\hline \hfill x_{02}& \hfill x_{11}& \hfill x_{12}& \hfill x_{21}& \hfill x_{22}\\ \hfill -1& \hfill 0& \hfill 1& \hfill 0& \hfill -1\\ \hfill 1& \hfill 1& \hfill 0& \hfill 1& \hfill -1\\ \hfill 0& \hfill 0& \hfill 0& \hfill -1& \hfill 1\\ \hfill 0& \hfill 0& \hfill 0& \hfill 0& \hfill -1\\ \hfill 0& \hfill 0& \hfill 0& \hfill 1& \hfill 0\\ \hfill -1& \hfill 1& \hfill -1& \hfill 1& \hfill 0\\ \hline\end{array}$$

That is, up to sign and translation by $x_{01}$ all generators of power integral bases of K are given by

$$\gamma =\omega x_{02}+(x_{11}+\omega x_{12})\alpha +(x_{21}+\omega x_{22}){\displaystyle \frac{{\alpha }^2(1+\omega )}{3}},$$

with the above-listed tuples $(x_{02},x_{11},x_{12,}{x}_{21},x_{22})$.

All calculations were performed in Maple and took just a few seconds.

7. Conclusions

There are several results on monogenity or non-monogenity of wide classes of number fields even for high degrees. However, to determine explicitly all generators of power integral bases in a specific number field turns out to be a much more complicated problem. We try to extend the methods of existing algorithms to some more general cases. In this paper, it is performed for sextic fields with a quadratic subfield, without the previous assumption of having a special type of relative integral basis.

We emphasize that, for our algorithm, it is crucial to choose the involved constants $c_1,c_2,c_3$ as accurately as possible. Also, an optimal choice of the H values in the reduction part produces a better reduced bound.

Note also that the same method can be applied for determining relative power integral bases in relative cubic and relative quartic extensions, in case the initial relative integral basis is not a power basis.