1 Introduction

There is an extensive literature of monogenity of number fields and power integral bases, see [4, 12]. A number field K of degree n is monogenic if its ring of integers \({\mathbb {Z}}_K\) is a simple ring extension of \({\mathbb {Z}}\), that is there exists \(\alpha \in {\mathbb {Z}}_K\) with \({\mathbb {Z}}_K={\mathbb {Z}}[\alpha ]\). In this case \((1,\alpha ,\ldots ,\alpha ^{n-1})\) is an integral basis of K, called power integral basis. We also call \(\alpha \) the generator of this power integral basis. The algebraic integer \(\alpha \) generates a power integral basis if and only if its index

$$\begin{aligned} I(\alpha )=\sqrt{\left| \frac{D(\alpha )}{D_K}\right| } \end{aligned}$$

is equal to 1, where \(D(\alpha )\) is the discriminant of \(\alpha \).

The calculation of generators of power integral bases can be reduced to the resolution of certain diophantine equations, called index form equations, cf. [4].

There exist general algorithms for solving index form equations in cubic, quartic, quintic, sextic fields, however the general algorithms for quintic and sextic fields are already quite tedious, see [4]. Therefore it is worthy to develop efficient methods for the resolution of special types of higher degree number fields, cf. [4, 8].

In some cases we considered monogenity in composites of fields, see [3, 9]. In these cases the index form factorizes that makes the resolution of the index form equation easier.

On the other hand, considering totally real relative Thue equations over imaginary quadratic fields, it has turned out, that the relative Thue equations can be reduced to absolute Thue equations (over \({\mathbb {Z}}\)), cf. [6]. A similar idea was used in [2, 5].

In the present paper we study composites \(K=LM\) of a totally real fields L and imaginary quadratic fields M. We show that in this case the resolution of the index form equation in K can be reduced to solving the index form equation in L. If L is of degree n, then the index form equation of L is of degree \(n(n-1)/2\) in \(n-1\) variables, but the index form equation in K is of degree \(2n(2n-1)/2\) in \(2n-1\) variables. Therefore our statement simplifies a lot the calculation of generators of power integral bases in fields of type K of degree 2n.

Surprisingly the proofs of our statements are quite simple. However, they provide a powerful tool. Our Theorem 2 immediately implies some results of [3] (see our remarks after the proof of Theorem 2).

The strengh and usefulness of our results is aslo demonstrated in the application given in Sect. 4, where we consider composites of the so called simplest quartic fields and imaginary quadratic fields.

2 Composites of totally real fields and imaginary quadratic fields

Let \(f(x)\in {\mathbb {Z}}[x]\) be a monic irreducible polynomial of degree n having all roots \(\xi =\xi ^{(1)},\ldots ,\xi ^{(n)}\) in \({\mathbb {R}}\). The field \(L={\mathbb {Q}}(\xi )\) is then totally real. Assume that L has integral basis \((\ell _1=1,\ell _2,\ldots ,\ell _n)\) and discriminant \(D_L\). We shall denote by \(\gamma ^{(j)}\) the conjugates of any \(\gamma \in L \; (j=1,\ldots , n)\). The index form corresponding to the basis \((\ell _1=1,\ell _2,\ldots ,\ell _n)\) is defined by

$$\begin{aligned}{} & {} I_L(x_2,\ldots ,x_n)\\ {}{} & {} \quad =\frac{1}{\sqrt{|D_L|}} \prod _{1\le j_1<j_2\le n} \left( x_2\left( \ell _2^{(j_1)}-\ell _2^{(j_2)}\right) +\ldots +x_n\left( \ell _n^{(j_1)}-\ell _n^{(j_2)}\right) \right) . \end{aligned}$$

As it is known, \(I_L(x_2,\ldots ,x_n)\in {\mathbb {Z}}[x_2\ldots ,x_n]\) and for any \(x_1,x_2,\ldots ,x_n\in {\mathbb {Z}}\) the algebraic integer element

$$\begin{aligned} \alpha =x_1+x_2\ell _2+\ldots x_n\ell _n \end{aligned}$$

generates a power integral basis \((1,\alpha ,\ldots ,\alpha ^{n-1})\) if and only if

$$\begin{aligned} I_L(x_2,\ldots ,x_n)=\pm 1. \end{aligned}$$

Let \(0<d\in {\mathbb {Z}}\) be a square-free integer, set

$$\begin{aligned} \omega = \left\{ \begin{array}{lr} i\sqrt{d}\;\;&{} \textrm{if}\; d\equiv 1,2\; (\bmod \; {4}),\\ \displaystyle {\frac{1+i\sqrt{d}}{2}}\;\; &{}\textrm{if} \; d \equiv 3\; (\bmod \; {4}). \end{array}\right. \end{aligned}$$

The conjugates of \(\omega \) are \(\omega ^{(1)}=\omega ,\omega ^{(2)}=\overline{\omega }\) (the complex conjugate of \(\omega \)). Let \(M={\mathbb {Q}}(i\sqrt{d})\) with discriminant \(D_M\).

Our purpose is to investigate the composite field \(K=L\cdot M\) of degree \(2\ell \) and discriminant \(D_K\). We assume \((D_L,D_M)=1\). Then \((1,\ell _2,\ldots ,\ell _n,\omega ,\omega \ell _2,\ldots ,\omega \ell _n)\) is an integral basis of K and \(D_K=D_M^n\cdot D_L^2\) (cf. [12]).

In the ring of integers \({\mathbb {Z}}_K\) of K any element \(\alpha \) can be represented as

$$\begin{aligned} \alpha =x_1+x_2\ell _2+\ldots x_n\ell _n+y_1\omega +y_2\omega \ell _2+\ldots +y_n\omega \ell _n =X_1+X_2\ell _2+\ldots +X_n\ell _n, \end{aligned}$$
(1)

with \(x_1,\ldots ,x_n,y_1,\ldots ,y_n\in {\mathbb {Z}}\) and \(X_j=x_j+\omega y_j\) are elements in the ring \({\mathbb {Z}}_M\) of algebraic integers of M.

We shall use the following consequence of Theorem 1 of [3]:

Lemma 1

If \(\alpha \) of (1) generates a power integral basis in K then

$$\begin{aligned} N_{M/Q}(I_L(X_2,\ldots ,X_n))=\pm 1 \end{aligned}$$
(2)

and

$$\begin{aligned} N_{L/Q}(y_1+y_2\ell _2+\ldots +y_n\ell _n)=\pm 1. \end{aligned}$$
(3)

The conjugates of \(\alpha \) of (1) are obtained obviously as

$$\begin{aligned} \alpha ^{(j,k)}=x_1+x_2\ell _2^{(j)}+\ldots x_n\ell _n^{(j)} +y_1\omega ^{(k)}+y_2\omega ^{(k)}\ell _2^{(j)}+\ldots +y_n \omega ^{(k)}\ell _n^{(j)}, \end{aligned}$$

for \(j=1,\ldots ,n,k=1,2\). Note that the index form corresponding to the basis

\((1,\ell _2,\ldots ,\ell _n,\omega ,\omega \ell _2,\ldots ,\omega \ell _n)\) has three factors, two of them are the polynomials on the left hand sides of equations (2), (3). The third factor is

$$\begin{aligned} F(x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n)= \prod _{1\le j_1,j_2\le n,j_1\ne j_2} (\alpha ^{(j_1,1)}-\alpha ^{(j_2,2)}), \end{aligned}$$
(4)

(cf. [3, 4]). This is also a polynomial with coefficients in \({\mathbb {Z}}\). Equations (2), (3) together with

$$\begin{aligned} F(x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n)=\pm 1 \end{aligned}$$
(5)

are already equivalent to \(\alpha \) generating a power integral basis.

In our main result we reduce the resolution of the relative index form equation (2) to the resolution of absolute equations, i.e. absolute inequalities. This makes the resolution of (2) much easier.

Theorem 2

Assume \(\alpha \) of (1) generates a power integral basis in K.

If \(d\equiv 1,2\; (\bmod \; {4})\), then

$$\begin{aligned} |I_L(x_2,\ldots ,x_n)|\le 1 \end{aligned}$$
(6)

and

$$\begin{aligned} |I_L(y_2,\ldots ,y_n)|\le \frac{1}{(\sqrt{d})^{n(n-1)/2}}. \end{aligned}$$
(7)

If \(d\equiv 3\; (\bmod \; {4})\), then

$$\begin{aligned} |I_L(2x_2+y_2,\ldots ,2x_n+y_n)|\le 2^{n(n-1)/2} \end{aligned}$$
(8)

and

$$\begin{aligned} |I_L(y_2,\ldots ,y_n)|\le \left( \frac{2}{\sqrt{d}}\right) ^{n(n-1)/2}. \end{aligned}$$
(9)

Proof of Theorem 2

According to the arguments in the proof of Theorem 1 of [3] we have

$$\begin{aligned} |N_{M/Q}(I_L(X_2,\ldots ,X_n))|=\prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}}\prod _{1\le j_1<j_2\le n} \left| \alpha ^{(j_1,k)}-\alpha ^{(j_2,k)}\right| \right) . \end{aligned}$$
(10)

If \(d\equiv 1,2\; (\bmod \; {4})\), then

$$\begin{aligned} \textrm{Re}(\alpha ^{(j_1,k)}-\alpha ^{(j_2,k)}) =x_2(\ell _2^{(j_1)}-\ell _2^{(j_2)})+\ldots +x_n(\ell _n^{(j_1)}-\ell _n^{(j_2)}) \end{aligned}$$

and

$$\begin{aligned} \textrm{Im}(\alpha ^{(j_1,k)}-\alpha ^{(j_2,k)}) =\sqrt{d}\cdot \left( y_2(\ell _2^{(j_1)}-\ell _2^{(j_2)})+\ldots +y_n(\ell _n^{(j_1)}-\ell _n^{(j_2)})\right) . \end{aligned}$$

We have

$$\begin{aligned}{} & {} |I_L(x_2,\ldots ,x_n)|^2 =|N_{M/Q}(I_L(x_2,\ldots ,x_n))| \\{} & {} \quad =\prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}} \prod _{1\le j_1<j_2\le n} \left| x_2(\ell _2^{(j_1)}-\ell _2^{(j_2)})+\ldots + x_n(\ell _n^{(j_1)}-\ell _n^{(j_2)})\right| \right) \\{} & {} \quad =\prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}}\prod _{1\le j_1<j_2\le n} \left| \textrm{Re}(\alpha ^{(j_1,k)}-\alpha ^{(j_2,k)})\right| \right) \\{} & {} \quad \le \prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}}\prod _{1\le j_1<j_2\le n} \left| \alpha ^{(j_1,k)}-\alpha ^{(j_2,k)}\right| \right) \\{} & {} \quad = |N_{M/Q}(I_L(X_2,\ldots ,X_n))|, \end{aligned}$$

whence by (2) we obtain (6). Similarly,

$$\begin{aligned}{} & {} \left( \sqrt{d}\right) ^{n(n-1)}|I_L(y_2,\ldots ,y_n)|^2 = \left( \sqrt{d}\right) ^{n(n-1)}|N_{M/Q}(I_L(y_2,\ldots ,y_n))| \\{} & {} \quad =\prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}} \prod _{1\le j_1<j_2\le n} \sqrt{d}\left| y_2(\ell _2^{(j_1)}-\ell _2^{(j_2)})+\ldots + y_n(\ell _n^{(j_1)}-\ell _n^{(j_2)}) \right| \right) \\{} & {} \quad =\prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}}\prod _{1\le j_1<j_2\le n} \left| \textrm{Im}(\alpha ^{(j_1,k)}-\alpha ^{(j_2,k)})\right| \right) \\{} & {} \quad \le \prod _{k=1}^2 \left( \frac{1}{\sqrt{|D_L|}}\prod _{1\le j_1<j_2\le n} \left| \alpha ^{(j_1,k)}-\alpha ^{(j_2,k)}\right| \right) \\{} & {} \quad = |N_{M/Q}(I_L(X_2,\ldots ,X_n))|, \end{aligned}$$

whence by (2) we obtain (7).

If \(d\equiv 3\; (\bmod \; {4})\), then

$$\begin{aligned} \textrm{Re}(\alpha ^{(j_1,k)}{-}\alpha ^{(j_2,k)}) {=}\frac{2x_2+y_2}{2}\cdot (\ell _2^{(j_1)}{-}\ell _2^{(j_2)})+\ldots +\frac{2x_n+y_n}{2}\cdot (\ell _n^{(j_1)}{-}\ell _n^{(j_2)}) \end{aligned}$$

and

$$\begin{aligned} \textrm{Im}(\alpha ^{(j_1,k)}-\alpha ^{(j_2,k)})=\sqrt{d} \cdot \left( \frac{y_2}{2}\cdot (\ell _2^{(j_1)}-\ell _2^{(j_2)}) +\ldots + \frac{y_n}{2}\cdot (\ell _n^{(j_1)}-\ell _n^{(j_2)})\right) . \end{aligned}$$

The above arguments lead us to

$$\begin{aligned} \left| I_L\left( \frac{2x_2+y_2}{2},\ldots ,\frac{2x_n+y_n}{2}\right) \right| \le 1 \end{aligned}$$

and

$$\begin{aligned} \left| I_L\left( \frac{y_2}{2},\ldots ,\frac{y_n}{2}\right) \right| \le \frac{1}{(\sqrt{d})^{n(n-1)/2}}, \end{aligned}$$

whence we obtain (8) and (9), accordingly. \(\square \)

Remark

Our Theorem 2 immediately implies the result of [3] on monogenity of composites of totally real cyclic fields of prime degree and imaginary quadratic fields, as well as on monogenity of composites of Lehmer’s quintics and imaginary quadratic fields. In the next section we show that our result implies that the above statements of [3] hold for any composite of a totally real number field and an imaginary quadratic field.

3 Applying Theorem 2

Let \(d\equiv 1,2\; (\bmod {4})\). If \(d=1\), then left hand side of both (6) and (7) is 1. Concerning inequality (6) this yields that either \(I_(x_2,\ldots ,x_n)=0\), that is \(x_2=\ldots =x_n=0\) or \(I_(x_2,\ldots ,x_n)=\pm 1\), that is \(\beta =x_2\ell _2+\ldots +x_n\ell _n\) generates a power integral basis in L. Similary, inequality (7) implies either \(y_2=\ldots =y_n=0\), or \(\gamma =y_2\ell _2+\ldots +y_n\ell _n\) generates a power integral basis in L. For given \(y_2,\ldots ,y_n\) we calculate \(y_1\) from (3). We test all these \(x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n\) in (5). Note that \(x_2=\ldots =x_n=0\), and simultaneously \(y_2=\ldots =y_n=0\) is not possible, since \(\omega \) does not generate K.

If \(d\ne 1\), then (7) gives \(y_2=\ldots =y_n=0\) and (3) gives \(y_1=\pm 1\). The values of \(x_2,\ldots ,x_n\) are obtained from (6). If \(I_L(x_2,\ldots ,x_n)=0\) then \(x_2=\ldots x_n=0\) and \(\alpha =\pm \omega \) which is again impossible, since \(\omega \) does not generate K. Hence we have to take those \(x_2,\ldots ,x_n\) for which \(I_L(x_2,\ldots ,x_n)=\pm 1\), that is \(\beta =x_2\ell _2+\ldots +x_n\ell _n\) generates a power integral basis in L. We test all these \(x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n\) in (5).

Let now \(d\equiv 3 \; (\bmod \; {4})\).

If \(d=3\), then the right hand side of (8) is \(2^{n(n-1)/2}\), the right hand side of (9) is \((2/\sqrt{3})^{n(n-1)/2}\). Hence we have to determine the elements \(\beta =z_2\ell _2+\ldots +z_n\ell _n\) having index \(\le 2^{n(n-1)/2}\) and to select from those the elements \(\gamma =y_2\ell _2+\ldots +y_n\ell _n\) having index \(\le (2/\sqrt{3})^{n(n-1)/2}\). We test if there exist \(x_i\in {\mathbb {Z}}\) with \(x_i=(z_i-y_i)/2\) for all \(i=2,\ldots ,n\). For given \(y_2,\ldots ,y_n\) we calculate \(y_1\) from (3). We test all possible \(x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n\) in (5).

Otherwise, if \(d\ne 3\), Eq (9) has left hand side \(<1\), therefore \(y_2=\ldots y_n=0\) and (3) gives \(y_1=\pm 1\). In this case inequality (8) implies \(|I_L(x_2,\ldots ,x_n)|\le 1\). \(I_L(x_2,\ldots ,x_n)=0\) implies \(x_2 =\ldots = x_n = 0\) which is not possible again. The equation \(I_L(x_2,\ldots ,x_n)=\pm 1\) yields that \(\beta =x_2\ell _2+\ldots +x_n\ell _n\) generates a power integral basis in L. We test all these \(x_2,\ldots ,x_n,y_1,y_2,\ldots ,y_n\) in (5).

This means that for calculating generators of power integral basis in K we only need the generators of power integral bases in L and, in case \(d=3\), elements of small indices of L. This is easy to calculate in lower degree (cubic, quartic) fields, therefore we obtain an efficient method to calculate generators of power integral bases in sextic, octic fields, that are composites of totally real cubic, quartic fields and imaginary quadratic fields.

As a consequence of the above arguments we have

Theorem 3

Under the above conditions, for \(d\ne 1,3\) the composite field \(K = LM\) can only be monogenic if the totally real field L is monogenic. Additionally, if \(d\ne 1,3\) then all generators of power integral bases of K are of the form

$$\begin{aligned} \alpha =x+\beta \pm \omega \end{aligned}$$

where \(\beta \) generates a power integral basis in L and \(x\in {\mathbb {Z}}\) is arbitrary.

Remark

Our Theorem 3 implies Theorem 2 of [3] on generators of power integral bases in composites of totally real cyclic fields of prime degree and imaginary quadratic fields.

4 Composites of the simplest quartic fields and imaginary quadratic fields

In this section we shall give an application to an infinite parametric family of number fields, that shows the strength of our method.

Let a be an integer with \(a\ne 0,\pm 3\). Let \(\xi \) be a root of

$$\begin{aligned} f(x)=x^4-ax^3-6x^2+ax+1. \end{aligned}$$

The parametric family of fields \(L={\mathbb {Q}}(\xi )\) is called simplest quartic fields, see Gras [10]. She showed that for \(a\ne 0,\pm 3\) the polynomial f is irreducible and \({\mathbb {Q}}(\xi ) = {\mathbb {Q}}(-\xi )\), so we can assume \(a > 0\) and \(a \ne 3\). In the following we shall also assume that \(a^2+16\) is not divisible by an odd square. This condition was needed by Kim and Lee [11] to determine an integral basis of L. (Gras [10] showed that \(a^2+16\) represents infinitely many square free integers.) Using the discriminant

$$\begin{aligned} D(f)=4 (a^2+16)^3 \end{aligned}$$

of the polynomial we obtain the discriminant of L. Let \(v_2(x)\) denote the exponent of 2 in the prime power decomposition of the integer x.

Lemma 4

Under the above assumptions on the parameter a an integral basis and the discriminant of L is given by

$$\begin{aligned} \begin{array}{rcl} \left( 1,\xi ,\xi ^2,\frac{1+\xi ^3}{2}\right) ,&{} D_L=(a^2+16)^3, &{} \textrm{if}\; v_2(a)=0,\\ \left( 1,\xi ,\frac{1+\xi ^2}{2},\frac{\xi +\xi ^3}{2}\right) ,&{} D_L=\frac{(a^2+16)^3}{4}, &{} \textrm{if}\; v_2(a)=1,\\ \left( 1,\xi ,\frac{1+\xi ^2}{2},\frac{1+\xi +\xi ^2+\xi ^3}{4}\right) ,&{} D_L=\frac{(a^2+16)^3}{16}, &{} \textrm{if}\; v_2(a)=2,\\ \left( 1,\xi ,\frac{1+2\xi -\xi ^2}{4},\frac{1+\xi +\xi ^2+\xi ^3}{4}\right) ,&{} D_L=\frac{(a^2+16)^3}{64}, &{} \textrm{if}\; v_2(a)\ge 3.\\ \end{array} \end{aligned}$$

Olajos [13] determined all generators of power integral bases (up to sign and translation by elements of \({\mathbb {Z}}\)).

Lemma 5

Under the above assumptions on the parameter a a power integral basis in L exists only for \(a=2\) and \(a=4\). All generators of power integral bases are given by

  • \(a = 2\), \(\alpha =x\xi +y\frac{1+\xi ^2}{2} +z\frac{\xi +\xi ^3}{2}\) where

    \((x, y, z) = (4, 2,-1), (-13,-9, 4), (-2, 1, 0), (1, 1, 0),\)

    \((-8,-3, 2), (-12,-4, 3), (0,-4, 1), (6, 5,-2), (-1, 1, 0), (0, 1, 0)\)

  • \(a = 4\), \(\alpha =x\xi +y\frac{1+\xi ^2}{2} +z\frac{1+\xi +\xi ^2+\xi ^3}{4}\) where

    \((x, y, z) = (3, 2,-1), (-2,-2, 1), (4, 8,-3), (-6,-7, 3),\)

    \((0, 3,-1), (1, 3,-1)\).

Note also that Gaál and Petrányi [7] calculated the minimal indices and all elements of minimal index for all parameters a.

Our purpose is to study composites of simplest quartic fields with imaginary quadratic fields.

Let d be a squarefree positive integer, \(M={\mathbb {Q}}(i\sqrt{d})\). Set

$$\begin{aligned} \omega =i\sqrt{d} \;\; \textrm{if} \;\; d\equiv 1,2\; (\bmod \; {4})\;\; \textrm{and}\;\; \omega =\frac{1+i\sqrt{d}}{4} \;\; \textrm{if} \;\; d\equiv 3\; (\bmod \; {4}). \end{aligned}$$

We assume that \((D_M,D_L)=1\). If \(v_2(a)\ge 1\) then \(D_L\) is even, hence in that case we must have \(d\equiv 3\; (\bmod \; {4})\), since \(D_M=-4d\) if \(d\equiv 1,2\; (\bmod \; {4})\).

We consider monogenity of the composite field \(K=LM\) of degree 8. This is an infinite parametric family of octic fields, depending on the parameters d, a.

We show:

Theorem 6

Under the above conditions for \(d\ne 3\) the field K is not monogenic.

Proof of Theorem 6

For a given parameter a denote by \((1,\ell _2,\ell _3,\ell _4)\) an integral basis of L. Denote by \(I_L(x_2,x_3,x_4)\) the index form corresponding to this integral basis. The condition \((D_L,D_M)=1\) implies that an integral basis of K is given by \((1,\ell _2,\ell _3,\ell _4,\omega ,\omega \ell _2,\omega \ell _3,\omega \ell _4)\). Represent any \(\alpha \in {\mathbb {Z}}_K\) in the form

$$\begin{aligned} \alpha =x_1+x_2\ell _2+x_3\ell _3+x_4\ell _4 +y_1\omega +y_2\omega \ell _2+y_3\omega \ell _3+y_4\omega \ell _4 \end{aligned}$$

with integer coefficients \(x_i,y_i\; (1\le i\le 4)\).

I. Assume \(d\ne 1\). In this case, by Theorem 3 and Lemma 5, the field K can only be monogenic if \(a = 2\) or \(a = 4\). Since in both cases we have \(v_2(a)\ge 1\), then we must have \(d\equiv 1 \; (\bmod \; 4)\). Further, by Theorem 3, if \(\alpha \) generates a power integral basis in K, then \(y_2 = y_3 = y_4 = 0, y_1 =\pm 1\) and \(\beta =x_2\ell _2 + x_3\ell _3 + x_4\ell _4\) generates a power integral basis in L. By Lemma 5, for \(a = 2\) and \(a = 4\) all possible \(x_2, x_3, x_4\) are known.

Now, we consider the third factor (5) of the index form equation of K for \(d \equiv 1 (\bmod \; 4)\), \(d \ne 1\). In both cases \(a = 2\) and \(a = 4\) we substitute the possible \(x_2, x_3, x_4\) and \(y_2 {=} y_3 {=} y_4 {=}0, y_1 {=} \pm 1\) into \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)\). In all cases we obtain a polynomial of degree 6 in \(d > 1\) with positive coefficients, implying that \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)=\pm \) is not possible.

II. Let now \(d=1\). As we showed at the beginning of Sect. 3, in this case all generators of power integral bases in K can be written in the form

$$\begin{aligned} \alpha = x_1 + y_1\omega +\varepsilon _1\beta +\varepsilon _2\gamma \omega \end{aligned}$$
(11)

where \(\beta = x_2\ell _2 + x_3\ell _3 + x_4\ell _4\) and \(\gamma =y_2\ell _2 + y_3\ell _3 + y_4\ell _4\) are generators of power integral bases in L, \(\varepsilon _1,\varepsilon _2\in \{0,1\}\), \((\varepsilon _1,\varepsilon _2)\ne (0,0)\) and \(x_1\in {\mathbb {Z}}\) is arbitrary. Therefore, by Lemma 5 the field K can be monogenic only if \(a = 2\) or \(a = 4\). Since, in both cases \(a = 2\) and \(a = 4\) all possible \(x_2, x_3, x_4\) and \(y_2, y_3, y_4\) are given in Lemma 5, we have to consider Eq (5) in the following three cases:

  1. a.

    If \((\varepsilon _1,\varepsilon _2)= (1,0)\) then the possible \(x_2, x_3, x_4\) corresponding to \(a = 2\) and \(a = 4\) are listed in Lemma 5, \(y_2=y_3=y_4=0\) and (3) gives \(y_1 =\pm 1\). We substitute the possible values of the variables into \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)\) and find that in all cases it takes huge values, not \(\pm 1\).

  2. b.

    If \((\varepsilon _1,\varepsilon _2)= (0,1)\) then \(x_2 = x_3 = x_4 = 0\) and the possible \(y_2, y_3, y_4\) corresponding to \(a = 2\) and \(a = 4\) are listed in Lemma 5. We substitute the possible values of the variables into \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)\) and obtain a polynomial of degree 12 in \(y_1\). Solving \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)=\pm 1\) in \(y_1\) we never get integer solutions for \(y_1\).

  3. c.

    If \((\varepsilon _1,\varepsilon _2)= (1,1)\) then \(x_2,x_3,x_4\) and \(y_2,y_3,y_4\) (corresponding to \(a = 2\) or \(a = 4\)) may run independently through the possible triplets listed in Lemma 5. In all cases, we substitute \(x_2,x_3,x_4\) and \(y_2,y_3,y_4\) into \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)\) and obtain a polynomial of degree 12 in \(y_1\). Solving \(F(x_2, x_3, x_4, y_1,y_2,y_3,y_4)=\pm 1\) in \(y_1\) we never get integer solutions for \(y_1\).

\(\square \)

Remark

The case \(a=3\) is not covered by Theorem 6. In this case we have

$$\begin{aligned} |I_L(2x_2+y_2,2x_3+y_3,2x_4+y_4)|\le 64 \end{aligned}$$

and

$$\begin{aligned} |I_L(y_2,y_3,y_4)|\le 2. \end{aligned}$$

According to [7] we have minimal index 2 in L for infinitely many parameters a. Especially the elements of index \(\le 64\) seems very difficult to determine. This could be the subject of a further research.

Remark

Note that in [9] we obtained conditions on the monogenity of composites of fields, among others of simplest quartic fields and quadratic fields. We did not assume that the discriminants are relative prime and involved also real quadratic fields. However we only obtained certain divisibility conditions on the parameters as a consequence of monogenity.

5 One more example

We also provide a positive example to show that such composite fields may happen to be monogenic. Consider the totally real quartic field L generated by a root \(\xi \) of the polynomial \(f(x)=x^4-4x^2-x+1\). In this field \((1,\xi ,\xi ^2,\xi ^2)\) is an integral basis and \(D_L=1957\). Let \(M={\mathbb {Q}}(i)\) with \(D_M=-4\), coprime to \(D_L\). The composite field \(K=LM\) can be generated e.g. by \(\alpha =i\xi \) having minimal polynomial \(g(x)=x^8+8x^6+18x^4+9x^2+1\). In K the element \(\alpha \) generates a power integral basis. Note that the generator \(\alpha = i\xi \) is of the form (11) with \((\varepsilon _1,\varepsilon _2)= (0,1)\).

6 Computational aspects

All calculations connected with the above examples were performed in Maple [1]. Our procedures were executed on an average laptop running under Windows. The CPU time took all together some seconds.