# On index divisors and monogenity of certain number fields defined by
trinomials $\displaystyle x^{7}+ax+b$

Lhoussain El Fadil<EMAIL_ADDRESS>
###### Abstract.

In this paper for every number field $\displaystyle K$ generated by a root
$\displaystyle\alpha$ of a trinomial $\displaystyle
x^{7}+ax+b\in\mathbb{Z}[x]$ and for every prime integer $\displaystyle p$, we
calculate $\displaystyle\nu_{p}(i(K))$, the highest power of $\displaystyle p$
dividing the index $\displaystyle i(K)$ of the field $\displaystyle K$. In
particular, we calculate the index $\displaystyle i(K)$. When the index of
$\displaystyle K$ is not trivial, namely $\displaystyle\nu_{p}(i(K))\neq q0$
for some prime integer $\displaystyle p$, then $\displaystyle K$ is not
monogenic.

###### Key words and phrases:

Power integral bases, theorem of Ore, prime ideal factorization, common index
divisor

###### 2010 Mathematics Subject Classification:

11R04,11Y40, 11R21

##  1\. Introduction

Let $\displaystyle K$ be a number field of degree $\displaystyle n$,
$\displaystyle\mathbb{Z}_{K}$ its ring of integers, and $\displaystyle d_{K}$
its absolute discriminant. It is well known that $\displaystyle\mathbb{Z}_{K}$
is a free abelian group of rank $\displaystyle n$ and by the The fundamental
theorem of finite abelian groups,
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\theta])$ for every primitive element
$\displaystyle\theta\in\mathbb{Z}_{K}$ of $\displaystyle K$. Let
$\displaystyle ind(\theta)=(\mathbb{Z}_{K}:\mathbb{Z}[\theta])$. Then
$\displaystyle ind(\theta)$ is called the index of $\displaystyle\theta$. The
index of the number field $\displaystyle K$ is defined by $\displaystyle
i(K)=GCD((\mathbb{Z}_{K}:\mathbb{Z}[\theta])\mid K=\mathbb{Q}(\theta)\mbox{
and }\theta\in\mathbb{Z}_{K})$. A rational prime integer $\displaystyle p$
dividing $\displaystyle i(K)$ is called a prime common index divisor of
$\displaystyle K$. The number field $\displaystyle K$ is called monogenic if
it admits a $\displaystyle\mathbb{Z}$ basis of type
$\displaystyle(1,\theta,\ldots,\theta^{n-1})$ for some
$\displaystyle\theta\in\mathbb{Z}_{K}$. Remark that if
$\displaystyle\mathbb{Z}_{K}$ has a power integral basis, then $\displaystyle
i(K)=1$. Therefore a field having a prime common index divisor is not
monogenic. Monogenity of number fields is a classical problem of algebraic
number theory, going back to Dedekind, Hasse and Hensel, see for instance [21,
22] and [17] for the present state of this area. It is called a problem of
Hasse to give an arithmetic characterization of those number fields which are
monogenic [21, 22, 19]. For any primitive element
$\displaystyle\theta\in\mathbb{Z}_{K}$ of $\displaystyle K$, we denote by

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

the index of $\displaystyle\theta$, that is the index of the
$\displaystyle\mathbb{Z}$-module $\displaystyle\mathbb{Z}[\theta]$ in the
free-$\displaystyle\mathbb{Z}$-module $\displaystyle\mathbb{Z}_{K}$ of rank
$\displaystyle n$. As it is known [17], we have

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

where $\displaystyle\triangle(\theta)$ is the discriminant of the minimal
polynomial of $\displaystyle\theta$ over $\displaystyle\mathbb{Q}$.

Clearly, $\displaystyle ind(\theta)=1$ for some primitive element
$\displaystyle\theta\in\mathbb{Z}_{K}$ of $\displaystyle K$ if and only if
$\displaystyle(1,\theta,\ldots,\theta^{n-1})$ is a power integral basis of
$\displaystyle\mathbb{Z}_{K}$.

The problem of testing the monogenity of number fields and constructing power
integral bases have been intensively studied during the last four decades
mainly by Gaál, Györy, Nakahara, Pohst and their collaborators (see for
instance [1, 14, 35]). In 1871, Dedekind was the first who gave an example of
a number field with non trivial index, he considered the cubic field
$\displaystyle K$ generated by a root of $\displaystyle x^{3}-x^{2}-2x-8$ and
showed that the rational prime $\displaystyle 2$ splits completely in
$\displaystyle K$ ([5, § 5, page 30]). According to a well known theorem of
Dedekind ([20, Chapter I, Proposition 8.3]), if we suppose that $\displaystyle
K$ is monogenic, then we would be able to find a cubic polynomial defining
$\displaystyle K$, that splits completely into distinct polynomials of degree
$\displaystyle 1$ in $\displaystyle\mathbb{F}_{2}[x]$. Since there is only two
distinct polynomials of degree $\displaystyle 1$ in
$\displaystyle\mathbb{F}_{2}[x]$, this is impossible. In $\displaystyle 1930$,
Engstrom was the first one who related the prime ideal factorization and the
index of a number field of degree less than $\displaystyle 8$ [11]. For any
number field $\displaystyle K$ of degree $\displaystyle n\leq 7$, he showed
that $\displaystyle\nu_{p}(i(K))$ is explicitly determined by the
factorization of $\displaystyle p\mathbb{Z}_{K}$ into powers of prime ideals
of $\displaystyle K$ for every positive rational prime integer $\displaystyle
p\leq n$. This motivated Narkiewicz to ask a very important question, stated
as problem 22 in Narkiewicz’s book ([34, Problem 22]), which asks for an
explicit formula of the highest power $\displaystyle\nu_{p}(i(K))$ for a given
rational prime $\displaystyle p$ dividing $\displaystyle i(K)$. In [33],
Nakahara studied the index of non-cyclic but abelian biquadratic number
fields. He showed that the field index of such fields is in the set
$\displaystyle\\{1,2,3,4,6,12\\}$. In [15] Gaál et al. characterized the field
indices of biquadratic number fields having Galois group $\displaystyle V_{4}$
and they proved that $\displaystyle i(K)\in\\{1,2,3,4,6,12\\}$. Recently, many
authors are interested on monogenity of number fields defined by trinomials.
In [26, 27], Khanduja et al. studied the integral closedness of some number
fields defined by trinomials. Their results are refined by Ibarra et al. (see
[25]). Recall that the results given in [26, 27], can only decide on the
integral closedness of $\displaystyle\mathbb{Z}[\alpha]$, but cannot test
whether the field is monogenic or not, where $\displaystyle\alpha$ is an
integral element over $\displaystyle\mathbb{Z}$. In [28, 29, 30, 31], Jones et
al. introduced monogenity of some irreducible polynomials. Recall that
according to Jones’ definition, if a polynomial $\displaystyle F(x)$ is
monogenic, then $\displaystyle\mathbb{Q}(\alpha)$ is monogenic, but the
converse is not true; it is possible that a number field generated by a root
$\displaystyle\alpha$ of a non monogenic polynomial can be monogenic (see
Example 2.2). Therefore Jones’ and Khanduja’s results cover partially the
study of monogenity of number fields defined by trinomials. Davis and Spearman
[6] studied the index of quartic number fields $\displaystyle K$ generated by
a root of such a quartic trinomial $\displaystyle
F(x)=x^{4}+ax+b\in\mathbb{Z}[x]$. They gave necessary and sufficient
conditions on $\displaystyle a$ and $\displaystyle b$ so that a prime
$\displaystyle p$ is a common index divisor of $\displaystyle K$ for
$\displaystyle p=2,3$. Their method is based on the calculation of the
$\displaystyle p$-index form of $\displaystyle K$, using $\displaystyle
p$-integral bases of $\displaystyle K$. El Fadil and Gaál [9] studied the
index of quartic number fields $\displaystyle K$ generated by a root of a
quadratic trinomial of the form $\displaystyle
F(x)=x^{4}+ax^{2}+b\in\mathbb{Z}[x]$. They gave necessary and sufficient
conditions on $\displaystyle a$ and $\displaystyle b$ so that a prime
$\displaystyle p$ is a common index divisor of $\displaystyle K$ for every
prime integer $\displaystyle p$. In [13], for a sextic number field
$\displaystyle K$ defined by a trinomial $\displaystyle
F(x)=x^{6}+ax^{3}+b\in\mathbb{Z}[x]$, Gaál studied the multi-monegenity of
$\displaystyle K$; he calculated all possible power integral bases of
$\displaystyle K$. In [7], El Fadil extended Gaál’s studies by providing some
cases where $\displaystyle K$ is not monogenic. Also in [8], for every prime
integer $\displaystyle p$, El Fadil gave necessary and sufficient conditions
on $\displaystyle a$ and $\displaystyle b$ so that $\displaystyle p$ is a
common index divisor of $\displaystyle K$, where $\displaystyle K$ is a number
field defined by an irreducible trinomial $\displaystyle
F(x)=x^{5}+ax^{2}+b\in\mathbb{Z}[x]$. In this paper, for a septic number field
generated by a root of a trinomial $\displaystyle
F(x)=x^{7}+ax+b\in\mathbb{Z}[x]$, we calculate $\displaystyle\nu_{p}(i(K))$,
the highest power of $\displaystyle p$ dividing the index $\displaystyle i(K)$
of the field $\displaystyle K$. When the index of $\displaystyle K$ is not
trivial, namely $\displaystyle\nu_{p}(i(K))\neq 0$ for some prime integer
$\displaystyle p$, then $\displaystyle K$ is not monogenic.

##  2\. Main Results

Throughout this section $\displaystyle K$ is a number field generated by a
root $\displaystyle\alpha$ of a trinomial $\displaystyle
F(x)=x^{7}+ax+b\in\mathbb{Z}[x]$ and we assume that for every rational prime
integer $\displaystyle p$, $\displaystyle\nu_{p}(a)\leq 5$ or
$\displaystyle\nu_{p}(b)\leq 6$. Along this paper, for every integer
$\displaystyle a\in\mathbb{Z}$ and a prime integer $\displaystyle p$, let
$\displaystyle a_{p}=\frac{a}{p^{\nu_{p}(a)}}$.

We start with the following theorem, which characterizes when is
$\displaystyle\mathbb{Z}[\alpha]$ integrally closed?

######  Theorem 2.1.

The ring $\displaystyle\mathbb{Z}[\alpha]$ is integrally closed if and only if
the following conditions hold:

  1. (1)

For every prime integer $\displaystyle p$, if $\displaystyle p$ divides
$\displaystyle a$ and $\displaystyle b$, then $\displaystyle\nu_{p}(b)=1$.

  2. (2)

If $\displaystyle 2$ divides $\displaystyle b$ and does not $\displaystyle a$,
then $\displaystyle a\equiv 3\ \mbox{\rm(mod }{4})$ and $\displaystyle b\equiv
2\ \mbox{\rm(mod }{4})$ or $\displaystyle a\equiv 1\ \mbox{\rm(mod }{4})$ and
$\displaystyle b\equiv 0\ \mbox{\rm(mod }{4})$ .

  3. (3)

If $\displaystyle 3$ divides $\displaystyle b$ and $\displaystyle a\equiv-1\
\mbox{\rm(mod }{3})$, then $\displaystyle\nu_{3}(b+1+a)=\nu_{3}(b-1-a)=1$.

  4. (4)

If $\displaystyle 3$ divides $\displaystyle b$ and $\displaystyle a\equiv 1\
\mbox{\rm(mod }{3})$, then $\displaystyle a\equiv 4\ \mbox{\rm(mod }{9})$ or
$\displaystyle a\equiv 7\ \mbox{\rm(mod }{9})$ or $\displaystyle\nu_{3}(b)=1$.

  5. (5)

If $\displaystyle 7$ divides $\displaystyle a$ and does not divide
$\displaystyle b$, then $\displaystyle\nu_{7}(1-a-b^{6})=1$.

  6. (6)

For every prime integer $\displaystyle p\not\in\\{2,3,7\\}$, if $\displaystyle
p$ does not divide both $\displaystyle a$ and $\displaystyle b$, then
$\displaystyle\nu_{p}(7^{7}b^{6}+6^{6}a^{7})\leq 1$.

The following example gives an infinite family of monogenic septic number
fields defined by non monogenic trinomials. In such a way, the results given
in [27, 28, 29, 30, 31, 25], cannot decide on monogenity of $\displaystyle K$.

######  Proposition 2.2.

Let $\displaystyle K$ be the number field generated by a root
$\displaystyle\alpha$ of $\displaystyle
F(x)=x^{7}+2^{u}ax+2^{v}b\in\mathbb{Z}[x]$, with $\displaystyle u\geq v-1$,
$\displaystyle 2\leq v\leq 6$, GCD$\displaystyle(6,b)=1$, $\displaystyle 7$
does not divide $\displaystyle a$, and for every odd prime integer
$\displaystyle p$, if $\displaystyle p$ does not divide $\displaystyle b$,
then $\displaystyle p^{2}$ does not divide $\displaystyle
7^{7}b^{6}+6^{6}a^{7}$. Then $\displaystyle F(x)$ is a non monogenic
polynomial and $\displaystyle K$ is a monogenic number field.

In the remainder of this section, for every prime integer $\displaystyle p$,
for every values of $\displaystyle a$ and $\displaystyle b$ and for every
prime integer $\displaystyle p$, we calculate $\displaystyle\nu_{p}(i(K))$. If
$\displaystyle\nu_{p}(i(K))\neq 0$ for some prime integer $\displaystyle p$,
then $\displaystyle K$ is not monogenic. For every integers $\displaystyle a$
and $\displaystyle b$, let $\displaystyle\triangle=-(6^{6}a^{7}+7^{7}b^{6})$
and for every prime integer $\displaystyle p$, let
$\displaystyle\triangle_{p}=\frac{\triangle}{p^{\nu_{p}(\triangle)}}$.

######  Theorem 2.3.

The following table provides the value of $\displaystyle\nu_{2}(i(K))$.

$\displaystyle\begin{array}[]{|c|c|c|}\hline\cr{\mbox{c}onditions}&\nu_{2}(i(K))\\\
\hline\cr a\equiv 28\ \mbox{\rm(mod }{32}){\mbox{ and }}b\equiv 0\
\mbox{\rm(mod }{32})&1\\\ \hline\cr a\equiv 112\ \mbox{\rm(mod }{128}){\mbox{
and }}b\equiv 0\ \mbox{\rm(mod }{128})&1\\\ \hline\cr a\equiv 1\ \mbox{\rm(mod
}{8}){\mbox{ and }}b\equiv 2\ \mbox{\rm(mod }{4})&\\\
\nu_{2}(\triangle){\mbox{ even and }}\triangle_{2}\equiv 3\ \mbox{\rm(mod
}{4})&1\\\ \hline\cr a\equiv 3\ \mbox{\rm(mod }{8}){\mbox{ and }}b\equiv 4\
\mbox{\rm(mod }{8})&1\\\ \hline\cr a\equiv 3\ \mbox{\rm(mod }{4}){\mbox{ and
}}b\equiv 0\ \mbox{\rm(mod }{8})&3\\\
\hline\cr(a,b)\in\\{(5,2),(5,6),(13,2),(13,14)\\}\ \mbox{\rm(mod }{16})&1\\\
\hline\cr{\mbox{O}therwise}&0\\\ \hline\cr\end{array}$

In particular, if $\displaystyle\nu_{2}(i(K))\geq 1$, then $\displaystyle K$
is not monogenic.

######  Theorem 2.4.

The following table provides the value of $\displaystyle\nu_{3}(i(K))$.

$\displaystyle\begin{array}[]{|c|c|c|}\hline\cr{\mbox{c}onditions}&\nu_{3}(i(K))\\\
\hline\cr a\equiv 5\ \mbox{\rm(mod }{9}){\mbox{ and }}b\in\\{3,6\\}\
\mbox{\rm(mod }{9})&1\\\ \hline\cr a\equiv 8\ \mbox{\rm(mod }{9}){\mbox{ and
}}b\equiv 0\ \mbox{\rm(mod }{9})&2\\\ \hline\cr a\equiv 2\ \mbox{\rm(mod
}{9}){\mbox{ and }}b\in\\{3,6\\}\ \mbox{\rm(mod }{9})&\\\
\nu_{3}(\triangle)=2k{\mbox{ and }}k\geq 5&1\\\ \hline\cr a\equiv 2\
\mbox{\rm(mod }{9}){\mbox{ and }}b\in\\{3,6\\}\ \mbox{\rm(mod }{9})&\\\
\nu_{3}(\triangle)=2k+1,\,k\geq 5{\mbox{ and }}\triangle_{3}\equiv 1\
\mbox{\rm(mod }{3})&2\\\ \hline\cr{\mbox{O}therwise}&0\\\
\hline\cr\end{array}$

In particular, if $\displaystyle\nu_{3}(i(K))\geq 1$, then $\displaystyle K$
is not monogenic.

######  Theorem 2.5.

For every prime integer $\displaystyle p\geq 5$ and for every integers
$\displaystyle a$ and $\displaystyle b$ such that $\displaystyle
F(x)=x^{7}+ax+b$ is irreducible over $\displaystyle\mathbb{Q}$, $\displaystyle
p$ does not divide $\displaystyle i(K)$, where $\displaystyle K$ is the number
field defined by $\displaystyle F(x)$.

######  Corollary 2.6.

For every integers $\displaystyle a$ and $\displaystyle b$ such that
$\displaystyle F(x)=x^{7}+ax+b$ is irreducible over $\displaystyle\mathbb{Q}$,
$\displaystyle i(K)\in\\{1,2,3,6,8,9,18,24,72\\}$, where $\displaystyle K$ is
the number field generated by a root of $\displaystyle F(x)$.

######  Remark 1.

  1. (1)

The field $\displaystyle K$ can be non monogenic even if the index
$\displaystyle i(K)=1$.

  2. (2)

The unique method which allows to test whether $\displaystyle K$ is monogenic
is to calculate the solutions of the index form equation of the field
$\displaystyle K$ (see for instance [16, 17]).

##  3\. A short introduction to prime ideal factorization based on Newton
polygons

In 1894, Hensel developed a powerful approach by showing that for every prime
integer $\displaystyle p$, the prime ideals of $\displaystyle\mathbb{Z}_{K}$
lying above $\displaystyle p$ are in one–one correspondence with monic
irreducible factors of $\displaystyle F(x)$ in
$\displaystyle\mathbb{Q}_{p}[x]$. For every prime ideal corresponding to any
irreducible factor in $\displaystyle\mathbb{Q}_{p}[x]$, the ramification index
and the residue degree together are the same as those of the local field
defined by the associated irreducible factor [24]. Since then, to factorize
$\displaystyle p\mathbb{Z}_{K}$, we need to factorize $\displaystyle F(x)$ in
$\displaystyle\mathbb{Q}_{p}[x]$. Newton’s polygon techniques can be used to
refine the factorization. This is a standard method which is rather technical
but very efficient to apply. We have introduced the corresponding concepts in
several former papers. Here we only give a brief introduction which makes our
proofs understandable. For a detailed description, we refer to Ore’s Paper
[37] and Guardia, Montes and Nart’s paper [18]. For every prime integer
$\displaystyle p$, let $\displaystyle\nu_{p}$ be the $\displaystyle p$-adic
valuation of $\displaystyle\mathbb{Q}_{p}$ and $\displaystyle\mathbb{Z}_{p}$
the ring of $\displaystyle p$-adic integers. Let $\displaystyle
F(x)\in\mathbb{Z}_{p}[x]$ be a monic polynomial and
$\displaystyle\phi\in\mathbb{Z}_{p}[x]$ a monic lift of an irreducible factor
of $\displaystyle\overline{F(x)}$ modulo $\displaystyle p$. Let $\displaystyle
F(x)=a_{0}(x)+a_{1}(x)\phi(x)+\cdots+a_{n}(x)\phi(x)^{l}$ be the
$\displaystyle\phi$-expansion of $\displaystyle F(x)$, $\displaystyle
N_{\phi}(F)$ the $\displaystyle\phi$-Newton polygon of $\displaystyle F(x)$
and $\displaystyle N_{\phi}^{+}(F)$ its principal part. Let
$\displaystyle\mathbb{F}_{\phi}$ be the field
$\displaystyle\mathbb{F}_{p}[x]/(\overline{\phi})$. For every side
$\displaystyle S$ of $\displaystyle N_{\phi}^{+}(F)$ with length
$\displaystyle l$ and initial point $\displaystyle(s,u_{s})$, for every
$\displaystyle i=0,\ldots,l$, let $\displaystyle c_{i}\in\mathbb{F}_{\phi}$ be
the residue coefficient, defined as follows:

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

Let $\displaystyle-\lambda=-h/e$ be the slope of $\displaystyle S$, where
$\displaystyle h$ and $\displaystyle e$ are two positive coprime integers.
Then $\displaystyle d=l/e$ is the degree of $\displaystyle S$. Let
$\displaystyle{R_{1}(F)(y)}=t_{d}y^{d}+t_{d-1}y^{d-1}+\cdots+t_{1}y+t_{0}\in\mathbb{F}_{\phi}[y]$,
called the residual polynomial of $\displaystyle F(x)$ associated to the side
$\displaystyle S$, where for every $\displaystyle i=0,\dots,d$, $\displaystyle
t_{i}=c_{ie}$. If $\displaystyle{R_{1}(F)(y)}$ is square free for each side of
the polygon $\displaystyle N_{\phi}^{+}(F)$, then we say that $\displaystyle
F(x)$ is $\displaystyle\phi$-regular.
Let $\displaystyle\overline{F(x)}=\prod_{i=1}^{r}\overline{\phi_{i}}^{l_{i}}$
be the factorization of $\displaystyle F(x)$ into powers of monic irreducible
coprime polynomials over $\displaystyle\mathbb{F}_{p}$, we say that the
polynomial $\displaystyle F(x)$ is $\displaystyle p$-regular if $\displaystyle
F(x)$ is a $\displaystyle\phi_{i}$-regular polynomial with respect to
$\displaystyle p$ for every $\displaystyle i=1,\dots,r$. Let $\displaystyle
N_{\phi_{i}}^{+}(F)=S_{i1}+\cdots+S_{ir_{i}}$ be the
$\displaystyle\phi_{i}$-principal Newton polygon of $\displaystyle F(x)$ with
respect to $\displaystyle p$. For every $\displaystyle j=1,\dots,r_{i}$, let
$\displaystyle R_{1_{ij}}(F)(y)=\prod_{s=1}^{s_{ij}}\psi_{ijs}^{a_{ijs}}(y)$
be the factorization of $\displaystyle R_{1_{ij}}(F)(y)$ in
$\displaystyle\mathbb{F}_{\phi_{i}}[y]$, where $\displaystyle
R_{1_{ij}}(F)(y)$ is the residual polynomial of $\displaystyle F(x)$ attached
to the side $\displaystyle S_{ij}$. Then we have the following theorem of
index of Ore:

######  Theorem 3.1.

$\displaystyle($[10, Theorem 1.7 and Theorem 1.9]$\displaystyle)$
Under the above hypothesis, we have the following:

  1. (1)

$\displaystyle\nu_{p}((\mathbb{Z}_{K}:\mathbb{Z}[\alpha]))\geq\sum_{i=1}^{r}\text{ind}_{\phi_{i}}(F).$

The equality holds if $\displaystyle F(x)\text{ is }p$-regular.

  2. (2)

If $\displaystyle F(x)\text{ is }p$-regular, then

$\displaystyle
p\mathbb{Z}_{K}=\prod_{i=1}^{r}\prod_{j=1}^{t_{i}}\prod_{s=1}^{s_{ij}}\mathfrak{p}_{ijs}^{e_{ij}}$

is the factorization of $\displaystyle p\mathbb{Z}_{K}$ into powers of prime
ideals of $\displaystyle\mathbb{Z}_{K}$, where $\displaystyle e_{ij}$ is the
smallest positive integer satisfying $\displaystyle
e_{ij}\lambda_{ij}\in\mathbb{Z}$ and the residue degree of
$\displaystyle\mathfrak{p}_{ijs}$ over $\displaystyle p$ is given by
$\displaystyle f_{ijs}=\deg(\phi_{i})\times\deg(\psi_{ijs})$ for every
$\displaystyle(i,j,s)$.

When the theorem of Ore fails, that is $\displaystyle F(x)$ is not
$\displaystyle p$-regular, then in order to complete the factorization of
$\displaystyle F(x)$, Guardia, Montes, and Nart introduced the notion of high
order Newton polygon. By analogous to the first order, for each order
$\displaystyle r$, the authors of [18] introduced the valuation
$\displaystyle\omega_{2}$ of order $\displaystyle r$, the key polynomial
$\displaystyle\phi_{2}(x)$ of such a valuation, $\displaystyle N_{r}(F)$ the
Newton polygon of any polynomial $\displaystyle F(x)$ with respect to
$\displaystyle\omega_{2}$ and $\displaystyle\phi_{2}(x)$, and for every side
of $\displaystyle N_{r}(F)$ the residual polynomial $\displaystyle R_{r}(F)$,
and the index of $\displaystyle F(x)$ in order $\displaystyle r$. For more
details, we refer to [18].

##  4\. proofs of our main results

Proof of Theorem 2.1.
Let $\displaystyle p$ be a prime integer. Since
$\displaystyle\triangle=-(6^{6}a^{7}+7^{7}b^{6})$ is the discriminant of
$\displaystyle F(x)$, then a necessary condition for $\displaystyle p$ to be a
candidate divisor of $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ is
that $\displaystyle p^{2}$ divides $\displaystyle 6^{6}a^{7}+7^{7}b^{6}$. We
conclude that if $\displaystyle p=2,3$, then $\displaystyle p$ divides
$\displaystyle b$ is a necessary condition for $\displaystyle p$ to be a
candidate divisor of $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. If
$\displaystyle p=7$, then $\displaystyle 7$ divides $\displaystyle a$ is a
necessary condition for $\displaystyle p$ to be a candidate divisor of
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. If $\displaystyle
p\not\in\\{2,3,7\\}$, then $\displaystyle p^{2}$ divides $\displaystyle
6^{6}a^{7}+7^{7}b^{6}$ is a necessary condition for $\displaystyle p$ to be a
candidate divisor of $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$.

  1. (1)

If $\displaystyle p$ divides $\displaystyle a$ and $\displaystyle b$, then
$\displaystyle p$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ if and only if
$\displaystyle\nu_{p}(b)=1$.

  2. (2)

For $\displaystyle p=2$,

    1. (a)

If $\displaystyle 2$ divides $\displaystyle b$ and does not $\displaystyle a$,
then $\displaystyle\overline{F(x)}=x(x-1)^{2}(x^{2}+x+1)^{2}$. Let
$\displaystyle\phi_{1}=x-1$ and $\displaystyle\phi_{2}=x^{2}+x+1$. Since
$\displaystyle F(x)=\dots-(4x-2)\phi_{2}+(a+1)x+b$ and $\displaystyle
F(x)=\dots+(a+7)\phi_{1}+(a+1+b)$, we conclude that $\displaystyle 2$ does not
divide $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ if and only if
$\displaystyle\nu_{2}((a+1)x+b)=1$ and $\displaystyle\nu_{2}(b+a+1)=1$, which
means $\displaystyle b\equiv 2\ \mbox{\rm(mod }{4})$ and $\displaystyle
a\equiv 3\ \mbox{\rm(mod }{4})$ or $\displaystyle b\equiv 0\ \mbox{\rm(mod
}{4})$ and $\displaystyle a\equiv 1\ \mbox{\rm(mod }{4})$.

  3. (3)

For $\displaystyle p=3$,

    1. (a)

If $\displaystyle 3$ divides $\displaystyle b$ and $\displaystyle a\equiv 1\
\mbox{\rm(mod }{3})$, then $\displaystyle\overline{F(x)}=x(x^{2}+1)^{3}$. Let
$\displaystyle\phi=x^{2}+1$. Since $\displaystyle
F(x)=x\phi^{3}-3x\phi^{2}+3x\phi+(a-1)x+b$, we conclude that $\displaystyle 3$
does not divide $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ if and only
if $\displaystyle\nu_{3}((a-1)x+b)=1$, which means that $\displaystyle
a\not\equiv 1\ \mbox{\rm(mod }{9})$ or $\displaystyle b\not\equiv 0\
\mbox{\rm(mod }{9})$.

    2. (b)

If $\displaystyle 3$ divides $\displaystyle b$ and $\displaystyle a\equiv-1\
\mbox{\rm(mod }{3})$, then $\displaystyle\overline{F(x)}=x(x-1)^{3}(x+1)^{2}$.
Let $\displaystyle\phi_{1}=x-1$ and $\displaystyle\phi_{2}=x+1$. Since
$\displaystyle
F(x)=\phi_{1}^{7}+7\phi_{1}^{6}+21\phi_{1}^{5}+35\phi_{1}^{4}+35\phi_{1}^{3}+21\phi_{1}^{2}+(a+7)\phi_{1}+(a+1+b)$
and $\displaystyle
F(x)=\phi_{1}^{7}-7\phi_{2}^{6}+21\phi_{2}^{5}-35\phi_{2}^{4}+35\phi_{2}^{3}-21\phi_{2}^{2}+(a+7)\phi_{2}+(b-a-1)$,
we conclude that $\displaystyle 3$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ if and only if
$\displaystyle\nu_{2}(a+1+b)=1$ and $\displaystyle\nu_{3}(b-a-1)=1$.

  4. (4)

For $\displaystyle p=7$, $\displaystyle 7$ divides $\displaystyle a$ and
$\displaystyle 7$ does not divide $\displaystyle b$, we have
$\displaystyle\overline{F(x)}=(x+b)^{7}$. Let $\displaystyle\phi=x+b$.
$\displaystyle
F(x)=\phi^{7}-7b\phi^{6}+21b^{2}\phi^{5}-35b^{3}\phi^{4}+35b^{4}\phi^{3}-21b^{5}\phi^{2}+(a+7b^{6})\phi+(b-ab-b^{7})$,
we conclude that $\displaystyle 7$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ if and only if
$\displaystyle\nu_{7}(1-a-b^{6})=1$.

  5. (5)

For $\displaystyle p\not\in\\{2,3,7\\}$ such that $\displaystyle p$ does not
divide both $\displaystyle a$ and $\displaystyle b$ and $\displaystyle p^{2}$
divides $\displaystyle 6^{6}a^{7}+7^{7}b^{6}$, let $\displaystyle t$ be an
integer such that $\displaystyle 6at\equiv-7b\ \mbox{\rm(mod }{p^{2}})$. Then
$\displaystyle(6a)^{6}F^{\prime}(t)=7(-7b)^{6}+6^{6}a^{7}\equiv 0\
\mbox{\rm(mod }{p^{2}})$ and $\displaystyle(6a)^{7}F(t)\equiv 0\ \mbox{\rm(mod
}{p^{2}})$. Thus by Dedekind’s criterion $\displaystyle p$ divides the index
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$.

For the proofs of Theorems 2.3 and 2.4, we need the following lemma, which
characterizes the prime common index divisors of $\displaystyle K$.

######  Lemma 4.1.

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

Proof of Theorem 2.3.
Since $\displaystyle\triangle=-(6^{6}a^{7}+7^{7}b^{6})$ is the discriminant of
$\displaystyle F(x)$, if $\displaystyle 2$ does not divide $\displaystyle b$,
then $\displaystyle 2$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. Thus
$\displaystyle\nu_{2}(i(K))=0$. Recall that according to Engstrom’s results,
in order to evaluate $\displaystyle\nu_{2}(i(K))$, we need to factorize
$\displaystyle 2\mathbb{Z}_{K}$ into powers of prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$. The following
table provides the factorization of $\displaystyle 2\mathbb{Z}_{K}$. Now we
assume that $\displaystyle 2$ divides $\displaystyle b$. Then we have the
following cases:

  1. (1)

If $\displaystyle 2$ divides $\displaystyle a$ and $\displaystyle b$, then for
$\displaystyle\phi=x$, we have $\displaystyle\overline{F(x)}=\phi^{7}$ in
$\displaystyle\mathbb{F}_{2}[x]$.

    1. (a)

If $\displaystyle N_{\phi}(F)=S$ has a single side, that is
$\displaystyle\nu_{2}(a)\geq\nu_{2}(b)$, then the side $\displaystyle S$ is of
degree $\displaystyle 1$. Thus there is a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$.

    2. (b)

If $\displaystyle N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,\nu_{2}(b))$, $\displaystyle(1,\nu_{2}(a))$, and
$\displaystyle(7,0)$, that is $\displaystyle\nu_{2}(a)+1\leq\nu_{2}(b)$, then
$\displaystyle S_{1}$ is of degree $\displaystyle 1$, and so it provides a
unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
2$ with residue degree $\displaystyle 1$. Let $\displaystyle d$ be the the
degree of $\displaystyle S_{2}$.

      1. (i)

If $\displaystyle\nu_{2}(a)\not\in\\{2,4\\}$, then $\displaystyle S_{2}$ is of
degree $\displaystyle 1$, and so there are exactly two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$ each.

      2. (ii)

If $\displaystyle\nu_{2}(a)=2$, then the slope of $\displaystyle S_{2}$ is
$\displaystyle\frac{-1}{3}$ and $\displaystyle R_{1}(F)(y)=(y+1)^{2}$ is the
residual polynomial of $\displaystyle F(x)$ attached to $\displaystyle S_{2}$.
Thus we have to use second order Newton polygon. Let
$\displaystyle\omega_{2}=3[\nu_{2},x,1/3]$ be the valuation of second order
Newton polygon and $\displaystyle\phi_{2}$ the key polynomial of
$\displaystyle\omega_{2}$, where $\displaystyle[\nu_{2},x,1/3]$ is the
augmented valuation of $\displaystyle\nu_{2}$ with respect to $\displaystyle
x$ and $\displaystyle 1/3$. Let $\displaystyle N_{2}(F)$ the
$\displaystyle\phi_{2}$-Newton polygon of $\displaystyle F(x)$ with respect to
the valuation $\displaystyle\omega_{2}$. It follows that:
If $\displaystyle\nu_{2}(b)=3$, then for $\displaystyle\phi_{2}=x^{3}+2x+2$,
we have $\displaystyle
F(x)=x\phi_{2}^{2}+(4-4x-4x^{2})\phi_{2}+8x^{2}+(a-4)x+b-8$. It follows that
if $\displaystyle\nu_{2}(a-4)=3$, then $\displaystyle N_{2}(F)=T$ has a single
side joining $\displaystyle(0,10)$ and $\displaystyle(2,7)$. Thus
$\displaystyle T$ is of degree $\displaystyle 1$ and so $\displaystyle S_{2}$
provides a unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$. If $\displaystyle\nu_{2}(a-4)\geq 4$ and
$\displaystyle\nu_{2}(b-8)\geq 4$, then $\displaystyle N_{2}(F)=T$ has a
single side joining $\displaystyle(0,11)$, $\displaystyle(1,9)$ and
$\displaystyle(2,7)$, with $\displaystyle R_{2}(F)(y)=y^{2}+y+1$, which is
irreducible over $\displaystyle\mathbb{F}_{2}=\mathbb{F}_{0}$. Thus
$\displaystyle S_{2}$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 2$. Hence $\displaystyle 2$ is not a common index
divisor of $\displaystyle K$.
If $\displaystyle\nu_{2}(b)\geq 4$ and $\displaystyle\nu_{2}(a+4)=3$, then for
$\displaystyle\phi_{2}=x^{3}+2$, we have $\displaystyle
F(x)=x\phi_{2}^{2}-4x\phi_{2}+(a+4)x+b$ is the
$\displaystyle\phi_{2}$-expansion of $\displaystyle F(x)$, and so
$\displaystyle N_{2}(F)=T$ has a single side joining $\displaystyle(0,10)$ and
$\displaystyle(2,7)$. In this case the side $\displaystyle T$ is of degree
$\displaystyle 1$ and $\displaystyle S_{2}$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$. If
$\displaystyle\nu_{2}(b)=4$ and $\displaystyle\nu_{2}(a+4)\geq 4$, then , then
for $\displaystyle\phi_{2}=x^{3}+2$, $\displaystyle N_{2}(F)=T$ has a single
side joining $\displaystyle(0,12)$ and $\displaystyle(2,7)$. Thus
$\displaystyle T$ is of degree $\displaystyle 1$ and $\displaystyle S_{2}$
provides a unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$.
If $\displaystyle\nu_{2}(b)\geq 5$ and $\displaystyle\nu_{2}(a+4)=4$, then for
$\displaystyle\phi_{2}=x^{3}+2$, we have $\displaystyle
F(x)=x\phi_{2}^{2}-4x\phi_{2}+(a+4)x+b$ be the
$\displaystyle\phi_{2}$-expansion of $\displaystyle F(x)$ and $\displaystyle
N_{2}(F)=T$ has a single side joining $\displaystyle(0,13)$,
$\displaystyle(1,10)$ and $\displaystyle(2,7)$. So $\displaystyle T$ is of
degree $\displaystyle 2$ with attached residual polynomial $\displaystyle
R_{2}(F)=y^{2}+y+1$ irreducible over
$\displaystyle\mathbb{F}_{2}=\mathbb{F}_{0}$. Thus $\displaystyle S_{2}$
provides a unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 2$.
If $\displaystyle\nu_{2}(b)\geq 5$ and $\displaystyle\nu_{2}(a+4)\geq 5$, then
for $\displaystyle\phi_{2}=x^{3}+2$, $\displaystyle N_{2}(F)=T_{1}+T_{2}$ has
two sides joining $\displaystyle(0,v)$ , $\displaystyle(1,10)$ and
$\displaystyle(2,7)$ with $\displaystyle v\geq 15$. So each $\displaystyle
T_{i}$ has degree $\displaystyle 1$ and $\displaystyle S_{2}$ provides two
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$
with residue degree $\displaystyle 1$ each. As $\displaystyle S_{1}$ provides
a prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$
with residue degree $\displaystyle 1$, we conclude that there are three prime
ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with
residue degree $\displaystyle 1$ each, and so $\displaystyle 2$ is a common
index divisor of $\displaystyle K$. In this last case, $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{111}\mathfrak{p}_{121}^{3}\mathfrak{p}_{131}^{3}$
with residue degree $\displaystyle 1$ each factor. Based on Engstrom’s result,
we conclude that $\displaystyle\nu_{2}(i(K))=1$.

      3. (iii)

Similarly for $\displaystyle\nu_{2}(a)=4$, $\displaystyle\nu_{2}(i(K))\geq 1$
If and only if $\displaystyle\nu_{2}(b)\geq 7$ and
$\displaystyle\nu_{2}(a+16)\geq 7$. In this case, $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{111}\mathfrak{p}_{121}^{3}\mathfrak{p}_{131}^{3}$
with residue degree $\displaystyle 1$ each factor. Based on Engstrom’s result,
we conclude that $\displaystyle\nu_{2}(i(K))=1$.

  2. (2)

$\displaystyle a\equiv 1\ \mbox{\rm(mod }{2})$ and $\displaystyle b\equiv 0\
\mbox{\rm(mod }{2})$. In this case
$\displaystyle\overline{F(x)}=x(x-1)^{2}(x^{2}+x+1)^{2}$ modulo $\displaystyle
2$. Let $\displaystyle\phi=x-1$, $\displaystyle g(x)=x^{2}+x+1$,
$\displaystyle F(x)=\dots-21\phi^{2}+(7+a)\phi+(b+a+1)$, and $\displaystyle
F(x)=(x-3)g^{3}+(5+3x)g^{2}-(4x+2)g+(a+1)x+b$. Since $\displaystyle x$
provides a unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$, we conclude that $\displaystyle 2$ is a common index
divisor of $\displaystyle K$ if and only if $\displaystyle\phi$ provides two
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ of
degree $\displaystyle 1$ each or $\displaystyle\phi$ provides a unique prime
ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ of degree
$\displaystyle 2$ and $\displaystyle g$ provides at least one prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ of degree
$\displaystyle 2$ or also $\displaystyle g$ provides two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ of degree
$\displaystyle 2$ each. That is if and only if one of the following conditions
holds:

    1. (a)

If $\displaystyle a\equiv 1\ \mbox{\rm(mod }{4})$ and $\displaystyle b\equiv
2\ \mbox{\rm(mod }{4})$, then $\displaystyle\nu_{2}(\triangle)\geq 7$ and
$\displaystyle N_{g}^{+}(F)$ has a single side of height $\displaystyle 1$,
and so $\displaystyle g$ provides a unique prime ideal
$\displaystyle\mathfrak{p}_{311}$ of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 2$. For $\displaystyle
N_{\phi}^{+}(F)$, let $\displaystyle u=\frac{-7b_{2}}{3a}$. Then
$\displaystyle u\in\mathbb{Z}_{2}$. Let $\displaystyle
F(x+u)=x^{7}+\dots+21u^{5}x^{2}+Ax+B$, where $\displaystyle
A=7u^{6}+a=\frac{-\triangle}{6^{6}a^{6}}$ and $\displaystyle
B=u^{7}+au+b=\frac{b\triangle}{6^{7}a^{7}}$. It follows that
$\displaystyle\nu_{2}(A)=\nu_{2}(B)=\nu_{2}(\triangle)-6$, and so
$\displaystyle N_{\phi}^{+}(F)=S_{1}$ has a single side joining
$\displaystyle(0,\nu_{2}(\triangle)-6)$ and $\displaystyle(2,0)$. Thus, if
$\displaystyle\nu_{2}(\triangle)$ is odd, then $\displaystyle\phi$ provides a
unique prime ideal $\displaystyle\mathfrak{p}_{211}$ of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$. If $\displaystyle\nu_{2}(\triangle)=2(k+3)$ for some
positive integer $\displaystyle k$, then let $\displaystyle
F(x+u+2^{k})=x^{7}+\dots+21(u+2^{k})^{5}x^{2}+A_{1}x+B_{1}$, where
$\displaystyle A_{1}=7u^{6}+a+3\cdot 2^{k+1}u^{5}+2^{2k}D=A+3\cdot
2^{k+1}u^{5}+2^{2k}D$ and $\displaystyle B_{1}=B+A\cdot 2^{k}+2^{2k}\cdot
21u^{5}+2^{3k}H=2^{2k}(\frac{b_{2}\triangle_{2}}{3^{7}a^{7}}+21u^{5})+2^{3k}H$
for some $\displaystyle D\in\mathbb{Z}_{2}$ and $\displaystyle
H\in\mathbb{Z}_{2}$. Thus, $\displaystyle B_{1}=2^{2k}(3\cdot a\cdot
b_{2}\triangle_{2}+3\cdot a\cdot b_{2})+2^{2k+3}L$ for some $\displaystyle
L\in\mathbb{Z}_{2}$. Hence if $\displaystyle k\geq 2$, then
$\displaystyle\nu_{2}(A_{1})=k+1$ and $\displaystyle\nu_{2}(B_{1})\geq 2k+1$.
More precisely, if $\displaystyle\triangle_{2}\equiv 1\ \mbox{\rm(mod }{4})$,
then $\displaystyle\nu_{2}(B_{1})=2k+1$, and so $\displaystyle\phi$ provides a
unique prime ideal $\displaystyle\mathfrak{p}_{211}$ of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$. If $\displaystyle\triangle_{2}\equiv 3\
\mbox{\rm(mod }{4})$, then $\displaystyle\nu_{2}(B_{1})\geq 2k+2$. It follows
that if $\displaystyle\nu_{2}(B_{1})=2k+2$, and so $\displaystyle\phi$
provides a unique prime ideal $\displaystyle\mathfrak{p}_{211}$ of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 2$. If $\displaystyle\nu_{2}(B_{1})\geq 2k+3$, then
$\displaystyle\nu_{2}(B_{1})\geq 2k+3$, and so $\displaystyle\phi$ provides
two prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
2$ with residue degree $\displaystyle 1$ each. In these last two cases, we
have $\displaystyle 2$ divides $\displaystyle i(K)$ and
$\displaystyle\nu_{2}(i(K))=1$.
For $\displaystyle k=1$, we have $\displaystyle\nu_{2}(\triangle)=8$ and
$\displaystyle a\equiv 5\ \mbox{\rm(mod }{8})$. In this case
$\displaystyle\overline{F(x)}=x(x-1)^{2}(x^{2}+x+1)^{2}$ modulo $\displaystyle
2$. Let $\displaystyle\phi=x-1$, $\displaystyle g(x)=x^{2}+x+1$,
$\displaystyle F(x)=\dots-21\phi^{2}+(7+a)\phi+(b+a+1)$, and $\displaystyle
F(x)=(x-3)g^{3}+(5+3x)g^{2}-(4x+2)g+(a+1)x+b$. Since $\displaystyle x$
provides a unique prime ideal of of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 1$ and $\displaystyle g$
provides a unique prime ideal of of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 2$, we conclude that
$\displaystyle\nu_{2}(i(K))\geq 1$ if and only if $\displaystyle\phi$ provides
a unique prime ideal of of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 2$ or $\displaystyle\phi$
provides two distinct prime ideals of of $\displaystyle\mathbb{Z}_{K}$ lying
above $\displaystyle 2$ with residue degree $\displaystyle 1$ each. If
$\displaystyle(a,b)\in\\{(5,10),(13,2)\\}\ \mbox{\rm(mod }{16})$, then
$\displaystyle\phi$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 2$ and so $\displaystyle\nu_{2}(i(K))=1$. If
$\displaystyle(a,b)\in\\{(5,10),(13,10)\\}\ \mbox{\rm(mod }{16})$, then
$\displaystyle\phi$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$ and so $\displaystyle\nu_{2}(i(K))=0$. For
$\displaystyle(a,b)\in\\{(5,6),(5,14),(13,6),(13,14)\\}\ \mbox{\rm(mod
}{16})$, let us replace $\displaystyle\phi$ by
$\displaystyle\phi^{\prime}=x-3$ and consider the
$\displaystyle\phi^{\prime}$-Newton polygon of $\displaystyle F(x)$ with
respect to $\displaystyle\nu_{2}$. It follows that If
$\displaystyle(a,b)\in\\{(5,6),(13,14)\\}\ \mbox{\rm(mod }{16})$, then
$\displaystyle\phi^{\prime}$ provides two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$ each and so $\displaystyle\nu_{2}(i(K))=1$. If
$\displaystyle(a,b)\in\\{(5,14),(13,6)\\}\ \mbox{\rm(mod }{16})$, then
$\displaystyle\phi^{\prime}$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$ and so $\displaystyle\nu_{2}(i(K))=0$.

    2. (b)

$\displaystyle a\equiv 3\ \mbox{\rm(mod }{4})$ and $\displaystyle
b\equiv-(a+1)\ \mbox{\rm(mod }{8})$ because $\displaystyle N_{\phi}^{+}(F)$
has two sides.

    3. (c)

If $\displaystyle a\equiv 3\ \mbox{\rm(mod }{8})$ and $\displaystyle b\equiv
0\ \mbox{\rm(mod }{8})$, then $\displaystyle\phi$ provides a unique prime
ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with
residue degree $\displaystyle 2$ and $\displaystyle g$ provides two prime
ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with
residue degree $\displaystyle 2$ each because $\displaystyle N_{g}^{+}(F)$ has
a single side of degree $\displaystyle 2$ with
$\displaystyle(1+x)y^{2}+y+x=(x+1)(y-1)(y-x)$ its attached residual polynomial
of $\displaystyle F(x)$. In this case $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{111}\mathfrak{p}_{211}\mathfrak{p}_{221}\mathfrak{p}_{311}\mathfrak{p}_{312}$
with residue degrees $\displaystyle f_{111}=1$ and $\displaystyle
f_{211}=f_{311}=f_{312}=2$, and so $\displaystyle\nu_{2}(i(K))=3$.

    4. (d)

$\displaystyle a\equiv 7\ \mbox{\rm(mod }{8})$ and $\displaystyle b\equiv 0\
\mbox{\rm(mod }{8})$. In this case $\displaystyle\phi$ provides a unique prime
ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with
residue degree $\displaystyle 2$ and $\displaystyle N_{g}^{+}(F)$ has two
sides. More precisely, $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{111}\mathfrak{p}_{211}\mathfrak{p}_{221}\mathfrak{p}_{311}\mathfrak{p}_{312}$
with residue degrees $\displaystyle f_{111}=f_{211}=f_{221}=1$ and
$\displaystyle f_{211}=f_{311}=f_{312}=2$, and so
$\displaystyle\nu_{2}(i(K))=3$.

    5. (e)

If $\displaystyle a\equiv 5\ \mbox{\rm(mod }{8})$ and $\displaystyle
b\equiv-(a+1)\ \mbox{\rm(mod }{16})$ because if $\displaystyle b\equiv-(a+1)\
\mbox{\rm(mod }{32})$, then $\displaystyle N_{\phi}^{+}(F)$ has two sides and
if $\displaystyle b\equiv-(a+1)+16\ \mbox{\rm(mod }{32})$, then $\displaystyle
N_{\phi}^{+}(F)$ has a single side of degree $\displaystyle 2$, which provides
a single prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 2$ and $\displaystyle
N_{g}^{+}(F)$ has a single side of degree $\displaystyle 1$. Thus there are
$\displaystyle 2$ prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 2$ with residue degree $\displaystyle 2$ each.

    6. (f)

If $\displaystyle a\equiv 5\ \mbox{\rm(mod }{8})$ and
$\displaystyle\nu_{2}(b+(a+1))=2$, then $\displaystyle\nu_{2}(b-(a+1))\geq 3$.
If $\displaystyle\nu_{2}(b-(a+1))=3$, then for $\displaystyle\phi=x+1$, we
have $\displaystyle N_{\phi}^{+}(F)$ has a single side of degree
$\displaystyle 1$. Since $\displaystyle\nu_{2}(a+1)=1$, then $\displaystyle
N^{+}_{g}(F)$ has a single side of height $\displaystyle 1$. Thus there are
two prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
2$ with residue degree $\displaystyle 1$ each and one prime ideal with residue
degree $\displaystyle 2$. If $\displaystyle\nu_{2}(b-(a+1))=4$, then for
$\displaystyle\phi=x+1$, we have $\displaystyle N_{\phi}^{+}(F)$ has a single
side of degree $\displaystyle 2$ and its attached residual polynomial of
$\displaystyle F$ is $\displaystyle R_{1}(F)(y)=y^{2}+y+1$. Since
$\displaystyle b\equiv 6\ \mbox{\rm(mod }{8})$, we conclude that
$\displaystyle N_{g}^{+}(F)$ has a single side of degree $\displaystyle 1$,
then there are $\displaystyle 2$ prime ideals of $\displaystyle\mathbb{Z}_{K}$
lying above $\displaystyle 2$ with residue degree $\displaystyle 2$ each, and
so $\displaystyle 2$ divides $\displaystyle i(K)$. If
$\displaystyle\nu_{2}(b-(a+1))\geq 5$, then for $\displaystyle\phi=x+1$, we
have $\displaystyle N_{\phi}^{+}(F)$ has two sides of degree $\displaystyle 1$
each, and so there are $\displaystyle 3$ prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$ each, and so $\displaystyle 2$ divides $\displaystyle
i(K)$.

Proof of Theorem 2.4.
Since $\displaystyle\triangle=-(6^{6}a^{7}+7^{7}b^{6})$ is the discriminant of
$\displaystyle F(x)$, by the formula
$\displaystyle\triangle=(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])^{2}d_{K}$, if
$\displaystyle 3$ does not divide $\displaystyle b$, then $\displaystyle 3$
does not divide $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, and so
$\displaystyle\nu_{3}(i(K))=0$. Now assume that $\displaystyle 3$ divides
$\displaystyle b$. Recall that according to Lemma 4.1,
$\displaystyle\nu_{3}(i(K))\geq 1$ if and only if there are at least four
primes ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 1$ each. We have the following cases:

  1. (1)

If $\displaystyle 3$ divide $\displaystyle a$ and $\displaystyle b$, then for
$\displaystyle\phi=x$, $\displaystyle\overline{F(x)}=\phi^{7}$ in
$\displaystyle\mathbb{F}_{3}[x]$. It follows that:

    1. (a)

If $\displaystyle\nu_{3}(a)\geq\nu_{3}(b)$, then $\displaystyle N_{\phi}(F)$
has a single side of degree $\displaystyle 1$, and so there is a unique prime
ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$.

    2. (b)

If $\displaystyle\nu_{3}(a)+1\leq\nu_{3}(b)$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,\nu_{3}(b))$,
$\displaystyle(1,\nu_{3}(a))$, and $\displaystyle(7,0)$. Since $\displaystyle
S_{1}$ is of degree $\displaystyle 1$, $\displaystyle S_{1}$ provides exactly
one prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 1$. Thus $\displaystyle\nu_{3}(i(K))\geq 1$
if and only if $\displaystyle S_{2}$ provides at least three primes ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each. If $\displaystyle\nu_{3}(a)\not\in\\{2,3,4\\}$,
then $\displaystyle S_{2}$ is of degree $\displaystyle 1$, and so
$\displaystyle S_{2}$ provides exactly one prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each. If $\displaystyle\nu_{3}(a)\in\\{2,4\\}$, then
$\displaystyle S_{2}$ is of degree $\displaystyle 2$, and so $\displaystyle
S_{2}$ provides at most two prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying
above $\displaystyle 3$. Hence $\displaystyle 3$ is not a common index divisor
of $\displaystyle K$. If $\displaystyle\nu_{3}(a)=3$, then $\displaystyle
S_{2}$ is of degree $\displaystyle 3$ and its attached residual polynomial of
$\displaystyle F(x)$ is $\displaystyle R_{1}(F)(y)=y^{3}+a_{3}=(y+a_{3})^{3}$.
So we have to use second order Newton polygon. Let
$\displaystyle\omega_{2}=2[\nu_{3},x,1/2]$ be the valuation of second order
Newton polygon, $\displaystyle\phi_{2}=x^{2}+3a_{3}$ the key polynomial of
$\displaystyle\omega_{2}$ and $\displaystyle N_{2}(F)$ the
$\displaystyle\phi_{2}$-Newton polygon of $\displaystyle F(x)$ with respect to
$\displaystyle\omega_{2}$. It follows that: If $\displaystyle a_{3}\equiv 1\
\mbox{\rm(mod }{3})$, then for $\displaystyle\phi_{2}=x^{2}+3$, we have
$\displaystyle F(x)=x\phi_{2}^{3}-9x\phi_{2}^{2}+27x\phi_{2}+(a-27)x+b$ is the
$\displaystyle\phi_{2}$-expansion of $\displaystyle F(x)$. We have the
following cases:

      1. (i)

If $\displaystyle\nu_{3}(b)=4$, then $\displaystyle N_{2}(F)=T$ has a single
side joining $\displaystyle(0,8)$ and $\displaystyle(3,7)$. Thus
$\displaystyle T$ is of degree $\displaystyle 1$ and $\displaystyle S_{2}$
provides a unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 3$ with residue degree $\displaystyle 1$.

      2. (ii)

If $\displaystyle\nu_{3}(b)\geq 5$ and $\displaystyle\nu_{3}(a-27)=4$, then
$\displaystyle N_{2}(F)=T$ has a single side joining $\displaystyle(0,9)$ and
$\displaystyle(3,7)$. Thus $\displaystyle T$ is of degree $\displaystyle 1$
and $\displaystyle S_{2}$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$.

      3. (iii)

If $\displaystyle\nu_{3}(b)=5$ and $\displaystyle\nu_{3}(a-27)\geq 5$, then
$\displaystyle N_{2}(F)=T$ has a single side joining $\displaystyle(0,10)$ and
$\displaystyle(3,7)$ and its attached residual polynomial of $\displaystyle F$
is $\displaystyle R_{2}(F)(y)=xy^{3}+xy+b_{3}$, which is irreducible over
$\displaystyle\mathbb{F}_{2}=\mathbb{F}_{\phi}$ because $\displaystyle\phi$ is
of degree $\displaystyle 1$. Thus $\displaystyle S_{2}$ provides a unique
prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 3$.

      4. (iv)

If $\displaystyle\nu_{3}(b)\geq 6$ and $\displaystyle\nu_{3}(a-27)\geq 5$,
then $\displaystyle N_{2}(F)=T_{1}+T_{2}$ has two sides joining
$\displaystyle(0,v)$, $\displaystyle(2,9)$ and $\displaystyle(3,7)$ with
$\displaystyle v\geq 11$. Thus $\displaystyle T_{1}$ is of degree
$\displaystyle 1$, $\displaystyle T_{2}$ of degree $\displaystyle 2$ and
$\displaystyle R_{2}(F)(y)=xy^{2}+x$ is its attached residual polynomial of
$\displaystyle F(x)$, which is irreducible over
$\displaystyle\mathbb{F}_{2}=\mathbb{F}_{\phi}$. Thus $\displaystyle S_{2}$
provides a unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 3$ with residue degree $\displaystyle 1$ and a unique prime
ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with
residue degree $\displaystyle 2$.

Similarly, for $\displaystyle a_{3}\equiv-1\ \mbox{\rm(mod }{3})$, let
$\displaystyle\phi_{2}=x^{2}-3$. Then $\displaystyle
F(x)=x\phi_{2}^{3}+9x\phi_{2}^{2}+27x\phi_{2}+(a+27)x+b$ is the
$\displaystyle\phi_{2}$-expansion of $\displaystyle F(x)$. By analogous to the
case $\displaystyle a_{3}\equiv 1\ \mbox{\rm(mod }{3})$, in every case
$\displaystyle 3$ does not divide $\displaystyle i(K)$. If $\displaystyle
a\equiv 1\ \mbox{\rm(mod }{3})$, then
$\displaystyle\overline{F(x)}=x(x^{2}+1)^{3}$ in
$\displaystyle\mathbb{F}_{3}[x]$. Then there are exactly a unique prime ideal
of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ and the others prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ are of residue
degrees at least $\displaystyle 2$ each. Hence $\displaystyle\nu_{3}(i(K))=0$.

    3. (c)

If $\displaystyle a\equiv-1\ \mbox{\rm(mod }{3})$, then
$\displaystyle\overline{F(x)}=x(x-1)^{3}(x+1)^{3}$ in
$\displaystyle\mathbb{F}_{3}[x]$. Let $\displaystyle\phi_{1}=x-1$,
$\displaystyle\phi_{2}=x+1$, $\displaystyle
F(x)=\phi_{1}^{7}+7\phi_{1}^{6}+21\phi_{1}^{5}+35\phi_{1}^{4}+35\phi_{1}^{3}+21\phi_{1}^{2}+(7+a)\phi_{1}+(b+a+1)$,
and $\displaystyle
F(x)=\phi_{2}^{7}-7\phi_{2}^{6}+21\phi_{2}^{5}-35\phi_{2}^{4}+35\phi_{2}^{3}-21\phi_{2}^{2}+(7+a)\phi_{2}+(b-(a+1))$.
It follows that:

      1. (i)

If $\displaystyle b\equiv-(1+a)\pm 3\ \mbox{\rm(mod }{9})$ and $\displaystyle
b\equiv(1+a)\pm 3\ \mbox{\rm(mod }{9})$, then for $\displaystyle
N_{\phi_{i}}^{+}(F)$ has a single side of height $\displaystyle 1$ for every
$\displaystyle i=1,2$. Thus there are exactly three prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each.

      2. (ii)

If $\displaystyle a\equiv 8\ \mbox{\rm(mod }{9})$ and $\displaystyle b\equiv
0\ \mbox{\rm(mod }{9})$, then $\displaystyle\nu_{3}(b+(1+a))\geq 2$ and
$\displaystyle\nu_{3}(b-(1+a))\geq 2$. Thus $\displaystyle x$ a provides a
unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
3$ with residue degree $\displaystyle 1$, and each of $\displaystyle\phi_{1}$
and $\displaystyle\phi_{2}$ provides two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each. In this two cases
$\displaystyle\nu_{3}(i(K))=2$.

      3. (iii)

If $\displaystyle a\equiv 5\ \mbox{\rm(mod }{9})$ and $\displaystyle b\equiv
3\ \mbox{\rm(mod }{9})$, then $\displaystyle\nu_{3}(b+(1+a))\geq 2$ and
$\displaystyle\nu_{3}(b-(1+a))=1$. Thus each of $\displaystyle x$ and
$\displaystyle\phi_{2}$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$, and $\displaystyle\phi_{1}$ provides two prime
ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with
residue degree $\displaystyle 1$ each. Similarly, if $\displaystyle a\equiv 5\
\mbox{\rm(mod }{9})$ and $\displaystyle b\equiv 6\ \mbox{\rm(mod }{9})$, then
$\displaystyle\nu_{3}(b-(1+a))\geq 2$ and $\displaystyle\nu_{3}(b+(1+a))=1$.
Thus each of $\displaystyle x$ and $\displaystyle\phi_{1}$ provides a unique
prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 1$, $\displaystyle\phi_{2}$ provides two
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 1$ each. In these two cases
$\displaystyle\nu_{3}(i(K))=1$.

      4. (iv)

If $\displaystyle a\equiv 2\ \mbox{\rm(mod }{9})$ and
$\displaystyle\nu_{3}(b)\geq 2$, then $\displaystyle\nu_{3}(b-(a+1))=1$ and
$\displaystyle\nu_{3}(b+(a+1))=1$. Thus $\displaystyle 3$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$.

      5. (v)

If $\displaystyle a\equiv 2\ \mbox{\rm(mod }{9})$ and $\displaystyle
b\equiv(1+a)\pm 9\ \mbox{\rm(mod }{27})$, then $\displaystyle
N_{\phi_{2}}^{+}(F)$ has a single side joining $\displaystyle(0,2)$ and
$\displaystyle(3,0)$ and $\displaystyle N_{\phi_{1}}^{+}(F)$ has a single side
joining $\displaystyle(0,1)$ and $\displaystyle(3,0)$. Thus there are
$\displaystyle 3$ prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 3$ with residue degree $\displaystyle 1$ each, and so
$\displaystyle\nu_{3}(i(K))=0$.

      6. (vi)

Similarly, if $\displaystyle a\equiv 2\ \mbox{\rm(mod }{9})$ and
$\displaystyle b\equiv-(1+a)\pm 9\ \mbox{\rm(mod }{27})$, then there are
$\displaystyle 3$ prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 3$ with residue degree $\displaystyle 1$ each, and so
$\displaystyle\nu_{3}(i(K))=0$.

      7. (vii)

If $\displaystyle a\equiv 2\ \mbox{\rm(mod }{9})$ and
$\displaystyle\nu_{3}(b)=1$, then $\displaystyle\nu_{3}(\triangle)\geq 8$. Let
$\displaystyle u=\frac{-7b_{3}}{2a}$. Then $\displaystyle u\in\mathbb{Z}_{3}$.
Let $\displaystyle\phi=x-u$ and $\displaystyle
F(x+u)=x^{7}+\dots+35u^{4}x^{3}+21u^{5}x^{2}+Ax+B$, where $\displaystyle
A=7u^{6}+a=\frac{-\triangle}{6^{6}a^{6}}$ and $\displaystyle
B=u^{7}+au+b=\frac{b\triangle}{6^{7}a^{7}}$. It follows that
$\displaystyle\nu_{3}(A)=\nu_{3}(B)=\nu_{3}(\triangle)-6$, and so
$\displaystyle N_{\phi}^{+}(F)=S_{1}$ has a single side joining
$\displaystyle(0,\nu_{3}(\triangle)-6)$ and $\displaystyle(2,0)$. Remark that
since $\displaystyle\nu_{3}(b)=1$ and $\displaystyle\nu_{3}(B)\geq 2$,
$\displaystyle\nu_{3}(-u^{7}-au+b)=1$, and so $\displaystyle(x+u)$ provides a
unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
3$ with residue degree $\displaystyle 1$. Thus $\displaystyle\nu_{3}(i(K))\geq
1$ if and only if $\displaystyle\phi$ provides at least two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each. Let $\displaystyle v=\nu_{3}(\triangle)$. We
have the following cases:

        1. (A)

If $\displaystyle v=8$, then $\displaystyle N_{\phi}^{+}(F)$ has a single side
of degree one, and so $\displaystyle\phi$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$.

        2. (B)

If $\displaystyle v=9$, then $\displaystyle N_{\phi}^{+}(F)=S$ has a single
side joining $\displaystyle(0,3)$ and $\displaystyle(3,0)$ with $\displaystyle
R_{1}(F)(y)=-u^{4}y^{3}+u^{5}y^{2}+B_{3}$ its attached residual polynomial of
$\displaystyle F(x)$. Since $\displaystyle a\equiv-1\ \mbox{\rm(mod }{3})$ and
$\displaystyle B=\frac{b\triangle}{6^{7}a^{7}}$, we have $\displaystyle
u\equiv-b_{3}\ \mbox{\rm(mod }{3})$ and $\displaystyle B_{3}\equiv
b_{3}\triangle_{3}\ \mbox{\rm(mod }{3})$. Thus $\displaystyle
R_{1}(F)(y)=-y^{3}-b_{3}y^{2}+b_{3}\triangle_{3}$. Since $\displaystyle
R_{1}(F)(y)$ is square free and $\displaystyle R_{1}(F)(0)\neq 0$, then
$\displaystyle R_{1}(F)(y)$ has at most one root in
$\displaystyle\mathbb{F}_{\phi}$. Thus $\displaystyle S$ provides at most a
unique prime ideal of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
3$ with residue degree $\displaystyle 1$. Therefore,
$\displaystyle\nu_{3}(i(K))=0$.

        3. (C)

If $\displaystyle v\geq 10$, then $\displaystyle N_{\phi}^{+}(F)=S_{1}+S_{2}$
has two sides joining $\displaystyle(0,v-6)$, $\displaystyle(1,v-6)$ and
$\displaystyle(3,1)$. It follows that Since $\displaystyle S_{2}$ is of degree
$\displaystyle 1$, it provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$. Moreover, if $\displaystyle v$ is even then
$\displaystyle S_{1}$ is of degree $\displaystyle 1$, and so
$\displaystyle\phi$ provides two prime ideals of $\displaystyle\mathbb{Z}_{K}$
lying above $\displaystyle 3$ with residue degree $\displaystyle 1$ each. In
this case $\displaystyle\nu_{3}(i(K))=1$. If $\displaystyle v=2(k+3)+1$, then
$\displaystyle S_{1}$ is of degree $\displaystyle 2$ with residual polynomial
$\displaystyle R_{1}(F)(y)=uy^{2}+b_{3}\triangle_{3}$. Since $\displaystyle
a\equiv-1\ \mbox{\rm(mod }{3})$, we have $\displaystyle 2a\equiv 1\
\mbox{\rm(mod }{3})$ and $\displaystyle u\equiv-b_{3}\ \mbox{\rm(mod }{3})$.
Thus $\displaystyle R_{1}(F)(y)=-b_{3}(y^{2}-\triangle_{3})$. It follows that
if $\displaystyle(\frac{\triangle_{3}}{3})=1$, then $\displaystyle
R_{1}(F)(y)$ has two different factors of degree $\displaystyle 1$ each, and
so $\displaystyle S_{1}$ provides two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each. In this case there are exactly five prime
ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with
residue degree $\displaystyle 1$ each and according to Engstrom’s results
$\displaystyle\nu_{3}(i(K))=2$. But if
$\displaystyle(\frac{\triangle_{3}}{3})=-1$, then $\displaystyle R_{1}(F)(y)$
is irreducible over $\displaystyle\mathbb{F}_{\phi}=\mathbb{F}_{3}$, and so
$\displaystyle S_{1}$ provides a unique prime ideal of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 2$. In this last case there are exactly three prime
ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with
residue degree $\displaystyle 1$ each, and so $\displaystyle\nu_{3}(i(K))=0$.

Proof of Theorem 2.5.

We start by showing that $\displaystyle 5$ does not divide $\displaystyle
i(K)$ for every integers of $\displaystyle a$ and $\displaystyle b$ such that
$\displaystyle x^{7}+ax+b$ is irreducible.

  1. (1)

If $\displaystyle 5^{2}$ does not divide $\displaystyle\triangle$, then
$\displaystyle 5$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. Consequently,
$\displaystyle\nu_{5}(i(K))=0$.

  2. (2)

Now assume that $\displaystyle 5^{2}$ divides
$\displaystyle\triangle=-(6^{6}a^{7}+7^{7}b^{6})$. Then $\displaystyle
6^{6}a^{7}+7^{7}b^{6}\equiv 0\ \mbox{\rm(mod }{5})$. Since $\displaystyle
a^{5}\equiv a\ \mbox{\rm(mod }{5})$ and $\displaystyle b^{5}\equiv b\
\mbox{\rm(mod }{5})$, then $\displaystyle a^{2}\equiv 3b\ \mbox{\rm(mod
}{5})$, which means $\displaystyle(a,b)\in\\{(0,0),(2,2),(3,2),(1,3),(4,3)\\}\
\mbox{\rm(mod }{5})$. In order to show that $\displaystyle\nu_{5}(i(K))=0$ it
suffices to show that for every value $\displaystyle(a,b)\in\mathbb{Z}^{2}$
such that $\displaystyle x^{7}+ax+b$ is irreducible and
$\displaystyle(a,b)\in\\{(0,0),(1,2),(-1,2),(2,3),(3,3)\\}\ \mbox{\rm(mod
}{5})$ there are at most four prime ideals of $\displaystyle\mathbb{Z}_{K}$
lying above $\displaystyle 5$ with residue degree $\displaystyle 1$, where
$\displaystyle K$ is the number field generated by a complex root of
$\displaystyle x^{7}+ax+b$ .

    1. (a)

For $\displaystyle(a,b)\equiv(0,0)\ \mbox{\rm(mod }{5})$, if
$\displaystyle\nu_{5}(a)\geq\nu_{5}(b)$, then $\displaystyle N_{\phi}(F)=S$
has a single side and it is of degree $\displaystyle 1$. Thus there is a
unique prime ideal $\displaystyle\mathfrak{p}$ of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 5$ with residue
degree $\displaystyle 1$. More precisely $\displaystyle
5\mathbb{Z}_{K}=\mathfrak{p}^{7}$.
If $\displaystyle\nu_{5}(a)+1\leq\nu_{5}(b)$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides. More precisely, $\displaystyle S_{1}$
is of degree $\displaystyle 1$. Let $\displaystyle d$ be degree of
$\displaystyle S_{2}$. Since $\displaystyle 6$ is the length of $\displaystyle
S_{2}$, then $\displaystyle d\in\\{1,2,3\\}$. Thus $\displaystyle S_{1}$
provides a unique prime ideal $\displaystyle\mathfrak{p}$ of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 5$ with residue
degree $\displaystyle 1$ and $\displaystyle S_{2}$ provides at most three
prime ideals $\displaystyle\mathfrak{p}$ of $\displaystyle\mathbb{Z}_{K}$
lying above $\displaystyle 5$ with residue degree $\displaystyle 1$ each.

    2. (b)

For $\displaystyle(a,b)\equiv(2,2)\ \mbox{\rm(mod }{5})$, since
$\displaystyle\overline{F(x)}=(x-1)(x-3)^{2}(x^{4}+2x^{3}+4x^{2}+2x+2)$ in
$\displaystyle\mathbb{F}_{5}[x]$, there is at most three prime ideals
$\displaystyle\mathfrak{p}$ of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 5$ with residue degree $\displaystyle 1$.

    3. (c)

For $\displaystyle(a,b)\in\\{(3,2),(4,3)\\}\ \mbox{\rm(mod }{5})$, since
$\displaystyle\overline{F(x)}$ is irreducible over
$\displaystyle\mathbb{F}_{5}$, there is a unique prime ideal
$\displaystyle\mathfrak{p}$ of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 5$ with residue degree $\displaystyle 7$.

    4. (d)

For $\displaystyle(a,b)\equiv(1,3)\ \mbox{\rm(mod }{5})$, since
$\displaystyle\overline{F(x)}=(x-1)(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+2)$ in
$\displaystyle\mathbb{F}_{5}[x]$, there is exactly a unique prime ideals
$\displaystyle\mathfrak{p}$ of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 5$ with residue degree $\displaystyle 1$.

We conclude that in all cases $\displaystyle\nu_{5}(i(K))=0$.
For $\displaystyle p\geq 7$, since the field $\displaystyle K$ is of degree
$\displaystyle 7$, there are at most $\displaystyle 7$ prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle p$. The fact that
there at least $\displaystyle p\geq 7$ monic irreducible polynomial of degree
$\displaystyle f$ in $\displaystyle\mathbb{F}_{p}[x]$ for every positive
integer $\displaystyle f\in\\{1,2,3\\}$, we conclude that $\displaystyle p$
does not divide $\displaystyle i(K)$.

Proof of Proposition 2.2.
First according to Theorem 2.1 and the hypotheses of Example 2.2,
$\displaystyle 2$ is the unique prime integer candidate to divide
$\displaystyle ind(\alpha)$. Let $\displaystyle\phi=x$. Then
$\displaystyle\overline{F(x)}=\phi^{7}$ in $\displaystyle\mathbb{F}_{2}[x]$
and $\displaystyle N_{\phi}(F)=S$ has a single side of degree
GCD$\displaystyle(7,v)=1$. Thus $\displaystyle F(x)$ is irreducible over
$\displaystyle\mathbb{Q}_{2}$. Let $\displaystyle K$ be the number field
generated by a root $\displaystyle\alpha$ of $\displaystyle F(x)$. Since
$\displaystyle F(x)$ is irreducible over $\displaystyle\mathbb{Q}_{2}$, there
is a unique valuation $\displaystyle\omega$ of $\displaystyle K$ extending
$\displaystyle\nu_{2}$. By Theorem 3.1, we have
$\displaystyle\nu_{2}(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])\geq ind_{\phi}(F)\geq
1$, and so $\displaystyle F(x)$ is not a monogenic polynomial. Let
$\displaystyle\theta=\frac{\alpha^{x}}{2^{y}}$, where $\displaystyle(x,y)$ is
the unique solution of integers of the Diophantine equation $\displaystyle
vx-7y=1$ and $\displaystyle 0\leq x\leq 6$. Then $\displaystyle\theta\in K$.
Since $\displaystyle v$ and $\displaystyle 7$ are coprime, we conclude that
$\displaystyle K=\mathbb{Q}(\theta)$. Let us show that
$\displaystyle\mathbb{Z}_{K}=\mathbb{Z}[\theta]$, and so $\displaystyle K$ is
monogenic. By [11, Corollary 3.1.4], in order to show that
$\displaystyle\theta\in\mathbb{Z}_{K}$, we need to show that
$\displaystyle\omega(\theta)\geq 0$, where $\displaystyle\omega$ is the unique
valuation of $\displaystyle K$ extending $\displaystyle\nu_{2}$. Since
$\displaystyle N_{\phi}(F)=S$ has a single side of slope $\displaystyle-v/7$,
we conclude that $\displaystyle\omega(\alpha)=v/7$, and so
$\displaystyle\omega(\theta)=\frac{xv}{7}-y=\frac{1}{7}$. Let $\displaystyle
g(x)$ be the minimal polynomial of $\displaystyle\theta$ over
$\displaystyle\mathbb{Q}$. By the formula relating roots and coefficients of a
monic polynomial, we conclude that $\displaystyle
g(x)=x^{7}+\sum_{i=1}^{7}(-1)^{i}s_{i}x^{7-i}$, where $\displaystyle
s_{i}=\displaystyle\sum_{k_{1}<\dots<k_{i}}\theta_{k_{1}}\cdots\theta_{k_{i}}$
and $\displaystyle\theta_{1},\dots,\theta_{7}$ are the
$\displaystyle\mathbb{Q}_{p}$-conjugates of $\displaystyle\theta$. Since there
is a unique valuation extending $\displaystyle\nu_{2}$ to any algebraic
extension of $\displaystyle\mathbb{Q}_{2}$, we conclude that
$\displaystyle\omega(\theta_{i})=1/7$ for every $\displaystyle i=1,\dots,7$.
Thus $\displaystyle\nu_{2}(s_{7})=\omega(\theta_{1}\cdots\theta_{7})=7\times
1/7=1$ and $\displaystyle\nu_{2}(s_{i})\geq i/7$ for every $\displaystyle
i=1,\dots,6$, which means that $\displaystyle g(x)$ is a $\displaystyle
2$-Eisenstein polynomial. Hence $\displaystyle 2$ does not divide the index
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\theta])$. As $\displaystyle 2$ is
the unique positive prime integer candidate to divide
$\displaystyle(\mathbb{Z}[\alpha]:\mathbb{Z}[\theta])$, we conclude that for
every prime integer $\displaystyle p$, $\displaystyle p$ does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\theta])$, which means that
$\displaystyle\mathbb{Z}_{K}=\mathbb{Z}[\theta]$.

##  5\. Examples

Let $\displaystyle F=x^{7}+ax+b\in\mathbb{Z}[x]$ be a monic irreducible
polynomial and $\displaystyle K$ a number field generated by a root
$\displaystyle\alpha$ of $\displaystyle F(x)$. In the following examples, we
calculate the index of the field $\displaystyle K$. First based on Theorem
2.5, $\displaystyle\nu_{p}(i(K))=0$ for every prime integer $\displaystyle
p\geq 5$. Thus we need only to calculate $\displaystyle\nu_{p}(i(K))$ for
$\displaystyle p=2,3$.

  1. (1)

For $\displaystyle a=6$ and $\displaystyle b=6$, since $\displaystyle{F(x)}$
is $\displaystyle p$-Eisenstein for every $\displaystyle p=2,3$, we conclude
that $\displaystyle F(x)$ is irreducible over $\displaystyle\mathbb{Q}$,
$\displaystyle 2$ (resp. $\displaystyle 3$) does not divide
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. Thus $\displaystyle 2$
(resp. $\displaystyle 3$) does not divide $\displaystyle i(K))$, and so
$\displaystyle i(K)=1$.

  2. (2)

For $\displaystyle a=28$ and $\displaystyle b=32$, since
$\displaystyle\overline{F(x)}$ is irreducible over
$\displaystyle\mathbb{F}_{5}$, $\displaystyle F(x)$ is irreducible over
$\displaystyle\mathbb{Q}$. By the first item of Theorem 2.3, we have
$\displaystyle\nu_{2}(i(K))=1$. By Theorem 2.4,
$\displaystyle\nu_{3}(i(K))=0$. Thus $\displaystyle i(K)=2$.

  3. (3)

For $\displaystyle a=3$ and $\displaystyle b=8$,
$\displaystyle\overline{F(x)}$ is irreducible over
$\displaystyle\mathbb{F}_{5}$, $\displaystyle F(x)$ is irreducible over
$\displaystyle\mathbb{Q}$. Again since $\displaystyle a\equiv 3\ \mbox{\rm(mod
}{4})$ and $\displaystyle b\equiv 0\ \mbox{\rm(mod }{8})$, by Theorem 2.3,
$\displaystyle\nu_{2}(i(K))=3$. By Theorem 2.4,
$\displaystyle\nu_{3}(i(K))=0$. Thus $\displaystyle i(K)=8$.

  4. (4)

For $\displaystyle a=-1$ and $\displaystyle b=9$, since
$\displaystyle\overline{F(x)}$ is irreducible over
$\displaystyle\mathbb{F}_{2}$, $\displaystyle F(x)$ is irreducible over
$\displaystyle\mathbb{Q}$. Since $\displaystyle 2\mathbb{Z}_{K}$ is a prime
ideal of $\displaystyle\mathbb{Z}_{K}$, $\displaystyle\nu_{2}(i(K))=0$. Also
since $\displaystyle a\equiv 8\ \mbox{\rm(mod }{9})$ and $\displaystyle
b\equiv 0\ \mbox{\rm(mod }{9})$, by Theorem 2.4,
$\displaystyle\nu_{3}(i(K))=2$. Thus $\displaystyle i(K)=9$.

  5. (5)

For $\displaystyle a=803$ and $\displaystyle b=2112$, since
$\displaystyle\overline{F(x)}$ is irreducible over
$\displaystyle\mathbb{F}_{5}$, $\displaystyle F(x)$ is irreducible over
$\displaystyle\mathbb{Q}$. Since $\displaystyle a\equiv 3\ \mbox{\rm(mod
}{4})$ and $\displaystyle b\equiv 0\ \mbox{\rm(mod }{8})$, by Theorem 2.3,
$\displaystyle\nu_{2}(i(K))=3$. Similarly since $\displaystyle a\equiv 5\
\mbox{\rm(mod }{9})$ and $\displaystyle b\equiv 6\ \mbox{\rm(mod }{9})$, by
Theorem 2.4, $\displaystyle\nu_{3}(i(K))=1$. Thus $\displaystyle i(K)=24$.

  6. (6)

For $\displaystyle a=35$ and $\displaystyle b=72$, since
$\displaystyle\overline{F(x)}$ is irreducible over
$\displaystyle\mathbb{F}_{11}$, $\displaystyle F(x)$ is irreducible over
$\displaystyle\mathbb{Q}$. Since $\displaystyle a\equiv 3\ \mbox{\rm(mod
}{4})$ and $\displaystyle b\equiv 0\ \mbox{\rm(mod }{8})$, by Theorem 2.3,
$\displaystyle\nu_{2}(i(K))=3$. Similarly since $\displaystyle a\equiv 8\
\mbox{\rm(mod }{9})$ and $\displaystyle b\equiv 0\ \mbox{\rm(mod }{9})$, by
Theorem 2.4, $\displaystyle\nu_{3}(i(K))=2$. Thus $\displaystyle i(K)=72$.

Data Availability Statement
Data sharing not applicable to this article as no datasets were generated or
analysed during the current study.

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# On the treewidth of generalized Kneser graphs

Klaus Metsch Justus-Liebig-Universität, Mathematisches Institut, Arndtstraße
2, D-35392 Gießen

###### Abstract

The generalized Kneser graph $K(n,k,t)$ for integers $k>t>0$ and $n>2k-t$ is
the graph whose vertices are the $k$-subsets of $\\{1,\dots,n\\}$ with two
vertices adjacent if and only if they share less than $t$ elements. We
determine the treewidth of the generalized Kneser graphs $K(n,k,t)$ when
$t\geq 2$ and $n$ is sufficiently large compared to $k$. The imposed bound on
$n$ is a significant improvement of a previously known bound. One consequence
of our result is the following. For each integer $c\geq 1$ there exists a
constant $K(c)\geq 2c$ such that $k\geq K(c)$ implies for $t=k-c$ that

$\operatorname{tw}(K(n,k,t))=\binom{n}{k}-\binom{n-t}{k-t}-1$

if and only if $n\geq(t+1)(k+1-t)$ .

Keywords: generalized Kneser graph, treewidth

MSC (2020): 51D05

##  1 Introduction

In this paper a _graph_ $\Gamma$ is a pair $(X,E)$ where $X$ is a finite non-
empty set and $E$ is a set of subsets of cardinality two of $X$. The elements
of $X$ are called _vertices_ and the elements of $E$ are called _edges_. We
write $X=V(\Gamma)$. $\Gamma$ is called empty if it has no edges.

A _tree decomposition_ of a graph $\Gamma$ is a pair $(T,B)$ where $T$ is a
tree and $B=(B_{t}:t\in V(T))$ is a collection of subsets $B_{t}$ of
$V(\Gamma)$, indexed by the vertices of $T$, such that

  1. 1.

every edge $\\{u,v\\}$ of $\Gamma$ is contained in $B_{t}$ for some $t\in
V(T)$, and

  2. 2.

for each $v\in V(\Gamma)$, the graph induced by $T$ on $\\{t\in V(T)\mid v\in
B_{t}\\}$ is connected and non-empty.

The _width_ of such a tree decomposition is the number $\max\\{|B_{t}|-1\mid
t\in V(T)\\}$, and the _treewidth_ $\operatorname{tw}(\Gamma)$ of a graph
$\Gamma$ is the smallest width of its tree decompositions. The treewidth of a
graph measures how treelike a graph is. For example, the treewidth of a non-
empty tree is one and the treewidth of a graph on $n$ vertices is at most
$n-1$ with equality if and only if the graph is complete. There is a vast
literature on the treewidth of graphs, see [1, 3, 4, 5, 6, 7, 9] for some
recent ones, and there are applications. A famous one is the use of treewidth
by Robertson and Seymour [8] in their minor theorem.

For integers $n,k,t$ with $k>t\geq 1$ and $n>2k-t$, the _generalized Kneser
graph_ $K(n,k,t)$ is the graph whose vertices are the $k$-element subsets of
the set $[n]:=\\{1,\dots,n\\}$ with two vertices $K$ and $K^{\prime}$ adjacent
if and only if $|K\cap K^{\prime}|<t$. The condition $n>2k-t$ ensures that the
graph is non-empty. If $t=1$, these graphs are called _Kneser graph_ s and are
denoted by $K(n,k)$.

It was proved by Harvey and Wood [2] that the treewidth of $K(n,k)$ is equal
to $\binom{n}{k}-\binom{n-1}{k-1}-1$ for $n\geq 4k^{2}-4k+1$ and $k\geq 3$.
More recently, Liu, Ciao and Lu proved the following.

######  Theorem 1.1 ([6]).

For integers $n,k,t$ with $k>t\geq 2$ and

$\displaystyle n\geq 2(k-t)(t+1)\binom{k}{t}+k+t+1$ (1)

we have $\operatorname{tw}(K(n,k,t))=\binom{n}{k}-\binom{n-t}{k-t}-1$.

The hard part in this theorem is to prove the lower bound for the treewidth.
For the upper bound, the authors of [6] also used (1) but in fact this holds
in general, which is our first result.

######  Theorem 1.2.

Let $n,k,t$ be positive integers with $n>2k-t$ and $k>t>0$.

  1. (a)

$\operatorname{tw}(K(n,k,t))\leq\binom{n}{k}-\binom{n-t}{k-t}-1$.

  2. (b)

If $n<(t+1)(k+1-t)$, then the bound in (a) is not tight.

The reason for (a) not to be tight when $n<(t+1)(k+1-t)$ is that in this
situation the generalized Kneser graph has independent sets that are larger
than the so called point pencils, see Section 2. Up to my knowledge, it is
open problem whether equality holds in (a) for all $n\geq(t+1)(k+1-t)$. We can
give an affirmative answer for some parameter sets as follows.

######  Theorem 1.3.

For each integer $c\geq 1$, there exists an integer $K(c)$ such that $k\geq
K(c)$ and $t=k-c$ implies that

$\displaystyle\operatorname{tw}(K(n,k,k-c))=\binom{n}{k}-\binom{n-t}{k-t}-1$
(2)

for all $n$ with $n\geq(t+1)(k+1-t)$.

We also give $K(c)$ more explicitly in Corollary 2.9. For general $k$ and $t$
we can improve the above result of Liu, Ciao and Lu as follows.

######  Theorem 1.4.

For integers $n,k,t$ with $k>t>1$ and

$\displaystyle n\geq\begin{cases}6k(k+1-t)(k-t)&\text{if $2\leq t\leq 16$,}\\\
\frac{t-1}{\ln(t)}k(k+1-t)(k-t)&\text{if $t\geq 17$}\end{cases}$ (3)

we have

$\displaystyle\operatorname{tw}(K(n,k,t))=\binom{n}{k}-\binom{n-t}{k-t}-1.$

The proofs in the present paper follows the lines of the proofs in [6] by
improving and simplifying their arguments. In particular, our proof avoids
results of the $t$-shadow of families of sets.

##  2 Proof

For every graph $\Gamma$ its maximum vertex degree is denoted by
$\Delta(\Gamma)$ and is called its _maximum degree_. The cardinality of a
largest independent set of $\Gamma$ is denoted by $\alpha(\Gamma)$ and is
called the _independence number_ of the graph. There is a connection between
treewidth, maximum degree and independence number.

To see this, consider a graph $\Gamma$ and an independent set $A$ of $\Gamma$
of size $\alpha(\Gamma)$. Put $S=V(\Gamma)\setminus A$. Then
$T:=(A\cup\\{S\\},\\{\\{a,S\\}\mid a\in A\\})$ is a tree (in fact a star) and
defining $B_{S}=S$ and $B_{a}=\\{a\\}\cup\Gamma_{1}(a)$ for $a\in A$ results
in the tree decomposition $(T,(B_{i})_{i\in A\cup\\{S\\}})$ of $\Gamma$. Since
$|B_{S}|=|S|=|V(\Gamma)|-\alpha(\Gamma)$ and $|B_{a}|\leq\Delta(\Gamma)+1$, it
follows that

$\displaystyle\operatorname{tw}(\Gamma)\leq\max\\{\Delta(\Gamma),|V(\Gamma)|-\alpha(\Gamma)-1\\}.$
(4)

This was proved in [2].

We will apply this to generalized Kneser graphs. The independence number of
the generalized Kneser graph $K(n,k,t)$ is at least $\binom{n-t}{k-t}$ since
the $k$-element subsets of an $n$-set containing a given $t$-set is an
independent set of this size. We now compare $\Delta(\Gamma)$ and
$|V(\Gamma)|-\alpha(\Gamma)-1$ for generalized Kneser graphs $\Gamma$.

######  Lemma 2.1.

For positive integers $n>k\geq t$ we have

$\displaystyle\binom{n-t}{k-t}+(k-t)t\binom{n-k}{k-t}+\sum_{i=0}^{t-1}\binom{k}{i}\binom{n-k}{k-i}\leq\binom{n}{k}.$

###### Proof.

The sum is equal to the cardinality of the set $T_{1}$ that consists of all
$k$-subsets of $[n]$ that have at most $t-1$ elements in $[k]$. Also
$\binom{n-t}{k-t}$ is equal to the cardinality of the set $T_{2}$ that
consists of all $k$-subsets of $[n]$ that contain $[t]$. Finally,
$(k-t)t\binom{n-k}{k-t}$ is the number of $k$-subsets of $[n]$ that contain
exactly $t-1$ elements of $[t]$ and one further element of $[k]$. As the three
sets $T_{i}$ are mutually disjoint, we have that $|T_{1}|+|T_{2}|+|T_{3}|$ is
at most the number $\binom{n}{k}$ of $k$-subsets of $[n]$. ∎

######  Corollary 2.2.

For a generalized Kneser graph $\Gamma=K(n,k,t)$ with $n>2k-t$ and $k>t>0$ we
have

$\displaystyle\Delta(\Gamma)\leq|V(\Gamma)|-\binom{n-t}{k-t}-(k-t)t\binom{n-k}{k-t}.$

###### Proof.

We have $\Delta(\Gamma)=\sum_{i=0}^{t-1}\binom{k}{i}\binom{n-k}{k-i}$. In fact
for every $k$-subset $K$ of $[n]$, this is the number of $k$-subsets of $[n]$
that meet $K$ in at most $t-1$ elements. The statement follows therefore from
the lemma. ∎

######  Proposition 2.3.

Let $n,k,t$ be positive integers with $n>2k-t$ and $k>t>0$.

  1. (a)

$\operatorname{tw}(K(n,k,t))\leq\binom{n}{k}-\binom{n-t}{k-t}-1$.

  2. (b)

$n<(t+1)(k+1-t)$, then the bound in (a) is not tight.

###### Proof.

Put $\Gamma=K(n,k,t)$. We will apply (4). Since $n>2k-t$ and $k>t>0$ we have
$\Delta(\Gamma)\leq\binom{n}{k}-\binom{n-t}{k-t}-2$ from Corollary 2.2.

We have $|V(K(n,k,t))|=\binom{n}{k}$ and $\alpha(\Gamma)\geq\binom{n-t}{k-t}$,
as noticed above. Hence
$|V(\Gamma)|-\alpha(\Gamma)\leq\binom{n}{k}-\binom{n-t}{k-t}$. Thus (a)
follows from (4).

Now suppose that $n<(t+1)(k+1-t)$. Let $A$ be the set consisting of all
$k$-subsets of $[n]$ that have at least $t+1$ elements in $[t+2]$. This is an
independent set of $\Gamma$ and thus $\alpha(\Gamma)\geq|A|$. Also, since
$n<(t+1)(k-1-t)$, it is easy to see that $|A|>\binom{n-t}{k-t}$. Hence
$|V(\Gamma)|-\alpha(\Gamma)<\binom{n}{k}-\binom{n-t}{k-t}$, and thus also (b)
follows from (4). ∎

It was proved in [6] that the upper bound is sharp when $n$ is sufficiently
large compared to $k$ and $t$. We will improve this result by weakening the
required bound on $n$ significantly. As in [6] and [2] we use a result of
Robertson and Seymour on separators. For a real number $p$ with
$\frac{2}{3}\leq p<1$ a _$p$ -separator_ of a graph $\Gamma$ is subset $X$ of
the vertex set $V(\Gamma)$ of $\Gamma$ such that every component of
$\Gamma\setminus X$ has at most $p|V(\Gamma)\setminus X)|$ vertices.

######  Result 2.4 ([8]).

Let $\Gamma$ be a finite graph and $p$ a real number with $\frac{2}{3}\leq
p<1$. Then $\Gamma$ has a $p$-separator $X$ with
$|X|\leq\operatorname{tw}(\Gamma)+1$.

A second ingredient of our proof is the result of Wilson on the independence
number of generalized Kneser graphs.

######  Result 2.5 ([10]).

For integers $k>t\geq 1$ and $n\geq(t+1)(k+1-t)$, we have
$\alpha(K(n,k,t))=\binom{n-t}{k-t}$.

As we have already noticed previously the independence number is larger when
$n<(t+1)(k+1-t)$.

######  Lemma 2.6.

Suppose $n,k,t$ are positive integers with $k>t$ and $n\geq
t+\frac{1}{2}(k+1-t)(k-t)$. Define the function

$f:\\{r\in\mathbb{Z}\mid 0\leq r\leq t-1\\}\to\mathbb{R},\
f(r):=\binom{k-r}{t-r}\binom{n-2t+r}{k-2t+r}.$

Then $f$ is monotone increasing.

###### Proof.

For integers $r$ with $0\leq r\leq t-2$, it is easy to see that $f(r)\leq
f(r+1)$ is equivalent to

$(k-r)(k-2t+r+1)\leq(n-2t+r+1)(t-r).$

This in turn can be written as $(n-t)(t-r)\geq(k-t)(k+1-t)$. In view of the
assumed lower bound on $n$ and $r\leq t-2$, this is true. ∎

######  Lemma 2.7.

Suppose $n,k,t$ are integers with $k>t>0$ and $p$ is a real number with
$\frac{2}{3}\leq p<1$ such that

$\displaystyle n$ $\displaystyle\geq(t+1)(k+1-t),$ $\displaystyle n$
$\displaystyle\geq t+\frac{1}{2}(k+1-t)(k-t),\ \text{and}$
$\displaystyle(1-p)\binom{n-t}{k-t}$
$\displaystyle\geq\sum_{s=1}^{t}\binom{t-1}{s-1}\binom{k+1-t}{s}\binom{k-t+s}{s}\binom{n-t-s}{k-t-s}.$

Then every $p$-separator of $K(n,k,t)$ has at least
$\binom{n}{k}-\binom{n-t}{k-t}$ elements.

###### Proof.

Assume on the contrary that there exits a $p$-separator $X$ with
$|X|<\binom{n}{k}-\binom{n-t}{k-t}$. Then $U:=V(K(n,k,t))\setminus X$
satisfies $|U|>\binom{n-t}{k-t}$. The components of $K(n,k,t)\setminus X$ have
each at most $p|U|$ elements, and hence there is a union ${\cal A}$ of
components with $(1-p)|U|\leq|{\cal A}|\leq p|U|$ (this is clear if some
component has at least $(1-p)|U|$ vertices and otherwise it follows from
$p\geq 2(1-p)$). Let ${\cal B}=U\setminus{\cal A}$ be the union of the
remaining components, so that also $(1-p)|U|\leq|{\cal B}|\leq p|U|$. As
${\cal A}$ and ${\cal B}$ are unions of components of the graph
$K(n,k,t)\setminus X$, we have $|A\cap B|\geq t$ for all $A\in{\cal A}$ and
$B\in{\cal B}$.

Since $n\geq(t+1)(k+1-t)$, Results 2.5 shows that $|U|>\alpha(K(n,k,t))$.
Hence $U$ contains adjacent vertices, that is there exists two elements
$A_{1},A_{2}\in U$ that intersect in less than $t$ elements. We may assume
that $A_{1},A_{2}\in A$. Put $s=|A_{1}\cap A_{2}|<t$. For every $t$-subset $Y$
of $A_{1}$ we put ${\cal B}_{Y}:=\\{B\in{\cal B}\mid Y\subseteq B\\}$. The
following observation is from [6].

Claim: If $Y$ is a $t$-subset of $A_{1}$ and if $r:=|Y\cap A_{2}|$, then
$|{\cal B}_{Y}|\leq f(r)$.
This can be seen as follows. If $B\in{\cal B}_{Y}$, then $|A_{2}\cap B|\geq t$
and hence $|(A_{2}\setminus Y)\cap B|\geq t-r$. On the other hand, for each of
the $\binom{k-r}{t-r}$ subsets $Z$ of size $t-r$ of $A_{2}\setminus Y$, there
are exactly $\binom{n-2t+r}{k-2t+r}$ $k$-subsets of $[n]$ that contain $Z\cup
Y$. It follows that $|{\cal
B}_{Y}|\leq\binom{k-r}{t-r}\binom{n-2t+r}{k-2t+r}=f(r)$ establishing the
claim.

Define $S:=A_{1}\cap A_{2}$, so that $s=|S|\leq t-1$. Since $|A_{1}\cap B|\geq
t$ for all $B\in{\cal B}$ we have

$\displaystyle\bigcup_{Y\in{A_{1}\choose t}}{\cal B}_{Y}={\cal B}.$ (5)

Fix any subset $T$ of $A_{1}$ with $S\subseteq T$ and $|T|=t-1$. Then

$\displaystyle|{\cal B}|$ $\displaystyle\leq\sum_{Y\in{A_{1}\choose t}}|{\cal
B}_{Y}|\leq\sum_{Y\in{A_{1}\choose t}}f(|Y\cap S|)$
$\displaystyle\leq\sum_{Y\in{A_{1}\choose t}}f(|Y\cap
T|)\leq\sum_{r=0}^{t-1}\binom{t-1}{r}\binom{k+1-t}{t-r}f(r)$

where the first inequality follows from (5), the second from the above claim,
the third from Lemma 2.6, and the fourth by counting for $0\leq r\leq t-1$ how
many $t$-subsets of $A_{1}$ share exactly $r$ elements with $T$. Using $|{\cal
B}|\geq(1-p)|U|$ and $|U|>\binom{n-t}{k-t}$ we find

$\displaystyle(1-p)\binom{n-t}{k-t}<\sum_{r=0}^{t-1}\binom{t-1}{r}\binom{k+1-t}{t-r}\binom{k-r}{t-r}\binom{n-2t+r}{k-2t+r}.$
(6)

If we substitute $r=t-s$ the resulting inequality contradicts the hypothesis
of the present lemma. ∎

######  Theorem 2.8.

Suppose $n,k,t$ are integers with $k>t>0$ such that

$\displaystyle n$ $\displaystyle\geq(t+1)(k+1-t),$ (7) $\displaystyle n$
$\displaystyle\geq t+\frac{1}{2}(k+1-t)(k-t),\ \text{and}$ (8)
$\displaystyle\frac{1}{3}\binom{n-t}{k-t}$
$\displaystyle\geq\sum_{s=1}^{\min\\{t,k-t\\}}\binom{t-1}{s-1}\binom{k+1-t}{s}\binom{k-t+s}{s}\binom{n-t-s}{k-t-s}.$
(9)

Then $\operatorname{tw}(K(n,k,t))=\binom{n}{k}-\binom{n-t}{k-t}-1$.

###### Proof.

By Lemma 2.7 every $\frac{2}{3}$-separator of $K(n,k,t)$ has at least
$\binom{n}{k}-\binom{n-t}{k-t}$ elements. Result 2.4 implies thus that
$\operatorname{tw}(K(n,k,t))\geq\binom{n}{k}-\binom{n-t}{k-t}-1$. Proposition
2.3 gives equality. ∎

It remains to analyze inequality (9). If $k-t$ is sufficiently small, we find
the following.

######  Corollary 2.9.

For integers $c\geq 1$ define

$\displaystyle
K(c)=c-1+3\sum_{s=1}^{c}\binom{c-1}{s-1}\binom{c+1}{s}\binom{c+s}{s}\frac{1}{c^{s-1}}.$

Then for all integers $k$ and $t$ with $k\geq K(c)$ and $t=k-c$ we have

$\displaystyle\operatorname{tw}(K(n,k,k-c))=\binom{n}{k}-\binom{n-t}{k-t}-1$
(10)

for all $n$ with $n\geq(t+1)(k+1-t)$.

###### Proof.

Consider integers $c,k,t,n$ with $c\geq 1$, $k\geq K(c)$, $t=k-c$ and

$\displaystyle n\geq(t+1)(k+1-t)=(t+1)(c+1).$ (11)

Since $K(c)>2c$, then $k>2c$ and $t=k-c\geq c+1\geq 2$. Also (11) implies
$n>2k-t$, so that $n,k,t$ are parameters of a generalized Kneser graph
$K(n,k,t)$. As $k>2c$, then (11) implies (8). Assume that (10) is not true.
Then Theorem 2.8 shows that

$\displaystyle\frac{1}{3}\binom{n-t}{c}$
$\displaystyle<\sum_{s=1}^{c}\binom{t-1}{s-1}\binom{c+1}{s}\binom{c+s}{s}\binom{n-t-s}{c-s}$
(12) $\displaystyle\Rightarrow\frac{1}{3}\frac{(n-t)!}{c!}$
$\displaystyle\leq\sum_{s=1}^{c}\frac{(t-1)!}{(t-s)!(s-1)!}\binom{c+1}{s}\binom{c+s}{s}\frac{(n-t-s)!}{(c-s)!}$
(13) $\displaystyle\Rightarrow n-t$
$\displaystyle\leq\sum_{s=1}^{c}\frac{3c!}{(c-s)!(s-1)!}\binom{c+1}{s}\binom{c+s}{s}\frac{(t-1)!}{(t-s)!}\frac{(n-t-s)!}{(n-t-1)!}$
(14)

From (11) we find $n-t\geq t$ and $n-t-1\geq(t+1)c$. This proves the
inequalities in

$\displaystyle\frac{(t-1)!}{(t-s)!}\frac{(n-t-s)!}{(n-t-1)!}=\prod_{i=1}^{s-1}\frac{t-i}{n-t-i}\leq\frac{(t-1)^{s-1}}{(n-t-1)^{s-1}}\leq\frac{1}{c^{s-1}}.$

On the left hand side of (14) we use that $n-t>(t+1)c=(k+1-c)c$ and find

$\displaystyle k+1-c$
$\displaystyle<\sum_{s=1}^{c}\frac{3(c-1)!}{(c-s)!(s-1)!}\binom{c+1}{s}\binom{c+s}{s}\frac{1}{c^{s-1}}$
(15)

and hence $k<K(c)$. Since we assumed $k\geq K(c)$, this is a contradiction
coming from the assumption that (10) is not true. Therefore (10) is true. ∎

Corollary 2.9 and Proposition 2.3 prove Theorem 1.3.

######  Remark 2.10.

The value $K(c)$ was chosen in such a way that (14) is not satisfied. For
fixed small $c$ one can determine $K^{\prime}(c)$ explicitly such that (14) is
not satisfied iff $k\geq K^{\prime}(c)$ and hence (10) is satisfied for all
$n,k,t$ with $k\geq K^{\prime}(c)$, $t=k-c$ and $n\geq(t+1)(k+1-t)$. For
example one finds $K^{\prime}(1)=12$, $K^{\prime}(2)=54$, $K^{\prime}(3)=195$
and $K^{\prime}(4)=626$. For $c=1$, a better result was proved in [6].

######  Corollary 2.11.

Suppose $n,k,t$ are integers with $k>t>1$ and

$\displaystyle n\geq\begin{cases}t+\frac{1}{\ln(t)}(t-1)k(k+1-t)(k-t)&\text{if
$t\geq 17$}\\\ t+6k(k+1-t)(k-t)&\text{if $2\leq t\leq 16$}\end{cases}$ (16)

Then $\operatorname{tw}(K(n,k,t))=\binom{n}{k}-\binom{n-t}{k-t}-1$.

###### Proof.

Define

$\displaystyle c:=\frac{(t-1)k(k-t)(k+1-t)}{n-t}.$ (17)

Assume the statement is wrong. Then Theorem 2.8 shows that

$\displaystyle\frac{1}{3}\binom{n-t}{k-t}\leq\sum_{s=1}^{t}\binom{t-1}{s-1}\binom{k+1-t}{s}\binom{k-t+s}{s}\binom{n-t-s}{k-t-s}.$
(18)

For every integer $s$ with $1\leq s\leq t$ we have

$\displaystyle\binom{n-t-s}{k-t-s}$
$\displaystyle=\binom{n-t}{k-t}\prod_{i=0}^{s-1}\frac{k-t-i}{n-t-i}\leq\left(\frac{k-t}{n-t}\right)^{s}\binom{n-t}{k-t}.$

Using this and $\binom{a}{b}\leq a^{b}/b!$ for binomial coefficients with
$a,b\geq 0$ in (18), we find

$\displaystyle\frac{1}{3}$
$\displaystyle\leq\sum_{s=1}^{t}\binom{t-1}{s-1}\binom{k+1-t}{s}\binom{k-t+s}{s}\left(\frac{k-t}{n-t}\right)^{s}$
$\displaystyle\leq\sum_{s=1}^{t}\frac{(t-1)^{s}(k+1-t)^{s}(k-t+s)^{s}}{(t-1)(s-1)!s!s!}\cdot\left(\frac{k-t}{n-t}\right)^{s}.$

Using the definition of $c$ in (17), this implies that

$\displaystyle\frac{t-1}{3}$
$\displaystyle\leq\sum_{s=1}^{t}\frac{c^{s}}{(s-1)!s!s!}.$ (19)

Case 1. We have $2\leq t\leq 16$. Then one can check for each possible value
of $t$ that (19) implies that $c>\frac{1}{6}(t-1)$. Using (17), this
contradicts (16).

Case 2. We have $t\geq 17$. By hypotheses and (17) we then have $c\leq\ln(t)$
and therefore

$\displaystyle\frac{t-1}{3}$
$\displaystyle\leq\sum_{s=1}^{t}\frac{c^{s}}{(s-1)!s!s!}$ $\displaystyle\leq
c+\frac{c^{2}}{4}+\frac{1}{2!3!}\sum_{s=3}^{\infty}\frac{c^{s}}{s!}$
$\displaystyle\leq c+\frac{c^{2}}{4}+\frac{1}{2!3!}(e^{c}-1)$
$\displaystyle\leq\ln t+\frac{1}{4}(\ln t)^{2}+\frac{1}{12}(t-1).$

This implies that

$\displaystyle t-1$ $\displaystyle\leq 4\ln t+(\ln t)^{2}.$

For $t=24$ and hence for all $t\geq 24$, this is a contradiction. For $17\leq
t\leq 23$ one can check easily that (19) implies that $c>\ln(t)$, which is a
contradiction. ∎

Corollary 2.11 proves Theorem 1.4.

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