# On integral bases and monogenity of pure octic number fields with non-square
free parameters

Lhoussain El Fadil and István Gaál  Faculty of Sciences Dhar El Mahraz, P.O.
Box 1796 Atlas-Fes, Sidi Mohamed ben Abdellah University, Morocco
<EMAIL_ADDRESS>University of Debrecen, Mathematical Institute, H-4002
Debrecen Pf.400, Hungary<EMAIL_ADDRESS>
###### Abstract.

In all available papers, on power integral bases of pure octic number fields
$K$, generated by a root $\alpha$ of a monic irreducible polynomial
$f(x)=x^{8}-m\in\mathbb{Z}[x]$, it was assumed that $m\neq\pm 1$ is square
free. In this paper, we investigate the monogenity of any pure octic number
field, without the condition that $m$ is square free. We start by calculating
an integral basis of $\mathbb{Z}_{K}$, the ring of integers of $K$. In
particular, we characterize when $\mathbb{Z}_{K}=\mathbb{Z}[\alpha]$. We give
sufficient conditions on $m$, which guarantee that $K$ is not monogenic. We
finish the paper by investigating the case when $m=a^{u}$, $u\in\\{1,3,5,7\\}$
and $a\neq\mp 1$ is a square free rational integer.

###### Key words and phrases:

Power integral basis, index, Theorem of Ore, prime ideal factorization

###### 2010 Mathematics Subject Classification:

11R04, 11R16, 11R21

##  1\. Introduction

Let $K=\mathbb{Q}(\alpha)$ be a number field generated by a root $\alpha$ of a
monic irreducible polynomial $f(x)\in\mathbb{Z}[x]$ of degree $n$, denote by
$\mathbb{Z}_{K}$ its ring of integers and $d_{K}$ its discriminant. The ring
$\mathbb{Z}_{K}$ is said to have a power integral basis if it has a
$\mathbb{Z}$-basis $(1,\theta,\cdots,\theta^{n-1})$ for some
$\theta\in\mathbb{Z}_{K}$. That means $\mathbb{Z}_{K}=\mathbb{Z}[\theta]$,
that is $\mathbb{Z}_{K}$ is mono-generated as a ring, with a single generator
$\theta$. In this case, the field $K$ is said to be monogenic and not
monogenic otherwise.

The problem of deciding the monogenity of a number field and constructing
power integral bases is a classical problem of algebraic number theory, going
back to Dedekind [6], Hensel [18] and Hasse [16]. This area is intensively
studied even nowadays, cf. [10] for the present state of this research.

It is well-known that $\mathbb{Z}_{K}$ is a free $\mathbb{Z}$-module of rank
$n$. For any primitive element $\theta\in\mathbb{Z}_{K}$ (that is
$K=\mathbb{Q}(\theta)$), $\mathbb{Z}[\theta]$ is a subgroup of
$\mathbb{Z}_{K}$ of finite index. We call

$\text{ind}(\theta)=(\mathbb{Z}_{K}:\mathbb{Z}[\theta])$

the index of $\theta$. Obviously, $\text{ind}(\theta)=1$ if and only if
$(1,\theta,\ldots,\theta^{n-1})$ is an integral basis of $\mathbb{Z}_{K}$. As
it is known [10], we have

$\triangle(\theta)=\text{ind}(\theta)^{2}\cdot d_{K}$

where $\triangle(\theta)$ is the discriminant of $\theta$.

The greatest common divisor $i(K)$ of the indices of the primitive elements
$\theta\in\mathbb{Z}_{K}$ of $K$ is called the (field) index of $K$. It is
clear that if $i(K)>1$, then $K$ is not monogenic. On the other hand, $i(K)=1$
does not imply that $K$ is monogenic. If a prime $p$ divides $i(K)$, then $p$
is called a prime common index divisor of $K$.

Let $(1,\omega_{1},\ldots,\omega_{n-1})$ be an integral basis of
$\mathbb{Z}_{K}$. The discriminant $\triangle(L(X_{1},\ldots,X_{n}))$ of the
linear form
$L(X_{1},\ldots,X_{n-1})=\omega_{1}X_{1}+\ldots+\omega_{n-1}X_{n-1}$ can be
written (cf. [10]) as

$\triangle(L(X_{1},\ldots,X_{n-1}))=(\text{ind}(X_{1},\ldots,X_{n-1}))^{2}\cdot
d_{K},$

where $\text{ind}(X_{1},\ldots,X_{n-1})$ is the index form corresponding to
the integer basis $(1,\omega_{1},\ldots,\omega_{n-1})$ having the property
that for any
$\theta=x_{0}+\omega_{1}x_{1}+\ldots+\omega_{n-1}x_{n-1}\in\mathbb{Z}_{K}$
(with $x_{0},x_{1},\ldots,x_{n-1}\in\mathbb{Z}$) we have
$\text{ind}(\theta)=|\text{ind}(x_{1},\ldots,x_{n-1})|$. Therefore $\theta$ is
a generator of a power integral basis if and only if
$x_{1},\ldots,x_{n-1}\in\mathbb{Z}$ is a solution of the index form equation
$\text{ind}(x_{1},\ldots,x_{n-1})=\pm 1$.

According to our notation $K$ is generated by a root $\alpha$ of a monic
irreducible polynomial $f\in\mathbb{Z}[x]$. The index

$\text{ind}(f)=(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$

is called the index of the polynomial $f$ and we also have

$\triangle(f)=\text{ind}(f)^{2}\cdot d_{K},$

$\triangle(f)$ denoting the discriminant of $f$.

The main tool of our proofs will be the theory of Newton polygons. If a prime
$p$ does not divide $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, then a well-known
theorem of Dedekind says that the factorization of $p\mathbb{Z}_{K}$ can be
derived directly from the factorization of $\overline{f(x)}$ in
$\mathbb{F}_{p}[x]$. In order to apply this theorem in an effective way, one
needs a criterion to test whether $p$ divides the index
$(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. Such a criterion was proved by Dedekind
in 1878, (see [5, Theorem 6.1.4]). When Dedekind’s criterion fails, then a
method of Ore $1928$, based on Newton polygon techniques, can be used in order
to evaluate the index $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, the absolute
discriminant of $K$, and the prime ideal factorization of the rational primes
into powers of prime ideals of $\mathbb{Z}_{K}$ (see [19, 21]). In case Ore’s
method also fails, then an algorithm developed by Guardia, Montes, and Nart
[13], based on higher order Newton polygons can be used. Such an algorithm
gives after a finite number of iterations a complete answer on the index
$(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, the absolute discriminant $d_{K}$ of
$K$, and the factorization of $p\mathbb{Z}_{K}$.

Among all types of number fields, most investigations deal with pure number
fields $K$ generated by a root of an irreducible polynomial $x^{n}-m$.
Assuming that $m$ is square free, Gaál and Remete [11] studied pure quartic
fields, Ahmad, Nakahara and Husnine [3, 1], Ahmad, Nakahara and Hameed [2], El
Fadil [8] pure sextic fields. Applying the index forms, Gaál and Remete [12]
investigated pure number fields of degrees $3\leq n\leq 9$.

The subject of our present paper is the monogenity of pure octic fields. For
square free $m$ Hameed and Nakahara [15], proved that if $m\equiv
1\;(\bmod\;16)$, then the octic number field generated by $m^{1/8}$ is not
monogenic, but if $m\equiv 2,3\;(\bmod\;4)$, then it is monogenic.

In all above quoted papers, the authors consider only pure octic number fields
generated by a root of a monic polynomial $f(x)=x^{8}-m$, where $m\neq\pm 1$
is a square free rational integer. Our purpose is to extend these results to
arbitrary parameters $m$, without assuming that $m$ is square free. We start
by calculating an integral basis of $\mathbb{Z}_{K}$ in Theorem 2.1. We give
sufficient conditions on $m$, which guarantee the non monogenity of $K$. We
conclude the paper by studying the case where $m=a^{u}$, $u\in\\{1,3,5,7\\}$,
and $a\neq\mp 1$ is a square free rational integer.

##  2\. Main results

Throughout this paper, $m\neq\pm 1$ is a rational integer such that the
polynomial $f(x)=x^{8}-m$ is irreducible over $\mathbb{Q}$. Let $\alpha$ be a
root of $f(x)$, let $K=\mathbb{Q}(\alpha)$ with ring of integers
$\mathbb{Z}_{K}$. Replacing $\alpha$ by $\frac{\alpha}{p}$ and $m$ by
$\frac{m}{p^{8}}$, and repeating this process until we get $\nu_{p}(m)<8$ for
every prime integer $p$, we can assume that
$m=a_{1}a_{2}^{2}a_{3}^{3}a_{4}^{4}a_{5}^{5}a_{6}^{6}a_{7}^{7}$, where
$a_{1},\dots,a_{7}$ are square free pairwise coprime rational integers. Set
$A_{2}=a_{4}a_{5}a_{6}a_{7}$, $A_{3}=a_{3}a_{4}a_{5}a_{6}^{2}a_{7}^{2}$,
$A_{4}=a_{2}a_{3}a_{4}^{2}a_{5}^{2}a_{6}^{3}a_{7}^{3}$,
$A_{5}=a_{2}a_{3}a_{4}^{2}a_{5}^{3}a_{6}^{3}a_{7}^{4}$,
$A_{6}=a_{2}a_{3}^{2}a_{4}^{3}a_{5}^{3}a_{6}^{4}a_{7}^{5}$, and
$A_{7}=a_{2}a_{3}^{2}a_{4}^{3}a_{5}^{4}a_{6}^{5}a_{7}^{6}$.

The following theorem gives explicitly an integral basis of $\mathbb{Z}_{K}$.

######  Theorem 2.1.

Using the above notations let $m_{2}=\frac{m}{2^{\nu_{2}(m)}}$ and let
$u\in\mathbb{Z}$ such that $m_{2}u\equiv 1\ \mbox{\rm(mod }{2^{6}})$. In the
following table $\bf{B}$ is an explicitly given integral basis of $K$.

Table A :  conditions | B
---|---
$\nu_{2}(m)$ is odd | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
$m\equiv 28\ \mbox{\rm(mod }{32})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{4A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{4A_{7}})$
|  $\phi_{2}(\alpha)=\alpha^{4}+2m_{2}u\alpha^{2}+2m_{2}u$
$m\equiv 12\ \mbox{\rm(mod }{32})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{4A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{4}+2m_{2}u\alpha^{2}+4m_{2}u\alpha+6m_{2}u$
$m\equiv 4\ \mbox{\rm(mod }{16})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{4}+2m_{2}u$
$m\equiv 48\ \mbox{\rm(mod }{64})$ |  $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha\phi_{2}(\alpha)^{2}}{2A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)^{2}}{2A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{2}+2m_{2}u$
$m\equiv 80\ \mbox{\rm(mod }{128})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha\phi_{2}(\alpha)}{2A_{3}},\frac{(\phi_{2}(\alpha))^{2}}{2A_{4}},\frac{\alpha(\phi_{2}(\alpha))^{2}}{2A_{5}},\frac{\alpha^{2}(\phi_{2}(\alpha))^{2}}{2A_{6}},\frac{\alpha(\phi_{2}(\alpha))^{3}}{4A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{2}+2m_{2}u$
$m\equiv 144\ \mbox{\rm(mod }{256})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)^{2}}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)^{2}}{2A_{5}},\frac{\theta}{4A_{6}},\frac{\alpha\cdot\theta}{4A_{7}})$
|  $\phi_{2}(\alpha)=\alpha^{2}+2m_{2}u$ and
$\theta=\phi_{2}^{3}(\alpha)-8m_{2}u\phi_{2}^{2}(\alpha)+24m_{2}u\phi_{2}(\alpha)-32m_{2}u$
$m\equiv 16\ \mbox{\rm(mod }{256})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)^{2}}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)^{2}}{2A_{5}},\frac{\theta}{4A_{6}},\frac{\alpha\cdot\theta}{8A_{7}})$
|  $\phi_{2}(\alpha)=\alpha^{2}+2m_{2}u$ and
$\theta=\phi_{2}^{3}(\alpha)-8m_{2}u\phi_{2}^{2}(\alpha)+24m_{2}u\phi_{2}(\alpha)-32m_{2}u$
$m\equiv 448\ \mbox{\rm(mod }{512})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{4A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{4}+4m_{2}u\alpha^{3}+12m_{2}u\alpha^{2}+24m_{2}u$
$m\equiv 192\ \mbox{\rm(mod }{512})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{4A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{4A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{4}+4m_{2}u\alpha^{2}+8m_{2}u$
$m\equiv 64\ \mbox{\rm(mod }{256})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
| $\phi_{2}(\alpha)=\alpha^{4}+8m_{2}u$
$m\equiv 3\ \mbox{\rm(mod }{4})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
$m\equiv 5\ \mbox{\rm(mod }{8})$ | $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{4}}{2A_{4}},\frac{\alpha^{5}+m^{4}\alpha}{2A_{5}},\frac{\alpha^{6}+m^{4}\alpha^{2}}{2A_{6}},\frac{\alpha^{7}+m^{4}\alpha^{3}}{2A_{7}})$
$m\equiv 9\ \mbox{\rm(mod }{16})$ |  $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{4}}{2A_{4}},\frac{\alpha^{5}+m^{4}\alpha}{2A_{5}},\frac{\alpha^{6}-2m\alpha^{5}-m^{2}\alpha^{4}+m^{2}\alpha^{2}+2m\alpha+3m^{2}}{4A_{6}}$,
|
$\frac{\alpha^{7}-m\alpha^{6}+m^{2}\alpha^{5}-m\alpha^{4}+m^{2}\alpha^{3}-m\alpha^{2}+(m^{2}+4m)\alpha+m}{4A_{7}})$
$m\equiv 1\ \mbox{\rm(mod }{16})$ |  $(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{2}}{2A_{4}},\frac{\alpha^{5}+m^{2}\alpha}{2A_{5}},\frac{\alpha^{6}-2m\alpha^{5}-m^{2}\alpha^{4}+m^{2}\alpha^{2}+2m\alpha+3m^{2}}{4A_{6}}$,
|
$\frac{\alpha^{7}-m\alpha^{6}+m^{2}\alpha^{5}-m\alpha^{4}+m^{2}\alpha^{3}-m\alpha^{2}+(m^{2}+4m)\alpha+m}{8A_{7}})$

As a consequence we obtain:

######  Corollary 2.2.

Let $K=\mathbb{Q}(\alpha)$ be a pure octic number field generated by a root
$\alpha$ of a monic irreducible polynomial $f(x)=x^{8}-m\in\mathbb{Z}[x]$.
Then $\mathbb{Z}_{K}=\mathbb{Z}[\alpha]$ if and only if $m$ is a square free
integer and $m\not\equiv 1\ \mbox{\rm(mod }{4})$.

In case of the integral basis
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$,
considering the explicit form of factors of the index form we conclude:

######  Theorem 2.3.

Keeping the notation of Section 2, if $\nu_{2}(m)$ is odd or $m\equiv 3\
\mbox{\rm(mod }{4})$, then

$8a_{1}a_{3}a_{5}a_{7}|(a_{2}^{2}a_{6}^{2}\pm 1)$

is a necessary condition for the monogenity of $K$.

By the $\pm$ sign we mean that for the monogenity of $K$ the divisibility
condition must hold with at least one of the signs. Note that similar
conditions can be derived also in the other cases of the integral basis, but
the calculation becomes far too complicated.

Our next main result gives sufficient conditions on $m$ for the non-monogenity
of $K$. It relaxes the condition $m$ is square free required in [15, 12].

######  Theorem 2.4.

Let $K=\mathbb{Q}(\alpha)$ be a pure octic number field generated by a root
$\alpha$ of a monic irreducible polynomial $f(x)=x^{8}-m\in\mathbb{Z}[x]$. If
one of the following conditions holds

  1. (1)

$m\equiv 1\ \mbox{\rm(mod }{32})$,

  2. (2)

$m\equiv 272\ \mbox{\rm(mod }{512})$,

  3. (3)

$\nu_{2}(m)$ is odd and $a_{2}a_{6}\ \mbox{\rm(mod }{8})\in\\{2,6\\}$,

then $K$ is not monogenic.

Finally, we consider monogenity of pure octic fields for $m$ of type $a^{u}$:

######  Theorem 2.5.

Assume that $m=a^{u}$ with $a\neq\pm 1$ a square free rational integer and
$u\in\\{3,5,7\\}$. Then

  1. (1)

If $a\not\equiv 1\ \mbox{\rm(mod }{4})$, then $K$ is monogenic and
$\mathbb{Z}_{K}$ is generated by $\theta=\frac{\alpha^{x}}{a^{y}}$, where
$(x,y)\in\mathbb{Z}^{2}$ is the unique solution in non negative integers of
the equation $ux-8y=1$ with $x<8$.

  2. (2)

If $a\equiv 1\ \mbox{\rm(mod }{4})$, then $K$ is not monogenic with the
exception of $a=-3$.

##  3\. Preliminaries

In order to show Theorem 2.1 and Theorem 2.4, we recall some fundamental facts
of Newton polygon techniques applied to algebraic number theory. Namely, the
theorems on the index and on the prime ideal factorization. For a detailed
presentation of this theory we refer to the paper of Guardia and Nart [14].

Let $f(x)\in\mathbb{Z}[x]$ be a monic irreducible polynomial with a root
$\alpha$, let $\overline{f(x)}=\prod_{i=1}^{r}\overline{\phi_{i}(x)}^{l_{i}}$
modulo $p$ be the factorization of $\overline{f(x)}$ into powers of monic
irreducible coprime polynomials of $\mathbb{F}_{p}[x]$. Recall that a well-
known theorem of Dedekind says that:

######  Theorem 3.1.

$($[22, Chapter I, Proposition 8.3]$)$

$\mbox{If }p\mbox{ does not divide the index
}(\mathbb{Z}_{K}:\mathbb{Z}[\alpha]),\mbox{ then
}p\mathbb{Z}_{K}=\prod_{i=1}^{r}\mathfrak{p}_{i}^{l_{i}},\mbox{ where
}\mathfrak{p}_{i}=p\mathbb{Z}_{K}+\phi_{i}(\alpha)\mathbb{Z}_{K}$

and the residue degree of $\mathfrak{p}_{i}$ is
$f(\mathfrak{p}_{i})={\mbox{deg}}(\phi_{i})$.

In order to apply this theorem, one needs a criterion to test whether $p$
divides the index $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. In $1878$, Dedekind
proved the following criterion:

######  Theorem 3.2.

$($Dedekind’s criterion [5, Theorem 6.1.4] and [6]$)$
For a number field $K$ generated by a root $\alpha$ of a monic irreducible
polynomial $f(x)\in\mathbb{Z}[x]$ and a rational prime integer $p$, let
$\overline{f}(x)=\prod_{i=1}^{r}\overline{\phi_{i}}^{l_{i}}(x)\ \mbox{\rm(mod
}{p})$ be the factorization of $\overline{f}(x)$ in $\mathbb{F}_{p}[x]$, where
the polynomials $\phi_{i}\in\mathbb{Z}[x]$ are monic with their reductions
irreducible over $\mathbb{F}_{p}$ and
$\gcd(\overline{\phi_{i}},\overline{\phi_{j}})=1$ for every $i\neq j$. If we
set $M(x)=\cfrac{f(x)-\prod_{i=1}^{r}{\phi_{i}}^{l_{i}}(x)}{p}$, then
$M(x)\in\mathbb{Z}[x]$ and the following statements are equivalent:

  1. 1.

$p$ does not divide the index $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$.

  2. 2.

For every $i=1,\dots,r$, either $l_{i}=1$ or $l_{i}\geq 2$ and
$\overline{\phi_{i}}(x)$ does not divide $\overline{M}(x)$ in
$\mathbb{F}_{p}[x]$.

When Dedekind’s criterion fails, that is, $p$ divides the index
$(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ for every primitive element
$\alpha\in\mathbb{Z}_{K}$ of $K$, then it is not possible to obtain the prime
ideal factorization of $p\mathbb{Z}_{K}$ by Dedekind’s theorem. In 1928, Ore
developed an alternative approach for obtaining the index
$(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, the absolute discriminant, and the
prime ideal factorization of the rational primes in a number field $K$ by
using Newton polygons (see [19, 21]). For more details on Newton polygon
techniques, we refer to [7, 13]. For any prime integer $p$, let $\nu_{p}$ be
the $p$-adic valuation of $\mathbb{Q}$, $\mathbb{Q}_{p}$ its $p$-adic
completion, and $\mathbb{Z}_{p}$ the ring of $p$-adic integers. Let $\nu_{p}$
be the Gauss’s extension of $\nu_{p}$ to $\mathbb{Q}_{p}(x)$;
$\nu_{p}(P)=\mbox{min}(\nu_{p}(a_{i}),\,i=0,\dots,n)$ for any polynomial
$P=\sum_{i=0}^{n}a_{i}x^{i}\in\mathbb{Q}_{p}[x]$ and extended by
$\nu_{p}(P/Q)=\nu_{p}(P)-\nu_{p}(Q)$ for every nonzero polynomials $P$ and $Q$
of $\mathbb{Q}_{p}[x]$. Let $\phi\in\mathbb{Z}_{p}[x]$ be a monic polynomial
whose reduction is irreducible in $\mathbb{F}_{p}[x]$, let $\mathbb{F}_{\phi}$
be the field $\frac{\mathbb{F}_{p}[x]}{(\overline{\phi})}$. For any monic
polynomial $f(x)\in\mathbb{Z}_{p}[x]$, upon the Euclidean division by
successive powers of $\phi$, we expand $f(x)$ as follows:
$f(x)=\sum_{i=0}^{l}a_{i}(x)\phi(x)^{i},$ called the $\phi$-expansion of
$f(x)$ (for every $i$, deg$(a_{i}(x))<$ deg$(\phi)$). The $\phi$-Newton
polygon of $f(x)$ with respect to $p$, is the lower boundary convex envelope
of the set of points $\\{(i,\nu_{p}(a_{i}(x))),\,a_{i}(x)\neq 0\\}$ in the
Euclidean plane, which we denote by $N_{\phi}(f)$. The $\phi$-Newton polygon
of $f$, is the process of joining the obtained edges $S_{1},\dots,S_{r}$
ordered by increasing slopes, which can be expressed as
$N_{\phi}(f)=S_{1}+\dots+S_{r}$. For every side $S_{i}$ of $N_{\phi}(f)$, the
length of $S_{i}$, denoted $l(S_{i})$ is the length of its projection to the
$x$-axis and its height , denoted $h(S_{i})$ is the length of its projection
to the $y$-axis. Let $d(S_{i})=\gcd(l(S_{i}),h(S_{i}))$ be the ramification
degree of $S$. The principal $\phi$-Newton polygon of ${f}$, denoted
$N_{\phi}^{+}(f)$, is the part of the polygon $N_{\phi}(f)$, which is
determined by joining all sides of negative slopes. For every side $S$ of
$N_{\phi}^{+}(f)$, with initial point $(s,u_{s})$ and length $l$, and for
every $0\leq i\leq l$, we attach the following residue coefficient
$c_{i}\in\mathbb{F}_{\phi}$ as follows:

$c_{i}=\displaystyle\left\\{\begin{array}[]{ll}0,&\mbox{ if }(s+i,{\it
u_{s+i}})\mbox{ lies strictly above }S,\\\ \left(\dfrac{a_{s+i}(x)}{p^{{\it
u_{s+i}}}}\right)\,\,\ \mbox{\rm(mod }{(p,\phi(x))}),&\mbox{ if }(s+i,{\it
u_{s+i}})\mbox{ lies on }S,\end{array}\right.$

where $(p,\phi(x))$ is the maximal ideal of $\mathbb{Z}_{p}[x]$ generated by
$p$ and $\phi$. Let $\lambda=-h/e$ be the slope of $S$, where $h$ and $e$ are
two positive coprime integers. Then $d=l/e$ is the degree of $S$. Notice that,
the points with integer coordinates lying on $S$ are exactly

$\displaystyle{(s,u_{s}),(s+e,u_{s}-h),\cdots,(s+de,u_{s}-dh)}.$

Thus, if $i$ is not a multiple of $e$, then $(s+i,u_{s+i})$ does not lie in
$S$, and so $c_{i}=0$. The polynomial

$f_{S}(y)=t_{d}y^{d}+t_{d-1}y^{d-1}+\cdots+t_{1}y+t_{0}\in\mathbb{F}_{\phi}[y],$

is called the residual polynomial of $f(x)$ associated to the side $S$, where
for every $i=0,\dots,d$, $t_{i}=c_{ie}$.

Let $N_{\phi}^{+}(f)=S_{1}+\dots+S_{r}$ be the principal $\phi$-Newton polygon
of $f$ with respect to $p$. We say that $f$ is a $\phi$-regular polynomial
with respect to $p$, if $f_{S_{i}}(y)$ is square free in
$\mathbb{F}_{\phi}[y]$ for every $i=1,\dots,r$. The polynomial $f$ is said to
be $p$-regular if $\overline{f(x)}=\prod_{i=1}^{r}\overline{\phi_{i}}^{l_{i}}$
for some monic polynomials $\phi_{1},\dots,\phi_{t}$ of $\mathbb{Z}[x]$ such
that $\overline{\phi_{1}},\dots,\overline{\phi_{t}}$ are irreducible coprime
polynomials over $\mathbb{F}_{p}$ and $f$ is a $\phi_{i}$-regular polynomial
with respect to $p$ for every $i=1,\dots,t$.

The theorem of Ore plays a fundamental role for proving our main Theorems:
Let $\phi\in\mathbb{Z}_{p}[x]$ be a monic polynomial, with
$\overline{\phi(x)}$ irreducible in $\mathbb{F}_{p}[x]$. As defined in [9,
Def. 1.3], the $\phi$-index of $f(x)$, denoted by $\text{ind}_{\phi}(f)$, is
deg$(\phi)$ times the number of points with natural integer coordinates that
lie below or on the polygon $N_{\phi}^{+}(f)$, strictly above the horizontal
axis, and strictly beyond the vertical axis (see $Figure\ 1$).

$0$$S_{1}$$S_{2}$$S_{3}$ Figure 1.  $N_{\phi}^{+}(f)$.

In the example of $Figure\ 1$, $\text{ind}_{\phi}(f)=9\times$deg$(\phi)$.

Now assume that $\overline{f(x)}=\prod_{i=1}^{r}\overline{\phi_{i}}^{l_{i}}$
is the factorization of $\overline{f(x)}$ in $\mathbb{F}_{p}[x]$, where every
$\phi_{i}\in\mathbb{Z}[x]$ is monic polynomial, with $\overline{\phi_{i}(x)}$
is irreducible in $\mathbb{F}_{p}[x]$, $\overline{\phi_{i}(x)}$ and
$\overline{\phi_{j}(x)}$ are coprime when $i\neq j$ and $i,j=1,\dots,t$. For
every $i=1,\dots,t$, let $N_{\phi_{i}}^{+}(f)=S_{i1}+\dots+S_{ir_{i}}$ be the
principal $\phi_{i}$-Newton polygon of $f$ with respect to $p$. For every
$j=1,\dots,r_{i}$, let
$f_{S_{ij}}(y)=\prod_{k=1}^{s_{ij}}\psi_{ijk}^{a_{ijk}}(y)$ be the
factorization of $f_{S_{ij}}(y)$ in $\mathbb{F}_{\phi_{i}}[y]$. Then we have
the following index theorem of Ore (see [9, Theorem 1.7 and Theorem 1.9], [7,
Theorem 3.9], [19, pp: 323–325], and [21]).

######  Theorem 3.3.

$($Theorem of Ore$)$

  1. (1)

$\nu_{p}(\text{ind}(f))\geq\sum_{i=1}^{r}\text{ind}_{\phi_{i}}(f).$

The equality holds if $f(x)$ is $p$-regular.

  2. (2)

If $f(x)$ is $p$-regular, then

$p\mathbb{Z}_{K}=\prod_{i=1}^{r}\prod_{j=1}^{r_{i}}\prod_{k=1}^{s_{ij}}\mathfrak{p}^{e_{ij}}_{ijk},$

is the factorization of $p\mathbb{Z}_{K}$ into powers of prime ideals of
$\mathbb{Z}_{K}$ lying above $p$, where $e_{ij}=l_{ij}/d_{ij}$, $l_{ij}$ is
the length of $S_{ij}$, $d_{ij}$ is the ramification degree of $S_{ij}$, and
$f_{ijk}=\mbox{deg}(\phi_{i})\times\mbox{deg}(\psi_{ijk})$ is the residue
degree of the prime ideal $\mathfrak{p}_{ijk}$ over $p$.

When Ore’s program fails; that is if $f(x)$ is not $p$-regular, then it may
happen that some factors of $f(x)$ provided by Hensel’s lemma and refined by
first order Newton polygon techniques are not irreducible over
$\mathbb{Q}_{p}$. In this case in order to complete the factorization of
$f(x)$ in $\mathbb{Q}_{p}[x]$, Guardia, Montes, and Nart introduced the notion
of high order Newton polygon. They showed, thanks to a theorem of index [13,
Theorem 4. 18], that after a finite number of iterations this process yields
all monic irreducible factors of $f(x)$ in $\mathbb{Q}_{p}[x]$, all prime
ideals of $\mathbb{Z}_{K}$ lying above a prime integer $p$, the $p$-valuation
of the index $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, and so up to a sign the
absolute discriminant of $K$. We recall here some fundamental techniques of
Newton polygons of high order. For more details, we refer to [13] and [14]. As
introduced in [13], a type of order $r-1$ is a data
${\bf{t}}=(g_{1}(x),-\lambda_{1},g_{2}(x),-\lambda_{2},\dots,g_{r-1}(x),-\lambda_{r-1},\psi_{r-1}(x))$,
where every $g_{i}(x)$ is a monic polynomial in $\mathbb{Z}_{p}[x]$,
$\lambda_{i}\in\mathbb{Q}^{+}$, and $\psi_{r-1}(y)$ is a polynomial over a
finite field of $\displaystyle p^{H}$ elements and
$H=\displaystyle\prod_{i=0}^{r-2}f_{i}$, with $f_{i}=\mbox{deg}(\psi_{i}(x))$,
satisfying the following recursive properties:

  1. (1)

$g_{1}(x)$ is irreducible modulo $p$, $\psi_{0}(y)\in\mathbb{F}[y]$
($\mathbb{F}_{0}=\mathbb{F}_{p}$) be the polynomial obtained by reduction of
$g_{1}(x)$ modulo $p$, and $\mathbb{F}_{1}:=\mathbb{F}_{0}[y]/(\psi_{0}(y)))$.

  2. (2)

For every $i=1,\dots,r-1$, the Newton polygon of $i^{th}$ order,
$N_{i}(g_{i+1}(x))$, has a single sided of slope $-\lambda_{i}$.

  3. (3)

For every $i=1,\dots,r-1$, the residual polynomial of $i^{th}$ order,
$R_{i}(g_{i+1})(y)$ is an irreducible polynomial in $\mathbb{F}_{i}[y]$. Let
$\psi_{i}(y)\in\mathbb{F}_{i}[y]$ be the monic polynomial determined by
$R_{i}(\phi_{i+1})(y)\simeq\psi_{i}(y)$ (are equal up to multiplication by a
nonzero element of $\mathbb{F}_{i}$), and
$\mathbb{F}_{i+1}=\mathbb{F}_{i}[y]/(\psi_{i}(y))$. Thus,
$\mathbb{F}_{0}\subset\mathbb{F}_{1}\subset\dots\subset\mathbb{F}_{r}$ is a
tower of finite fields.

  4. (4)

For every $i=1,\dots,r-1$, $g_{i+1}(x)$ has minimal degree among all monic
polynomials in $\mathbb{Z}_{p}[x]$ satisfying $(2)$ and $(3)$.

  5. (5)

$\psi_{r-1}(y)\in\mathbb{F}_{r-1}[y]$ is a monic irreducible polynomial,
$\psi_{r-1}(y)\neq y$, and
$\mathbb{F}_{r}=\mathbb{F}_{r-1}[Y]/(\psi_{r-1}(y))$.

Here the field $\mathbb{F}_{i}$ should not be confused with the finite field
of $i$ elements.
As for every $i=1,\dots,r-1$, the residual polynomial of the $i^{th}$ order,
$R_{i}(g_{i+1})(y)$ is an irreducible polynomial in $\mathbb{F}_{i}[y]$, by
theorem of the product in order $i$, the polynomial $g_{i}(x)$ is irreducible
in $\mathbb{Z}_{p}[x]$. Let $\omega_{0}=[\nu_{p},x,0]$ be the Gauss’s
extension of $\nu_{p}$ to $\mathbb{Q}_{p}(x)$. As for every $i=1,\dots,r-1$,
the residual polynomial of the $i^{th}$ order, $R_{i}(g_{i+1})(y)$ is an
irreducible polynomial in $\mathbb{F}_{i}[y]$, then according to MacLane’s
notations and definitions [20], $g_{i+1}(x)$ induces a valuation on
$\mathbb{Q}_{p}(x)$, denoted by
$\omega_{i+1}=e_{i}[\omega_{i},g_{i+1},\lambda_{i+1}]$, where
$\lambda_{i}=h_{i}/e_{i}$, $e_{i}$ and $h_{i}$ are positive coprime integers.
The valuation $\omega_{i+1}$ is called the augmented valuation of $\omega_{i}$
with respect to $g_{i+1}$ and $\lambda_{i+1}$ is defined over
$\mathbb{Q}_{p}[x]$ as follows:

$\omega_{i+1}(f(x))=\mbox{min}\\{e_{i+1}\omega_{i}(a_{j}^{i+1}(x))+jh_{i+1},\,j=0,\dots,n_{i+1}\\},$

where $f(x)=\displaystyle\sum_{j=0}^{n_{i+1}}a_{j}^{i+1}(x)g_{i+1}^{j}(x)$ is
the $g_{i+1}(x)$-expansion of $f(x)$. According to the terminology in [13],
the valuation $\omega_{r}$ is called the $r^{th}$-order valuation associated
to the data ${\bf{t}}$. For every order $r\geq 1$, the $g_{r}$-Newton polygon
of $f(x)$, with respect to the valuation $\omega_{r}$ is the lower boundary of
the convex envelope of the set of points $\\{(i,\mu_{i}),i=0,\dots,n_{r}\\}$
in the Euclidean plane, where $\mu_{i}=\omega_{r}(a^{r}_{i}(x)g_{r}^{i}(x))$.
The following are the relevant theorems from Guardia-Montes-Nart’s work (high
order Newton polygon):

######  Theorem 3.4.

$($[13, Theorem 3.1]$)$
Let $f(x)\in\mathbb{Z}_{p}[x]$ be a monic polynomial such that
$\overline{f(x)}$ is a positive power of $\overline{\phi(x)}$ for some monic
polynomial $\phi(x)$ such that $\overline{\phi(x)}$ is irreducible over
$\mathbb{F}_{0}$. If $N_{r}(f)=S_{1}+\dots+S_{g}$ has $g$ sides, then we can
split $f(x)=f_{1}(x)\times\dots\times f_{g}(x)$ in $\mathbb{Z}_{p}[x]$, such
that $N_{r}(f_{i})=S_{i}$ and $R_{r}({f_{i}})(y)=R_{r}({f})(y)$ up to
multiplication by a nonzero element of $\mathbb{F}_{r}$ for every
$i=1,\cdots,g$.

######  Theorem 3.5.

$($[13, Theorem 3.7]$)$
Let $f(x)\in\mathbb{Z}_{p}[x]$ be a monic polynomial such that $N_{r}(f)=S$
has a single side of finite slope $-\lambda_{r}$. If
$R_{r}({f})(y)=\displaystyle\prod_{i=1}^{t}\psi_{i}(y)^{a_{i}}$ is the
factorization in $\mathbb{F}_{r}[y]$, then $f(x)$ splits as
$f(x)=f_{1}(x)\times\cdots\times f_{t}(x)$ in $\mathbb{Z}_{p}[x]$ such that
$N_{r}({f_{i}})=S$ has a single side of slope $-\lambda_{r}$ and
$R_{r}({f_{i}})(y)=\psi_{i}(y)^{a_{i}}$ up to multiplication by a nonzero
element of $\mathbb{F}_{r}$ for every $i=1,\cdots,t$.

In [13, Definition 4.15], the authors introduced the notion of $r^{th}$-order
index of a monic polynomial $f(x)\in\mathbb{Z}[x]$ as follows: For a fixed
data

${\bf{t}}=(g_{1}(x),-\lambda_{1},g_{2}(x),-\lambda_{2},\dots,g_{r-1}(x),-\lambda_{r-1},\psi_{r-1}(x)),$

let $N_{r}(f)$ be the Newton polygon of $r^{th}$-order with respect to the
data t and $\text{ind}_{t}(f)=f_{0}\cdots f_{r-1}\text{ind}(N_{r}(f))$, where
$\text{ind}(N_{r}(f))$ is the index of the polygon $N_{r}(f)$; the number of
points with natural integer coordinates that lie below or on the polygon
$N_{\phi}^{+}(f)$, strictly above the horizontal line of equation
$y=\omega_{r}(f)$, and strictly beyond the vertical axis. They showed the
following theorem on the index which generalizes the theorem of Ore on the
index:

######  Theorem 3.6.

$($[13, Theorem 4.18]$)$

$\nu_{p}(\text{ind}(f))\geq\text{ind}_{1}(f)+\dots+\text{ind}_{r}(f).$

The equality holds if and only if $\text{ind}_{r+1}(f)=0$.

Recall that by definition $\text{ind}(N_{r+1}(f))=0$ if and only if
$N_{r+1}(f)$ has a single side of length $1$ or height $1$. By [13, Lemma
2.17] $(2)$, if $R_{r}(f)$ is square free, then the length of $N_{r}(f)$ is
$1$. Thus if $R_{r}(f)$ is square free, then $\text{ind}_{r+1}(f)=0$, and so
the equality
$\nu_{p}(\text{ind}(f))=\text{ind}_{1}(f)+\dots+\text{ind}_{r}(f)$ holds.

##  4\. Proofs of main results

Proof of Theorem 2.1.
During this proof, $\mathbb{F}_{i}$ is the $i^{th}$ field of the tower
provided by Montes algorithm. Since $\triangle(f)=\mp 8^{8}m^{7}$ is the
discriminant of $f$, thanks to the formula
$\triangle(f)=(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])^{2}d_{K}$, the prime
candidates to divide the index $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ are those
dividing $2\cdot m$.

  1. (1)

Let $p$ be a prime dividing $m$. Then $\overline{f(x)}=x^{8}\ \mbox{\rm(mod
}{p})$. For $\phi=x$, $N_{\phi}(f)=S$ has a single side joining the points
$(0,\nu_{p}(m))$ and $(8,0)$. Then
$\text{ind}_{1}(f)=\text{ind}_{\phi}(f)=\nu_{p}(\prod_{i=2}^{7}A_{i})$. Also
by [9, Theorem 2.7],
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
is a free $\mathbb{Z}$-submodule of $\mathbb{Z}_{K}$. Now, let
$d=$gcd$(\nu_{p}(m),8)$. Then $R_{1}(f)(y)=y^{d}-\overline{m_{p}}$, where
$m_{p}=m/p^{\nu_{p}(m)}$. If $2$ does not divide $\nu_{p}(m)$, then $d=1$,
$R_{1}(f)(y)$ is irreducible over $\mathbb{F}_{1}$, and so by Theorem 3.3,
$\text{ind}(f)=\text{ind}_{\phi}(f)$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
is a $p$-integral basis of $\mathbb{Z}_{K}$. If $2$ divides $d$, then
$d\in\\{2,4\\}$. In this case if $p\neq 2$, then
$R_{1}(f)(y)=y^{d}-\overline{m_{p}}$ is square free over
$\mathbb{F}_{1}\simeq\mathbb{F}_{0}$ (because $\phi=x$, and so
deg$(\phi)=1=[\mathbb{F}_{1}:\mathbb{F}_{0}]$). Thus by Theorem 3.3,
$\nu_{p}(\text{ind}(f))=\text{ind}_{\phi}(f)$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
is a $p$-integral basis of $\mathbb{Z}_{K}$. It follows that if $2$ divides
$m$ and $\nu_{2}(m)$ is odd, then
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.
If $d\in\\{2,4\\}$ and $p=2$, then $R_{1}(f)(y)=(y-\overline{m_{2}})^{d}$ is
not square free and we have to use second order Newton polygon techniques.

    1. (a)

If $\nu_{2}(m)=2$, then for $\phi=x$, we have $\overline{f(x)}=\phi^{8}\
\mbox{\rm(mod }{2})$, $N_{\phi}(f)=S$ has a single side of slope
$-\lambda_{1}=-1/4$, $e_{1}=4$, and $R_{1}(f)(y)=y^{2}+1=(y+1)^{2}$. According
to the definitions and notations of [13, 20], let
$\omega_{2}=e_{1}[\nu_{2},\lambda_{1}]$ be the valuation of second order
Newton polygon defined by $\omega_{2}(a)=e_{1}\nu_{2}(a)=4\nu_{2}(a)$ for
every $a\in\mathbb{Q}_{2}$ and $\omega_{2}(x)=e_{1}\lambda_{1}=1$,
$\phi_{2}=x^{4}+2$ a key polynomial of $\omega_{2}$, and
$f(x)=\phi_{2}^{2}-4\phi_{2}+(4-m)$ the $\phi_{2}$-expansion. As
$\omega_{2}(\phi_{2})=4$, $\omega_{2}(\phi_{2}^{2})=8$, and
$\omega_{2}(4\phi_{2})=12$. It follows that:

      1. (i)

If $\nu_{2}(4-m)=3$; $m\equiv 12\ \mbox{\rm(mod }{16})$, then there are $2$
cases :
If $\nu_{2}(12-m)=4$; $m\equiv 28\ \mbox{\rm(mod }{32})$, then for
$\phi_{2}=x^{4}+2x^{2}+2$, we have $f(x)=\phi_{2}^{2}-4x^{2}\phi_{2}-(4+m)$ is
the $\phi_{2}$-expansion of $f(x)$. As $\omega_{2}(\phi_{2}^{2})=8$,
$\omega_{2}((-4x^{2})\phi_{2})=14$, and $\omega_{2}(4+m)\geq 20$. It follows
that if $\omega_{2}(4+m)=20$, then $N_{2}(f)$ has a single side and
$\text{ind}_{2}(f)=6$ and with $R_{2}(f)(y)=y^{2}+y+1$. If
$\omega_{2}(4+m)>20$, then $N_{2}(f)$ has two sides of degree $1$ each. In
both cases $\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+6$. Based on the polygon
$N_{1}(f)$, we conclude that $V(\alpha)=1/4$. Similarly, based on the polygon
$N_{2}(f)$, we conclude that $V(\phi_{2}(\alpha))\geq 5/2$, and so
$(1,\alpha,{\alpha^{2}},{\alpha^{3}},\frac{\phi_{2}(\alpha)}{4},\frac{\alpha\phi_{2}(\alpha)}{4},\frac{\alpha^{2}\phi_{2}(\alpha)}{8},\frac{\alpha^{3}\phi_{2}(\alpha)}{8})$
is a $2$-integral basis of $\mathbb{Z}_{K}$. Since the Montes algorithm is
local; the algorithm $p$-integral elements, sometimes we have to replace
$\phi_{2}(x)$ by an equivalent polynomial. For example, in our case, if
$\nu_{p}(A_{i})\geq 1$ for some $i=2,\dots,7$ and for some odd prime integer
$p$, then $\frac{\alpha^{3}\phi_{2}(\alpha)}{8A_{7}}$ is not $p$-integral
($V(\frac{\alpha^{3}\phi_{2}(\alpha)}{8A_{7}})<0$ for some valuation $V$ of
$K$ extending $\nu_{p}$). So, we have to replace $\phi_{2}(x)$ by
$g(x)=x^{4}+2m_{2}ux^{2}+2m_{2}u$ with $u$ an integer which satisfies
$um_{2}\equiv 1\ \mbox{\rm(mod }{32})$ and show that
$\mathcal{B}=(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{g(\alpha)}{2A_{4}},\frac{\alpha
g(\alpha)}{2A_{5}},\frac{\alpha^{2}g(\alpha)}{4A_{6}},\frac{\alpha^{3}g(\alpha)}{4A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$. Since for every prime integer $p$,
$\nu_{p}(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])=\nu_{p}(2^{6}\prod_{i=2}^{7}A_{i})$,
we need only to show that every element of $\mathcal{B}$ is integral over
$\mathbb{Z}$. By the definition of $g(x)$ and by the first point of this proof
the $V$-valuation of each element of $\mathcal{B}$ is greater or equal than
$0$ for every valuation $V$ of $K$ extending $\nu_{p}$ for every odd prime
integer $p$. Let us show the same result for $p=2$. For this reason we need to
gave a lower bound of $V(\alpha)$ and $V(\phi_{2}(\alpha))$ for every
valuation $V$ of $K$ extending $\nu_{2}$. Let $V$ be a valuation of $K$
extending $\nu_{2}$. Since $V(\alpha)=1/4$, $V(\phi_{2}(\alpha))\geq 5/2$,
$\nu_{2}(g(x)-\phi_{2}(x))\geq 5$, and $\alpha\in\mathbb{Z}_{K}$, we conclude
that $V(g(\alpha)-\phi_{2}(\alpha))\geq 5$. Thus $V(g(\alpha))\geq 5/2$, and
so by a simple verification, the $V$-valuation of each element of
$\mathcal{B}$ is greater or equal than $0$. Hence
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{4A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{4A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.
In the remainder of this proof, this technique will be repeated. So in every
case, we give an adequate $\phi_{2}(x)$ for which $f(x)$ is regular with
respect to $\omega_{2}$, we give the $\phi_{2}$-expansion of $f(x)$, and a
lower bound of $V(\phi_{2}(\alpha))$ for every valuation $V$ of $K$ extending
$\nu_{2}$.
If $\nu_{2}(12-m)\geq 5$ ($m\equiv 12\ \mbox{\rm(mod }{32})$), then for
$\phi_{2}=x^{4}+2x^{2}+4x+6$,
$f(x)=\phi_{2}^{2}+(-8-4x^{2}-8x)\phi_{2}+(12-m+16x^{3}+32x^{2}+32x)$ is the
$\phi_{2}$-expansion of $f(x)$. As $\omega_{2}(\phi_{2}^{2})=8$,
$\omega_{2}((-8-4x^{2}-8x)\phi_{2})=14$, and
$\omega_{2}(12-m+16x^{3}+32x^{2}+32x)=\omega_{2}(16x^{3})=19$, we conclude
that $N_{2}(f)=S$ has a single side joining $(0,19)$ and $(2,8)$. Therefore
$\text{ind}_{2}(f)=5$, the side $S$ is of degree $1$, and
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+5$, and so
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{4A_{7}})$
is a $2$-integral basis of $\mathbb{Z}_{K}$. $\omega_{2}(\phi_{2}^{2})=8$,
$\omega_{2}((-8-4x^{2}-8x)\phi_{2})=14$, and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{4A_{7}})$
is a $2$-integral basis of $\mathbb{Z}_{K}$. Now by replacing $\phi_{2}(x)$ by
$g(x)=x^{4}+2m_{2}ux^{2}+4m_{2}ux+6m_{2}u$ with $u$ an integer which satisfies
$um_{2}\equiv 1\ \mbox{\rm(mod }{16})$, we conclude that
$\mathcal{B}=(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{g(\alpha)}{2A_{4}},\frac{\alpha
g(\alpha)}{2A_{5}},\frac{\alpha^{2}g(\alpha)}{2A_{6}},\frac{\alpha^{3}g(\alpha)}{4A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

      2. (ii)

If $\nu_{2}(4-m)=4$; $m\equiv 20\ \mbox{\rm(mod }{32})$, then
$\omega_{2}(4-m)=16$ and $N_{2}(f)=T$ has a single side joining the points
$(0,16)$, $(1,12)$, and $(2,8)$. Thus, $\text{ind}_{1}(f)=4$ and
$R_{2}(f)(y)=y^{2}+y+1$ is irreducible over
$\mathbb{F}_{2}=\mathbb{F}_{1}=\mathbb{F}_{0}$ (because
deg$(\phi)=$deg$(\psi_{1})=1$). Hence by Theorem 3.6
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+4$. Based on the polygon $N_{2}(f)$,
we conclude that $V(\phi_{2}(\alpha))=2$. Replacing $\phi_{2}(x)$ by
$x^{4}+2m_{2}u$ with an integer $u$ satisfying $um_{2}\equiv 1\ \mbox{\rm(mod
}{16})$, we conclude that
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

      3. (iii)

If $\nu_{2}(4-m)\geq 5$; $m\equiv 4\ \mbox{\rm(mod }{32})$, then
$\omega_{2}(4-m)\geq 20$ and $N_{2}(f)=T_{1}+T_{2}$ has two sides joining
$(0,v)$, $(1,12)$, and $(2,8)$ with $v\geq 20$. Thus each side is of degree
$1$, and so
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+\text{ind}_{2}(f)=\text{ind}_{1}(f)+4$.
Based on $N_{2}(f)$, we have $V(\phi_{2}(\alpha))\geq 2$, and so by replacing
$\phi_{2}(x)$ by $x^{4}+2m_{2}u$ with an integer $u$ satisfying $um_{2}\equiv
1\ \mbox{\rm(mod }{16})$, we conclude that
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+2m_{2}u}{2A_{4}},\frac{\alpha^{5}+2m_{2}u\alpha}{2A_{5}},\frac{\alpha^{6}+2m_{2}u\alpha^{2}}{2A_{6}},\frac{\alpha^{7}+2m_{2}u\alpha^{3}}{2A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.
Note that the two cases $(i)$ and $(ii)$ could combined into one case only,
namely $\nu_{2}(4-m)\geq 4$.

    2. (b)

If $\nu_{2}(m)=4$; $m\equiv 16\ \mbox{\rm(mod }{32})$, then for $\phi=x$,
$\overline{f(x)}=\phi^{8}\ \mbox{\rm(mod }{2})$, $N_{\phi}(f)=S$ has a single
side of slope $-1/2$, $R_{1}(f)(y)=y^{4}+1=(y+1)^{4}$. Let $\omega_{2}$ be the
valuation of second order Newton polygon defined by
$\omega_{2}(a)=2\nu_{2}(a)$ for every $a\in\mathbb{Q}_{2}$ and
$\omega_{2}(x)=1$. Let $\phi_{2}=x^{2}+2$ and
$f(x)=\phi_{2}^{4}-8\phi_{2}^{3}+24\phi_{2}^{2}-32\phi_{2}+(16-m)$ the
$\phi_{2}$-expansion of $f(x)$. Since $m\equiv 16\ \mbox{\rm(mod }{32})$, then
$\nu_{2}(m-16)\geq 5$. It follows that:

      1. (i)

If $\nu_{2}(16-m)=5$; $m\equiv 48\ \mbox{\rm(mod }{64})$, then $N_{2}(f)=T$
has a single side joining the points $(0,10)$ and $(4,8)$ with slope $-1/2$,
$\text{ind}_{2}(f)=2$, and attached residual polynomial
$R_{2}(f)(y)=(y+1)^{2}$. Let us use the third order Newton polygon associated
to the data $t=(x,1/2,\phi_{2},1/2,\phi_{3})$, where
$\phi_{3}=\phi_{2}^{2}(x)+4x=x^{4}+4x^{2}+4x+4$ is a key polynomial of
$\omega_{2}$ and let $\omega_{3}=2[\omega_{2},1/2]$ be the valuation of third
order Newton polygon; $\omega_{3}(a)=4\nu_{2}(a)$ for every
$a\in\mathbb{Q}_{2}$, $\omega_{3}(x)=2$, and
$\omega_{3}(\phi_{2})=2(2+1/2)=5$. Let
$f(x)=\phi_{3}^{2}+(8-8x^{2}-8x)\phi_{3}+(-m-48-16x^{2}-32x+32x^{3})$ be the
$\phi_{3}$-expansion of $f(x)$. Since $m=48+64k$ for some integer $k$,
$-m-48-16x^{2}-32x+32x^{3}=-16\phi_{2}(x)-64+64k-32x+32x^{3}$, and so
$\omega_{3}(-m-48-16x^{2}-32x+32x^{3})=\omega_{3}(16)+\omega_{3}(\phi_{2})=21$.
Thus $N_{3}(f)=T$ has a single side joining the points $(0,21)$ and $(2,20)$.
Thus its height is $1$, and so $\text{ind}_{3}(f)=0$. By Theorem 3.6
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+\text{ind}_{2}(f)=\text{ind}_{1}(f)+2$.
Based on $N_{1}(f)$, we have $V(\alpha)=1/2$ and based on $N_{2}(f)$, we get
$V(\phi_{2}(\alpha))=5/4$. Replacing $\phi_{2}(x)$ by by $x^{2}+2m_{2}u$ with
$um_{2}\equiv 1\ \mbox{\rm(mod }{16})$, we get
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha\phi_{2}^{2}(\alpha)}{2A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{3}\phi_{2}^{2}(\alpha)}{2A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

      2. (ii)

If $\nu_{2}(16-m)=6$ ($m\equiv 80\ \mbox{\rm(mod }{128})$), then
$N_{2}(f)=T_{2}$ has a single side joining the points $(0,12)$ and $(4,8)$.
Thus $\text{ind}_{2}(f)=6$. As $T$ is of degree $4$ and
$R_{2}(f)(y)=y^{4}+y^{2}+1=(y^{2}+y+1)^{2}$, we have to use third order Newton
polygon techniques. Let $\omega_{3}$ be the valuation of third order Newton
polygon, $\phi_{3}=\phi_{2}^{2}+2x\phi_{2}+4x^{2}$ and
$f(x)=\phi_{3}^{2}+(-(4x+12)\phi_{2}+16+24x)\phi_{3}+(-(128x+32)\phi_{2}+192x-m+80)$.
Since $\omega_{3}(\phi_{2})=3$, we conclude that $N_{3}(f)=T_{3}$ has a single
side joining the points $(0,12)$ and $(2,13)$, which is of degree $1$. Thus
$\text{ind}_{3}(f)=0$ and $\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+6$. Based
on $N_{1}(f)$ and $N_{2}(f)$, we get $V(\alpha)=1/2$ and
$V(\phi_{2}(\alpha))=3/2$. Replacing $\phi_{2}(x)$ by $x^{2}+2m_{2}u$ by
$um_{2}\equiv 1\ \mbox{\rm(mod }{16})$, $V(\phi_{2}(\alpha))=3/2$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha\phi_{2}(\alpha)}{2A_{3}},\frac{\phi_{2}^{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}^{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}^{2}(\alpha)}{2A_{6}},\frac{\alpha\phi_{2}^{3}(\alpha)}{4A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

      3. (iii)

If $\nu_{2}(16-m)=7$; $m\equiv 144\ \mbox{\rm(mod }{256})$, then
$\omega_{2}(16-m)=14$ and $N_{2}(f)=T_{1}+T_{2}$ has $2$ sides joining the
points $(0,14)$, $(1,12)$, $(2,10)$, and $(4,8)$. Thus $\text{ind}_{2}(f)=7$.
Since the attached residual polynomials of $f(x)$ are $R_{12}(f)(y)=y^{2}+y+1$
and $R_{22}(f)(y)=(y+1)^{2}$, in order to complete the calculation of the
index $\nu_{2}(\text{ind}(f))$, we have to use third order Newton polygon. Let
$\phi_{3}=\phi_{2}+4x=x^{2}+4x+2$ be the key polynomial of $\omega_{2}$,
$\omega_{3}=[\omega_{2},1]$ the valuation of third order Newton polygon;
$\omega_{3}(a)=2\nu_{2}(a)$ for every $a\in\mathbb{Q}_{2}$, $\omega_{3}(x)=1$,
and $\omega_{3}(\phi_{2})=3$. Let
$f(x)=\phi_{3}^{4}+(88-16x)\phi_{3}^{3}+(728-544x)\phi_{3}^{2}+(96-3776x)\phi_{3}-({6528x}+m+3824)$
be the $\phi_{3}$-expansion of $f(x)$. As
$\omega_{3}((728-544x)\phi_{3}^{2})=12$, $\omega_{3}((96-3776x)\phi_{3})=13$,
and $\omega_{3}(6528x+m+3824)=13$, $N_{3}^{+}(f)=T$ has a single side joining
the points $(0,13)$ and $(2,12)$. Thus $T$ is of height $1$, and so
$\text{ind}_{3}(f)=0$. Therefore
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+\text{ind}_{2}(f)=\text{ind}_{1}(f)+7$.
Based on $N_{1}(f)$, $N_{2}(f)$, and $N_{3}(f)$, we get $V(\alpha)=1/2$ and
$V(\phi_{2}(\alpha))\geq 3/2$. Let
$\theta=\phi_{2}^{3}(\alpha)-8\phi_{2}^{2}(\alpha)+24\phi_{2}(\alpha)-32$. We
need to show that $\frac{\theta}{2^{5}}\in\mathbb{Z}_{K}$. For this reason we
have to show that $V(\theta)\geq 5$ for every valuation $V$ of $K$ extending
$\nu_{2}$. Since $N_{2}(f)$ has two sides of slopes $-2$ and $-1$, by Theorem
3.4, $f(x)=f_{1}(x)\times f_{2}(x)$ in $\mathbb{Z}_{2}[x]$ and there are two
distinct valuations $V_{1}$ and $V_{2}$ of $K$ extending $\nu_{2}$ which
satisfy $V_{1}(\phi_{2}(\alpha))=2$ and $V_{2}(\phi_{2}(\alpha))=3/2$. If
$V(\phi_{2}(\alpha))=2$, then a simple verification shows that $V(\theta)\geq
0$. If $V(\phi_{2}(\alpha))=3/2$, as $f(\alpha)=0$, we have
$\theta=\frac{m-16}{\phi_{2}(\alpha)}$, and so $V(\theta)=7-3/2\geq 5$. Hence
$\frac{\theta}{2^{5}}\in\mathbb{Z}_{K}$ and
$(1,\alpha,\frac{\alpha^{2})}{2},\frac{\alpha\phi_{2}(\alpha)}{4},\frac{\phi_{2}^{2}(\alpha)}{8},\frac{\alpha\phi_{2}^{2}(\alpha)}{8},\frac{\theta}{32},\frac{\alpha\theta}{32})$
is a $2$-integral basis of $\mathbb{Z}_{K}$. By replacing $\phi_{2}(x)$ by
$x^{2}+2m_{2}u$ with $m_{2}u\equiv 1\ \mbox{\rm(mod }{32})$, we get
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha\phi_{2}(\alpha)}{2A_{3}},\frac{\phi_{2}^{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}^{2}(\alpha)}{2A_{5}},\frac{\theta}{4A_{6}},\frac{\alpha\theta}{4A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$, where
$\theta=\phi_{2}^{3}(\alpha)-8m_{2}u\phi_{2}^{2}(\alpha)+24m_{2}u\phi_{2}(\alpha)-32m_{2}u$.

      4. (iv)

If $\nu_{2}(16-m)\geq 8$, then $m\equiv 16\ \mbox{\rm(mod }{256})$. Let
$v=\omega_{2}(16-m)$, $\phi_{2}=x^{2}-2x+2$ and
$f(x)=\phi_{2}^{4}+(16+8x)\phi_{2}^{3}+(-40+32x)\phi_{2}^{2}-32x\phi_{2}+(16-m)$
be the $\phi_{2}$-expansion of $f(x)$. If $v\geq 9$, then
$N_{2}(f)=T_{1}+T_{2}+T_{3}$ has $3$ sides joining the points $(0,v)$,
$(1,13)$, $(2,10)$, and $(4,8)$ with $v\geq 18$. If $v=8$, then
$N_{2}(f)=T_{1}+T_{2}$ has $2$ sides joining the points $(0,16)$, $(1,13)$,
$(2,10)$, and $(4,8)$. In both cases, $\text{ind}_{2}(f)=8$. Since the
attached residual polynomial of the last side is $R_{2}(f)(y)=(y+1)^{2}$, we
have to use third order Newton polygon techniques. Let
$\phi_{3}=\phi_{2}+2x=x^{2}+4x+2$ be the key polynomial of $\omega_{2}$,
$\omega_{3}$ the valuation of third order Newton polygon, and
$f(x)=\phi_{3}^{4}+(88-16x)\phi_{3}^{3}+(728-544x)\phi_{3}^{2}+(96-3776x)\phi_{3}-(6528x+m+3824)$
the $\phi_{3}$-expansion of $f(x)$. As
$\omega_{3}((728-544x)\phi_{3}^{2})=12$, $\omega_{3}((96-3776x)\phi_{3})=13$,
and $\omega_{3}(6528x+m+3824)=13$, $N_{3}^{+}(f)=T$ has a single side joining
the points $(2,13)$ and $(2,12)$ with height $1$. Hence $\text{ind}_{3}(f)=0$,
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+\text{ind}_{2}(f)=\text{ind}_{1}(f)+8$.
Let $V$ be a valuation of $K$ extending $\nu_{2}$. Based on $N_{1}(f)$ and
$N_{2}(f)$, we get $V(\alpha)=1/2$ and $V(\phi_{2}(\alpha))\geq 3/2$. If
$v=8$, then $N_{2}(f)$ has two sides of slopes $-3$ and $-1$. So by Theorem
3.4, $f(x)=f_{1}(x)\times f_{2}(x)$ in $\mathbb{Z}_{2}[x]$ and there are two
distinct valuations $V_{1}$ and $V_{2}$ of $K$ extending $\nu_{2}$ which
satisfy $V_{1}(\phi_{2}(\alpha))=5/2$ and $V_{2}(\phi_{2}(\alpha))=3/2$. If
$v\geq 9$, then $N_{2}(f)$ has three sides of slopes $-t$, $-3$ and $-1$. So
by Theorem 3.4, $f(x)=f_{1}(x)\times f_{2}(x)\times f_{3}(x)$ in
$\mathbb{Z}_{2}[x]$ and there are three distinct valuations $V_{1}$, $V_{2}$,
and $V_{3}$ of $K$ extending $\nu_{2}$ which satisfy
$V_{1}(\phi_{2}(\alpha))\geq 3$, $V_{1}(\phi_{2}(\alpha))=5/2$, and
$V_{3}(\phi_{2}(\alpha))=3/2$. If $V(\phi_{2}(\alpha))\geq 5/2$, then
$V(\phi_{2}^{3}(\alpha)+(16+8\alpha)\phi_{2}^{2}(\alpha)+(-40+32\alpha)\phi_{2}(\alpha)-32\alpha)\geq
11/2$, and so $V(\frac{\alpha\theta}{2^{6}}\geq 0$, where
$\theta=\phi_{2}^{3}(\alpha)+(16+8\alpha)\phi_{2}^{2}(\alpha)+(-40+32\alpha)\phi_{2}(\alpha)-32\alpha$.
If $V(\phi_{2}(\alpha))=3/2$, then
$V(\frac{\phi_{2}^{3}(\alpha)+(16+8\alpha)\phi_{2}^{2}(\alpha)+(-40+32\alpha)\phi_{2}(\alpha)-32\alpha}{2^{5}})=V(\frac{16-m}{2^{5}\phi_{2}(\alpha)})\geq
8-5$. Thus $V(\frac{\theta}{2^{5}})\geq 0$ and
$V(\frac{\alpha\theta}{2^{6}})\geq 0$ for every valuation $V$ of $K$ extending
$\nu_{2}$. Thus
$(1,\alpha,\frac{\alpha^{2})}{2},\frac{\alpha\phi_{2}(\alpha)}{4},\frac{\phi_{2}^{2}(\alpha)}{8},\frac{\alpha\phi_{2}^{2}(\alpha)}{8},\frac{\theta}{32},\frac{\alpha\theta}{64})$
is a $2$-integral basis of $\mathbb{Z}_{K}$. Replace $\phi_{2}(x)$ by
$x^{2}-2m_{2}ux+2m_{2}u$ with $m_{2}u\equiv 1\ \mbox{\rm(mod }{64})$, we
conclude that
$(1,\alpha,\frac{\alpha^{2})}{A_{2}},\frac{\alpha\phi_{2}(\alpha)}{2A_{3}},\frac{\phi_{2}^{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}^{2}(\alpha)}{2A_{5}},\frac{\theta}{4A_{6}},\frac{\alpha\theta}{8A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$, where
$\theta=\phi_{2}^{3}(\alpha)+(16m_{2}u+8m_{2}u\alpha)\phi_{2}^{2}(\alpha)+(-40m_{2}u+32m_{2}u\alpha)\phi_{2}(\alpha)-32m_{2}u\alpha$.

    3. (c)

If $\nu_{2}(m)=6$; $m\equiv 64\ \mbox{\rm(mod }{128})$, then for $\phi=x$, we
have $\overline{f(x)}=\phi^{8}\ \mbox{\rm(mod }{2})$, $N_{\phi}(f)=S$ has a
single side of slope $-\lambda_{1}=-3/4$, $e_{1}=4$, and
$R_{1}(f)(y)=y^{2}+1=(y+1)^{2}$. Let $\omega_{2}$ be the valuation of second
order Newton polygon defined by $\omega_{2}(a)=e_{1}\nu_{2}(a)=4\nu_{2}(a)$
for every $a\in\mathbb{Q}_{2}$ and $\omega_{2}(x)=e_{1}\lambda_{1}=3$. It
follows that:

      1. (i)

If $\nu_{2}(m-64)=7$ ($m\equiv 192\ \mbox{\rm(mod }{256})$), then there are
two case:
If $\nu_{2}(m-192)=8$ ($m\equiv 448\ \mbox{\rm(mod }{512})$), then for
$\phi_{2}=x^{4}+4x^{3}+12x^{2}+24$, we have
$f(x)=\phi_{2}^{2}+(-224-8x^{3}-8x^{2}+32x)\phi_{2}+(4800-m+512x^{3}+2304x^{2}-768x)$
is the $\phi_{2}$-expansion of $f(x)$. As $\omega_{2}(\phi_{2}^{2})=24$,
$\omega_{2}((-224-8x^{3}-8x^{2}+32x)\phi_{2})=30$, and
$\omega_{2}(4800-m+512x^{3}+2304x^{2}-768x)=\omega_{2}(768x)=35$, we conclude
that $N_{2}(f)$ has a single side of degree $1$. Thus, $\text{ind}_{2}(f)=5$
and $\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+5$. Since $V(\alpha)=3/4$ and
$V(\phi_{2}(\alpha))=35/8$. Replacing $\phi_{2}(x)$ by
$x^{4}+4um_{2}x^{3}+12um_{2}x^{2}+24um_{2}$ with $m_{2}u\equiv 1\
\mbox{\rm(mod }{32})$, we get
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{4A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.
If $\nu_{2}(m-192)\geq 9$ ($m\equiv 192\ \mbox{\rm(mod }{512})$), then for
$\phi_{2}=x^{4}+4x^{2}+8$, we have $f(x)=\phi_{2}^{2}-8x^{2}\phi_{2}-(64+m)$
is the $\phi_{2}$-expansion of $f(x)$. As $\omega_{2}(\phi_{2}^{2})=24$,
$\omega_{2}(-8x^{2}\phi_{2})=30$, and $\omega_{2}(64+m)=\omega_{2}(m-192)\geq
36$, then $N_{2}(f)$ has a single side with $R_{2}(f)(y)=y^{2}+y+1$ or $2$
sides of degree $1$ each. Thus, $\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)+6$.
Since $V(\phi_{2}(\alpha))\geq 9/2$ and $V(\alpha)=3/4$, if we replace
$\phi_{2}(x)$ by $x^{4}+4m_{2}ux+8m_{2}u$ with $um_{2}\equiv 1\ \mbox{\rm(mod
}{32})$, then
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{4A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{4A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

      2. (ii)

If $\nu_{2}(m-64)\geq 8$ ( $m\equiv 64\ \mbox{\rm(mod }{256})$), then for
$\phi_{2}=x^{4}+8$, we have $f(x)=\phi_{2}^{2}-16\phi_{2}+(64-m)$ is the
$\phi_{2}$-expansion of $f(x)$. Since $\omega_{2}(\phi_{2}^{2})=24$,
$\omega_{2}(16\phi_{2})=28$, and $\omega_{2}(64-m)\geq 32$, it follows that if
$\omega_{2}(64-m)=32$, then $N_{2}(f)$ has a single side with
$R_{2}(f)(y)=y^{2}+y+1$. So $\nu_{2}(\text{ind}(f))=\text{ind}_{2}(f)+4$. If
$\omega_{2}(64-m)>32$, then $N_{2}(f)$ has two side with degree $1$ each.
Thus, $\nu_{2}(\text{ind}(f))=\text{ind}_{2}(f)+4$. In both cases we have
$V(\alpha)=3/2$ and $V(\phi_{2}(\alpha))\geq 4$. Replacing $\phi_{2}$ by
$\phi_{2}=x^{4}+8um_{2}$ with $um_{2}\equiv 1\ \mbox{\rm(mod }{32})$, we get
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\phi_{2}(\alpha)}{2A_{4}},\frac{\alpha\phi_{2}(\alpha)}{2A_{5}},\frac{\alpha^{2}\phi_{2}(\alpha)}{2A_{6}},\frac{\alpha^{3}\phi_{2}(\alpha)}{2A_{7}})$
an integral basis of $\mathbb{Z}_{K}$.

  2. (2)

If $m\equiv 1\ \mbox{\rm(mod }{2})$, then $f(x)\equiv(x-1)^{8}\ \mbox{\rm(mod
}{2})$. Let $\phi=x-1$,
$f(x)=\phi^{8}+8\phi^{7}+28\phi^{6}+56\phi^{5}+70\phi^{4}+56\phi^{3}+28\phi^{2}+8\phi+1-m$.
It follows that:

    1. (a)

If $\nu_{2}(1-m)=1$; $m\equiv 3\ \mbox{\rm(mod }{4})$, then $N_{\phi}(f)=S$
has a single side joining $(0,1)$ and $(8,0)$. Thus
$\nu_{2}(\text{ind}(f))=\text{ind}_{\phi}(f)=0$. It follows that
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha\phi_{2}(\alpha)}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

    2. (b)

If $\nu_{2}(1-m)=2$; $m\equiv 5\ \mbox{\rm(mod }{8})$, then
$N_{\phi}(f)=S_{1}$ has a single side joining $(0,2)$, $(4,1)$ and $(8,0)$
with residual polynomial $R_{1}(f)(y)=y^{2}+y+1$, which is irreducible over
$\mathbb{F}_{1}=\mathbb{F}_{0}$. Thus
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)=4$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{4}}{2A_{4}},\frac{\alpha^{5}+m^{4}\alpha}{2A_{5}},\frac{\alpha^{6}+m^{4}\alpha^{2}}{2A_{6}},\frac{\alpha^{7}+m^{4}\alpha^{3}}{2A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

    3. (c)

If $\nu_{2}(1-m)=3$; $m\equiv 9\ \mbox{\rm(mod }{16})$, then
$N_{\phi}(f)=S_{1}+S_{2}$ has $2$ sides joining $(0,3)$, $(2,2)$, $(4,1)$, and
$(8,0)$. Thus $R_{11}(f)(y)=y^{2}+y+1$ and $R_{21}(f)(y)=y+1$. It follows by
Theorem 3.3 that $\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)=6$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{4}}{2A_{4}},\frac{\alpha^{5}+m^{4}\alpha}{2A_{5}},\frac{\alpha^{6}+2m\alpha^{5}+3m^{2}\alpha^{4}+m\alpha^{2}+2m\alpha+3m^{2}}{4A_{6}},$
$\frac{\alpha^{7}+2m\alpha^{6}+3m^{2}\alpha^{5}+m\alpha^{3}+2m\alpha^{2}+3m^{2}\alpha}{4A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

    4. (d)

If $\nu_{2}(1-m)=4$; $m\equiv 17\ \mbox{\rm(mod }{32})$, then
$N_{\phi}(f)=S_{1}+S_{2}+S_{3}$ has $3$ sides joining $(0,4)$, $(1,3)$,
$(2,2)$, $(4,1)$, and $(8,0)$. Thus $R_{11}(f)(y)=y^{2}+y+1$ and
$R_{i1}(f)(y)=y+1$ for every $i=2,3$. It follows by Theorem 3.3 that
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)=7$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{4}}{2A_{4}},\frac{\alpha^{5}+m^{4}\alpha}{2A_{5}},\frac{\alpha^{6}-2m\alpha^{5}-m^{2}\alpha^{4}+m^{2}\alpha^{2}+2m\alpha+3m^{2}}{4A_{6}},\\\
\frac{\alpha^{7}-m\alpha^{6}+m^{2}\alpha^{5}-m\alpha^{4}+m^{2}\alpha^{3}-m\alpha^{2}+(m^{2}+4m)\alpha+m}{8A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

    5. (e)

If $\nu_{2}(1-m)\geq 5$; $m\equiv 1\ \mbox{\rm(mod }{32})$, then
$N_{\phi}(f)=S_{1}+S_{2}+S_{3}+S_{4}$ has $4$ sides joining $(0,v)$, $(1,3)$,
$(2,2)$, $(4,1)$, and $(8,0)$ with $v=\nu_{2}(1-m)\geq 5$. Thus
$R_{i1}(f)(y)=y+1$ for every $i=1,2,3,4$. It follows by Theorem 3.3 that
$\nu_{2}(\text{ind}(f))=\text{ind}_{1}(f)=7$ and
$(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}+m^{4}}{2A_{4}},\frac{\alpha^{5}+m^{4}\alpha}{2A_{5}},\frac{\alpha^{6}-2m\alpha^{5}-m^{2}\alpha^{4}+m^{2}\alpha^{2}+2m\alpha+3m^{2}}{4A_{6}},\\\
\frac{\alpha^{7}-m\alpha^{6}+m^{2}\alpha^{5}-m\alpha^{4}+m^{2}\alpha^{3}-m\alpha^{2}+(m^{2}+4m)\alpha+m}{8A_{7}})$
is an integral basis of $\mathbb{Z}_{K}$.

$\Box$

Proof of Theorem 2.3.
Let $\alpha^{(i)}=\zeta^{i}\alpha$ ($i=0,\dots,7$) be the roots of
$f(x)=x^{8}-m$, where $\zeta$ is a primitive eighth root of unity and
$\alpha=.\alpha^{(0)}$. Let $L(x)=\alpha
x_{1}+\alpha^{2}x_{2}+\alpha^{3}x_{3}+\alpha^{4}x_{4}+\alpha^{5}x_{5}+\alpha^{6}x_{6}+\alpha^{7}x_{7}$
and denote by $L^{(i)}(x)$ its conjugates, corresponding to $\alpha^{(i)}$.
Let

$H_{ij}(x)=\frac{L^{(i)}(x)-L^{(j)}(x)}{\alpha^{(i)}-\alpha^{(j)}}$

for $1\leq i<j\leq 8$. The authomorphism $\sigma(x)=\zeta x$ brings
$\alpha^{(i)}$ into $\alpha^{(i+1)}$ for $1\leq i\leq 7$. Set

$\displaystyle F_{4}(x)$ $\displaystyle=$ $\displaystyle H_{15}(x)\cdot
H_{26}(x)\cdot H_{37}(x)\cdot H_{48}(x),$ $\displaystyle F_{8}(x)$
$\displaystyle=$ $\displaystyle H_{12}(x)\cdot H_{23}(x)\cdot H_{34}(x)\cdot
H_{45}(x)\cdot H_{56}(x)\cdot H_{67}(x)\cdot H_{78}(x)\cdot H_{81}(x).$

Since these are complete norms, they are polynomials with integer
coefficients, that divide $\prod_{1\leq i<j\leq 8}H_{ij}(x)$ which is also of
integer coefficients. Denote by $F_{16}(x)$ the third factor of the above
product. ($F_{16}(x)$ is of degree 16, since the product has 28 factors.)
Using symmetric polynomials these factors can be explicitely calculated by
using Maple [4]. Therefore we obtain

$\prod_{1\leq i<j\leq
8}\frac{L^{(i)}(x)-L^{(j)}(x)}{\alpha^{(i)}-\alpha^{(j)}}=F_{4}(x)\cdot
F_{8}(x)\cdot F_{16}(x),$

where the factors $F_{4}(x),F_{8}(x),F_{16}(x)$ have integer coefficients, of
degrees $4$, $8$, $16$, having $10$, $169$, $9185$ terms, respectively.

If $\nu_{2}(m)$ is odd or $m\equiv 3\ \mbox{\rm(mod }{4})$, then
$K=\mathbb{Q}(\sqrt[8]{m})$ has integral basis

$\left(1,\alpha,\frac{\alpha^{2}}{A_{2}},\frac{\alpha^{3}}{A_{3}},\frac{\alpha^{4}}{A_{4}},\frac{\alpha^{5}}{A_{5}},\frac{\alpha^{6}}{A_{6}},\frac{\alpha^{7}}{A_{7}}\right).$

Therefore

$\sqrt{|D_{K}|}=\frac{\sqrt{|D(\alpha)|}}{A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}}=\frac{\prod_{1\leq
i<j\leq 8}|\alpha^{(i)}-\alpha^{(j)}|}{A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}}.$

Any algebraic integer $\vartheta\in\mathbb{Z}_{K}$ can be written as

$\vartheta=\sum_{k=0}^{7}y_{k}\frac{\alpha^{k}}{A_{k}}=\frac{1}{A_{7}}\sum_{k=0}^{7}\frac{A_{7}}{A_{k}}y_{k}\alpha^{k}=\frac{1}{A_{7}}\sum_{k=0}^{7}z_{k}\alpha^{k},$

where $A_{0}=1$, $A_{7}/A_{k}$ and $y_{k}$ are integers,
$z_{k}=\frac{A_{7}}{A_{k}}y_{k},\;\;(k=0,1,\ldots,7)$. Hence

$I(\vartheta)=\frac{1}{\sqrt{|D_{K}|}}\prod_{1\leq i<j\leq
8}|\vartheta^{(i)}-\vartheta^{(j)}|$
$=\frac{A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}}{\prod_{1\leq i<j\leq
8}|\alpha^{(i)}-\alpha^{(j)}|}\cdot\frac{1}{A_{7}^{28}}\cdot\prod_{1\leq
i<j\leq 8}(L^{(i)}(z)-L^{(j)}(z))$
$=\frac{A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}}{A_{7}^{28}}\prod_{1\leq i<j\leq
8}\frac{L^{(i)}(z)-L^{(j)}(z)}{\alpha^{(i)}-\alpha^{(j)}}.$

Hence

$I(\vartheta)=\frac{A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}}{A_{7}^{28}}\cdot
F_{4}(z)\cdot F_{8}(z)\cdot F_{16}(z).$

In $F_{4}(z),F_{8}(z),F_{16}(z)$ we substitute the representation of
$A_{2},\ldots,A_{7}$ and $m$ in terms of $a_{1},\ldots,a_{7}$. We extract the
gcd $W_{4},W_{8},W_{16}$ of the coefficients of $F_{4}(z),F_{8}(z),F_{16}(z)$,
respectively, then we obtain

$F_{4}(z)=W_{4}\cdot G_{4}(z),\;F_{8}(z)=W_{8}\cdot
G_{8}(z),\;F_{16}(z)=W_{16}\cdot G_{16}(z),$

where $G_{4}(z),G_{8}(z),G_{16}(z)$ are polynomials of $z_{1},\ldots,z_{7}$
with integer coefficients, and

$W_{4}W_{8}W_{16}=\frac{A_{7}^{28}}{A_{2}A_{3}A_{4}A_{5}A_{6}A_{7}}.$

Therefore

$I(\vartheta)=G_{4}(z)\cdot G_{8}(z)\cdot G_{16}(z).$

Calculating explicitly $G_{16}(z)-a_{2}^{2}a_{6}^{2}G_{8}^{2}(z)$ we obtain

$8a_{1}a_{3}a_{5}a_{7}|(G_{16}(z)-a_{2}^{2}a_{6}^{2}G_{8}^{2}(z)).$

If $K$ is monogenic, then for some $z_{1},\ldots,z_{7}\in\mathbb{Z}$ we have
$I(\vartheta)=1$, hence $G_{4}(z)=\pm 1,G_{8}(z)=\pm 1,G_{16}(z)=\pm 1$, that
is the above divisibility relation implies

$8a_{1}a_{3}a_{5}a_{7}|(a_{2}^{2}a_{6}^{2}\pm 1).$

with at least one of the signs. $\Box$

The existence of prime common index divisors was first established in $1871$
by Dedekind who exhibited examples in cubic number fields. For example, he
considered the cubic field $K$ generated by a root of $x^{3}-x^{2}-2x-8$ and
he showed that the prime $2$ splits completely in $\mathbb{Z}_{K}$. So, if we
suppose that $K$ is monogenic, then we would be able to find a cubic
polynomial generating $K$, that splits completely into distinct polynomials of
degree $1$ in $\mathbb{F}_{2}[x]$. Since there are only $2$ distinct
polynomials of degree $1$ in $\mathbb{F}_{2}[x]$, this is impossible. Based on
these ideas and using Kronecker’s theory of algebraic numbers, Hensel gave a
necessary and sufficient condition on the so-called ”index divisors” for any
prime integer $p$ to be a prime common index divisor [17].

In order to prove Theorem 2.4, we need the following two lemmas. The first one
is an immediate consequence of Dedekind’s theorem. The second one follows from
[13, Corollary 3.8].

######  Lemma 4.1.

Let $p$ be rational prime integer and $K$ a number field. For every positive
integer $f$, let $P_{f}$ be the number of distinct prime ideals of
$\mathbb{Z}_{K}$ lying above $p$ with residue degree $f$ and $N_{f}$ the
number of monic irreducible polynomials of $\mathbb{F}_{p}[x]$ of degree $f$.
If $P_{f}>N_{f}$ for some positive integer $f$, then $p$ is a common index
divisor of $K$.

######  Lemma 4.2.

Let $p$ be a prime integer, $f(x)\in\mathbb{Z}_{p}[x]$ a monic polynomial such
that $\overline{f(x)}$ is a power of $\overline{\phi(x)}$ for some monic
polynomial $\phi\in\mathbb{Z}_{p}[x]$, whose reduction is irreducible over
$\mathbb{F}_{p}$, $N_{i}({f})=S_{i}$ has a single side of slope $-\lambda_{i}$
and $R_{i}(f)(y)=\psi_{i}^{a_{i}}(y)$ for some monic irreducible polynomial
$\psi_{i}\in\mathbb{F}_{i}[y]$ for every $i=1,\dots,r-1$. Let $e_{i}$ be the
smallest positive integer satisfying $e_{i}\lambda_{i}\in\mathbb{Z}$ for every
$i=1,\dots,r-1$. If $N_{r}(f)=T$ has a single side of degree $1$, then $f(x)$
is irreducible over $\mathbb{Q}_{p}$. Let $\mathfrak{p}$ be the unique prime
ideal of $\mathbb{Q}_{p}(\beta)$ lying above $p$, where $\beta$ is a root of
$f(x)$. Then $e_{1}\times\dots\times e_{r}$ is the ramification index of
$\mathfrak{p}$ and
$f=\mbox{deg}(\phi)\times\prod_{i=1}^{r-1}\mbox{deg}(\psi_{i})$ is its residue
degree.

Proof of Theorem 2.4

  1. (1)

If $m\equiv 1\ \mbox{\rm(mod }{32})$, then
$\overline{f(x)}=\overline{\phi(x)}^{8}\ \mbox{\rm(mod }{2})$ and
$N_{\phi}(f)=S_{1}+S_{2}+S_{3}+S_{4}$ has $4$ sides joining $(0,v)$, $(1,3)$,
$(2,2)$, $(4,1)$, and $(8,0)$, where $v=\nu_{2}(1-m)$. Thus each side is of
degree $1$, and so there are $4$ prime ideals of $\mathbb{Z}_{K}$ lying above
$2$. Since there are only $2$ monic irreducible polynomial of degree modulo
$2$. By Lemma 4.1, $2$ divides $i(K)$, and so $K$ is not monogenic.

  2. (2)

If $\nu_{2}(16-m)\geq 8$; $m\equiv 272\ \mbox{\rm(mod }{512})$, then
$N_{1}(f)=S$ has a single side of slope $-\lambda_{1}=-1/2$ (see Figure 2).
Then $e_{1}=1$ is the ramification index of $S$. Also
$N_{2}(f)=T_{1}+T_{2}+T_{3}$ has $3$ sides joining the points $(0,v)$,
$(1,12)$, $(2,10)$, and $(2,8)$ (see Figue 3) with $v=\omega_{2}(16-m)\geq
16$, residual polynomials $R_{i2}(f)(y)=y+1$ for $i=1,2$, and
$R_{32}(f)(y)=(y+1)^{2}$. By Theorem 3.4, $f(x)=f_{1}(x)\times f_{2}(x)\times
f_{3}(x)$ in $\mathbb{Z}_{2}[x]$ where $f_{i}(x)$ is monic irreducible over
$\mathbb{Q}_{2}$ for $i=1,2$ and $f_{3}(x)$ is monic. In order to complete the
factorization of $f_{3}(x)$, we have to use third order Newton polygon. For
$\phi_{3}=\phi_{2}+4x=x^{2}+4x+2$, we get $N_{3}^{+}(f)=T$ has a single side
of degree $1$ (see Figure 4). Thus $f_{3}(x)$ is irreducible over
$\mathbb{Q}_{2}$. By Hensel’s correspondence, there are exactly $3$ prime
ideals $\mathfrak{p}_{1},\mathfrak{p}_{2},\mathfrak{p}_{3}$ of
$\mathbb{Z}_{K}$ lying above $2$ (each $\mathfrak{p}_{i}$ is associated to the
factor $f_{i}$). By Lemma 4.2, $e(\mathfrak{p}_{i})=e_{1}\times e_{2}=2$ for
$i=1,2$ and $e(\mathfrak{p}_{3})=e_{1}\times e_{2}\times e_{3}=4$. As
$\sum_{i=1}^{3}e(\mathfrak{p}_{i})f(\mathfrak{p}_{i})=8$, we conclude that the
residue degree $f(\mathfrak{p}_{i})=1$ for every $i=1,2,3$. As there are only
$2$ monic irreducible polynomial in $\mathbb{F}_{0}[x]$, $2$ divides $i(K)$,
and so $K$ is not monogenic.

$0$$S$ Figure 2.  $N_{1}({f})$. $8$$4$$2$$1$$T_{1}$$T_{2}$$T_{3}$ Figure 3.
$N_{2}({f})$. $12$$13$$4$$2$$1$$T$ Figure 4.  $N_{3}^{+}({f})$.

  3. (3)

If $\nu_{2}(m)$ is odd, then by Theorem 2.3,

$8a_{1}a_{3}a_{5}a_{7}|(a_{2}^{2}a_{6}^{2}\pm 1)$

is a necessary condition for monogenity of $K$ with at least one of the signs.
Thus if $a_{2}a_{6}\equiv 2\ \mbox{\rm(mod }{8})$ or $a_{2}a_{6}\equiv 6\
\mbox{\rm(mod }{8})$, then $8a_{1}a_{3}a_{5}a_{7}$ does not divide
$(a_{2}a_{6})^{2}\pm 1$ with neither of the signs. Thus $K$ is not monogenic.

$\Box$

Proof of of Theorem 2.5
Assume that $m=a^{t}$ with $a\neq\pm 1$ is a square free rational integer and
$t\in\\{3,5,7\\}$. Since gcd$(t,8)=1$, let $(u,v)$ be two non-negative
integers solution of $tu-8v=1$ and $u<8$ and
$\theta=\frac{\alpha^{u}}{a^{v}}$. Since
$\theta^{8}=\frac{\alpha^{8u}}{a^{8v}}=\frac{m^{u}}{a^{8v}}=\frac{a^{tu}}{a^{8v}}=a$
and $x^{8}-a$ is irreducible over $\mathbb{Q}$ (because it is an Eisenstein
polynomial), hence $\mathbb{Q}(\theta)=\mathbb{Q}(\alpha)=K$. Thus $K$ is the
octic number field generated by a root $\theta$ of $g(x)=x^{8}-a$ with a
square free integer $a$. Applying [12, Theorem 8], we get the following
results:

  1. (1)

If $a\not\equiv 1\ \mbox{\rm(mod }{4})$, then $K$ is monogenic and
$\mathbb{Z}_{K}$ is generated by $\theta$.

  2. (2)

If $a\equiv 1\ \mbox{\rm(mod }{4})$, then $K$ is not monogenic, except for
$a=-3$.

$\Box$

## Acknowledgments

The authors are deeply grateful to the anonymous referee whose valuable
comments and suggestions have tremendously improved the quality of this paper.
Also, the first author is deeply grateful to Professor Enric Nart who
introduced him to Newton polygon techniques when he was a post-doc at CRM of
Barcelona (2007–2008).

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# Cr-doping effect on the orbital fluctuation of
heavily doped Nd1-xSrxMnO3 ($x$ $\approx$ 0.625)

R. Tasaki<EMAIL_ADDRESS>S. Fukushima  M. Akaki  D. Akahoshi  H.
Kuwahara  Department of Physics, Sophia University
Chiyoda-ku, Tokyo 102-8554, JAPAN

###### Abstract

We have investigated the Cr-doping effect of Nd0.375Sr0.625MnO3 near the phase
boundary between the $x^{2}-y^{2}$ and $3z^{2}-r^{2}$ orbital ordered states,
where a ferromagnetic correlation and concomitant large magnetoresistance are
observed owing to orbital fluctuation. Cr-doping steeply suppresses the
ferromagnetic correlation and magnetoresistance in Nd0.375Sr0.625Mn1-yCryO3
with $0\leq y\leq 0.05$, while they reappear in $0.05<y\leq 0.10$. Such a
reentrant behavior implies that a phase boundary is located at $y=0.05$, or a
phase crossover occurs across $y=0.05$.

Perovskite manganites, Colossal magnetoresistance (CMR) , Orbital Fluctuation,
Impurity Effect

###### pacs:

75.47.Gk, 75.30.Kz, 75.47.Lx

††preprint: AP-06

##  .1 Introduction

Mn oxides with a perovskite structure have attracted much attention because of
the colossal magnetoresistance (CMR) effectDagotto_PR_344 ; Tokura_RPP_69 .
Since the magnetic and transport properties of the perovskite manganites are
strongly affected by ordering patterns of $x^{2}-y^{2}$ and/or $3z^{2}-r^{2}$
orbitals, a detailed investigation of the orbital-ordered (OO) states is
significant for understanding the CMR effect. In heavily doped Nd1-xSrxMnO3
(NSMO), there exist two types of OO states, which exhibit highly anisotropic
magnetic and transport propertiesKajimoto_PRB_60 . One is the $x^{2}-y^{2}$ OO
state ($0.53\leq x<0.63$), accompanying the $A$-type antiferromagnetic (AF)
order in which the $x^{2}-y^{2}$ electrons are conducting within the
ferromagnetic (F) planeKuwahara_PRB_82 . The other is the $3z^{2}-r^{2}$ OO
state ($0.63\leq x\leq 0.80$), accompanying the $C$-type AF order. These two
OO states compete with each other in a bicritical manner at
$x=0.625$Kajimoto_PRB_60 . Near the bicritical region, competition between the
two OO states causes spatial orbital fluctuation on nanometer scale, which
gives rise to the F correlation and concomitant large magnetoresistance
(MR)Akahoshi_PRB_77 ; Nagao_JPCM_19 .

It is well-known that the presence of quenched disorder in a bicritical (or
multicritical) region where a ferromagnetic metallic (FM) and AF insulating
states meet often causes phase separation phenomena, which are essential for
the CMR. In Nd0.5Ca0.5MnO3, for example, Cr-substitution on Mn-sites turns the
charge- and orbital-ordered (CO/OO) state into the FM oneKimura_PRB_62 ;
Kimura_PRL_83 .

Therefore, it can be expected that Cr-doping into NSMO near the bicritical
region ($x=0.625$) induces phase separation and/or enhances the orbital
fluctuation, which might lead to nontrivial phenomena such as the CMR. In this
study, we have investigated the Cr-doping effect of
Nd0.375Sr0.625Mn1-yCryO3(NSMCO) ($0\leq y\leq 0.10$).

##  .2 Experiment

NSMCO crystals with $0\leq y\leq 0.10$ were prepared using the floating zone
method. We confirmed that all synthesized crystals are of single phase by the
powder X-ray diffraction method. Magnetic and transport properties were
measured using a Quantum Design physical property measurement system (PPMS).
We randomly cut the synthesized crystals with the size larger than twin-domain
size for measurements of magnetic and transport properties.

##  .3 Results and discussion

Figure 1:  (Color online) Temperature ($T$) dependence of (a) magnetization
($M$) and (b) magnetoresistance [MR(80 kOe)] of
Nd0.375Sr0.625Mn1-yCryO3(NSMCO) with $y$ = 0, 0.03, and 0.05. ZFC represents
zero field cooling process. MR(80 kOe) is defined as MR(80 kOe)
$\equiv\rho$(80 kOe) / $\rho$(0 Oe).

First, we show in Figs. 1(a) and 1(b) the temperature ($T$) dependence of the
magnetization ($M$) and MR(80 kOe) of NSMCO with $0\leq y\leq 0.05$,
respectively. Here MR(80 kOe) is defined as MR(80 kOe) $\equiv\rho$(80 kOe) /
$\rho$(0 Oe), where $\rho$(0 Oe) and $\rho$(80 kOe) are resistivities measured
in $H$ = 0 Oe and 80 kOe, respectively. In $y$ = 0, the F correlation is
observed due to the orbital fluctuation below 65 K, and the MR(80 kOe) at 5 K
is below 0.01: the resistivity drops more than two orders of magnitude by
applying a magnetic field of $H$ = 80 kOe. This result is consistent with our
previous reportAkahoshi_PRB_77 . Cr-doping steeply suppresses the F
correlation and concomitant MR, which are most suppressed at $y$ = 0.05.

Figure 2:  (Color online) $T$ dependence of (a) $M$ and (b) MR(80 kOe) of
NSMCO with $y$= 0.05, 0.07, and 0.10.

Then, we exhibit the $T$ dependence of the $M$ and MR(80 kOe) of NSMCO with
$0.05\leq y\leq 0.10$ in Figs. 2(a) and 2(b). In $y>0.05$, the F correlation
and MR reappear and are evolving with an increase of $y$ from 0.05 to 0.10.
Note that the $M$ and MR of $y=0.10$ are quite similar to those of $y=0$, as
clearly seen from Figs. 1 and 2, indicating that the F correlation and MR of
NSMCO show a reentrant behavior with a change of Cr-concentration $y$.

Figure 3:  Cr-doping concentration $y$ dependence of (a) $M$ under $H$=1 kOe
and (b) MR(80 kOe) at 15 K.

We plot the $M$ under $H$ = 1 kOe and MR(80 kOe) at 15K as a function of $y$
in Figs. 3(a) and 3(b), respectively. These figures, as mentioned above,
demonstrate that the F correlation and MR are most suppressed at $y$ = 0.05
and show the reentrant behavior, implying that a phase boundary is located at
$y=0.05$, or a phase crossover occurs across $y=0.05$.

Let us discuss the origin of the reentrant behavior. In $0\leq y\leq 0.05$,
the F correlation and MR are systematically reduced with increasing $y$. This
behavior is quite similar to that observed in NSMO with $x=0.625$, the F
correlation and MR of which are also suppressed with increasing hole-
concentration $x$ from 0.625Akahoshi_PRB_77 . Therefore, we interpret that Cr-
and hole-doping have almost the same effect on the magnetic and transport
properties of NSMO with $x=0.625$ in $0\leq y\leq 0.05$. This is probably
because Cr3+ and Mn4+ have the same electronic configuration of
$t_{2g}^{3}e_{g}^{0}$. In $y>0.05$, the F correlation and MR reappear; the F
correlation is developing with further increasing $y$ from 0.05. This reminds
us the fact that Cr-doping into CO/OO Nd0.5Ca0.5MnO3 produces the FM clusters
embedded in the CO/OO matrixKimura_PRB_62 ; Kimura_PRL_83 . In NSMO with
$x=0.625$ as well as Nd0.5Ca0.5MnO3, the F correlation is probably induced
around Cr3+. However, the F correlation is not so strong compared with that of
Cr-doped Nd0.5Ca0.5MnO3, the reason for which might be explained by the fact
that NSMO with $x=0.625$ is apart from the FM state, which is often found in
the low-doped ($x\leq 0.5$) perovskite manganites. The reentrant behavior
observed in Cr-doped NSMO with $x=0.625$ is perhaps due to competition between
the hole-doping effect and the FM cluster effect caused by Cr-doping. In
$0\leq y\leq 0.05$, the number (or the size) of the FM clusters is so small
that the hole-doping effect is dominant. With increasing $y$, the number (or
the size) of the FM clusters is becoming large, and the FM cluster effect
finally overcomes the hole-doping effect in $y>0.05$. As a result, the F
correlation is macroscopically observed again in the magnetic and transport
properties of NSMCO with $y>0.05$. The detailed mechanism of the reentrant
behavior is now under investigation.

##  .4 Acknowledgment

We thank Y. Izuchi for her help in growing single crystals and the
measurements using PPMS. This work was partly supported by the Mazda
Foundation, the Asahi Glass Foundation, and Grant-in-Aid for scientific
research (C) from the Japan Society for Promotion of Science.

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