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

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, [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 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" requires very often too long CPU time. On the other hand there are some very fast methods for determining generators of power integral bases with "small" coefficients, say, 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, certainly all generators that can be used in practice for further calculations. It is usual to apply such algorithms also if we need to solve a large number of equation (cf. [2]).

In sextic fields with a quadratic subfield we developed some efficient methods for calculating "small" solutions of the index form equation, see [4,5]. For simplicity, in these 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 apriori 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 [6], where they consider sextic trinomials of the form $f\left(x\right)=x^6+a{x}^3+b$. Considering the sextic fields in [6], 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 worthy 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

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

(1)

where $A,B,C,D,E\in {\mathbb{Z}}_M$, $0<k,\mathcal{l}\in \mathbb{Z}$. Note that if K is (absolute) monogenic, then it is also relative 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

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

(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 say, $C={10}^{100}$, 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

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

(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

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

Therefore we obtain

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

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

(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

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

(7)

where

$$Z_1=\left(k\mathcal{l}\right)Y_1,Z_2=\left(k\mathcal{l}\right)Y_2\in {\mathbb{Z}}_M.$$

(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 $\pm {\left(k\mathcal{l}\right)}^5$. 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

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

(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 $c_1={\left|k\mathcal{l}\right|}^{5/6}$, 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), (8) we have to consider all $(z_{11},z_{12},z_{21},z_{22})$ with

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

(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). Similary $max\left(\right|z_{21}|,|z_{22}\left|\right)\le 2|Z_2^{\left(1\right)}|$, therefore

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

By (10) and (12) we obtain

$$\left|Z_1^{\left(1\right)}-{\delta }^{(1,j_0)}{Z}_2^{\left(1\right)}\right|\le {\displaystyle \frac{{\left(k\mathcal{l}\right)}^{5/2}}{c_2^2}}{\left|Z_2^{\left(1\right)}\right|}^{-2}\le c_4{A}^{-2},$$

with

$$c_4={\displaystyle \frac{{\left(k\mathcal{l}\right)}^{5/2}}{4c_2^2{(0.1+\overline{\left|\delta \right|})}^2}},$$

whence

$$\left|z_{11}+{\omega }^{\left(1\right)}{z}_{12}-{\delta }^{(1,j_0)}{z}_{21}-{\delta }^{(1,j_0)}{\omega }^{\left(1\right)}{z}_{22}\right|\le c_4{A}^{-2}.$$

(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, 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 bases on (14)), if $A\le A_0$ and H is large enough to have

$$|b_1|\ge \sqrt{40}·A_0,$$

(15)

then

$$A\le {\left({\displaystyle \frac{c_4·H}{A_0}}\right)}^{1/2}.$$

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:

$$\begin{array}{ccccc}A& {10}^{100}& 1.5805·{10}^{51}& 6.2833·{10}^{26}& 1.2528·{10}^{15}\\ H& {10}^{202}& 2.4979·{10}^{104}& 3.9481·{10}^{55}& 1.5695·{10}^{32}\\ \mathrm{new}A& 1.5805·{10}^{51}& 6.2833·{10}^{26}& 1.2528·{10}^{15}& 5.5942·{10}^8\\ \end{array}$$

$$\begin{array}{ccccc}A& 5.5942·{10}^8& 3.7382·{10}^5& 8663& 1553\\ H& 3.1295·{10}^{19}& 1.3974·{10}^{13}& 9.3381·{10}^9& 2.4139·{10}^8\\ \mathrm{new}A& 3.7382·{10}^5& 8663& 1553& 622\\ \end{array}$$

$$\begin{array}{ccccc}A& 622& 394& 313& 280\\ H& 3.8810·{10}^7& 1.5562·{10}^7& 9.8543·{10}^6& 7.8416·{10}^6\\ \mathrm{new}A& 394& 313& 280& 264\\ \end{array}$$

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

The reduction procedure was executed with 250 digits accuracy and took only a few seconds. It has to be performed for each 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 $|z_{11}|,|z_{12}|$, 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, 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 [6] 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 $x^2+3x+9=0$. It’s root is $\beta =(-3+3i\sqrt{3})/2$, 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, $\delta =\alpha ,k=1,\mathcal{l}=3$,

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

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

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

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, 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 calculate $Y_1,Y_2$, 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:

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

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 done for sextic fields with a quadratic subfield, without the previous assumption of having a special type of relative integral basis.