#  A GENERALIZATION OF SIMPLEST NUMBER FIELDS AND THEIR INTEGRAL BASIS

L. REMETE
University of Debrecen, Mathematical Institute
H–4002 Debrecen Pf.400., Hungary,
e–mail<EMAIL_ADDRESS>Supported through the ÚNKP-19-3 New
National Excellence Program of the Ministry for Innovation and Technology.

###### Abstract

An integral basis of the simplest number fields of degree 3,4 and 6 over
${\mathbb{Q}}$ are well-known, and widely investigated. We generalize the
simplest number fields to any degree, and show that an integral basis of these
fields is repeating periodically.

##  1 Introduction

Let $a,b,c,d\in{\mathbb{Q}}$ and

$\sigma:{\mathbb{C}}\mapsto{\mathbb{C}},\qquad\sigma:z\mapsto\frac{az+b}{cz+d}.$

Assume, that $f(X)\in{\mathbb{Z}}[X]$ is a polynomial with real roots, such
that $\sigma$ transitively permutes the roots of $f(X)$. In this case, if
$\beta$ is a root of $f(X)$, then ${\mathbb{Q}}(\beta)$ is a totally real
cyclic number field. This requires, that the matrix

$M=\left(\begin{array}[]{rr}a&b\\\ c&d\end{array}\right)\in
PGL_{2}({\mathbb{Q}}),$

is of finite order. It can be shown that each non-trivial torsion element of
$PGL_{2}({\mathbb{Q}})$ has order 2,3,4 or 6, whence, there exist such
polynomials and number fields only of degrees 2,3,4 and 6.
According to D.Shanks [27], A.J.Lazarus [18], G.Lettl, A.Pethő and P.Voutier
[21] and A.Hoshi [14] the polynomials of degrees 3,4 and 6 with these
properties are called simplest polynomials, and the corresponding number
fields are called simplest number fields. These fields have an extensive
literature.
First, the simplest cubic fields are investigated by H.Cohn [1] and D.Shanks
[27], because they have easily computable and relatively large class numbers.
M.N.Gras [9], [10], V.Ennola [2], [3] and A.J.Lazarus [18] investigated the
unit group of the simplest fields. K.Foster [4] obtained the simplest
parametric polynomials, just by using a special identity of units in cyclic
extensions. It shows, that these fields have some other unique and interesting
properties.
E.Thomas [28], M.Mignotte [22], G.Lettl, A.Pethő and P.Voutier [21], [20] and
G.Lettl and A.Pethő [19], I.Gaál [5] solved Thue equations corresponding to
the simplest polynomials in absolute case, and C.Heuberger [11], I.Gaál,
B.Jadrijević and L.Remete [6] in certain relative cases.
A.Hoshi [12], [13], [14] gave a correspondence between solutions of a family
of Thue equations and the isomorphism classes of the simplest number fields.
He extended his results to a family of polynomials of degree 12, which has
similar properties as the simplest polynomials, but over
${\mathbb{Q}}(\sqrt{-3})$. This generalization provided the main motivation of
our approach.
Assume that $\beta$ is a root of a simplest polynomial
$f(X)\in{\mathbb{Z}}[X]$ of degree $n$, and the Möbius transformation $\sigma$
permutes its roots transitively.
Then the conjugates of $\beta$ are:

$\\{\beta,\sigma(\beta),\sigma^{2}(\beta),\ldots,\sigma^{n-1}(\beta)\\}.$

Let $m$ denote

$m=\frac{\beta+\sigma(\beta)+\sigma^{2}(\beta)+\ldots+\sigma^{n-1}(\beta)}{n}.$

If $n=3$ or $n=6$, and

$\sigma:{\mathbb{C}}\mapsto{\mathbb{C}},\qquad\sigma:z\mapsto\frac{az-1}{z+a+1}.$

where $a=0$ or $-1$ for degree 3, and $a=1$ or $-2$ for degree 6, then we can
write the minimal polynomial of $\beta$ over ${\mathbb{Q}}$ in the form:

$f^{(3)}_{m}(X)=X^{3}-3mX^{2}-3(m+1)X-1,$
$f^{(6)}_{m}(X)=X^{6}-6mX^{5}-15(m+1)X^{4}-20X^{3}+15mX^{2}+6(m+1)X+1.$

Therefore, if $m$ is a rational parameter, then the polynomials above are
cyclic, they have rational coefficients and

$\sigma:z\mapsto\frac{az-1}{z+a+1}$

permutes their roots transitively.
Although these polynomials have nice properties for any $m\in{\mathbb{Q}}$, we
say that their root generates a simplest number field, if they are
irreducible, and have integer coefficients.
In Section 2 we will generalize these polynomials to any degree. We keep the
property that the coefficients are rational numbers, and a Möbius
transformation

$\sigma:z\mapsto\frac{\alpha z-1}{z+\alpha+1}$

permutes the roots transitively, where $\alpha\in{\mathbb{C}}$ is an algebraic
integer, such that

$M=\left(\begin{array}[]{rr}\alpha&-1\\\ 1&\alpha+1\end{array}\right)\in
PGL_{2}({\mathbb{C}}),$

has finite order. In such way, we will obtain infinite parametric families of
polynomials of degree $n$, such that their Galois groups are cyclic over
${\mathbb{Q}}(\alpha)$.
We get some well-known families as special cases.

  * •

Simplest cubic fields: $n=3$, $m=\frac{t}{3}$, $t\in{\mathbb{Z}}$ (c.f. [27])

  * •

Simplest sextic fields: $n=6$, $m=\frac{t}{3}$,
$t\in{\mathbb{Z}}\setminus\\{-8,-3,0,5\\}$ (c.f. [10],[8])

  * •

Family of polynomials of degree 12 investigated by A.Hoshi: $n=12$,
$m=\frac{t}{3}$,
$t\in{\mathbb{Z}}\setminus\\{-8,-3,0,5\\}$ (c.f. [15])

We call the number fields generated by a root of our polynomials generalized
simplest number fields. As the above examples show, these are indeed
generalizations of the simplest fields of degrees 3 and 6, but we remark, that
for degree 4, our family of number fields is different from the simplest
quartic fields.
In Section 2.1 and 2.2, we describe the discriminants of these polynomials,
and give a sufficient condition for the irreducibility.

Explicit integral bases of the simplest fields are given by D.Shanks [27],
H.K.Kim and J.H.Lee [16], G.Lettl, A.Pethő, P.Voutier [20] and I.Gaál and
L.Remete [8]. In each cases an integral basis of these fields is repeating
periodically in $m$. In Section 3, we will show, that in our generalization
the same phenomena occurs. Let $t$ be an integer. If $t$ is congruent to 1 or
2 modulo 3, then we set $m=t$, otherwise we set $m=\frac{t}{3}$. In such way
we obtain families of parametric polynomials, which have integer coefficients
for any integer parameter $t$. We prove, that if the $3$-free part of
$t^{2}+t+1$ (or $t^{2}+3t+9$, respectively) is square-free, then an integral
basis of the fields generated by a root $\beta_{t}$ of $f^{(n)}_{t}(X)$ is
repeating periodically.
We say that an integral basis is repeating periodically modulo $n_{0}$, if for
$r=0,\ldots,n_{0}-1$ there are polynomials $h^{(r)}_{i}(X)\in{\mathbb{Q}}[X]$,
$(i=0,\ldots n-1)$, such that, if $t\equiv r\mod{n_{0}}$ and $f^{(n)}_{t}(X)$
is irreducible, then

$\left(h^{(r)}_{0}(\beta_{t}),h^{(r)}_{1}(\beta_{t}),h^{(r)}_{2}(\beta_{t}),\ldots,h^{(r)}_{n-1}(\beta_{t})\right),$

is an integral basis of ${\mathbb{Q}}(\beta_{t})$. (See [7],[8],[26]).
We give a general upper bound for the period length for any
$n\in{\mathbb{N}}$, and for $n\leq 12$ we significantly improve it. Using
these better bounds, we are able to determine the smallest period lengths for
$n=2,3,4,5,6,8,9,12$.

##  2 Generalized simplest number fields

In this section we define the parametric family $f^{(n)}_{m}$ of polynomials
of degrees $n>1$, such that there exists a Möbius transformation

$\sigma:z\mapsto\frac{\alpha z-1}{z+\alpha+1},\qquad\alpha\in{\mathbb{C}}$

which permutes the roots of $f^{(n)}_{m}(X)$ transitively for any parameter
$m$. All along this section let $m$ be a rational parameter.
Let $g,h:{\mathbb{N}}\mapsto{\mathbb{Q}}$ be defined by

$g(i):=\left\\{\begin{array}[]{lr}1,&\mbox{ if }i\equiv 0\mod{6},\\\
-m,&\mbox{ if }i\equiv 1\mod{6},\\\ -m-1,&\mbox{ if }i\equiv 2\mod{6},\\\
-1,&\mbox{ if }i\equiv 3\mod{6},\\\ m,&\mbox{ if }i\equiv 4\mod{6},\\\
m+1,&\mbox{ if }i\equiv 5\mod{6}.\end{array}\right.\qquad\mbox{and}\qquad
h(i):=\left\\{\begin{array}[]{lr}0,&\mbox{ if }i\equiv 0\mod{6},\\\ -1,&\mbox{
if }i\equiv 1\mod{6},\\\ -1,&\mbox{ if }i\equiv 2\mod{6},\\\ 0,&\mbox{ if
}i\equiv 3\mod{6},\\\ 1,&\mbox{ if }i\equiv 4\mod{6},\\\ 1,&\mbox{ if }i\equiv
5\mod{6}.\end{array}\right.$

For $n\geq 0$, let

$f^{(n)}_{m}(X):=\sum_{i=0}^{n}\binom{n}{i}\cdot X^{i}\cdot
g(n-i)\qquad\in{\mathbb{Q}}[m,X],$

and

$r^{(n)}(X):=\sum_{i=0}^{n}\binom{n}{i}\cdot X^{i}\cdot
h(n-i)\qquad\in{\mathbb{Z}}[X].$

These polynomials satisfy the following recursion formulas:

######  Lemma 1.

For $n\geq 0$,

(1) $f^{(n+1)}_{m}(X)=(X-m)\cdot f^{(n)}_{m}(X)+(m^{2}+m+1)\cdot r^{(n)}(X),$

and

(2) $r^{(n+1)}(X)=(X+m+1)\cdot r^{(n)}(X)-f^{(n)}_{m}(X).$

###### Proof.

We prove only (2), the other identity can be proved in the same way. We
compare the coefficients of $X^{k}$ in the two side of the equation.
The coefficient of $X^{k}$ on the left hand side of (2) is

$\binom{n+1}{k}\cdot h(n+1-k)$

and on the right side of (2) is

$\binom{n}{k-1}\cdot h(n+1-k)+(m+1)\cdot\binom{n}{k}\cdot
h(n-k)-\binom{n}{k}\cdot g(n-k).$

It is easy to check that for any remainder of $n-k\mod{6}$, these coefficients
are the same. ∎

By the definition of $f^{(n)}_{m}$ and $r^{(n)}$, it is easy to determine
their derivatives.

######  Lemma 2.

For $n\geq 0$,

$\left(f^{(n+1)}_{m}\right)^{\prime}(X)=\sum_{i=1}^{n+1}\binom{n+1}{i}\cdot
i\cdot X^{i-1}\cdot g(n+1-i)=(n+1)\sum_{i=1}^{n+1}\binom{n}{i-1}\cdot
X^{i-1}\cdot g(n+1-i)=$ $=(n+1)\cdot f^{(n)}_{m}(X),$

and similarly

$\left(r^{(n+1)}\right)^{\prime}(X)=(n+1)\cdot r^{(n)}(X).$

Now we prove a nice variable transformation identity for $f^{(n)}_{m}(X)$.

######  Theorem 3.

For any $\alpha\in{\mathbb{C}}$ and $n\geq 1$,

(3) $(X+\alpha+1)^{n}\cdot f^{(n)}_{m}\left(\frac{\alpha
X-1}{X+\alpha+1}\right)=f^{(n)}_{m}(\alpha)\cdot
f^{(n)}_{m}(X)-(m^{2}+m+1)\cdot r^{(n)}(\alpha)\cdot r^{(n)}(X).$

###### Proof.

If $n=1$, then $f^{(1)}_{m}(X)=X-m$, $r^{(1)}(X)=-1$, and

$(X+\alpha+1)\cdot\left(\frac{\alpha
X-1}{X+\alpha+1}-m\right)=(\alpha-m)\cdot(X-m)-(m^{2}+m+1)\cdot(-1)\cdot(-1).$

Hence, (3) is true for $n=1$.
Assume, that (3) is true for $n$:

(4) $(X+\alpha+1)^{n}\cdot f^{(n)}_{m}\left(\frac{\alpha
X-1}{X+\alpha+1}\right)=f^{(n)}_{m}(\alpha)\cdot
f^{(n)}_{m}(X)-(m^{2}+m+1)\cdot r^{(n)}(\alpha)\cdot r^{(n)}(X).$

Let

$F_{n+1}(X):=\frac{f^{(n+1)}_{m}(\alpha)\cdot
f^{(n+1)}_{m}(X)-(m^{2}+m+1)\cdot r^{(n+1)}(\alpha)\cdot
r^{(n+1)}(X)}{(X+\alpha+1)^{n+1}}.$

Using Lemma 2, the derivative of $F_{n+1}$ with respect to $X$ is:

$F^{\prime}_{n+1}(X)=\frac{n+1}{(X+\alpha+1)^{n+2}}\cdot\left[\left(f^{(n+1)}_{m}(\alpha)\cdot
f^{(n)}_{m}(X)-(m^{2}+m+1)\cdot r^{(n+1)}(\alpha)\cdot
r^{(n)}(X)\right)\cdot(X+\alpha+1)-\right.$ $\left.f^{(n+1)}_{m}(\alpha)\cdot
f^{(n+1)}_{m}(X)+(m^{2}+m+1)\cdot r^{(n+1)}(\alpha)\cdot r^{(n+1)}(X)\right].$

By Lemma 1, substitute

$r^{(n+1)}(X)=(X+m+1)\cdot r^{(n)}(X)-f^{(n)}_{m}(X),$
$f^{(n+1)}_{m}(X)=(X-m)\cdot f^{(n)}_{m}(X)+(m^{2}+m+1)\cdot r^{(n)}(X),$
$r^{(n+1)}(\alpha)=(\alpha+m+1)\cdot r^{(n)}(\alpha)-f^{(n)}_{m}(\alpha),$
$f^{(n+1)}_{m}(\alpha)=(\alpha-m)\cdot f^{(n)}_{m}(\alpha)+(m^{2}+m+1)\cdot
r^{(n)}(\alpha).$

After simplification we get

$F^{\prime}_{n+1}(X)=\frac{n+1}{(X+\alpha+1)^{n+2}}\cdot(\alpha^{2}+\alpha+1)\cdot\left(f^{(n)}_{m}(X)\cdot
f^{(n)}_{m}(\alpha)-(m^{2}+m+1)\cdot r^{(n)}(X)\cdot r^{(n)}(\alpha)\right).$

The induction hypothesis (4) implies

$F^{\prime}_{n+1}(X)=\frac{n+1}{(X+\alpha+1)^{n+2}}\cdot(\alpha^{2}+\alpha+1)\cdot(X+\alpha+1)^{n}\cdot
f^{(n)}_{m}\left(\frac{\alpha X-1}{X+\alpha+1}\right)=$
$=\frac{n+1}{(X+\alpha+1)^{2}}\cdot(\alpha^{2}+\alpha+1)\cdot
f^{(n)}_{m}\left(\frac{\alpha
X-1}{X+\alpha+1}\right)=\left(f^{(n+1)}_{m}\right)^{\prime}\left(\frac{\alpha
X-1}{X+\alpha+1}\right).$

In the last step we applied Lemma 2, and the derivation rules. This shows,
that

$F_{n+1}(X)=f^{(n+1)}_{m}\left(\frac{\alpha X-1}{X+\alpha+1}\right)+c,$

with an appropriate constant $c$. We have to show that $c=0$. In order to
prove this, multiply both sides of this equation by $(X+\alpha+1)^{n+1}$:

$f^{(n+1)}_{m}(\alpha)\cdot f^{(n+1)}_{m}(X)-(m^{2}+m+1)\cdot
r^{(n+1)}(\alpha)\cdot r^{(n+1)}(X)=$
$(X+\alpha+1)^{n+1}\cdot\left(f^{(n+1)}_{m}\left(\frac{\alpha
X-1}{X+\alpha+1}\right)+c\right)$

Thus, both sides are polynomials in $X$ of degree $n+1$. The leading
coefficient of $X$ in the left hand side is equal to

$f^{(n+1)}_{m}(\alpha)\cdot g(0)-(m^{2}+m+1)\cdot r^{(n+1)}(\alpha)\cdot
h(0)=f^{(n+1)}_{m}(\alpha).$

The right hand side is

$(X+\alpha+1)^{n+1}\cdot\left(f^{(n+1)}_{m}\left(\frac{\alpha
X-1}{X+\alpha+1}\right)+c\right)=$
$\left(\sum_{i=0}^{n+1}\binom{n+1}{i}\cdot(\alpha
X-1)^{i}\cdot(X+\alpha+1)^{n+1-i}\cdot g(n+1-i)\right)+c(X+\alpha+1)^{n+1},$

and the leading coefficient of $X$ in this polynomial is equal to

$\left(\sum_{i=0}^{n+1}\binom{n+1}{i}\cdot\alpha^{i}\cdot
g(n+1-i)\right)+c=f^{(n+1)}_{m}(\alpha)+c.$

The leading coefficients have to be the same, which implies that $c=0$, and
therefore by mathematical induction, the equation (3) is true for any
$n\in{\mathbb{N}}$. ∎

We will use the following interesting lemma about the periodic alternating
multisection of sums of binomial coefficients.

######  Lemma 4.

Let

$p(i):=\left\\{\begin{array}[]{lr}a,&\mbox{ if }i\equiv 0\mod{6},\\\ b,&\mbox{
if }i\equiv 1\mod{6},\\\ c,&\mbox{ if }i\equiv 2\mod{6},\\\ -a,&\mbox{ if
}i\equiv 3\mod{6},\\\ -b,&\mbox{ if }i\equiv 4\mod{6},\\\ -c,&\mbox{ if
}i\equiv 5\mod{6}.\end{array}\right.$

with arbitrary complex numbers $a,b,c$, and let

$t_{n}:=\sum_{i=0}^{n}\binom{n}{i}\cdot p(i).$

Then

$t_{n+12}=3^{6}\cdot t_{n}.$

###### Proof.

We will use the formula of multisections of sums of binomial coefficients (see
e.g. [25],[17]). Let $0\leq u<v$ be integers, then

$S_{n}(u,v):=\binom{n}{u}+\binom{n}{u+v}+\binom{n}{u+2v}+\ldots=\frac{1}{v}\sum_{i=0}^{v-1}\left(2\cos\frac{\pi
i}{v}\right)^{n}\cdot\cos\frac{\pi(n-2u)i}{v}.$

By this formula,

(5)
$t_{n}=a\cdot(S_{n}(0,6)-S_{n}(3,6))+b\cdot(S_{n}(1,6)-S_{n}(4,6))+c\cdot(S_{n}(2,6)-S_{n}(5,6)).$

The coefficient of $a$ is:

$S_{n}(0,6)-S_{n}(3,6)=\frac{1}{6}\sum_{i=0}^{5}\left(2\cos\frac{\pi
i}{6}\right)^{n}\cdot\left(\cos\frac{\pi
ni}{6}-\cos\frac{\pi(n-6)i}{6}\right)=$
$=\frac{1}{6}\sum_{i=0}^{5}\left(2\cos\frac{\pi
i}{6}\right)^{n}\cdot\cos\frac{\pi ni}{6}\left(1-\cos\pi i\right).$

For $i=3$, $\cos\frac{\pi i}{6}=0$;
for $i=0,2,4$, $\left(1-\cos\pi i\right)=0$;
for $i=1,5$, $\left(1-\cos\pi i\right)=2$, so

$S_{n}(0,6)-S_{n}(3,6)=\frac{1}{3}\left(\left(2\cos\frac{\pi}{6}\right)^{n}\cdot\cos\frac{\pi
n}{6}+\left(2\cos\frac{5\pi}{6}\right)^{n}\cdot\cos\frac{5\pi n}{6}\right)=$
$=\frac{1}{3}\sqrt{3}^{n}\left(\cos\frac{\pi n}{6}+(-1)^{n}\cdot\cos\frac{5\pi
n}{6}\right).$

Hence,

$S_{n+12}(0,6)-S_{n+12}(3,6)=\frac{1}{3}\sqrt{3}^{n+12}\left(\cos\frac{\pi(n+12)}{6}+(-1)^{n+12}\cdot\cos\frac{5\pi(n+12)}{6}\right)=$
$=3^{6}\cdot(S_{n}(0,6)-S_{n}(3,6)).$

Similarly:

$S_{n+12}(1,6)-S_{n+12}(4,6)=3^{6}\cdot(S_{n}(1,6)-S_{n}(4,6)),$
$S_{n+12}(2,6)-S_{n+12}(5,6)=3^{6}\cdot(S_{n}(2,6)-S_{n}(5,6)).$

Finally by (5)

$t_{n+12}=3^{6}\cdot t_{n}.$

∎

By this lemma, we are able to calculate $r^{(n)}(\varepsilon_{3})$, where
$\varepsilon_{3}$ is a primitive cube root of unity.
Since

$r^{(n)}(\varepsilon_{3})=\sum_{i=0}^{n}\binom{n}{i}\cdot\varepsilon_{3}^{i}\cdot
h(n-i),$

we can use Lemma 4 with

$p(i)=\varepsilon_{3}^{i}\cdot h(n-i).$

It is easy to see, that $p(i)$ is of the form we prescribed in Lemma 4, since

$p(i+3)=\varepsilon_{3}^{i+3}\cdot h(n-i-3)=-\varepsilon_{3}^{i}\cdot
h(n-i)=-p(i).$

Therefore we get

$r^{(n+12)}(\varepsilon_{3})=3^{6}\cdot r^{(n)}(\varepsilon_{3}).$

By calculating $r^{(n)}(\varepsilon_{3})$ for $n=1,\ldots,12$, we get

(6) $r^{(n)}(\varepsilon_{3})=-(-I\sqrt{3})^{n-1}.$

As $3^{6}=(-I\sqrt{3})^{12}$, (6) is true for any $n\in{\mathbb{N}}$.
An important consequence is that $\varepsilon_{3}$ is not a root of
$r^{(n)}(X)$.
In the following we show, that there is an interesting connection between the
roots of $f_{m}^{(n)}(X)$ and $r^{(n)}(X)$.

######  Corollary 5.

Let $n\geq 2$. If $\beta$ is a root of $f^{(n)}_{m}(X)$ and $\alpha$ is a root
of $r^{(n)}(X)$, then

$\frac{\alpha\beta-1}{\beta+\alpha+1}$

is also a root of $f^{(n)}_{m}(X)$.

###### Proof.

First we show that if $\beta$ is a root of $f^{(n)}_{m}(X)$ and $\alpha$ is a
root of $r^{(n)}(X)$, then $\beta+\alpha+1\neq 0$. We will use, that

(7) $r^{(n)}(X)=(-1)^{n-1}\cdot r^{(n)}(-X-1).$

This is trivially true for $n=1$. Assume, that it is true for $n$, then

$\left(r^{(n+1)}(X)\right)^{\prime}=(n+1)\cdot
r^{(n)}(X)=(n+1)\cdot(-1)^{n-1}\cdot r^{(n)}(-X-1)=\left((-1)^{n}\cdot
r^{(n+1)}(-X-1)\right)^{\prime}.$

The leading coefficients of $r^{(n+1)}(X)$ and $(-1)^{n}\cdot r^{(n+1)}(-X-1)$
are the same, so (7) is true for any $n\geq 1$.
The second step is to show that $f^{(n)}_{m}(X)$ and $r^{(n)}(X)$ can not have
a common root. By Lemma 1, if $\alpha$ was a common root of $f^{(n)}_{m}(X)$
and $r^{(n)}(X)$, and the determinant of

$\left(\begin{array}[]{rr}\alpha-m&m^{2}+m+1\\\
-1&\alpha+m+1\end{array}\right)$

is not zero, then $\alpha$ would also be a common root of $f^{(n-1)}_{m}(X)$
and $r^{(n-1)}(X)$. However, the determinant of this matrix is
$\alpha^{2}+\alpha+1$, which is not zero, since $\alpha$ is a root of
$r^{(n)}(X)$, and by (6), the cube roots of unity are not roots of
$r^{(n)}(X)$.
Iteratively using this correspondence, we get, that $\alpha$ is a common root
of $f^{(1)}_{m}(X)$ and $r^{(1)}(X)$, but $r^{(1)}(X)=-1$ does not have any
root, which is contradiction.
If $n\geq 2$ and $\alpha$ is a root of $r^{(n)}(X)$, then $-\alpha-1$ is also
a root of $r^{(n)}(X)$. Therefore, as $f^{(n)}_{m}(X)$ and $r^{(n)}(X)$ don’t
have a common root, $\beta\neq-\alpha-1$, hence $\beta+\alpha+1\neq 0$.
Finally, by Theorem 3:

$(\beta+\alpha+1)^{n}\cdot
f^{(n)}_{m}\left(\frac{\alpha\beta-1}{\beta+\alpha+1}\right)=f^{(n)}_{m}(\alpha)\cdot
f^{(n)}_{m}(\beta)-(m^{2}+m+1)\cdot r^{(n)}(\alpha)\cdot r^{(n)}(\beta)=$
$=f^{(n)}_{m}(\alpha)\cdot 0-(m^{2}+m+1)\cdot 0\cdot r^{(n)}(\beta)=0,$

whence

$\frac{\alpha\beta-1}{\beta+\alpha+1}$

is also a root of $f^{(n)}_{m}(X)$. ∎

Now we will prove, that if $n\geq 2$, then there exists a root $\alpha$ of
$r^{(n)}(X)$, such that the order of the matrix

$M=\left(\begin{array}[]{rr}\alpha&-1\\\ 1&\alpha+1\end{array}\right)\in
PGL_{2}({\mathbb{Q}}(\alpha))$

is equal to $n$. Let $\varepsilon_{6}$ be a primitive sixth root of unity, and
$\varepsilon_{n}$ be a primitive $n$-th root of unity. Set

(8)
$\alpha=\varepsilon_{6}\cdot\frac{\varepsilon_{6}+\varepsilon_{n}}{1-\varepsilon_{n}}.$

First we show, that $\alpha$ is a root of $r^{(n)}(X)$. It is easy to see,
that

(9)
$\frac{\alpha+\varepsilon_{6}^{5}}{\alpha+\varepsilon_{6}}=\varepsilon_{n}.$

Hence

$\sum_{i=0}^{n-1}(\alpha+\varepsilon_{6}^{5})^{n-1-i}(\alpha+\varepsilon_{6})^{i}=0,$

and $\alpha+\varepsilon_{6}$ is a root of

$R^{(n)}(X):=\sum_{i=0}^{n-1}(X+\varepsilon_{6}^{5}-\varepsilon_{6})^{n-1-i}X^{i}=\sum_{i=0}^{n-1}(X-I\sqrt{3})^{n-1-i}X^{i}.$

We will show that

$R^{(n)}(X)=-r^{(n)}(X-\varepsilon_{6}).$

For $k=0,\ldots,n-1$, the coefficient of $X^{k}$ in $R^{(n)}(X)$ is:

$\sum_{i=0}^{k}\binom{n-1-i}{k-i}\cdot(-I\sqrt{3})^{n-1-i-(k-i)}=(-I\sqrt{3})^{n-1-k}\cdot\sum_{i=0}^{k}\binom{n-1-i}{k-i}=$
$=(-I\sqrt{3})^{n-1-k}\cdot\binom{n}{k}.$

The coefficient of $X^{k}$ in $r^{(n)}(X-\varepsilon_{6})$ is:

$\sum_{i=k}^{n}\binom{n}{i}\cdot\binom{i}{k}\cdot(-\varepsilon_{6})^{i-k}\cdot
h(n-i)=\sum_{i=k}^{n}\binom{n}{k}\cdot\binom{n-k}{i-k}\cdot(-\varepsilon_{6})^{i-k}\cdot
h(n-i)=$
$=\binom{n}{k}\cdot\sum_{j=0}^{n-k}\binom{n-k}{j}\cdot(-\varepsilon_{6})^{j}\cdot
h(n-k-j)=\binom{n}{k}\cdot
r_{n-k}(-\varepsilon_{6})=-\binom{n}{k}\cdot(-I\sqrt{3})^{n-k-1}.$

In the last step we used equation (6), and that $-\varepsilon_{6}$ is a
primitive cube root of unity. Since all coefficients of $R^{(n)}(X)$ are the
negative of the corresponding coefficients of $r^{(n)}(X-\varepsilon_{6})$, we
obtain

$R^{(n)}(X)=-r^{(n)}(X-\varepsilon_{6}).$

Thus $\alpha+\varepsilon_{6}$ is a root of $r^{(n)}(X-\varepsilon_{6})$, and
therefore $\alpha$ is a root of $r^{(n)}(X)$.
Due to the special form of $\alpha$, it is also easy to determine the order of

$M=\left(\begin{array}[]{rr}\alpha&-1\\\ 1&\alpha+1\end{array}\right)\in
PGL_{2}({\mathbb{Q}}(\alpha)).$

The eigenvalues of this matrix are $\alpha+\varepsilon_{6}$ and
$\alpha+\varepsilon_{6}^{5}$, and by (9) the quotient of the eigenvalues of
$M$ is $\varepsilon_{n}$, therefore $M\in PGL_{2}({\mathbb{Q}}(\alpha))$ is a
matrix of order $n$.
This allows us to determine the Galois group of $f^{(n)}_{m}(X)$ over
${\mathbb{Q}}(\alpha)$.

######  Theorem 6.

Let $n\geq 2$. If $\alpha$ is defined as in (8), and $m$ is a rational number,
such that $f^{(n)}_{m}(X)$ is irreducible over ${\mathbb{Q}}(\alpha)$, then
the Galois group of $f^{(n)}_{m}(X)$ over ${\mathbb{Q}}(\alpha)$ is the cyclic
group of order $n$.

###### Proof.

By Corollary 5, if $\beta$ is a root of $f^{(n)}_{m}(X)$, then

$\frac{\alpha\beta-1}{\beta+\alpha+1}$

is also a root of $f^{(n)}_{m}(X)$. As the order of the matrix

$M=\left(\begin{array}[]{rr}\alpha&-1\\\ 1&\alpha+1\end{array}\right)\in
PGL_{2}({\mathbb{Q}}(\alpha))$

is $n$, $\\{\sigma^{i}(\beta)|i=0,\ldots,n-1\\}$ is the complete set of roots
of $f^{(n)}_{m}(X)$, where

$\sigma:z\mapsto\frac{\alpha z-1}{z+\alpha+1}.$

Finally $\sigma^{i}(\beta)\in{\mathbb{Q}}(\alpha,\beta)$ for all
$i=0,\ldots,n-1$, hence the splitting field of $f^{(n)}_{m}(X)$ over
${\mathbb{Q}}(\alpha)$ is ${\mathbb{Q}}(\alpha,\beta)$. Therefore,
${\mathbb{Q}}(\alpha,\beta)/{\mathbb{Q}}(\alpha)$, is a cyclic Galois
extension of degree $n$. ∎

###  2.1 The discriminant of $f^{(n)}_{m}(X)$

In this section, we determine the discriminants of the polynomials
$f^{(n)}_{m}(X)$.
We will use the following properties of the resultant
$\operatorname{res}(A,B)$ of polynomials

$A(X)=a_{n}X^{n}+a_{n-1}X^{n-1}+\ldots+a_{1}X+a_{0},$
$B(X)=b_{m}X^{m}+b_{m-1}X^{m-1}+\ldots+b_{1}X+b_{0}.$

(see for e.g. Chapter 2.3.3. [24]).

(10) $\operatorname{res}(A,B)=(-1)^{nm}\cdot\operatorname{res}(B,A).$

If $n\geq m$ and $Q,R$ are polynomials such that $A=QB+R$, where $\deg(R)=r$,
then

(11) $\operatorname{res}(A,B)=b_{m}^{n-r}\cdot\operatorname{res}(R,B).$

For any monic polynomial $Q$,

(12)
$\operatorname{res}(AQ,B)=\operatorname{res}(A,B)\cdot\operatorname{res}(Q,B).$

If $\lambda\in{\mathbb{R}}$, then

(13) $\displaystyle\operatorname{res}(\lambda\cdot
A,B)=\lambda^{m}\cdot\operatorname{res}(A,B),$ (14)
$\displaystyle\operatorname{res}(A,\lambda\cdot
B)=\lambda^{n}\cdot\operatorname{res}(A,B).$

######  Lemma 7.

For $n\geq 2$ the resultant of $f^{(n)}_{m}(X)$ and $f^{(n-1)}_{m}(X)$ with
respect to $X$ is

$\operatorname{res}\left(f^{(n)}_{m}(X),f^{(n-1)}_{m}(X)\right)=(m^{2}+m+1)^{n-1}\cdot
3^{\frac{(n-1)(n-2)}{2}}\cdot(-1)^{\frac{n(n-1)}{2}}.$

###### Proof.

First we determine the resultant of $r^{(n)}(X)$ and $X^{2}+X+1$.
The remainder of $r^{(n)}(X)$ divided by $X^{2}+X+1$ is

$T_{n}(X)=\sum_{i=0}^{n}\binom{n}{i}\cdot G(i)\cdot h(n-i),$

where

$G(i)=\left\\{\begin{array}[]{lr}1,&\mbox{ if }i\equiv 0\mod{3},\\\ X,&\mbox{
if }i\equiv 1\mod{3},\\\ -X-1,&\mbox{ if }i\equiv 2\mod{3}.\end{array}\right.$

Thus, we can write $T_{n}(X)=A+B\cdot X$, with

$A=\sum_{i=0}^{n}\binom{n}{i}\cdot G_{0}(i)\cdot h(n-i),$
$B=\sum_{i=0}^{n}\binom{n}{i}\cdot G_{1}(i)\cdot h(n-i),$
$G_{0}(i):=\left\\{\begin{array}[]{lr}1,&\mbox{ if }i\equiv 0\mod{3},\\\
0,&\mbox{ if }i\equiv 1\mod{3},\\\ -1,&\mbox{ if }i\equiv 2\mod{3}.\\\
\end{array}\right.\qquad\mbox{and}\qquad
G_{1}(i):=\left\\{\begin{array}[]{lr}0,&\mbox{ if }i\equiv 0\mod{3},\\\
1,&\mbox{ if }i\equiv 1\mod{3},\\\ -1,&\mbox{ if }i\equiv 2\mod{3}.\\\
\end{array}\right.$

By applying Lemma 4 to the coefficients of $T_{n}(X)$, we obtain,
$T_{n+12}(X)=3^{6}\cdot T_{n}(X)$, whence using the identities (11) and (13),

$\operatorname{res}\left(r^{(n+12)}(X),X^{2}+X+1\right)=\operatorname{res}\left(T_{n+12}(X),X^{2}+X+1\right)=$
$\operatorname{res}\left(3^{6}\cdot
T_{n}(X),X^{2}+X+1\right)=3^{12}\cdot\operatorname{res}\left(T_{n}(X),X^{2}+X+1\right)=$
$3^{12}\cdot\operatorname{res}\left(r^{(n)}(X),X^{2}+X+1\right).$

By calculating $\operatorname{res}\left(r^{(n)}(X),X^{2}+X+1\right)$ for
$n=1,\ldots,12$, we get that

$\operatorname{res}\left(r^{(n)}(X),X^{2}+X+1\right)=3^{n-1},$

for any $n\geq 1$, and by (10),

(15) $\operatorname{res}\left(X^{2}+X+1,r^{(n)}(X)\right)=(-1)^{2(n-1)}\cdot
3^{n-1}=3^{n-1}.$

The next step is to determine the resultant of $r^{(n)}(X)$ and
$f^{(n)}_{m}(X)$ with respect to $X$.
By (1) and (2), we have

$f^{(n)}_{m}(X)=-(X-m)\cdot r^{(n)}(X)+(X^{2}+X+1)\cdot r^{(n-1)}(X).$

Therefore, by (10), (11), (12) and (15),

$\operatorname{res}\left(r^{(n)}(X),f^{(n)}_{m}(X)\right)=(-1)^{n(n-1)}\cdot\operatorname{res}\left(f^{(n)}_{m}(X),r^{(n)}(X)\right)=$
$n^{0}\cdot\operatorname{res}\left((X^{2}+X+1)\cdot
r^{(n-1)}(X),r^{(n)}(X)\right)=$
$\operatorname{res}\left(X^{2}+X+1,r^{(n)}(X)\right)\cdot\operatorname{res}\left(r^{(n-1)}(X),r^{(n)}(X)\right)=3^{n-1}\operatorname{res}\left(r^{(n-1)}(X),r^{(n)}(X)\right).$

By (2), we have

$r^{(n)}(X)=(X+m+1)\cdot r^{(n-1)}(X)-f^{(n-1)}_{m}(X),$

whence by (11), (12) and (10)

$\operatorname{res}\left(r^{(n)}(X),r^{(n-1)}(X)\right)=(n-1)^{0}\cdot\operatorname{res}\left(-f^{(n-1)}_{m}(X),r^{(n-1)}(X)\right)=$
$(-1)^{n-2}\cdot\operatorname{res}\left(f^{(n-1)}_{m}(X),r^{(n-1)}(X)\right)=(-1)^{(n-1)(n-2)+n-2}\cdot\operatorname{res}\left(r^{(n-1)}(X),f^{(n-1)}_{m}(X)\right)=$
$(-1)^{n}\cdot\operatorname{res}\left(r^{(n-1)}(X),f^{(n-1)}_{m}(X)\right).$

Hence,

$\operatorname{res}\left(r^{(n)}(X),f^{(n)}_{m}(X)\right)=(-1)^{n}\cdot
3^{n-1}\cdot\operatorname{res}\left(r^{(n-1)}(X),f^{(n-1)}_{m}(X)\right).$

By iterating this equation, and using that
$\operatorname{res}(r^{(1)}(X),f^{(1)}_{m}(X))=-1$, we obtain

(16)
$\operatorname{res}\left(r^{(n)}(X),f^{(n)}_{m}(X)\right)=3^{\frac{n(n-1)}{2}}\cdot(-1)^{\frac{n(n+1)}{2}}.$

Finally, we obtain the resultant of $f^{(n)}_{m}(X)$ and $f^{(n-1)}_{m}(X)$
with respect to $X$.
By (1), we have

$f^{(n)}_{m}(X)=(X-m)\cdot f^{(n-1)}_{m}(X)+(m^{2}+m+1)\cdot r^{(n-1)}(X).$

Therefore, by (11), (13) and (16)

$\operatorname{res}\left(f^{(n)}_{m}(X),f^{(n-1)}_{m}(X)\right)=\operatorname{res}\left((m^{2}+m+1)\cdot
r^{(n-1)}(X),f^{(n-1)}_{m}(X)\right)=$
$(m^{2}+m+1)^{n-1}\cdot\operatorname{res}\left(r^{(n-1)}(X),f^{(n-1)}_{m}(X)\right)=$
$(m^{2}+m+1)^{n-1}\cdot 3^{\frac{(n-1)(n-2)}{2}}\cdot(-1)^{\frac{n(n-1)}{2}}.$

∎

######  Theorem 8.

For $n\geq 2$ the discriminant of $f^{(n)}_{m}(X)$ with respect to $X$ is

(17) $D_{f^{(n)}_{m}}=3^{\frac{(n-1)(n-2)}{2}}\cdot
n^{n}\cdot(m^{2}+m+1)^{n-1}.$

###### Proof.

The discriminant of $f^{(n)}_{m}(X)$ with respect to $X$, is

$D_{f^{(n)}_{m}}=(-1)^{\frac{n(n-1)}{2}}\cdot\operatorname{res}\left(f^{(n)}_{m}(X),\left(f^{(n)}_{m}\right)^{\prime}(X)\right).$

By Lemma 2 and (13):

$D_{f^{(n)}_{m}}=(-1)^{\frac{n(n-1)}{2}}\cdot\operatorname{res}\left(f^{(n)}_{m}(X),n\cdot
f^{(n-1)}_{m}(X)\right)=n^{n}\cdot(-1)^{\frac{n(n-1)}{2}}\cdot\operatorname{res}\left(f^{(n)}_{m}(X),f^{(n-1)}_{m}(X)\right).$

Thus by Lemma 7,

$D_{f^{(n)}_{m}}=n^{n}\cdot(m^{2}+m+1)^{n-1}\cdot(-1)^{\frac{n(n-1)}{2}}\cdot\operatorname{res}\left(r^{(n-1)}(X),f^{(n-1)}_{m}(X)\right)=$
$=3^{\frac{(n-1)(n-2)}{2}}\cdot n^{n}\cdot(m^{2}+m+1)^{n-1}.$

∎

###  2.2 Irreducibility of $f^{(n)}_{m}(X)$

In this section, we will give a sufficient condition for the irreducibility of
$f^{(n)}_{m}(X)$ over ${\mathbb{Q}}$. For the investigation of an integral
basis of these polynomials, we will use exactly the same condition, and
therefore we do not need the necessary condition of the irreducibility.
Here $v_{p}(x)$ denotes the $p$-adic valuation of a rational number $x$.

######  Lemma 9.

Let $n\geq 2$. If there exist a prime $p\neq 3$, such that
$v_{p}(m^{2}+m+1)=1$, then $v_{p}\left(f^{(n)}_{m}(m)\right)=1$.

###### Proof.

By Lemma 1, if $n\geq 2$ then

$r^{(n)}(m)=(2m+1)\cdot r^{(n-1)}(m)-f^{(n-1)}_{m}(m)$

and

$f^{(n)}_{m}(m)=(m^{2}+m+1)\cdot r^{(n-1)}(m),$

thus

$r^{(n)}(m)=(2m+1)\cdot r^{(n-1)}(m)-(m^{2}+m+1)\cdot r^{(n-2)}(m).$

We have $v_{p}(2m+1)=0$, since $4(m^{2}+m+1)-(2m+1)^{2}=3$ and $p\neq 3$.
Now assume that
$v_{p}\left(r^{(n-1)}(m)\right)=v_{p}\left(r^{(n-2)}(m)\right)$, then

$v_{p}\left((2m+1)\cdot r^{(n-1)}(m)\right)=0+v_{p}\left(r^{(n-1)}(m)\right),$
$v_{p}\left((m^{2}+m+1)\cdot
r^{(n-2)}(m)\right)=1+v_{p}\left(r^{(n-1)}(m)\right),$

whence

$v_{p}\left(r^{(n)}(m)\right)=\min(v_{p}\left(r^{(n-1)}(m)\right),v_{p}\left(r^{(n-1)}(m)\right)+1)=v_{p}\left(r^{(n-1)}(m)\right).$

Therefore, $v_{p}\left(r^{(1)}(m)\right)=v_{p}(-1)=0$ and
$v_{p}\left(r^{(2)}(m)\right)=v_{p}(-2m-1)=0$ implies

$v_{p}\left(r^{(n)}(m)\right)=0$

and

$v_{p}\left(f^{(n)}_{m}(m)\right)=v_{p}\left(m^{2}+m+1\right)+v_{p}\left(r^{(n-1)}(m)\right)=1.$

∎

######  Corollary 10.

Let $n\geq 2$. If there exist a prime $p\neq 3$, such that
$v_{p}(m^{2}+m+1)=1$, then $f^{(n)}_{m}(X)$ is irreducible over
${\mathbb{Q}}$.

###### Proof.

The Taylor series of $f^{(n)}_{m}(X)$ at $m$ is:

$f^{(n)}_{m}(X)=\sum_{i=0}^{n}\frac{\left({f^{(n)}_{m}}\right)^{(i)}(m)}{i!}\cdot(X-m)^{i}.$

Thus, by shifting the polynomial by $m$:

$f^{(n)}_{m}(X+m)=\sum_{i=0}^{n}\frac{\left({f^{(n)}_{m}}\right)^{(i)}(m)}{i!}\cdot
X^{i}.$

By Lemma 2

$\left({f^{(n)}_{m}}\right)^{(i)}(X)=\frac{n!}{(n-i)!}\cdot f^{(n-i)}_{m}(X),$

whence

$f^{(n)}_{m}(X+m)=\sum_{i=0}^{n}\binom{n}{i}\cdot f^{(n-i)}_{m}(m)\cdot
X^{i}.$

With $f^{(0)}_{m}(m)=1$ and $f^{(1)}_{m}(m)=0$, Lemma 9 shows that if $n\geq
2$, then $f^{(n)}_{m}(X+m)$ is Eisenstein with respect to $p$, so it is
irreducible, and consequently $f^{(n)}_{m}(X)$ is also irreducible. ∎

##  3 Periodically repeating integral bases

In this section we will prove, that an integral basis of the generalized
simplest fields generated by a root of $f^{(n)}_{m}(X)$ $(n\geq 2)$ is
repeating periodically, and we give an upper bound for the period length
$n_{0}$, in terms of $n$.

The periodic property of an integral basis makes sense if we investigate the
integral bases of a parametric family of number fields of type
${\mathbb{Q}}(\beta_{t})$, where $\beta_{t}$ is a root of a polynomial
$f^{(n)}_{t}(X)\in{\mathbb{Z}}[t][X]$ of degree $n$, with integer parameter
$t\in{\mathbb{Z}}$. In this case we determine an integral basis of
${\mathbb{Q}}(\beta_{t})$ of the form

$\left(h_{0}(\beta_{t}),h_{1}(\beta_{t}),h_{2}(\beta_{t}),\ldots,h_{n-1}(\beta_{t})\right),$

where $h_{i}(X)\in{\mathbb{Q}}[X]$, $(i=0,\ldots,n-1)$, are polynomials with
$\deg(h_{i}(X))=i$.

In order to construct a parametric family of polynomials with integer
coefficients from $f_{m}^{(n)}$, let

  * •

$m=t$, if $n\equiv 1,2\mod{3},$

  * •

$m=\frac{t}{3}$, if $n\equiv 0\mod{3}$.

It is easy to check, that for any $t\in{\mathbb{Z}}$, the generalized simplest
polynomials $f^{(n)}_{m}(X)$ have integer coefficients. In the following we
write $f^{(n)}_{t}(X)$, and it refers to the definition above.
In this case the modified results of the previous section are the following.

The discriminant of $f_{t}^{(n)}(X)$ with respect to $X$ is

(18) $D_{f_{t}^{(n)}}=\left\\{\begin{array}[]{lr}3^{\frac{(n-1)(n-2)}{2}}\cdot
n^{n}\cdot(t^{2}+t+1)^{n-1},&$if $n\equiv 1,2\mod{3},\\\
3^{\frac{(n-1)(n-6)}{2}}\cdot n^{n}\cdot(t^{2}+3t+9)^{n-1},&$if $n\equiv
0\mod{3}.\end{array}\right.$

If $t\in{\mathbb{Z}}$ is an integer parameter, such that there exists a prime
$p\neq 3$, for which

  * •

$v_{p}(t^{2}+t+1)=1$, if $n\equiv 1,2\mod{3},$

  * •

$v_{p}(t^{2}+3t+9)=1$, if $n\equiv 0\mod{3},$

then $f_{t}^{(n)}(X)$ is irreducible over ${\mathbb{Q}}$.

Let $\beta_{t}$ be a root of $f^{(n)}_{t}(X)$. We recall that an integral
basis of the fields ${\mathbb{Q}}(\beta_{t})$ is repeating periodically modulo
$n_{0}$, if for each residue class $r$ modulo $n_{0}$, there exist polynomials
$h^{(r)}_{i}(X)\in{\mathbb{Q}}[X]$, $(i=0,\ldots n-1)$, such that, if $t\equiv
r\mod{n_{0}}$, and $f^{(n)}_{t}(X)$ is irreducible then

$\left(h^{(r)}_{0}(\beta_{t}),h^{(r)}_{1}(\beta_{t}),h^{(r)}_{2}(\beta_{t}),\ldots,h^{(r)}_{n-1}(\beta_{t})\right),$

is an integral basis of ${\mathbb{Q}}(\beta_{t})$. (See [7],[8],[26]).
There is an equivalent definition of a periodically repeating integral basis
which is more convenient to use in practice.

######  Lemma 11.

Let $f_{t}^{(n)}(X)\in{\mathbb{Z}}[t,X]$ be a parametric family of
polynomials, where $t$ is an integer parameter such that $f_{t}^{(n)}(X)$ is
irreducible, and let $\beta_{t}$ be a root of $f_{t}^{(n)}(X)$.
An integral basis of the fields ${\mathbb{Q}}(\beta_{t})$ is repeating
periodically modulo $n_{0}$, if and only if for any $k(X)\in{\mathbb{Q}}[X]$,
$\deg(k(X))\leq n-1$, the following property is satisfied:
for any parameters $t,s\in{\mathbb{Z}}$ for which $f_{t}^{(n)}(X)$ and
$f_{s}^{(n)}(X)$ are irreducible and $n_{0}\mid(t-s)$, $k(\beta_{t})$ is an
algebraic integer if and only if $k(\beta_{s})$ is an algebraic integer.

###### Proof.

First assume, that an integral basis of ${\mathbb{Q}}(\beta_{t})$ is repeating
periodically modulo $n_{0}$. Let $V=\sum_{i=0}^{n-1}{\mathbb{Q}}X^{i}$. Then
the ${\mathbb{Q}}$-linear map $\varphi:X^{i}\mapsto\beta_{t}^{i}$ gives an
isomorphism of the ${\mathbb{Q}}$-vector spaces $V$ and
${\mathbb{Q}}(\beta_{t})$ for any $t\in{\mathbb{N}}$. Let $M$ be the
${\mathbb{Z}}$-module generated by

$h^{(r)}_{0}(X),h^{(r)}_{1}(X),h^{(r)}_{2}(X),\ldots,h^{(r)}_{n-1}(X),$

where $t\equiv r\mod{n_{0}}$. If we restrict $\varphi$ to $M$ then we obtain
an isomorphism of $M$ and the ring of integers of ${\mathbb{Q}}(\beta_{t})$
for any $t\equiv r\mod{n_{0}}$. This means that if $t\equiv r\mod{n_{0}}$,
then for any $k(X)\in{\mathbb{Q}}[X]$ with $\deg(k(X))\leq n-1$,
$k(\beta_{t})$ is an algebraic integer if and only if $k(X)\in M$. Therefore
if $t\equiv s\equiv r\mod{n_{0}}$, then $k(\beta_{t})$ is an algebraic integer
if and only if $k(\beta_{s})$ is an algebraic integer.
On the other hand, assume that for any polynomial $k(X)\in{\mathbb{Q}}[X]$
with $\deg(k(X))\leq n-1$, $k(\beta_{t})$ is an algebraic integer if and only
if $k(\beta_{s})$ is an algebraic integer. Let $r\equiv t\equiv s\mod{n_{0}}$
and $h^{(r)}_{i}(X)\in{\mathbb{Q}}[X]$, $(i=0,\ldots n-1)$ be polynomials,
such that

$\left(h^{(r)}_{0}(\beta_{t}),h^{(r)}_{1}(\beta_{t}),h^{(r)}_{2}(\beta_{t}),\ldots,h^{(r)}_{n-1}(\beta_{t})\right)$

is an integral basis of ${\mathbb{Q}}(\beta_{t})$, where
$\deg(h^{(r)}_{i}(X))=i$ (one can use the Hermite normal form representation
of any integral basis of ${\mathbb{Q}}(\beta_{t})$ with respect to the
${\mathbb{Q}}$-basis
$\left\\{1,\beta_{t},\beta_{t}^{2},\ldots,\beta_{t}^{n-1}\right\\}$ to obtain
such $h^{(r)}_{i}(X)$ polynomials). Since all elements of an integral basis
are algebraic integers, by our assumption we have that
$h^{(r)}_{i}(\beta_{s})$, is algebraic integer for any $i=0,\ldots n-1$. We
should prove, that the elements $h^{(r)}_{i}(\beta_{s})$ $(i=0,\ldots n-1)$
form an integral basis of ${\mathbb{Q}}(\beta_{s})$.
Let $k(X)\in{\mathbb{Q}}[X]$, $\deg(k(X))\leq n-1$, be a polynomial, such that
$k(\beta_{s})$ is an algebraic integer. By our assumption $k(\beta_{t})$ is
also an algebraic integer, thus we can represent it in the integral basis

$\left(h^{(r)}_{0}(\beta_{t}),h^{(r)}_{1}(\beta_{t}),h^{(r)}_{2}(\beta_{t}),\ldots,h^{(r)}_{n-1}(\beta_{t})\right)$

with integer coefficients. Therefore we can write

$k(X)=k_{0}\cdot h^{(r)}_{0}(X)+k_{1}\cdot h^{(r)}_{1}(X)+\ldots+k_{n-1}\cdot
h^{(r)}_{n-1}(X),\qquad k_{0},k_{1},\ldots,k_{n-1}\in{\mathbb{Z}}.$

By substituting $\beta_{s}$ in place of $\beta_{t}$ in the equation above, we
obtain that we can represent $k(\beta_{s})$ as an integer linear combination
of the elements $h^{(r)}_{i}(\beta_{s})$, $(i=0,\ldots n-1)$. Therefore, if
these elements are linearly independent over ${\mathbb{Q}}$, then

$\left(h^{(r)}_{0}(\beta_{s}),h^{(r)}_{1}(\beta_{s}),h^{(r)}_{2}(\beta_{s}),\ldots,h^{(r)}_{n-1}(\beta_{s})\right)$

is an integral basis of ${\mathbb{Q}}(\beta_{s})$, and an integral basis of
the fields ${\mathbb{Q}}(\beta_{t})$ is repeating periodically modulo $n_{0}$.
However, it is trivial, since $\deg(h^{(r)}_{i}(X))=i$, $(i=0,\ldots n-1)$,
and the minimal polynomial of $\beta_{s}$ is the irreducible polynomial
$f_{s}^{(n)}(X)$ of degree $n$. ∎

###  3.1 Upper bound for the period length

In this section we prove, that if the 3-free part of $t^{2}+t+1$ (or
$t^{2}+3t+9$ respectively) is square-free, then an integral basis of
${\mathbb{Q}}(\beta_{t})$ is repeating periodically modulo $n_{0}$, where

$n_{0}^{2}\mid\left(3^{\frac{n^{2}}{2}}\cdot n^{n}\right)^{n}.$

First we show, that if the 3-free part of $t^{2}+t+1$ (or $t^{2}+3t+9$
respectively) is square-free, then the index $\operatorname{ind}(\beta_{t})$
of $\beta_{t}$ is independent of $t$. Therefore, for any $n\in{\mathbb{N}}$
there exists an integer $C_{n}$, such that for any algebraic integer
$\gamma_{t}\in{\mathbb{Q}}(\beta_{t})$:

$C_{n}\cdot\gamma_{t}\in{\mathbb{Z}}[\beta_{t}].$

It means, that we can represent $\gamma_{t}$ in the ${\mathbb{Q}}$-basis

$\left\\{1,\beta_{t},\beta_{t}^{2},\ldots,\beta_{t}^{n-1}\right\\},$

with rational coefficients having denominator $d\mid C_{n}$.
Finally, we prove that if such $C_{n}$ exists, then an integral basis of these
fields is repeating periodically modulo $n_{0}=C_{n}^{n}$.

The index of an algebraic integer $\beta_{t}$ is defined by

$\operatorname{ind}(\beta_{t}):=({\cal
O}_{{\mathbb{Q}}(\beta_{t})}^{+}:{\mathbb{Z}}[\beta_{t}]^{+}),$

where ${\cal O}_{{\mathbb{Q}}(\beta_{t})}$ is the ring of integers of
${\mathbb{Q}}(\beta_{t})$. That is the index of the additive group of the
polynomial ring ${\mathbb{Z}}[\beta_{t}]$ in the additive group of ${\cal
O}_{{\mathbb{Q}}(\beta_{t})}$. (for more details see for e.g. I.Gaál [5],
Section 1.2).
This is in fact, the product of the denominators of the elements of an
integral basis of ${\mathbb{Q}}(\beta_{t})$, represented in the
${\mathbb{Q}}$-basis $\left(1,\beta_{t},\ldots,\beta_{t}^{n-1}\right)$.
If $D(\beta_{t})$ denotes the discriminant of the minimal polynomial of
$\beta_{t}$ (which is equal to the discriminant of the basis
$\left(1,\beta_{t},\ldots,\beta_{t}^{n-1}\right)$), and $D_{M}$ denotes the
discriminant of the field $M={\mathbb{Q}}(\beta_{t})$, then

(19) $D(\beta_{t})=\operatorname{ind}(\beta_{t})^{2}\cdot D_{M}.$

######  Lemma 12.

([23] Chapter II., Lemma 2.17) Let $M$ be a number field of degree $n$, and
let $\beta\in M$ be a nonzero algebraic integer of degree $n$. Suppose that
the minimal polynomial of $\beta$ is Eisenstein with respect to the prime $p$.
Then $p\nmid\operatorname{ind}(\beta_{t})$.

By this lemma, we can exclude certain prime divisors of the discriminant
$D(\beta_{t})$.

######  Theorem 13.

Let $t$ be an integer and $\beta_{t}$ be a root of $f^{(n)}_{t}(X)$. If $p\neq
3$ is a prime, for which

$v_{p}(t^{2}+t+1)=1,\qquad\mbox{ if }n\equiv 1,2\mod{3},$
$v_{p}(t^{2}+3t+9)=1,\qquad\mbox{ if }n\equiv 0\mod{3},$

then $p\nmid\operatorname{ind}(\beta_{t})$.

###### Proof.

If $n\equiv 1$ or $2\mod{3}$, then by Corollary 10, the minimal polynomial of
$\beta_{t}-t$ is Eisenstein with respect to $p$, so by Lemma 12,
$p\nmid\operatorname{ind}(\beta_{t}-t)$. Since the index of an algebraic
integer is invariant under translation, we obtain
$p\nmid\operatorname{ind}(\beta_{t})$.
If $n\equiv 0\mod{3}$, then by Corollary 10, the minimal polynomial of
$3\beta_{t}-t$ is Eisenstein with respect to $p$, so by Lemma 12,
$p\nmid\operatorname{ind}(3\beta_{t}-t)$, and therefore
$p\nmid\operatorname{ind}(3\beta_{t})$. If $p\neq 3$, then
$v_{p}(\operatorname{ind}(3\beta_{t}))=v_{p}(\operatorname{ind}(\beta_{t}))$,
and thus we have $p\nmid\operatorname{ind}(\beta_{t})$ in this case, too. ∎

By (18) and (19)

$\operatorname{ind}(\beta_{t})^{2}\mid 3^{\frac{(n-1)(n-2)}{2}}\cdot
n^{n}\cdot(t^{2}+t+1)^{n-1},\qquad\mbox{ if }n\equiv 1,2\mod{3},$
$\operatorname{ind}(\beta_{t})^{2}\mid 3^{\frac{(n-1)(n-6)}{2}}\cdot
n^{n}\cdot(t^{2}+3t+9)^{n-1},\qquad\mbox{ if }n\equiv 0\mod{3}.$

Hence, by Theorem 13, if the 3-free part of $t^{2}+t+1$ (or $t^{2}+3t+9$,
respectively) is square-free, then

$\operatorname{ind}(\beta_{t})^{2}\mid
3^{\frac{(n-1)(n-2)}{2}+v_{3}(t^{2}+t+1)}\cdot n^{n},\qquad\mbox{ if }n\equiv
1,2\mod{3},$ $\operatorname{ind}(\beta_{t})^{2}\mid
3^{\frac{(n-1)(n-6)}{2}+v_{3}(t^{2}+3t+9)}\cdot n^{n},\qquad\mbox{ if }n\equiv
0\mod{3}.$

But it is easy to check, that for any $t\in{\mathbb{Z}}$,
$v_{3}(t^{2}+t+1)\leq 1$ and $v_{3}(t^{2}+3t+9)\leq 3$. Thus we get the
following corollary.

######  Corollary 14.

Let $\beta_{t}$ be a root of $f^{(n)}_{t}(X)$, and $C_{n}$ be the greatest
integer, such that

$C_{n}^{2}\mid 3^{\frac{n^{2}-3n+4}{2}}\cdot n^{n},\qquad\mbox{ if }n\equiv
1,2\mod{3},$ $C_{n}^{2}\mid 3^{\frac{n^{2}-7n+12}{2}}\cdot n^{n},\qquad\mbox{
if }n\equiv 0\mod{3}.$

If $t^{2}+t+1$ (or $t^{2}+3t+9$, respectively) is square-free, then for any
algebraic integer $\gamma_{t}\in{\mathbb{Q}}(\beta_{t})$,

$C_{n}\cdot\gamma_{t}\in{\mathbb{Z}}[\beta_{t}].$

Therefore, in both cases

$C_{n}^{2}\mid 3^{\frac{n^{2}}{2}}\cdot n^{n}.$

Following similar approach as in Theorem 2 of [7], we show, that it implies
that an integral basis of these fields is repeating periodically.

######  Lemma 15.

Let $\beta_{t}$ be a root of $f_{t}^{(n)}(X)$. If there exists an integer
$C_{n}$, such that for any algebraic integer
$\gamma_{t}\in{\mathbb{Q}}(\beta_{t})$,

$C_{n}\cdot\gamma_{t}\in{\mathbb{Z}}[\beta_{t}],$

then an integral basis of the fields ${\mathbb{Q}}(\beta_{t})$ is repeating
periodically modulo $n_{0}=C_{n}^{n}$.

###### Proof.

Let $\gamma_{t}\in{\mathbb{Q}}(\beta_{t})$ be an algebraic integer and
represent it the ${\mathbb{Q}}$-basis
$\left\\{1,\beta_{t},\beta_{t}^{2},\ldots,\beta_{t}^{n-1}\right\\}$:

$\gamma_{t}=\frac{a_{0}+a_{1}\beta_{t}+a_{2}\beta_{t}^{2}+\ldots+a_{n-1}\beta_{t}^{n-1}}{d},\qquad
a_{i}\in{\mathbb{Z}},d\mid C_{n}.$

Denote the conjugates of $\gamma_{t}$ by $\gamma_{t}^{(j)}$, $j=1,\ldots,n$.
The defining polynomial of $\gamma_{t}$ is

$\prod_{j=1}^{n}(X-\gamma_{t}^{(j)})=\frac{1}{d^{n}}\prod_{j=1}^{n}(dX-
a_{0}-a_{1}\beta_{t}^{(j)}-a_{2}(\beta_{t}^{(j)})^{2}-\ldots-
a_{n-1}(\beta_{t}^{(j)})^{n-1}).$

The product is a symmetric polynomial of
$\beta_{t}^{(1)},\beta_{t}^{(2)},\ldots,\beta_{t}^{(n)}$, hence its
coefficients can be expressed as integer polynomials of the coefficients of
the defining polynomial of $\beta_{t}$, that is $f_{t}^{(n)}(X)$. The
coefficients of $f_{t}^{(n)}(X)$ are integer polynomials of $t$, hence there
exist polynomials $P_{0},\ldots,P_{n-1}\in{\mathbb{Z}}[X]$, such that

$\prod_{j=1}^{n}(X-\gamma_{t}^{(j)})=\frac{1}{d^{n}}\left(d^{n}X^{n}+P_{n-1}(t)(dX)^{n-1}+\ldots+P_{1}(t)(dX)+P_{0}(t)\right).$

Therefore the element $\gamma_{t}$ is an algebraic integer if $d^{n}\mid
d^{j}P_{j}(t)$ for all $j=1,\ldots,n-1$.
Now let $s\in{\mathbb{Z}}$ be an integer, such that $C_{n}^{n}\mid(t-s)$. Then
$C_{n}^{n}\mid P_{j}(t)-P_{j}(s)$, and therefore $d^{n}\mid d^{j}P_{j}(t)$ if
and only if $d^{n}\mid d^{j}P_{j}(s)$. It implies that for any
$k(X)\in{\mathbb{Q}}[X]$, $(\deg(k(X))\leq n-1)$, $\gamma_{t}=k(\beta_{t})$ is
algebraic integer if and only if $\gamma_{s}=k(\beta_{s})$ is an algebraic
integer, thus by Lemma 11 an integral basis of ${\mathbb{Q}}(\beta_{t})$ is
repeating periodically modulo $n_{0}=C_{n}^{n}$. ∎

By Corollary 14 and Lemma 15, we obtain the main result of this section.

######  Theorem 16.

Let $\beta_{t}$ be a root of $f^{(n)}_{t}(X)$, and $n_{0}$ be a greatest
integer, such that

$n_{0}^{2}\mid\left(3^{\frac{n^{2}}{2}}\cdot n^{n}\right)^{n}.$

If $t^{2}+t+1$ (or $t^{2}+3t+9$, respectively) is square-free, then, an
integral basis of the fields ${\mathbb{Q}}(\beta_{t})$ is repeating
periodically modulo $n_{0}$.

###  3.2 Special cases

In this section we give better upper bounds for the period lengths of the
integral bases of the fields ${\mathbb{Q}}(\beta_{t})$ of degree $n\leq 12$.
We will obtain the better upper bounds for the period length by using the dual
bases of certain rational bases of these fields. For the advantage of the dual
basis see for e.g. [24].

Let $M$ be an algebraic number field of degree $n$. If
$(\beta_{0},\beta_{1},\ldots,\beta_{n-1})$ is a ${\mathbb{Q}}$-basis of $M$,
then its dual basis is $(\gamma_{0},\gamma_{1},\ldots,\gamma_{n-1})$, where
$\gamma_{i}\in M$, such that

$Tr_{M/{\mathbb{Q}}}(\gamma_{i}\cdot\beta_{j})=\delta_{ij}=\left\\{\begin{array}[]{ll}1,&$if
$i=j,\\\ 0,&$otherwise.$\end{array}\right.$

The dual basis of any ${\mathbb{Q}}$-basis of $M$ always uniquely exists.
Let $\left(\beta_{0},\beta_{1},\ldots,\beta_{n-1}\right)$ be a
${\mathbb{Q}}$-basis of $M$ containing algebraic integers, and
$\left(\gamma_{0},\gamma_{1},\ldots,\gamma_{n-1}\right)$ its dual basis. Then
it is easy to see, that we can represent any algebraic integer $\gamma\in M$
in the ${\mathbb{Q}}$-basis
$\left(\gamma_{0},\gamma_{1},\ldots,\gamma_{n-1}\right)$ with integer
coefficients.
Let $\gamma\in M$ be an algebraic integer, and represent is in the
${\mathbb{Q}}$-basis $\left(\gamma_{0},\gamma_{1},\ldots,\gamma_{n-1}\right)$:

$\gamma=f_{0}\gamma_{0},+f_{1}\gamma_{1}+\ldots+f_{n-1}\gamma_{n-1}.$

By the definition of the dual basis,
$f_{i}=Tr_{M/{\mathbb{Q}}}(\gamma\cdot\beta_{i})$, which is a rational
integer, since $\gamma$ and $\beta_{i}$ both algebraic integers.
Let $\beta_{t}$ be a root of $f^{(n)}_{t}(X)$, and
$\left(\gamma_{0},\gamma_{1},\ldots,\gamma_{n-1}\right)$ be the dual basis of
$\left(1,\beta_{t},\beta_{t}^{2}\ldots,\beta_{t}^{n-1}\right)$ in
${\mathbb{Q}}(\beta_{t})$. Represent the elements $\gamma_{i}$ of the dual
basis in the ${\mathbb{Q}}$-basis
$\left(1,\beta_{t},\beta_{t}^{2}\ldots,\beta_{t}^{n-1}\right)$:

$\gamma_{i}=\sum_{j=0}^{n-1}c_{ij}\beta_{t}^{j}.$

Let $d$ be the least common multiple of the denominators of the rational
numbers $c_{ij}$, $(i,j=0,\ldots,n-1)$. Then we can represent any algebraic
integer $\gamma\in{\mathbb{Q}}(\beta_{t})$ in the ${\mathbb{Q}}$-basis
$\left(\gamma_{0},\gamma_{1},\ldots,\gamma_{n-1}\right)$ with integer
coefficients, thus in the ${\mathbb{Q}}$-basis
$\left(1,\beta_{t},\beta_{t}^{2}\ldots,\beta_{t}^{n-1}\right)$, we can
represent $\gamma$ with rational coefficients having denominator $d$. By the
results of the previous section, an integral basis of the fields
${\mathbb{Q}}(\beta_{t})$ is repeating periodically modulo $d^{n}$.
We explicitly calculated these dual bases for $n=2,\ldots,12$, and obtained
Theorem 17.
For instance, if $n=4$, the dual basis of
$\left(1,\beta_{t},\beta_{t}^{2},\beta_{t}^{3}\right)$ is

$\displaystyle\left(\begin{array}[]{rrrr}\frac{8t^{2}+3t+12}{12(t^{2}+t+1)}&+\frac{-4t^{2}+7t+18}{12(t^{2}+t+1)}\cdot\beta_{t}&+\frac{-8t^{2}+10t}{12(t^{2}+t+1)}\cdot\beta_{t}^{2}&+\frac{2t-3}{12(t^{2}+t+1)}\cdot\beta_{t}^{3},\\\
&&&\\\
\frac{-4t^{2}+7t+18}{12(t^{2}+t+1)}&+\frac{40t^{2}+92t+62}{12(t^{2}+t+1)}\cdot\beta_{t}&+\frac{32t^{2}+46t+5}{12(t^{2}+t+1)}\cdot\beta_{t}^{2}&+\frac{-8t-11}{12(t^{2}+t+1)}\cdot\beta_{t}^{3},\\\
&&&\\\
\frac{-8t^{2}+10t}{12(t^{2}+t+1)}&+\frac{32t^{2}+46t+5}{12(t^{2}+t+1)}\cdot\beta_{t}&+\frac{32t^{2}+8t+1}{12(t^{2}+t+1)}\cdot\beta_{t}^{2}&+\frac{-8t-1}{12(t^{2}+t+1)}\cdot\beta_{t}^{3},\\\
&&&\\\
\frac{2t-3}{12(t^{2}+t+1)}&+\frac{-8t-11}{12(t^{2}+t+1)}\cdot\beta_{t}&+\frac{-8t-1}{12(t^{2}+t+1)}\cdot\beta_{t}^{2}&+\frac{2}{12(t^{2}+t+1)}\cdot\beta_{t}^{3}\end{array}\right)$

######  Theorem 17.

With the notation above, if $n\not\equiv 0\mod{3}$ and $t^{2}+t+1$ is square-
free, else $t^{2}+3t+9$ is square-free, then an integral basis of the fields
${\mathbb{Q}}(\beta_{t})$ is repeating periodically modulo $n_{0}$ given in
the following table:

$\scriptsize\begin{array}[]{c|c|c|c|c|c|c|c|c|c|c|c}n&2&3&4&5&6&7&8&9&10&11&12\\\
\hline\cr n_{0}&2^{2}&1^{3}&(3\cdot 4)^{4}&(3^{3}\cdot 5)^{5}&(3^{2}\cdot
6)^{6}&(3^{4}\cdot 7)^{7}&(3^{5}\cdot 8)^{8}&(3^{4}\cdot 9)^{9}&(3^{6}\cdot
10)^{10}&(3^{9}\cdot 11)^{11}&(3^{8}\cdot 12)^{12}\end{array}$

###### Proof.

In all cases, the least common multiple $d$ of the denominators of the
elements of the dual basis represented in the ${\mathbb{Q}}$-basis
$\left(1,\beta_{t},\beta_{t}^{2}\ldots,\beta_{t}^{n-1}\right)$, has the form

$d=3^{\delta_{n}}\cdot n\cdot(t^{2}+t+1),\qquad\mbox{ if }n\not\equiv
0\mod{3},$

or

$d=3^{\delta_{n}}\cdot n\cdot(t^{2}+3t+9),\qquad\mbox{ if }n\equiv 0\mod{3},$

with

$\begin{array}[]{c|c|c|c|c|c|c|c|c|c|c|c}n&2&3&4&5&6&7&8&9&10&11&12\\\
\hline\cr\delta_{n}&0&0&1&3&2&4&5&4&6&9&8\end{array}$

By Corollary 14 we can represent any algebraic integer in
${\mathbb{Q}}(\beta_{t})$ in the ${\mathbb{Q}}$-basis
$\left(1,\beta_{t},\beta_{t}^{2}\ldots,\beta_{t}^{n-1}\right)$, having
denominator $C_{n}$, for which,

$C_{n}^{2}\mid 3^{\frac{n^{2}}{2}}\cdot n^{n}.$

Thus, we can represent any algebraic integer
$\gamma\in{\mathbb{Q}}(\beta_{t})$ in the ${\mathbb{Q}}$-basis
$(1,\beta_{t},\beta_{t}^{2}\ldots,\beta_{t}^{n-1})$, with rational coefficient
having denominator

$\gcd(3^{\delta_{n}}\cdot n,C_{n}),$

whence by Lemma 15, an integral basis of these fields repeating periodically
modulo $n_{0}$ given in the table. ∎

However these period lengths are much smaller than we give in Theorem 16, the
smallest period lengths are even smaller. It was shown in [8], that an
integral basis of the simplest sextic fields is repeating periodically modulo
$36$ instead of $54^{6}$. In order to find the smallest period lengths, we
have to calculate an integral basis of the fields belong to the parameters
less than the period length given in the table above, and detect the
$h_{0}^{(r)},\ldots,h_{n-1}^{(r)}$ polynomials, that are repeated. If the
period length given in the table is too large for this calculations, we can
use the method described in [7] Theorem 3. For this we first have to make a
guess for the smallest period length $n_{0}$, then we have to show, that there
is no algebraic integer represented in the initial basis belonging to the
residue class of our parameter, with rational coefficient having prime
denominator $p$. We have to test it for all residue classes of $n_{0}$ and for
all possible primes, which divides the initial period length given in the
above table. For all of these primes this calculation requires to calculate an
integral basis of $n_{0}\cdot p^{n}$ number fields.
We can manage it for $n=2,3,4,5,6,8,9$ and $12$, but for $n=7,10,11$, the
$7^{7},5^{10},11^{11}$ tests would take too much time.
Therefore, we obtained the following result.

######  Theorem 18.

With the notation above, if $n\not\equiv 0\mod{3}$ and $t^{2}+t+1$ is square-
free, else $t^{2}+3t+9$ is square-free, then an integral basis of the fields
${\mathbb{Q}}(\beta_{t})$ is repeating periodically modulo $n_{0}$ given in
the following table:

$\begin{array}[]{c|c|c|c|c|c|c|c|c}n&2&3&4&5&6&8&9&12\\\ \hline\cr
n_{0}&4&1&24&75&36&432&1&1944\end{array}$

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