# The index of certain nonic number fields defined by $\displaystyle
x^{9}+ax^{2}+b$

Lhoussain El Fadil  and  Omar Kchit  Faculty of Sciences Dhar El Mahraz, P.O.
Box 1796 Atlas-Fes, Sidi Mohamed ben Abdellah University, Morocco
<EMAIL_ADDRESS>orcid: 0000.0003.4175.8064
<EMAIL_ADDRESS>orcid: 0000.0002.0844.5034

###### Abstract.

In this paper, for any nonic number field $\displaystyle K$ defined by a monic
irreducible trinomial $\displaystyle F(x)=x^{9}+ax^{2}+b\in\mathbb{Z}[x]$, we
calculate $\displaystyle\nu_{p}(i(K))$ for every rational prime $\displaystyle
p$. In particular, we characterize the index $\displaystyle i(K)$ of this
family of number fields. As an application of our results, if $\displaystyle
i(K)\neq 1$, then $\displaystyle K$ is not monogenic. We illustrate our
results by some computational examples.

###### Key words and phrases:

Theorem of Dedekind, Theorem of Ore, prime ideal factorization, Newton
polygon, Index of a number field, Power integral basis, Monogenic

###### 2010 Mathematics Subject Classification:

11R04, 11Y40, 11R21

##  1\. Introduction

Let $\displaystyle K$ be a number field of degree $\displaystyle n$ with ring
of integers $\displaystyle\mathbb{Z}_{K}$. Let
$\displaystyle\alpha\in\mathbb{Z}_{K}$ be a primitive element of
$\displaystyle K$. The index of $\displaystyle\alpha$, denoted by
$\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, is the index the Abelian
group $\displaystyle\mathbb{Z}[\alpha]$ in $\displaystyle\mathbb{Z}_{K}$, and
the formula linking $\displaystyle(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$,
$\displaystyle\Delta(\alpha)$, and $\displaystyle d_{K}$ is given by:

(1.1) $\Delta(\alpha)=\pm(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])^{2}\cdot d_{K},$

where $\displaystyle d_{K}$ is the absolute discriminant of $\displaystyle K$
and $\displaystyle\Delta(\alpha)$ is the discriminant of the minimal
polynomial of $\displaystyle\alpha$ over $\displaystyle\mathbb{Q}$. The index
of $\displaystyle K$, denoted by $\displaystyle i(K)$, is the greatest common
divisor of the indices of all primitive integers of $\displaystyle K$. Say
$\displaystyle i(K)=\gcd\
\\{(\mathbb{Z}_{K}:\mathbb{Z}[\theta])\,|\,{K=\mathbb{Q}(\theta)\text{ and
}\theta\in\mathbb{Z}_{K}}\\}$. A rational prime $\displaystyle p$ dividing
$\displaystyle i(K)$ is called a prime common index divisor of $\displaystyle
K$. If $\displaystyle K$ is monogenic, then $\displaystyle\mathbb{Z}_{K}$ has
a power integral basis, i.e., a $\displaystyle\mathbb{Z}$-basis of the form
$\displaystyle(1,\theta,\dots,\theta^{n-1})$, and the index of $\displaystyle
K$ is trivial, say $\displaystyle i(K)=1$. Therefore a field having a prime
common index divisor is not monogenic. The first number field with non trivial
index was given by Dedekind in $\displaystyle 1871$, who exhibited the example
of the cubic number field $\displaystyle K$ defined by $\displaystyle
x^{3}-x^{2}-2x-8$ and showed that the index of any primitive integer of
$\displaystyle K$ is even. In $\displaystyle 1930$, Engstrom [8] was the first
one who studied the prime power decomposition of the index of a number field.
He showed that, for number fields of degree $\displaystyle n\leq 7$,
$\displaystyle\nu_{p}(i(K))$ is determined by the form of the factorization of
$\displaystyle p\mathbb{Z}_{K}$, where $\displaystyle\nu_{p}$ is the
$\displaystyle p$-adic valuation of $\displaystyle\mathbb{Q}$. This motivated
a very important question, stated as problem $\displaystyle 22$ in
Narkiewicz’s book ([19]), which asks for an explicit formula of the highest
power $\displaystyle\nu_{p}(i(K))$ for a given rational prime p dividing
$\displaystyle i(K)$. In [23], Śliwa showed that, if $\displaystyle p$ is a
non-ramified ideal in $\displaystyle K$, then $\displaystyle\nu_{p}(i(K))$ is
determined by the factorization of $\displaystyle p\mathbb{Z}_{K}$. These
results were generalized by Nart ([20]), who developed a $\displaystyle
p$-adic characterization of the index of a number field. In [18], Nakahara
studied the index of non-cyclic but abelian biquadratic number fields. In [10]
Gaál et al. characterized the field indices of biquadratic number fields
having Galois group $\displaystyle V_{4}$. In [5], we characterized the index
and studied the monogenity of any sextic number field defined by
$\displaystyle x^{6}+ax^{5}+b$. In [6], for every rational prime
$\displaystyle p$, we characterized $\displaystyle\nu_{p}(i(K))$ for any
septic number field defined by a trinomial $\displaystyle x^{7}+ax^{3}+b$. In
[4], El Fadil characterized the index of any quintic number field defined by
$\displaystyle x^{5}+ax^{3}+b$. In [15], Kchit studied the index of any nonic
number field defined by $\displaystyle x^{9}+ax+b$. In this paper, for any
nonic number field $\displaystyle K$ defined by a monic irreducible trinomial
$\displaystyle x^{9}+ax^{2}+b\in\mathbb{Z}[x]$, we calculate
$\displaystyle\nu_{p}(i(K))$ for every rational prime $\displaystyle p$. In
particular, we show that $\displaystyle i(K)\in\\{0,1,2,3,6\\}$.

##  2\. Main Results

Throughout this section, $\displaystyle K$ is a number field generated by a
complex root $\displaystyle\alpha$ of a monic irreducible trinomial
$\displaystyle F(x)=x^{9}+ax^{2}+b\in\mathbb{Z}[x]$. Without loss of
generality, we assume that for every rational prime $\displaystyle p$,
$\displaystyle\nu_{p}(a)\leq 6$ or $\displaystyle\nu_{p}(b)\leq 8$.
Along this paper, for every integer $\displaystyle m\in\mathbb{Z}$ and a
rational prime $\displaystyle p$, let $\displaystyle
m_{p}=\cfrac{m}{p^{\nu_{p}(m)}}$. We denote $\displaystyle\Delta=\Delta(F)$
the discriminant of $\displaystyle F(x)$.
The following theorem characterizes when $\displaystyle 2$ divides
$\displaystyle i(K)$.

######  Theorem 2.1.

The following table provides $\displaystyle\nu_{2}(i(K))$:

Table 1. $\displaystyle\nu_{2}(i(K))$ Conditions | $\displaystyle\quad\nu_{2}(i(K))\quad$
---|---
$\displaystyle\nu_{2}(a)=0$ and $\displaystyle\nu_{2}(b)=2k$ | $\displaystyle b_{2}\equiv-a\ \mbox{\rm(mod }{8})$ | $\displaystyle 1$
$\displaystyle a\equiv 2\ \mbox{\rm(mod }{4})$ and $\displaystyle\nu_{2}(b)=2k+1$ | $\displaystyle b_{2}\equiv-a_{2}\ \mbox{\rm(mod }{8})$
$\displaystyle a\equiv 4\ \mbox{\rm(mod }{8})$ and
$\displaystyle\nu_{2}(b)=2k+2$
$\displaystyle a\equiv 8\ \mbox{\rm(mod }{16})$ and
$\displaystyle\nu_{2}(b)=2k+3$
$\displaystyle a\equiv 16\ \mbox{\rm(mod }{32})$ and
$\displaystyle\nu_{2}(b)=2k+4$
$\displaystyle a\equiv 32\ \mbox{\rm(mod }{64})$ and
$\displaystyle\nu_{2}(b)=2k+7$
$\displaystyle a\equiv 64\ \mbox{\rm(mod }{128})$ and
$\displaystyle\nu_{2}(b)=2k+8$
$\displaystyle a\equiv 32\ \mbox{\rm(mod }{64})$ and $\displaystyle b\equiv 128\ \mbox{\rm(mod }{256})$ | $\displaystyle a_{2}+b_{2}\equiv 4\ \mbox{\rm(mod }{8})$
$\displaystyle a\equiv 64\ \mbox{\rm(mod }{128})$ and $\displaystyle b\equiv 256\ \mbox{\rm(mod }{512})$ |  $\displaystyle\nu_{2}(\Delta)=2k$ and $\displaystyle\Delta_{2}\equiv a_{2}b_{2}\ \mbox{\rm(mod }{8})$
Otherwise | $\displaystyle 0$

The following theorem characterizes when $\displaystyle 3$ divides
$\displaystyle i(K)$.

######  Theorem 2.2.

The following table provides $\displaystyle\nu_{3}(i(K))$:

Table 2. $\displaystyle\nu_{3}(i(K))$ Conditions | $\displaystyle\nu_{3}(i(K))$
---|---
$\displaystyle(a,b)\in\\{(9,71),(36,44)\\}\ \mbox{\rm(mod }{81})$ and $\displaystyle a+b\equiv-1\ \mbox{\rm(mod }{243})$ | $\displaystyle\nu_{3}(\Delta)=2k$ | $\displaystyle 1$
$\displaystyle(a,b)\equiv(63,17)\ \mbox{\rm(mod }{81})$ and $\displaystyle a+b\equiv 80\ \mbox{\rm(mod }{243})$ | and
$\displaystyle(a,b)\equiv(18,64)\ \mbox{\rm(mod }{81})$ and $\displaystyle
a+b\equiv 163\ \mbox{\rm(mod }{243})$
$\displaystyle(a,b)\in\\{(45,37),(72,10)\\}\ \mbox{\rm(mod }{81})$ and $\displaystyle a+b\equiv 1\ \mbox{\rm(mod }{243})$ | $\displaystyle\Delta_{3}\equiv-1\ \mbox{\rm(mod }{3})$
Otherwise | $\displaystyle 0$

######  Theorem 2.3.

For every rational prime $\displaystyle p\geq 5$ and for every
$\displaystyle(a,b)\in\mathbb{Z}^{2}$ such that $\displaystyle
F(x)=x^{9}+ax^{2}+b$ is irreducible, $\displaystyle p$ does not divide the
index $\displaystyle i(K)$, where $\displaystyle K$ is the number field
defined by $\displaystyle F(x)$.

######  Corollary 2.4.

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

##  3\. Preliminaries

Our proofs are based on Newton polygon techniques applied on prime ideal
factorization, which 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 the theorem of index of Ore which plays a key
role for proving our main results.
Let $\displaystyle K=\mathbb{Q}(\alpha)$ be a number field generated by a
complex root $\displaystyle\alpha$ of a monic irreducible polynomial
$\displaystyle F(x)\in\mathbb{Z}[x]$. We shall use Dedekind’s theorem [21,
Chapter I, Proposition 8.3] and Dedekind’s criterion [1, Theorem 6.1.4]. Let
$\displaystyle\phi\in\mathbb{Z}_{p}[x]$ be a monic lift to an irreducible
factor of $\displaystyle F(x)$ modulo $\displaystyle p$, $\displaystyle
F(x)=a_{0}(x)+a_{1}(x)\phi(x)+\cdots+a_{l}(x)\phi(x)^{l}$ the
$\displaystyle\phi$-expansion of $\displaystyle F(x)$, and $\displaystyle
N_{\phi}^{+}(F)$ the principal $\displaystyle\phi$-Newton polygon of
$\displaystyle F(x)$. Let $\displaystyle\mathbb{F}_{\phi}$ be the field
$\displaystyle\mathbb{F}_{p}[x]/(\overline{\phi})$, then to every side
$\displaystyle S$ of $\displaystyle N_{\phi}^{+}(F)$ with initial point
$\displaystyle(i,u_{i})$, and every $\displaystyle i=0,\ldots,l$, let the
residue coefficient $\displaystyle c_{i}\in\mathbb{F}_{\phi}$ 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 and
$\displaystyle l=l(S)$ its length. Then $\displaystyle d=l/e$ is the degree of
$\displaystyle S$. Hence, if $\displaystyle i$ is not a multiple of
$\displaystyle e$, then $\displaystyle(s+i,u_{s+i})$ does not lie on
$\displaystyle S$, and so $\displaystyle c_{i}=0$. Let
$\displaystyle{R_{\lambda}(F)(y)}=t_{d}y^{d}+t_{d-1}y^{d-1}+\cdots+t_{1}y+t_{0}\in\mathbb{F}_{\phi}[y]$,
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 $\displaystyle{R_{\lambda}(F)(y)}$ 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
$\displaystyle{R_{\lambda_{ij}}(F)(y)}=\prod_{s=1}^{s_{ij}}\psi_{ijs}^{a_{ijs}}(y)$
be the factorization of $\displaystyle R_{\lambda_{ij}}(F)(y)$ in
$\displaystyle\mathbb{F}_{\phi_{i}}[y]$. Then we have the following theorem of
index of Ore:

######  Theorem 3.1.

$\displaystyle($[7, 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
$\displaystyle f_{ijs}=\deg(\phi_{i})\cdot\deg(\psi_{ijs})$ for every
$\displaystyle(i,j,s)$.

If the theorem of Ore fails, that is, $\displaystyle F(x)$ is not
$\displaystyle p$-regular, then in order to complete the factorization of
$\displaystyle F(x)$, Guàrdia, Montes, and Nart introduced the notion of high
order Newton polygon ([11]). Similar to first order, for each order
$\displaystyle r$, they introduced the valuation $\displaystyle\omega_{r}$,
the key polynomial $\displaystyle\phi_{r}(x)$ for this valuation, the Newton
polygon $\displaystyle N_{r}(F)$ of any polynomial $\displaystyle F(x)$ with
respect to $\displaystyle\omega_{r}$ and $\displaystyle\phi_{r}(x)$, and for
each side $\displaystyle T_{i}$ of $\displaystyle N_{r}(F)$, the residual
polynomial $\displaystyle R_{r}(F)(y)$, and the index of $\displaystyle F(x)$
in order $\displaystyle r$. For more details, we refer to [11].

For the proof of our results, we need the following lemma, which characterizes
the prime common index divisors of $\displaystyle K$.

######  Lemma 3.2.

$\displaystyle($[8]$\displaystyle)$
Let $\displaystyle p$ be a rational prime and $\displaystyle K$ 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 $\displaystyle p$ divides the index $\displaystyle
i(K)$ if and only if $\displaystyle\mathcal{P}_{f}>\mathcal{N}_{f}$ for some
positive integer $\displaystyle f$.

##  4\. Proofs of Main Results

Throughout this section, if $\displaystyle
p\mathbb{Z}_{K}=\prod_{i=1}^{r}\prod_{j=1}^{t_{i}}\prod_{s=1}^{s_{ij}}\mathfrak{p}_{ijs}^{e_{ij}}$,
then $\displaystyle e_{ij}$ denotes the ramification index of
$\displaystyle\mathfrak{p}_{ijs}$ and $\displaystyle f_{ijs}$ denotes its
residue degree for every $\displaystyle(i,j,s)$.

Proof of Theorem 2.1.
If $\displaystyle\nu_{2}(b)=0$, then since
$\displaystyle\Delta=b(3^{18}b^{7}+2^{2}\cdot 7^{7}a^{9})$ is the discriminant
of $\displaystyle F(x)$, thanks to the index formula $\displaystyle(1.1)$,
$\displaystyle\nu_{2}((\mathbb{Z}_{K}:\mathbb{Z}[\alpha]))=0$ and so
$\displaystyle\nu_{2}(i(K))=0$. Assume that $\displaystyle 2$ divides
$\displaystyle b$. Then we have the following cases:

  1. (1)

If $\displaystyle\nu_{2}(a)=0$, then $\displaystyle F(x)\equiv
x^{2}(x-1)(x^{3}+x+1)(x^{3}+x^{2}+1)\ \mbox{\rm(mod }{2})$. Then
$\displaystyle 2$ divides the index $\displaystyle i(K)$ if and only if
$\displaystyle\phi=x$ provides two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 2$ with residue
degree $\displaystyle 1$ each ideal factor. That is
$\displaystyle\nu_{2}(b)=2k$ and $\displaystyle b_{2}+a\equiv 0\ \mbox{\rm(mod
}{8})$. In this case, we replace $\displaystyle\phi$ by $\displaystyle
g=x-2^{k}$. Then $\displaystyle
F(x)=\cdots+(a+36(2^{7k}))g^{2}+(2^{k+1}a+9(2^{8k}))g+b+2^{2k}a+2^{9k}$. Thus
$\displaystyle N_{g}^{+}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,w)$, $\displaystyle(1,k+1)$, and $\displaystyle(2,0)$ with
$\displaystyle w\geq 2k+3$. Hence the degree of each side is $\displaystyle
1$, and so $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{11}\mathfrak{p}_{12}\mathfrak{p}_{21}\mathfrak{p}_{31}\mathfrak{p}_{41}$
with $\displaystyle f_{11}=f_{12}=f_{2}=1$ and $\displaystyle f_{3}=f_{4}=3$.
Applying [8, Theorem 4], we get $\displaystyle\nu_{2}(i(K))=1$.

  2. (2)

If $\displaystyle\nu_{2}(a)\geq 1$, then $\displaystyle F(x)=x^{9}\
\mbox{\rm(mod }{2})$. Let $\displaystyle\phi=x$. By assumption,
$\displaystyle\nu_{2}(a)\leq 6$ or $\displaystyle\nu_{2}(b)\leq 8$.
If $\displaystyle\nu_{2}(b)<\cfrac{9}{7}\nu_{2}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}$ has a single side joining $\displaystyle(0,\nu_{2}(b))$ and
$\displaystyle(9,0)$ with $\displaystyle d(S_{1})\in\\{1,3\\}$ since
$\displaystyle\nu_{2}(b)\leq 8$ by assumption. Let $\displaystyle d=d(S_{1})$,
then we have the following cases:

    1. (i)

If $\displaystyle d=1$; $\displaystyle\nu_{2}(b)\in\\{1,2,4,5,7,8\\}$, then
$\displaystyle 2\mathbb{Z}_{K}=\mathfrak{p}_{1}^{9}$ with residue degree
$\displaystyle 1$. Hence $\displaystyle\nu_{2}(i(K))=0$.

    2. (ii)

If $\displaystyle d=3$; $\displaystyle\nu_{2}(b)\in\\{3,6\\}$, then
$\displaystyle
R_{\lambda}(F)(y)=y^{3}+1=(y+1)(y^{2}+y+1)\in\mathbb{F}_{\phi}[y]$. Thus
$\displaystyle 2\mathbb{Z}_{K}=\mathfrak{p}_{11}^{3}\mathfrak{p}_{12}^{3}$
with $\displaystyle f_{11}=1$ and $\displaystyle f_{12}=2$. Hence
$\displaystyle\nu_{2}(i(K))=0$.

If $\displaystyle\nu_{2}(b)>\cfrac{9}{7}\nu_{2}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,\nu_{2}(b))$,
$\displaystyle(2,\nu_{2}(a))$, and $\displaystyle(9,0)$ with $\displaystyle
d(S_{1})\in\\{1,2\\}$ and $\displaystyle d(S_{2})=1$ since
$\displaystyle\nu_{2}(a)\leq 6$ by assumption.

    1. (i)

If $\displaystyle\nu_{2}(b)\not\equiv\nu_{2}(a)\ \mbox{\rm(mod }{2})$, then
$\displaystyle d(S_{1})=1$. Thus $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}^{7}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{2}(i(K))=0$.

    2. (ii)

If $\displaystyle\nu_{2}(b)=2k+\nu_{2}(a)$, then $\displaystyle d(S_{1})=2$
with $\displaystyle R_{\lambda_{1}}(F)(y)=(y+1)^{2}\in\mathbb{F}_{\phi}[y]$.
Let us replace $\displaystyle\phi$ by $\displaystyle g=x-2^{k}$. Then
$\displaystyle
F(x)=\cdots+(a+36(2^{7k}))g^{2}+(2^{k+1}a+9(2^{8k}))g+b+2^{2k}a+2^{9k}$. Then
$\displaystyle\nu_{2}(a+36(2^{7k}))=\nu_{2}(a)$.
If $\displaystyle k\geq 2$ or $\displaystyle k=1$ and
$\displaystyle\nu_{2}(a)\in\\{1,2,3,4\\}$, then
$\displaystyle\nu_{2}(2^{k+1}a+9(2^{8k}))=\nu_{2}(a)+k+1$, and
$\displaystyle\nu_{2}(b+2^{2k}a+2^{9k})\geq\nu_{2}(a)+2k+1$.

      1. (a)

If $\displaystyle b_{2}+a_{2}\equiv 2\ \mbox{\rm(mod }{4})$, then
$\displaystyle N_{g}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,2k+1+\nu_{2}(a))$, $\displaystyle(2,\nu_{2}(a))$, and
$\displaystyle(9,0)$. Thus the degree of each side is $\displaystyle 1$, and
so $\displaystyle 2\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}^{7}$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{2}(i(K))=0$.

      2. (b)

If $\displaystyle b_{2}+a_{2}\equiv 4\ \mbox{\rm(mod }{8})$, then
$\displaystyle N_{g}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,2k+2+\nu_{2}(a))$, $\displaystyle(2,\nu_{2}(a))$, and
$\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$ and $\displaystyle
R_{\lambda_{1}}(F)=y^{2}+y+1$, which is irreducible over
$\displaystyle\mathbb{F}_{g}$. Thus $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{7}$ with $\displaystyle
f_{1}=2$ and $\displaystyle f_{2}=1$. Hence $\displaystyle\nu_{2}(i(K))=0$.

      3. (c)

If $\displaystyle b_{2}+a_{2}\equiv 0\ \mbox{\rm(mod }{8})$, then
$\displaystyle N_{g}(F)=S_{1}+S_{2}+S_{3}$ has three sides joining
$\displaystyle(0,w)$, $\displaystyle(1,k+1+\nu_{2}(a))$,
$\displaystyle(2,\nu_{2}(a))$, and $\displaystyle(9,0)$ with $\displaystyle
w\geq 2k+3+\nu_{2}(a)$. Thus the degree of each side is $\displaystyle 1$, and
so $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{7}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence $\displaystyle 2$
divides $\displaystyle i(K)$. Applying [8, Corollary, p: 230], we get
$\displaystyle\nu_{2}(i(K))=1$.

If $\displaystyle k=1$ and $\displaystyle\nu_{2}(a)=5$; $\displaystyle a\equiv
32\ \mbox{\rm(mod }{64})$ and $\displaystyle b\equiv 128\ \mbox{\rm(mod
}{256})$, then $\displaystyle\nu_{2}(4a+9(2^{8})=7$, and
$\displaystyle\nu_{2}(b+4a+2^{9})\geq 8$.

      1. (a)

If $\displaystyle b_{2}+a_{2}\equiv 2\ \mbox{\rm(mod }{4})$, then
$\displaystyle N_{g}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,8)$, $\displaystyle(2,5)$, and $\displaystyle(9,0)$. Thus the
degree of each side is $\displaystyle 1$, and so $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}^{7}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{2}(i(K))=0$.

      2. (b)

If $\displaystyle b_{2}+a_{2}\equiv 4\ \mbox{\rm(mod }{8})$, then
$\displaystyle N_{g}(F)=S_{1}+S_{2}+S_{3}$ has three sides joining
$\displaystyle(0,w)$, $\displaystyle(1,7)$, $\displaystyle(2,5)$, and
$\displaystyle(9,0)$ with $\displaystyle w\geq 10$. Thus the degree of each
side is $\displaystyle 1$, and so $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{7}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence $\displaystyle 2$
divides $\displaystyle i(K)$. Applying [8, Corollary, p: 230], we get
$\displaystyle\nu_{2}(i(K))=1$.

      3. (c)

If $\displaystyle b_{2}+a_{2}\equiv 0\ \mbox{\rm(mod }{8})$, then
$\displaystyle N_{g}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,9)$, $\displaystyle(2,5)$, and $\displaystyle(9,0)$ with
$\displaystyle d(S_{2})=1$ and $\displaystyle R_{\lambda_{1}}(F)=y^{2}+y+1$,
which is irreducible over $\displaystyle\mathbb{F}_{g}$. Thus $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{7}$ with $\displaystyle
f_{1}=2$ and $\displaystyle f_{2}=1$. Hence $\displaystyle\nu_{2}(i(K))=0$.

If $\displaystyle k=1$ and $\displaystyle\nu_{2}(a)=6$; $\displaystyle a\equiv
64\ \mbox{\rm(mod }{128})$ and $\displaystyle b\equiv 256\ \mbox{\rm(mod
}{512})$, then $\displaystyle\nu_{2}(\Delta)\geq 65$. Let $\displaystyle
u=\cfrac{2\cdot 7^{3}a_{2}^{4}}{3^{8}b_{2}^{3}}$. Since
$\displaystyle\nu_{2}=(3^{8}b_{2}^{3})=0$, then $\displaystyle
u\in\mathbb{Z}_{2}$. Let $\displaystyle\phi_{1}=x-u$. Then $\displaystyle
F(x)=\cdots+(36u^{7}+a)\phi_{1}^{2}+A\phi_{1}+B$, where
$\displaystyle\nu_{2}(36u^{7}+a)=6$,

$\displaystyle\begin{array}[]{lllll}A&=&9u^{8}+2au&=&\cfrac{3^{7}a_{2}^{5}\Delta^{3}}{2^{184}\cdot
3^{62}b_{2}^{27}}-\cfrac{3^{7}a_{2}^{5}\Delta^{2}}{2^{120}\cdot
3^{43}b_{2}^{19}}+\cfrac{3^{7}a_{2}^{5}\Delta}{2^{56}\cdot
3^{25}b_{2}^{11}}~{}~{},\text{ and}\\\ &&&&\\\
B&=&u^{9}+au^{2}+b&=&\cfrac{\Delta^{4}}{2^{246}\cdot 3^{72}\cdot
7b_{2}^{31}}-\cfrac{\Delta^{3}}{2^{181}\cdot 3^{54}\cdot
7b_{2}^{23}}+\cfrac{\Delta^{2}}{2^{118}\cdot 3^{35}\cdot
7b_{2}^{15}}+\cfrac{\Delta}{2^{56}\cdot 3^{18}\cdot
7b_{2}^{7}}~{}~{}.\end{array}$

Thus $\displaystyle\nu_{2}(A)=\nu_{2}(B)=\nu_{2}(\Delta)-56$. It follows that
$\displaystyle N_{\phi_{1}}(F)=S_{1}+S_{2}$ has two side joining
$\displaystyle(0,\nu_{2}(\Delta)-56)$, $\displaystyle(2,6)$, and
$\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$. If
$\displaystyle\nu_{2}(\Delta)$ is odd, then $\displaystyle d(S_{1})=1$, and so
$\displaystyle 2\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}^{7}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{2}(i(K))=0$. If $\displaystyle\nu_{2}(\Delta)=2k+62$
$\displaystyle(k\geq 2)$, then $\displaystyle d(S_{1})=2$ with $\displaystyle
R_{\lambda_{1}}(F)(y)=(y+1)^{2}\in\mathbb{F}_{\phi_{1}}[y]$. Let us replace
$\displaystyle\phi_{1}$ by $\displaystyle x-(u+2^{k})$. Then $\displaystyle
F(x)=\cdots+A_{2}(x-(u+2^{k}))^{2}+A_{1}(x-(u+2^{k}))+A_{0}$, where
$\displaystyle\nu_{2}(A_{2})=6$,

$\displaystyle\begin{array}[]{lll}A_{1}&=&A+2^{k+1}a+2^{k}(72u^{7})+2^{2k}(252u^{6})+2^{3k}(504u^{5})+2^{4k}(630u^{4})+2^{5k}(504u^{3})\\\
&&+2^{6k}(252u^{2})+2^{7k}(72u)+2^{8k}(9)~{}~{},\text{ and}\\\
A_{0}&=&B+2^{k}A+2^{2k}(a+36u^{7})+2^{3k}(84u^{6})+2^{4k}(126u^{5})+2^{5k}(126u^{4})+2^{6k}(84u^{3})\\\
&&+2^{7k}(36u^{2})+2^{8k}(9u)+2^{9k}~{}~{}.\end{array}$

Thus $\displaystyle\nu_{2}(A_{1})=k+7$ and $\displaystyle\nu_{2}(A_{0})\geq
2k+7$. We have

$\displaystyle\begin{array}[]{lllll}A_{0}&\equiv&B+2^{2k}a\ \mbox{\rm(mod
}{2^{2k+9}})&\equiv&\cfrac{2^{2k+6}}{3^{18}\cdot
7b_{2}^{7}}\left(\Delta_{2}+3^{18}\cdot 7a_{2}b_{2}^{7}\right)\ \mbox{\rm(mod
}{2^{2k+9}}).\end{array}$

Since $\displaystyle\Delta_{2}+3^{18}\cdot
7a_{2}b_{2}^{7}\equiv\Delta_{2}-a_{2}b_{2}\ \mbox{\rm(mod }{8})$, then three
cases arise:

    3. -

If $\displaystyle\Delta_{2}\equiv a_{2}b_{2}+2\ \mbox{\rm(mod }{4})$, then
$\displaystyle\nu_{2}(A_{0})=2k+7$. Thus $\displaystyle
N_{x-(u+2^{k})}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,2k+7)$,
$\displaystyle(2,6)$, and $\displaystyle(9,0)$. It follows that $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}^{7}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{2}(i(K))=0$.

    4. -

If $\displaystyle\Delta_{2}\equiv a_{2}b_{2}+4\ \mbox{\rm(mod }{8})$, then
$\displaystyle\nu_{2}(A_{0})=2k+8$ . Thus $\displaystyle
N_{x-(u+2^{k})}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,2k+8)$,
$\displaystyle(2,6)$, and $\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$
and $\displaystyle R_{\lambda_{1}}(F)(y)=y^{2}+y+1$, which is irreducible over
$\displaystyle\mathbb{F}_{x-(u+2^{k})}$. It follows that $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{7}$ with $\displaystyle
f_{1}=2$ and $\displaystyle f_{2}=1$. Hence $\displaystyle\nu_{2}(i(K))=0$.

    5. -

If $\displaystyle\Delta_{2}\equiv a_{2}b_{2}\ \mbox{\rm(mod }{8})$, then
$\displaystyle\nu_{2}(A_{0})\geq 2k+9$. Thus $\displaystyle
N_{x-(u+2^{k})}(F)=S_{1}+S_{2}+S_{3}$ has three sides joining
$\displaystyle(0,\nu_{2}(A_{0}))$, $\displaystyle(1,k+7)$,
$\displaystyle(2,6)$, and $\displaystyle(9,0)$. It follows that $\displaystyle
2\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{7}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence $\displaystyle 2$
divides $\displaystyle i(K)$. Applying [8, Corollary, p: 230], we get
$\displaystyle\nu_{2}(i(K))=1$.

$\displaystyle\square$

Proof of Theorem 2.2.
If $\displaystyle\nu_{3}(ab)=0$, then since
$\displaystyle\Delta=b(3^{18}b^{7}+2^{2}\cdot 7^{7}a^{9})$ is the discriminant
of $\displaystyle F(x)$, thanks to the index formula $\displaystyle(1.1)$,
$\displaystyle\nu_{3}((\mathbb{Z}_{K}:\mathbb{Z}[\alpha]))=0$ and so
$\displaystyle\nu_{3}(i(K))=0$. Now assume that $\displaystyle 3$ divides
$\displaystyle ab$. Then we have the following cases:

  1. (1)

If $\displaystyle a\equiv-1\ \mbox{\rm(mod }{3})$ and
$\displaystyle\nu_{3}(b)\geq 1$, then $\displaystyle F(x)\equiv
x^{2}(x-1)(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1)\ \mbox{\rm(mod }{3})$. Thus
there are two prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 3$ with residue degrees $\displaystyle 1$ and $\displaystyle 6$
provided by $\displaystyle x-1$ and $\displaystyle
x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$ respectively. On the other hand,
$\displaystyle x$ can provides at most two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$ with residue
degree $\displaystyle 1$ each, or one prime ideal with residue degree
$\displaystyle 2$. Thus the possible factorizations of $\displaystyle
3\mathbb{Z}_{K}$ are
$\displaystyle\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}$ with
$\displaystyle f_{1}=2$, $\displaystyle f_{2}=1$, and $\displaystyle f_{3}=6$,
or $\displaystyle\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}\mathfrak{p}_{3}$ with
$\displaystyle f_{1}=f_{2}=1$ and $\displaystyle f_{3}=6$, or
$\displaystyle\mathfrak{p}_{11}\mathfrak{p}_{12}\mathfrak{p}_{21}\mathfrak{p}_{31}$
with $\displaystyle f_{11}=f_{12}=f_{21}=1$, and $\displaystyle f_{31}=6$. In
each case, $\displaystyle\nu_{3}(i(K))=0$.

  2. (2)

If $\displaystyle a\equiv 1\ \mbox{\rm(mod }{3})$ and
$\displaystyle\nu_{3}(b)\geq 1$, then $\displaystyle F(x)\equiv
x^{2}(x+1)(x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1)\ \mbox{\rm(mod }{3})$. Similar
to the case above, $\displaystyle\nu_{3}(i(K))=0$.

  3. (3)

If $\displaystyle\nu_{3}(a)\geq 1$ and $\displaystyle b\equiv-1\ \mbox{\rm(mod
}{3})$, then $\displaystyle F(x)\equiv(x-1)^{9}\ \mbox{\rm(mod }{3})$. Let
$\displaystyle\phi=x-1$. Then $\displaystyle
F(x)=\phi^{9}+9\phi^{8}+36\phi^{7}+84\phi^{6}+126\phi^{5}+126\phi^{4}+84\phi^{3}+(a+36)\phi^{2}+(2a+9)\phi+a+b+1$.

    1. (i)

If $\displaystyle(a,b)\in\\{(0,2),(0,5),(3,-1),(3,2),(6,-1),(6,5)\\}\
\mbox{\rm(mod }{9})$, then $\displaystyle N_{\phi}(F)=S_{1}$ has a single side
of height $\displaystyle 1$. Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{9}$ with residue degree $\displaystyle 1$.
Hence $\displaystyle\nu_{3}(i(K))=0$.

    2. (ii)

If $\displaystyle(a,b)\in\\{(3,5),(6,2)\\}\ \mbox{\rm(mod }{9})$, then
$\displaystyle N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining
$\displaystyle(0,w)$, $\displaystyle(1,1)$, and $\displaystyle(9,0)$ with
$\displaystyle w\geq 2$. Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{8}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{3}(i(K))=0$.

    3. (iii)

If $\displaystyle(a,b)\in\\{(0,8),(0,17),(9,-1),(9,8),(18,-1),(18,17)\\}\
\mbox{\rm(mod }{27})$, then $\displaystyle N_{\phi}(F)=S_{1}+S_{2}$ has two
sides joining $\displaystyle(0,2)$, $\displaystyle(3,1)$, and
$\displaystyle(9,0)$. Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{3}\mathfrak{p}_{2}^{6}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{3}(i(K))=0$.

    4. (iv)

If $\displaystyle(a,b)\in\\{(0,-1),(18,8)\\}\ \mbox{\rm(mod }{27})$, then
$\displaystyle N_{\phi}(F)=S_{1}+S_{2}+S_{3}$ has three sides joining
$\displaystyle(0,w)$, $\displaystyle(1,2)$, $\displaystyle(3,1)$, and
$\displaystyle(9,0)$ with $\displaystyle w\geq 3$. Thus the degree of each
side $\displaystyle 1$, and so $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{2}\mathfrak{p}_{3}^{6}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

    5. (v)

If $\displaystyle(a,b)\in\\{(9,17),(9,44),(36,17),(36,71),(63,44),(63,71)\\}\
\mbox{\rm(mod }{81})$, then $\displaystyle N_{\phi}(F)=S_{1}+S_{2}$ has two
sides joining $\displaystyle(0,3)$, $\displaystyle(3,1)$, and
$\displaystyle(9,0)$. Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{3}\mathfrak{p}_{2}^{6}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{3}(i(K))=0$.

    6. (vi)

If $\displaystyle(a,b)\equiv(9,71)\ \mbox{\rm(mod }{81})$, then we have the
following sub-cases:

      1. (a)

If $\displaystyle a+b+1\equiv 81\ \mbox{\rm(mod }{243})$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,4)$,
$\displaystyle(3,1)$, and $\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$
and $\displaystyle R_{\lambda_{1}}(F)(y)=y^{3}-y^{2}+y+1$, which is
irreducible over $\displaystyle\mathbb{F}_{\phi}$. It follows that
$\displaystyle 3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{6}$ with
$\displaystyle f_{1}=3$ and $\displaystyle f_{2}=1$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      2. (b)

If $\displaystyle a+b+1\equiv 162\ \mbox{\rm(mod }{243})$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,4)$,
$\displaystyle(3,1)$, and $\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$
and $\displaystyle
R_{\lambda_{1}}(F)(y)=y^{3}-y^{2}+y-1=(y-1)(y^{2}+1)\in\mathbb{F}_{\phi}[y]$.
Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{11}\mathfrak{p}_{12}\mathfrak{p}_{21}^{6}$ with
$\displaystyle f_{11}=f_{21}=1$ and $\displaystyle f_{12}=2$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      3. (c)

If $\displaystyle a+b+1\equiv 0\ \mbox{\rm(mod }{243})$, then
$\displaystyle\nu_{3}(\Delta)\geq 23$. Let $\displaystyle a_{3}=\cfrac{a}{9}$
and $\displaystyle u=\cfrac{2\cdot 7^{3}a_{3}^{4}}{b^{3}}\in\mathbb{Q}$. Since
$\displaystyle\nu_{3}(b^{3})=0$, then $\displaystyle u\in\mathbb{Z}_{3}$. Let
$\displaystyle\phi=x-u$. Then $\displaystyle
F(x)=\cdots+84u^{6}\phi^{3}+(36u^{7}+a)\phi_{2}^{2}+A\phi_{2}+B$, with
$\displaystyle\nu_{3}(36u^{7}+a)=2$,

$\displaystyle\begin{array}[]{lllllll}A&=&9u^{8}+2au&=&\cfrac{2^{2}\cdot
7^{3}a_{3}^{5}\Delta^{3}}{3^{52}b^{27}}-\cfrac{2^{2}\cdot
7^{3}a_{3}^{5}\Delta^{2}}{3^{33}b^{19}}+\cfrac{2^{2}\cdot
7^{3}a_{3}^{5}\Delta}{3^{15}b^{11}}~{}~{},{\text{and}}\\\ &&&&\\\
B&=&u^{9}+au^{2}+b&=&\cfrac{2\Delta^{4}}{3^{71}\cdot
7b^{31}}-\cfrac{8\Delta^{3}}{3^{54}\cdot
7b^{23}}+\cfrac{4\Delta^{2}}{3^{35}\cdot 7b^{15}}+\cfrac{\Delta}{3^{18}\cdot
7b^{7}}~{}~{}.\end{array}$

Obviously, $\displaystyle\nu_{3}(A)=\nu_{3}(\Delta)-15$ and
$\displaystyle\nu_{3}(B)=\nu_{3}(\Delta)-18$. It follows that $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}+S_{3}$ has three sides joining
$\displaystyle(\nu_{3}(\Delta)-18)$, $\displaystyle(2,2)$,
$\displaystyle(3,1)$, and $\displaystyle(9,0)$ with $\displaystyle
d(S_{2})=d(S_{3})=1$. If $\displaystyle\nu_{3}(\Delta)$ is odd, then
$\displaystyle d(S_{1})=1$, and so $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.
If $\displaystyle\nu_{3}(\Delta)$ is even, then $\displaystyle d(S_{1})=2$
with $\displaystyle R_{\lambda_{1}}(F)(y)=-y^{2}+B_{3}\in\mathbb{F}_{\phi}[y]$
since $\displaystyle(36u^{7}+a)/9\equiv-1\ \mbox{\rm(mod }{3})$.

      4. -

If $\displaystyle\Delta_{3}\equiv-b\ \mbox{\rm(mod }{3})$;
$\displaystyle\Delta\equiv 1\ \mbox{\rm(mod }{3})$, then $\displaystyle
R_{\lambda_{1}}(F)(y)=-y^{2}-1$, which is irreducible over
$\displaystyle\mathbb{F}_{\phi}$. Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
$\displaystyle f_{1}=2$ and $\displaystyle f_{2}=f_{3}=1$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      5. -

If $\displaystyle\Delta_{3}\equiv-1\ \mbox{\rm(mod }{3})$, then $\displaystyle
R_{\lambda_{1}}(F)(y)=-y^{2}+1=-(y-1)(y+1)\in\mathbb{F}_{\phi}[y]$. Thus
$\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{11}\mathfrak{p}_{12}\mathfrak{p}_{21}\mathfrak{p}_{31}^{6}$
with residue degree $\displaystyle 1$ each ideal factor. Hence $\displaystyle
3$ divides $\displaystyle i(K)$. Applying [8, Corollary, p: 230], we get
$\displaystyle\nu_{3}(i(K))=1$.

    7. (vii)

If $\displaystyle(a,b)\equiv(36,44)\ \mbox{\rm(mod }{81})$, then we have the
following cases:

      1. (a)

If $\displaystyle a+b+1\equiv 81\ \mbox{\rm(mod }{243})$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,4)$,
$\displaystyle(3,1)$, and $\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$
and $\displaystyle R_{\lambda_{1}}(F)(y)=y^{3}-y^{2}+1$, which is irreducible
over $\displaystyle\mathbb{F}_{\phi}$. It follows that $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{6}$ with $\displaystyle
f_{1}=3$ and $\displaystyle f_{2}=1$. Hence $\displaystyle\nu_{3}(i(K))=0$.

      2. (b)

If $\displaystyle a+b+1\equiv 162\ \mbox{\rm(mod }{243})$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,4)$,
$\displaystyle(3,1)$, and $\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$
and $\displaystyle
R_{\lambda_{1}}(F)(y)=y^{3}-y^{2}-1=(y+1)(y^{2}+y-1)\in\mathbb{F}_{\phi}[y]$.
It follows that $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{11}\mathfrak{p}_{12}\mathfrak{p}_{21}^{6}$ with
$\displaystyle f_{11}=f_{21}=1$ and $\displaystyle f_{12}=2$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      3. (c)

If $\displaystyle a+b+1\equiv 0\ \mbox{\rm(mod }{243})$, then
$\displaystyle\nu_{3}(\Delta)\geq 23$. Let $\displaystyle a_{3}=\cfrac{a}{9}$
and $\displaystyle u=\cfrac{b}{20a_{3}^{2}}\in\mathbb{Q}$. Similar to the case
(vi). (c) above, $\displaystyle 3$ divides $\displaystyle i(K)$ if and only if
$\displaystyle\nu_{3}(\Delta)$ is even and $\displaystyle\Delta_{3}\equiv-1\
\mbox{\rm(mod }{3})$. Also in this case, we have
$\displaystyle\nu_{3}(i(K))=1$.

    8. (viii)

If $\displaystyle(a,b)\equiv(63,17)\ \mbox{\rm(mod }{81})$, then we have the
following sub-cases:

      1. (a)

If $\displaystyle a+b+1\equiv 81\ \mbox{\rm(mod }{243})$, then
$\displaystyle\nu_{3}(\Delta)\geq 23$. Let $\displaystyle a_{3}=\cfrac{a}{9}$
and $\displaystyle u=\cfrac{b}{20a_{3}^{2}}\in\mathbb{Q}$. Let
$\displaystyle\phi=x-u$. Similar to the case (vi). (c) above, $\displaystyle
3$ divides $\displaystyle i(K)$ if and only if $\displaystyle\nu_{3}(\Delta)$
is even and $\displaystyle\Delta_{3}\equiv-1\ \mbox{\rm(mod }{3})$. Also in
this case, we have $\displaystyle\nu_{3}(i(K))=1$.

      2. (b)

If $\displaystyle a+b+1\equiv 162\ \mbox{\rm(mod }{243})$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,4)$,
$\displaystyle(3,1)$, and $\displaystyle(9,0)$ with $\displaystyle d(S_{2})=1$
and $\displaystyle R_{\lambda_{1}}(F)(y)=y^{3}-y^{2}-y-1$, which is
irreducible over $\displaystyle\mathbb{F}_{\phi}$. Thus $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{6}$ with $\displaystyle
f_{1}=3$ and $\displaystyle f_{12}=2$. Hence $\displaystyle\nu_{3}(i(K))=0$.

      3. (c)

If $\displaystyle a+b+1\equiv 0\ \mbox{\rm(mod }{243})$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}+S_{3}$ has three sides joining $\displaystyle(0,w)$,
$\displaystyle(1,3)$, $\displaystyle(3,1)$, and $\displaystyle(9,0)$ with
$\displaystyle w\geq 5$ and $\displaystyle R_{\lambda_{2}}(F)(y)=y^{2}-y-1$,
which is irreducible over $\displaystyle\mathbb{F}_{\phi}$. Thus
$\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
$\displaystyle f_{1}=f_{3}=1$ and $\displaystyle f_{2}=2$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

  4. (4)

If $\displaystyle\nu_{3}(a)\geq 1$ and $\displaystyle b\equiv 1\ \mbox{\rm(mod
}{3})$, then $\displaystyle F(x)\equiv(x+1)^{9}\ \mbox{\rm(mod }{3})$. Let
$\displaystyle\phi=x+1$. Then $\displaystyle
F(x)=\phi^{9}-9\phi^{8}+36\phi^{7}-84\phi^{6}+126\phi^{5}-126\phi^{4}+84\phi^{3}+(a-36)\phi^{2}+(-2a+9)\phi+a+b-1$.
The treatment of this case is similar to the case (3) above
$\displaystyle(\nu_{3}(a)\geq 1$ and $\displaystyle b\equiv-1\ \mbox{\rm(mod
}{3}))$. The following are the obtained results:

    1. (i)

If $\displaystyle(a,b)\in\\{(0,4),(0,7),(3,1),(3,4),(6,1),(6,7)\\}\
\mbox{\rm(mod }{9})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{9}$ with residue degree $\displaystyle 1$.
Hence $\displaystyle\nu_{3}(i(K))=0$.

    2. (ii)

If $\displaystyle(a,b)\in\\{(3,7),(6,4)\\}\ \mbox{\rm(mod }{9})$, then
$\displaystyle 3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{8}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

    3. (iii)

If $\displaystyle(a,b)\in\\{(0,10),(0,19),(9,1),(9,10),(18,1),(18,19)\\}\
\mbox{\rm(mod }{27})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{3}\mathfrak{p}_{2}^{6}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{3}(i(K))=0$.

    4. (iv)

If $\displaystyle(a,b)\in\\{(0,1),(9,19)\\}\ \mbox{\rm(mod }{27})$, then
$\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{2}\mathfrak{p}_{3}^{6}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

    5. (v)

If
$\displaystyle(a,b)\in\\{(18,10),(18,37),(45,10),(45,64),(72,37),(72,64)\\}\
\mbox{\rm(mod }{81})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{3}\mathfrak{p}_{2}^{6}$ with residue degree
$\displaystyle 1$ each ideal factor. Hence $\displaystyle\nu_{3}(i(K))=0$.

    6. (vi)

If $\displaystyle(a,b)\equiv(18,64)\ \mbox{\rm(mod }{81})$, then we have the
following sub-cases:

      1. (a)

If $\displaystyle a+b\equiv 82\ \mbox{\rm(mod }{243})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}^{6}$ with $\displaystyle
f_{1}=3$ and $\displaystyle f_{2}=1$. Hence $\displaystyle\nu_{3}(i(K))=0$.

      2. (b)

If $\displaystyle a+b\equiv 1\ \mbox{\rm(mod }{243})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
$\displaystyle f_{1}=f_{3}=1$ and $\displaystyle f_{2}=2$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      3. (c)

If $\displaystyle a+b\equiv 163\ \mbox{\rm(mod }{243})$ and
$\displaystyle\nu_{3}(\Delta)$ is odd, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      4. (d)

If $\displaystyle a+b\equiv 163\ \mbox{\rm(mod }{243})$,
$\displaystyle\nu_{3}(\Delta)$ is even, and $\displaystyle\Delta_{3}\equiv 1\
\mbox{\rm(mod }{3})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
$\displaystyle f_{1}=2$ and $\displaystyle f_{2}=f_{3}=1$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      5. (e)

If $\displaystyle a+b\equiv 163\ \mbox{\rm(mod }{243})$,
$\displaystyle\nu_{3}(\Delta)$ is even, and $\displaystyle\Delta_{3}\equiv-1\
\mbox{\rm(mod }{3})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}\mathfrak{p}_{3}^{6}$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=1$.

    7. (vii)

If $\displaystyle(a,b)\in\\{(45,37),(72,10)\\}\ \mbox{\rm(mod }{81})$, then we
have the following sub-cases:

      1. (a)

If $\displaystyle a+b\equiv 82\ \mbox{\rm(mod }{243})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
$\displaystyle f_{1}=f_{3}=1$ and $\displaystyle f_{2}=2$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      2. (b)

If $\displaystyle a+b\equiv 163\ \mbox{\rm(mod }{243})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{3}^{6}$ with $\displaystyle
f_{1}=3$ and $\displaystyle f_{2}=1$. Hence $\displaystyle\nu_{3}(i(K))=0$.

      3. (c)

If $\displaystyle a+b\equiv 1\ \mbox{\rm(mod }{243})$ and
$\displaystyle\nu_{3}(\Delta)$ is odd, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}^{2}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      4. (d)

If $\displaystyle a+b\equiv 1\ \mbox{\rm(mod }{243})$,
$\displaystyle\nu_{3}(\Delta)$ is even, and $\displaystyle\Delta_{3}\equiv 1\
\mbox{\rm(mod }{3})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}^{6}$ with
$\displaystyle f_{1}=2$ and $\displaystyle f_{2}=f_{3}=1$. Hence
$\displaystyle\nu_{3}(i(K))=0$.

      5. (e)

If $\displaystyle a+b\equiv 1\ \mbox{\rm(mod }{243})$,
$\displaystyle\nu_{3}(\Delta)$ is even, and $\displaystyle\Delta_{3}\equiv-1\
\mbox{\rm(mod }{3})$, then $\displaystyle
3\mathbb{Z}_{K}=\mathfrak{p}_{1}\mathfrak{p}_{2}\mathfrak{p}_{3}\mathfrak{p}_{3}^{6}$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=1$.

  5. (5)

If $\displaystyle\nu_{3}(a)\geq 1$ and $\displaystyle\nu_{3}(b)\geq 1$, then
$\displaystyle F(x)\equiv x^{9}\ \mbox{\rm(mod }{3})$. Let
$\displaystyle\phi=x$. By assumption, $\displaystyle\nu_{3}(a)\leq 6$ or
$\displaystyle\nu_{3}(b)\leq 8$.

    1. (i)

If $\displaystyle\nu_{3}(b)<\cfrac{9}{7}\nu_{3}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}$ has a single side joining $\displaystyle(0,\nu_{3}(b))$ and
$\displaystyle(9,0)$ with $\displaystyle d(S_{1})\in\\{1,3\\}$ since
$\displaystyle\nu_{3}(b)\leq 8$ by assumption. Thus there are at most three
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

    2. (ii)

If $\displaystyle\nu_{3}(b)>\cfrac{9}{7}\nu_{3}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,\nu_{3}(b))$,
$\displaystyle(2,\nu_{3}(a))$, and $\displaystyle(9,0)$ with $\displaystyle
d(S_{1})\in\\{1,2\\}$ and $\displaystyle d(S_{2})=1$ since
$\displaystyle\nu_{3}(a)\leq 6$ by assumption. Thus there are at most three
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 3$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{3}(i(K))=0$.

$\displaystyle\square$

Proof of Theorem 2.3.
For $\displaystyle p=5$. Since $\displaystyle\Delta=b(3^{18}b^{7}+2^{2}\cdot
7^{7}a^{9})$, then $\displaystyle\nu_{5}(\Delta)\geq 1$ if and only if
$\displaystyle\nu_{5}(b)\geq 1$ or
$\displaystyle(a,b)\in\\{(1,3),(2,4),(3,1),(4,2)\\}\ \mbox{\rm(mod }{5})$.

  1. (1)

If $\displaystyle(a,b)\in\\{(1,3),(2,4),(3,1),(4,2)\\}\ \mbox{\rm(mod }{5})$
or $\displaystyle\nu_{5}(b)\geq 1$ and $\displaystyle\nu_{5}(a)=0$, then one
can check easily that $\displaystyle
F(x)\equiv\phi_{i}^{2}\cdot\phi_{j}\cdot\phi_{k}\ \mbox{\rm(mod }{5})$ with
$\displaystyle\deg(\phi_{i})=\deg(\phi_{j})=1$,
$\displaystyle\deg(\phi_{k})=6$, and the polynomials $\displaystyle\phi_{i}$,
$\displaystyle\phi_{j}$, and $\displaystyle\phi_{k}$ are coprime irreducible
over $\displaystyle\mathbb{F}_{5}$ for every $\displaystyle i,j,k$. Thus there
are at most three prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above
$\displaystyle 5$ with residue degree $\displaystyle 1$ each ideal factor.
Hence $\displaystyle\nu_{5}(i(K))=0$.

  2. (2)

If $\displaystyle\nu_{5}(b)\geq 1$, and $\displaystyle\nu_{5}(a)\geq 1$ then
$\displaystyle F(x)\equiv x^{9}\ \mbox{\rm(mod }{5})$. Let
$\displaystyle\phi=x$. By assumption, $\displaystyle\nu_{5}(a)\leq 6$ or
$\displaystyle\nu_{5}(b)\leq 8$.

    1. (i)

If $\displaystyle\nu_{5}(b)<\cfrac{9}{7}\nu_{5}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}$ has a single side joining $\displaystyle(0,\nu_{5}(b))$ and
$\displaystyle(9,0)$ with $\displaystyle d(S_{1})\in\\{1,3\\}$ since
$\displaystyle\nu_{5}(b)\leq 8$ by assumption. Thus there are at most three
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 5$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{5}(i(K))=0$.

    2. (ii)

If $\displaystyle\nu_{5}(b)>\cfrac{9}{7}\nu_{5}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,\nu_{5}(b))$,
$\displaystyle(5,\nu_{5}(a))$, and $\displaystyle(9,0)$ with $\displaystyle
d(S_{1})\in\\{1,2\\}$ and $\displaystyle d(S_{2})=1$ since
$\displaystyle\nu_{5}(a)\leq 6$ by assumption. Thus there are at most three
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 5$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{5}(i(K))=0$.

For $\displaystyle p=7$. Since $\displaystyle\Delta=b(3^{18}b^{7}+2^{2}\cdot
7^{7}a^{9})$, thanks to the index formula, if $\displaystyle\nu_{7}(b)=0$,
then $\displaystyle\nu_{7}((\mathbb{Z}_{K}:\mathbb{Z}[\alpha]))=0$, and so
$\displaystyle\nu_{7}(i(K))=0$. Assume that $\displaystyle 7$ divides
$\displaystyle b$. Then we have the following cases:

  1. (1)

If $\displaystyle\nu_{7}(a)=0$, then $\displaystyle F(x)\equiv x^{2}(x+a)^{7}\
\mbox{\rm(mod }{7})$. Let $\displaystyle\phi_{1}=x$ and
$\displaystyle\phi_{2}=x+a$. Then

$\displaystyle\begin{array}[]{lll}F(x)&=&\phi_{1}^{9}+a\phi_{1}^{2}+b,\\\
&=&\cdots+36a^{2}\phi_{2}^{7}-84a^{3}\phi_{2}^{6}+126a^{4}\phi_{2}^{5}-126a^{5}\phi_{2}^{4}+84a^{6}\phi_{2}^{3}+(-36a^{7}+a)\phi_{2}^{2}\\\
&&+(-2a^{2}+9a^{8})\phi_{2}+b+a^{3}-a^{9}.\end{array}$

Obviously, $\displaystyle\phi_{1}$ can provides at most two prime ideals of
$\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 7$ with residue
degree $\displaystyle 1$ each ideal factor. On the other hand,
$\displaystyle\nu_{7}(84a^{6})=1$, then $\displaystyle\phi_{2}$ provides a
prime ideal $\displaystyle\mathfrak{p}$ of $\displaystyle\mathbb{Z}_{K}$ lying
above $\displaystyle 7$ with ramification index $\displaystyle
e(\mathfrak{p}/7)\geq 4$. Thus $\displaystyle\phi_{2}$ can provides at most
four prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle
7$ with residue degree $\displaystyle 1$ each ideal factor. We conclude that
there are at most six prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying
above $\displaystyle 7$ with residue degree $\displaystyle 1$ each ideal
factor. Hence $\displaystyle\nu_{7}(i(K))=0$.

  2. (2)

If $\displaystyle\nu_{7}(a)\geq 1$, then $\displaystyle F(x)\equiv x^{9}\
\mbox{\rm(mod }{7})$. Let $\displaystyle\phi=x$. By assumption,
$\displaystyle\nu_{7}(a)\leq 6$ or $\displaystyle\nu_{7}(b)\leq 8$.

    1. (i)

If $\displaystyle\nu_{7}(b)<\cfrac{9}{7}\nu_{7}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}$ has a single side joining $\displaystyle(0,\nu_{7}(b))$ and
$\displaystyle(9,0)$ with $\displaystyle d(S_{1})\in\\{1,3\\}$ since
$\displaystyle\nu_{7}(b)\leq 8$ by assumption. Thus there are at most three
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 7$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{7}(i(K))=0$.

    2. (ii)

If $\displaystyle\nu_{7}(b)>\cfrac{9}{7}\nu_{7}(a)$, then $\displaystyle
N_{\phi}(F)=S_{1}+S_{2}$ has two sides joining $\displaystyle(0,\nu_{7}(b))$,
$\displaystyle(2,\nu_{7}(a))$, and $\displaystyle(9,0)$ with $\displaystyle
d(S_{1})\in\\{1,2\\}$ and $\displaystyle d(S_{2})=1$ since
$\displaystyle\nu_{7}(a)\leq 6$ by assumption. Thus there are at most three
prime ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle 7$
with residue degree $\displaystyle 1$ each ideal factor. Hence
$\displaystyle\nu_{7}(i(K))=0$.

For $\displaystyle p\geq 11$, since there is at most $\displaystyle 9$ prime
ideals of $\displaystyle\mathbb{Z}_{K}$ lying above $\displaystyle p$ with
residue degree $\displaystyle 1$ each ideal factor, and there is at least
$\displaystyle p\geq 11$ monic irreducible polynomial of degree $\displaystyle
f$ in $\displaystyle\mathbb{F}_{p}[x]$ for every positive integer
$\displaystyle f$, we conclude that $\displaystyle p$ does not divide
$\displaystyle i(K)$.

$\displaystyle\square$

##  5\. Examples

Let $\displaystyle F(x)=x^{9}+ax^{2}+b\in\mathbb{Z}[x]$ be a monic irreducible
polynomial and $\displaystyle K$ the nonic number field generated by a complex
root of $\displaystyle F(x)$.

  1. (1)

For $\displaystyle a=54$ and $\displaystyle b=87$, we have
$\displaystyle\nu_{2}(b)=0$, then by Theorem 2.1,
$\displaystyle\nu_{2}(i(K))=0$. On the other hand,
$\displaystyle(a,b)\equiv(54,6)\ \mbox{\rm(mod }{81})$, then by Theorem 2.2,
$\displaystyle\nu_{3}(i(K))=0$. We conclude that $\displaystyle i(K)=1$.

  2. (2)

For $\displaystyle a=64$ and $\displaystyle b=256$, we have $\displaystyle
a\equiv 64\ \mbox{\rm(mod }{128})$, $\displaystyle b\equiv 256\ \mbox{\rm(mod
}{256})$, $\displaystyle\nu_{2}(\Delta)=70$, and
$\displaystyle\Delta_{2}\equiv a_{2}b_{2}\equiv 1\ \mbox{\rm(mod }{8})$, then
by Theorem 2.1, $\displaystyle\nu_{2}(i(K))=1$. On the other hand,
$\displaystyle(a,b)\equiv(64,13)\ \mbox{\rm(mod }{81})$, then by Theorem 2.2,
$\displaystyle\nu_{3}(i(K))=0$. Hence $\displaystyle i(K)=2$.

  3. (3)

For $\displaystyle a=90$ and $\displaystyle b=19835$, we have
$\displaystyle\nu_{2}(b)=0$, by Theorem 2.1, $\displaystyle\nu_{2}(i(K))=0$.
On the other hand, $\displaystyle(a,b)\equiv(9,71)\ \mbox{\rm(mod }{81})$,
$\displaystyle a+b\equiv-\ \mbox{\rm(mod }{243})$,
$\displaystyle\nu_{3}(\Delta)=26$, and $\displaystyle\Delta_{3}\equiv-1\
\mbox{\rm(mod }{3})$, by Theorem 2.2, $\displaystyle\nu_{3}(i(K))=1$. Hence
$\displaystyle i(K)=3$.

  4. (4)

For $\displaystyle a=99$ and $\displaystyle b=8055028$, we have
$\displaystyle\nu_{2}(a)=0$, $\displaystyle\nu_{2}(b)=2$, and $\displaystyle
b_{2}\equiv-a\equiv 5\ \mbox{\rm(mod }{8})$, then by Theorem 2.1,
$\displaystyle\nu_{2}(i(K))=1$. On the other hand,
$\displaystyle(a,b)\equiv(9,50)\ \mbox{\rm(mod }{81})$, $\displaystyle
a+b\equiv 163\ \mbox{\rm(mod }{243})$, $\displaystyle\nu_{3}(\Delta)=28$, and
$\displaystyle\Delta_{3}\equiv-1\ \mbox{\rm(mod }{3})$, then by Theorem 2.2,
$\displaystyle\nu_{3}(i(K))=1$. Hence $\displaystyle i(K)=6$.

Conflict of interest
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Author Contribution and Funding Statement
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