On Indices of Septic Number Fields Defined by Trinomials x⁷ + ax + b

Abstract

Let ${\displaystyle K}$ be a septic number field generated by a root, ${\displaystyle \alpha }$, of an irreducible trinomial, ${\displaystyle x^7+ax+b\in \mathbb{Z}\left[x\right]}$. In this paper, for every prime integer, ${\displaystyle p}$, we calculate ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$; the highest power of ${\displaystyle p}$ dividing the index, ${\displaystyle i\left(K\right)}$, of the number field, ${\displaystyle K}$. In particular, we calculate the index, ${\displaystyle i\left(K\right)}$. In application, when the index of ${\displaystyle K}$ is not trivial, then ${\displaystyle K}$ is not monogenic.

1. Introduction

Let ${\displaystyle K}$ be a number field of degree ${\displaystyle n}$ and ${\displaystyle {\mathbb{Z}}_K}$ its ring of integers. It is well known that ${\displaystyle {\mathbb{Z}}_K}$ is a free abelian group of rank ${\displaystyle n}$, and so, for every primitive integral element, ${\displaystyle \theta \in {\mathbb{Z}}_K}$, of ${\displaystyle K}$, the quotient group, ${\displaystyle {\mathbb{Z}}_K/\mathbb{Z}\left[\theta \right]}$, is finite. Let ${\displaystyle \theta \in {\mathbb{Z}}_K}$ be a primitive element of ${\displaystyle K}$ and ${\displaystyle ind\left(\theta \right)=({\mathbb{Z}}_K:\mathbb{Z}\left[\theta \right])}$ the cardinal order of ${\displaystyle {\mathbb{Z}}_K/\mathbb{Z}\left[\theta \right]}$. The index ${\displaystyle ind\left(\theta \right)}$ is called the index of ${\displaystyle \theta }$ . The index of the number field, ${\displaystyle K}$, denoted ${\displaystyle i\left(K\right)}$, is the greatest common divisor of all indices of all integral primitive elements of ${\displaystyle K}$. Namely, ${\displaystyle i\left(K\right)=GCD(({\mathbb{Z}}_K:\mathbb{Z}\left[\theta \right])\mid K=\mathbb{Q}\left(\theta \right) \mathrm{and} \theta \in {\mathbb{Z}}_K)}$. A rational prime integer ${\displaystyle p}$ dividing ${\displaystyle i\left(K\right)}$ is called a prime common index divisor of ${\displaystyle K}$. If, for some ${\displaystyle \theta \in {\mathbb{Z}}_K}$, ${\displaystyle (1,\theta ,\dots ,{\theta }^{n-1})}$ is a ${\displaystyle \mathbb{Z}}$ basis of ${\displaystyle {\mathbb{Z}}_K}$, then the number field, ${\displaystyle K}$, is said to be monogenic and ${\displaystyle \theta }$ is a generator of a power integral basis of ${\displaystyle K}$. Otherwise, the number field, ${\displaystyle K}$, is said to be non-monogenic. Remark that, if ${\displaystyle K}$ is monogenic, then ${\displaystyle i\left(K\right)=1}$. It follows that, if a number field has a prime common index divisor, then it is non-monogenic. Monogenity of number fields is a classical problem of algebraic number fields, going back to Dedekind, Hasse and Hensel; see for instance [1,2,3] for the present state of this area. Let ${\displaystyle \theta \in {\mathbb{Z}}_K}$ be a primitive element of ${\displaystyle K}$, ${\displaystyle d_K}$ the absolute discriminant of ${\displaystyle K}$, and ${\displaystyle ▵\left(\theta \right)}$ the discriminant of the minimal polynomial of ${\displaystyle \theta }$ over ${\displaystyle \mathbb{Q}}$. A well-known relation linking the discriminant and the index says that:

$$|▵\left(\theta \right)|=ind{\left(\theta \right)}^2·\left|d_K\right|$$

(see [3]).

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

The problem of studying monogenity of a number field and constructing all possible generators of power integral bases have been intensively studied during the last decades, mainly by Gaál, Nakahara, Pohst, and their collaborators (see for instance [4,5,6]). In 1871, Dedekind gave an example of a number field with a non-trivial index. He considered the cubic number field, ${\displaystyle K}$, generated by a root of ${\displaystyle x^3-x^2-2x-8}$, and he showed that the rational prime integer ${\displaystyle 2}$ splits completely in ${\displaystyle K}$ ([7], § 5, page 30). In ${\displaystyle 1930}$, for every number field, ${\displaystyle K}$, of a degree less than ${\displaystyle 8}$ and for every prime integer ${\displaystyle p\le n}$, Engstrom linked the ${\displaystyle p}$-index, ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$, of the number field, ${\displaystyle K}$, and the prime ideal factorization of ${\displaystyle p{\mathbb{Z}}_K}$ [8]. He gave an an explicit formula linking the value of ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$ and the factorization of ${\displaystyle p{\mathbb{Z}}_K}$ into powers of prime ideals of ${\displaystyle K}$ for every positive rational prime integer ${\displaystyle p\le n}$. This motivated Narkiewicz to ask a very important question, stated as problem 22 in Narkiewicz’s book ([9], Problem 22), which asks for an explicit formula of the highest power, ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$, for a given rational prime ${\displaystyle p}$ dividing ${\displaystyle i\left(K\right)}$. In [10], Nakahara characterized the index of abelian biquadratic number fields. Also in [11], Gaál et al. studied the indices of biquadratic number fields. Recently, many authors have focused on monogenity of number fields defined by trinomials. Based on the calculation of the ${\displaystyle p}$-index form of ${\displaystyle K}$ using ${\displaystyle p}$-integral bases of ${\displaystyle K}$ given in [12], Davis and Spearman [13] characterized the index of any quartic number field, ${\displaystyle K}$, generated by a root of an irreducible quartic irreducible trinomial, ${\displaystyle F\left(x\right)=x^4+ax+b\in \mathbb{Z}\left[x\right]}$. In [14], based on Newton’s polygon techniques, we reformulated Davis’ and Spearman’s results. In [15], El Fadil and Gaál characterized the index of quartic number fields, ${\displaystyle K}$, generated by a root of a quadratic irreducible trinomial of the form ${\displaystyle F\left(x\right)=x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]}$. They gave necessary and sufficient conditions on ${\displaystyle a}$ and ${\displaystyle b}$, which characterize when a prime, ${\displaystyle p\in \{2,3\}}$, is a common index divisor of ${\displaystyle K}$. In [16], for a sextic number field, ${\displaystyle K}$, defined by a trinomial, ${\displaystyle F\left(x\right)=x^6+a{x}^3+b\in \mathbb{Z}\left[x\right]}$, Gaál calculated all possible generators of power integral bases of ${\displaystyle K}$. In [17], we extended Gaál’s studies by providing all cases where ${\displaystyle K}$ is not monogenic. Also in [18], we characterized when a prime integer, ${\displaystyle p\in \{2,3\}}$, is a common index divisor of ${\displaystyle K}$, where ${\displaystyle K}$ is a number field defined by an irreducible trinomial, ${\displaystyle F\left(x\right)=x^5+a{x}^2+b\in \mathbb{Z}\left[x\right]}$. In [19], for a number field defined by an irreducible trinomial, ${\displaystyle F\left(x\right)=x^n+a{x}^m+b\in \mathbb{Z}\left[x\right]}$, we provided some cases where ${\displaystyle i\left(K\right)}$ is not trivial, and so ${\displaystyle K}$ is not monogenic. In this paper, for a septic number field generated by a root of an irreducible trinomial, ${\displaystyle F\left(x\right)=x^7+ax+b\in \mathbb{Z}\left[x\right]}$, and for every prime integer, ${\displaystyle p}$, we calculate ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$, the highest power of ${\displaystyle p}$ dividing the index ${\displaystyle i\left(K\right)}$ of the septic field, ${\displaystyle K}$. Our study key is based on prime ideal factorization in number fields, which is carried out in [20,21] and in the thesis of Montes (1999). The author warmly thanks Professor Enric Nart who provided him with a copy of Montes’ thesis.

2. Main Results

Let ${\displaystyle K}$ be a septic number field generated by a root, ${\displaystyle \alpha }$, of an irreducible trinomial, ${\displaystyle F\left(x\right)=x^7+ax+b\in \mathbb{Z}\left[x\right]}$. Without loss of generality, we can assume that for every rational prime integer, ${\displaystyle p}$, ${\displaystyle {\nu }_p\left(a\right)\le 5}$ or ${\displaystyle {\nu }_p\left(b\right)\le 6}$. Let ${\displaystyle a\in \mathbb{Z}}$ be a rational integer and a prime integer, ${\displaystyle p}$. Along this paper, let ${\displaystyle a_p=\frac{a}{p^{{\nu }_p\left(a\right)}}}$.

Our first main result characterizes when is ${\displaystyle \mathbb{Z}\left[\alpha \right]}$ the ring of integers of ${\displaystyle K}$?

Theorem 1.

The ring, ${\displaystyle \mathbb{Z}\left[\alpha \right]}$, is the ring of integers of ${\displaystyle K}$ if and only if every prime integer, ${\displaystyle p}$, satisfies one of the following conditions:

1. 

If ${\displaystyle p|a}$ and ${\displaystyle p|b}$, then ${\displaystyle {\nu }_p\left(b\right)=1}$.

2. 

If ${\displaystyle p=2}$, ${\displaystyle p}$ divides ${\displaystyle b}$ and does not divide ${\displaystyle a}$, then ${\displaystyle (a,b)\in \left\{\right(1,0),(3,2\left)\right\}(\mathrm{mod}4)}$.

3. 

If ${\displaystyle p=3}$, ${\displaystyle p}$ divides ${\displaystyle b}$ and ${\displaystyle a\equiv -1(\mathrm{mod}3)}$, then ${\displaystyle (a,b)\in \left\{\right(2,0),(8,3),(8,6\left)\right\}(\mathrm{mod}9)}$.

4. 

If ${\displaystyle p=3}$, ${\displaystyle p}$ divides ${\displaystyle b}$ and ${\displaystyle a\equiv 1(\mathrm{mod}3)}$, then ${\displaystyle (a,b) \equiv (1,0)(\mathrm{mod}9)}$.

5. 

If ${\displaystyle p=7}$, ${\displaystyle p}$ divides ${\displaystyle a}$ and does not divide ${\displaystyle b}$, then ${\displaystyle {\nu }_7(1-a-b^6)=1}$.

6. 

If ${\displaystyle p\notin \{2,3,7\}}$, ${\displaystyle p}$ does not divide both ${\displaystyle a}$ and ${\displaystyle b}$, then ${\displaystyle {\nu }_p(7^7{b}^6+6^6{a}^7)\le 1}$.

The following is an example of a monogenic septic number field defined by a non-monogenic irreducible trinomial.

Example 1.

Let ${\displaystyle K}$ be the number field generated by a root, ${\displaystyle \alpha }$, of ${\displaystyle F\left(x\right)=x^7+16x+16\in \mathbb{Z}\left[x\right]}$. Then ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$, which is a non-monogenic polynomial, and ${\displaystyle K}$ is a monogenic number field.

Indeed, since ${\displaystyle ▵\left(F\right)=-2^{24}·977·1607}$ is the discriminant of ${\displaystyle F\left(x\right)}$, and ${\displaystyle \overline{F\left(x\right)}=x^7}$ in ${\displaystyle {\mathbb{F}}_2\left[x\right]}$, then according to the notations of Section 3, let ${\displaystyle \varphi =x}$ and ${\displaystyle N_{\varphi }\left(F\right)}$ be the ${\displaystyle \varphi }$-Newton polygon of ${\displaystyle F\left(x\right)}$ with respect to ${\displaystyle p=2}$. Then, ${\displaystyle N_{\varphi }\left(F\right)=S}$ has a single side of degree ${\displaystyle 1}$. Thus by Remark 2 (2) of [22], ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$. Let ${\displaystyle \theta }$ be a root of ${\displaystyle F\left(x\right)}$. Since the height ${\displaystyle h\left(S\right)=4}$, by Theorem 5, ${\displaystyle {\nu }_2\left(ind\left(\theta \right)\right)\ge in{d}_{\varphi }\left(F\right)=9}$. Therefore, ${\displaystyle F\left(x\right)}$ is a non-monogenic polynomial. Let ${\displaystyle \alpha =\frac{{\theta }^2}{2}}$. Then ${\displaystyle G\left(x\right)=x^7+4x^4+4x-2}$ is the minimal polynomial of ${\displaystyle \alpha }$ over ${\displaystyle \mathbb{Q}}$. So ${\displaystyle \alpha }$ is a primitive integral element of ${\displaystyle K}$. Therefore, in order to show that ${\displaystyle {\mathbb{Z}}_K=\mathbb{Z}\left[\alpha \right]}$, we need to show that ${\displaystyle {\nu }_2\left(ind\left(\alpha \right)\right)=0}$. Since ${\displaystyle ▵\left(F\right)=-2^6·977·1607}$ is the discriminant of ${\displaystyle G\left(x\right)}$, then thanks to the formula

$$|▵\left(\theta \right)|=ind{\left(\theta \right)}^2·\left|d_K\right|,$$

the unique prime candidate to divide ${\displaystyle ind\left(\alpha \right)}$ is ${\displaystyle 2}$. Let ${\displaystyle N_{\varphi }\left(G\right)}$ be the ${\displaystyle \varphi }$-Newton polygon of ${\displaystyle G\left(x\right)}$ with respect to ${\displaystyle p=2}$. Then ${\displaystyle N_{\varphi }\left(G\right)=T}$ has a single side of height ${\displaystyle 1}$, by Theorem 6, ${\displaystyle {\nu }_2\left(ind\left(\alpha \right)\right)=0}$. Thus ${\displaystyle K}$ is monogenic and ${\displaystyle \alpha }$ generates a power integral basis of ${\displaystyle K}$.

For every ${\displaystyle (a,b)\in {\mathbb{Z}}^2}$, such that ${\displaystyle x^7+ax+b}$ is irreducible over ${\displaystyle \mathbb{Q}}$, let ${\displaystyle ▵=-(6^6{a}^7+7^7{b}^6)}$ be the discriminant of ${\displaystyle F\left(x\right)}$. Let ${\displaystyle p}$ be a prime integer and ${\displaystyle {▵}_p=\frac{▵}{p^{{\nu }_p(▵)}}}$. In the remainder of this section, for every prime integer, ${\displaystyle p}$, and for every value of ${\displaystyle a}$ and ${\displaystyle b}$, we calculate ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$.

Theorem 2.

The following table provides the value of ${\displaystyle {\nu }_2\left(i\left(K\right)\right)}$.

$$\begin{gathered} \begin{array}{|cc|}\hline conditions& {\nu }_2\left(i\left(K\right)\right)\\ a\equiv 28(\mathrm{mod}32) and b\equiv 0(\mathrm{mod}32)& 1\\ a\equiv 112(\mathrm{mod}128) and b\equiv 0(\mathrm{mod}128)& 1\\ a\equiv 1(\mathrm{mod}8) and b\equiv 2(\mathrm{mod}4) \\ {\nu }_2(▵)even and {▵}_2\equiv 3(\mathrm{mod}4)& 1\\ a\equiv 3(\mathrm{mod}8) and b\equiv 4(\mathrm{mod}8)& 1\\ a\equiv 3(\mathrm{mod}4) and b\equiv 0(\mathrm{mod}8)& 3\\ (a,b)\in \left\{\right(5,2),(5,6),(13,2),(13,14\left)\right\}(\mathrm{mod}16)& 1\\ Otherwise& 0\\ \hline\end{array} \end{gathered}$$

Theorem 3.

The following table provides the value of ${\displaystyle {\nu }_3\left(i\left(K\right)\right)}$.

$$\begin{gathered} \begin{array}{|cc|}\hline conditions& {\nu }_3\left(i\left(K\right)\right)\\ a\equiv 5(\mathrm{mod}9) and b\in \{3,6\left\}(\mathrm{mod}9\right)& 1\\ a\equiv 8(\mathrm{mod}9) and b\equiv 0(\mathrm{mod}9)& 2\\ a\equiv 2(\mathrm{mod}9) and b\in \{3,6\left\}(\mathrm{mod}9\right) \\ {\nu }_3(▵)=2k and k\ge 5& 1\\ a\equiv 2(\mathrm{mod}9) and b\in \{3,6\left\}(\mathrm{mod}9\right) \\ {\nu }_3(▵)=2k+1,k\ge 5 and {▵}_3\equiv 1(\mathrm{mod}3)& 2\\ Otherwise& 0\\ \hline\end{array} \end{gathered}$$

Theorem 4.

Let ${\displaystyle p\ge 5}$ be a prime integer and ${\displaystyle (a,b)\in {\mathbb{Z}}^2}$, such that ${\displaystyle F\left(x\right)=x^7+ax+b}$ is irreducible over ${\displaystyle \mathbb{Q}}$. Let ${\displaystyle K}$ be the number field generated by a root, ${\displaystyle \alpha }$, of ${\displaystyle F\left(x\right)}$. Then ${\displaystyle p}$ does not divide ${\displaystyle i\left(K\right)}$.

Corollary 1.

Let ${\displaystyle (a,b)\in {\mathbb{Z}}^2}$, such that ${\displaystyle F\left(x\right)=x^7+ax+b}$ is an irreducible polynomial over ${\displaystyle \mathbb{Q}}$. Then ${\displaystyle i\left(K\right)\in \{1,2,3,6,8,9,18,24,72\}}$.

Remark 1.

(1) 

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

(2) 

The unique method that characterises the monogenity of ${\displaystyle K}$ is to calculate the solutions of the index form equation of ${\displaystyle K}$ (see for instance [3,23]).

3. A Short Introduction to Newton’s Polygon Techniques Applied on Prime Ideal Factorization

For a number field, ${\displaystyle K}$, generated by a root of a monic irreducible polynomial, ${\displaystyle F\left(x\right)\in \mathbb{Q}\left[x\right]}$, and for every prime integer, ${\displaystyle p}$, in 1894, Hensel developed a powerful technique by establishing a one–one correspondence between maximal ideals of ${\displaystyle {\mathbb{Z}}_K}$ containing ${\displaystyle p}$ and monic irreducible factors of ${\displaystyle F\left(x\right)}$ with a coefficient in ${\displaystyle {\mathbb{Q}}_p}$. For every prime ideal corresponding to any irreducible factor, 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 ${\displaystyle p{\mathbb{Z}}_K}$, one needs to factorize ${\displaystyle F\left(x\right)}$ in ${\displaystyle {\mathbb{Q}}_p\left[x\right]}$. Newton’s polygon techniques can be used to refine the factorization of ${\displaystyle F\left(x\right)}$. 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 [25] and Guardia, Montes, and Nart’s paper [26]. 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 ${\displaystyle F\left(x\right)\in {\mathbb{Z}}_p\left[x\right]}$ be a monic polynomial and ${\displaystyle \varphi \in {\mathbb{Z}}_p\left[x\right]}$ a monic lift of an irreducible factor of ${\displaystyle \overline{F\left(x\right)}}$ modulo ${\displaystyle p}$. Let ${\displaystyle F\left(x\right)=a_0\left(x\right)+a_1\left(x\right)\varphi \left(x\right)+\cdots +a_n\left(x\right)\varphi {\left(x\right)}^l}$ be the ${\displaystyle \varphi }$-expansion of ${\displaystyle F\left(x\right)}$, ${\displaystyle N_{\varphi }\left(F\right)}$ the ${\displaystyle \varphi }$-Newton polygon of ${\displaystyle F\left(x\right)}$, and ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ its principal part. Let ${\displaystyle {\mathbb{F}}_{\varphi }}$ be the field ${\displaystyle {\mathbb{F}}_p\left[x\right]/\left(\overline{\varphi }\right)}$. For every side, ${\displaystyle S}$, of ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ with length ${\displaystyle l}$ and initial point ${\displaystyle (s,u_s)}$, for every ${\displaystyle i=0,\dots ,l}$, let ${\displaystyle c_i\in {\mathbb{F}}_{\varphi }}$ be the residue coefficient, defined as follows:

$$c_i=\left\{\begin{array}{cc}0,\hfill & \mathrm{if} (s+i,{\mathit{u}}_{\mathit{s}+\mathit{i}})\mathrm{lies} \mathrm{strictly} \mathrm{above} S,\hfill \\ \left({\displaystyle \frac{a_{s+i}\left(x\right)}{p^{{\mathit{u}}_{\mathit{s}+\mathit{i}}}}}\right)mod(p,\varphi \left(x\right)),\hfill & \mathrm{if} (s+i,{\mathit{u}}_{\mathit{s}+\mathit{i}})\mathrm{lies} \mathrm{on} S.\hfill \end{array}\right.$$

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 ${\displaystyle R_1\left(F\right)\left(y\right)=t_d{y}^d+t_{d-1}{y}^{d-1}+\cdots +t_{1}y+t_0\in {\mathbb{F}}_{\varphi }\left[y\right]}$ be the residual polynomial of ${\displaystyle F\left(x\right)}$ attached to the side ${\displaystyle S}$, where for every ${\displaystyle i=0,\cdots ,d}$, ${\displaystyle t_i=c_{ie}}$. If ${\displaystyle R_1\left(F\right)\left(y\right)}$ is square free for each side of the polygon, ${\displaystyle N_{\varphi }^{+}\left(F\right)}$, then we say that ${\displaystyle F\left(x\right)}$ is ${\displaystyle \varphi }$-regular.

Let ${\displaystyle \overline{F\left(x\right)}=\prod _{i=1}^r{\overline{{\varphi }_i}}^{l_i}}$ be the factorization of ${\displaystyle F\left(x\right)}$ into powers of monic irreducible coprime polynomials over ${\displaystyle {\mathbb{F}}_p}$, and we say that the polynomial ${\displaystyle F\left(x\right)}$ is ${\displaystyle p}$-regular if ${\displaystyle F\left(x\right)}$ is a ${\displaystyle {\varphi }_i}$-regular polynomial with respect to ${\displaystyle p}$ for every ${\displaystyle i=1,\cdots ,r}$. Let ${\displaystyle N_{{\varphi }_i}^{+}\left(F\right)=S_{i1}+\cdots +S_{i{r}_i}}$ be the ${\displaystyle {\varphi }_i}$-principal Newton polygon of ${\displaystyle F\left(x\right)}$ with respect to ${\displaystyle p}$. For every ${\displaystyle j=1,\dots ,r_i}$, let ${\displaystyle R_{1_{ij}}\left(F\right)\left(y\right)=\prod _{s=1}^{s_{ij}}{\psi }_{ijs}^{a_{ijs}}\left(y\right)}$ be the factorization of ${\displaystyle R_{1_{ij}}\left(F\right)\left(y\right)}$ in ${\displaystyle {\mathbb{F}}_{{\varphi }_i}\left[y\right]}$, where ${\displaystyle R_{1_{ij}}\left(F\right)\left(y\right)}$ is the residual polynomial of ${\displaystyle F\left(x\right)}$ attached to the side ${\displaystyle S_{ij}}$. Then we have the following theorem of index of Ore:

Theorem 5

(Theorems 1.7 and 1.9 of [12]).

1. 

$${\nu }_p\left(({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])\right)\ge \sum _{i=1}^r{ind}_{{\varphi }_i}\left(F\right).$$

The equality holds if ${\displaystyle F\left(x\right)isp}$-regular.

2. 

If ${\displaystyle F\left(x\right)isp}$-regular, then

$$p{\mathbb{Z}}_K=\prod _{i=1}^r\prod _{j=1}^{t_i}\prod _{s=1}^{s_{ij}}{\mathfrak{p}}_{ijs}^{e_{ij}},$$

where ${\displaystyle {\mathfrak{p}}_{ijs}}$ is a prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle p}$, ${\displaystyle e_{ij}}$ 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 ${\displaystyle f_{ijs}=deg\left({\varphi }_i\right)\times deg\left({\psi }_{ijs}\right)}$ for every ${\displaystyle (i,j,s)}$.

The Dedekind’s criterion can be reformulated as follows:

Theorem 6

(Theorem 1.1 of [27]). Let ${\displaystyle \overline{F\left(x\right)}=\prod _{i=1}^r{\overline{{\varphi }_i}}^{l_i}}$ be the factorization of ${\displaystyle F\left(x\right)}$ into powers of monic irreducible coprime polynomials over ${\displaystyle {\mathbb{F}}_p}$. For every ${\displaystyle i=1,\cdots ,r}$, let ${\displaystyle R_i\left(x\right)}$ be the remainder of the Euclidean division of ${\displaystyle F\left(x\right)}$ by ${\displaystyle {\varphi }_i\left(x\right)}$. Then ${\displaystyle {\nu }_p\left(ind\left(\alpha \right)\right)=0}$ if and only if ${\displaystyle l_i=1}$ or ${\displaystyle {\nu }_p\left(R_i\left(x\right)\right)=1}$ for every ${\displaystyle i=1,\dots ,r}$.

When the theorem of Ore fails, that is, ${\displaystyle F\left(x\right)}$ is not ${\displaystyle p}$-regular, then in order to complete the factorization of ${\displaystyle F\left(x\right)}$, Guardia, Montes, and Nart introduced the notion of higher order Newton’s polygons. Analogous to the first order, for each order, ${\displaystyle r}$, the authors of [26] introduced the valuation, ${\displaystyle {\omega }_r}$, of order ${\displaystyle r}$, the key polynomial, ${\displaystyle {\varphi }_r\left(x\right)}$, of such a valuation, ${\displaystyle N_r\left(F\right)}$ the Newton polygon of any polynomial, ${\displaystyle F\left(x\right)}$, with respect to ${\displaystyle {\omega }_r}$ and ${\displaystyle {\varphi }_r\left(x\right)}$, and for every side of ${\displaystyle N_r\left(F\right)}$ the residual polynomial ${\displaystyle R_r\left(F\right)}$, and the index of ${\displaystyle F\left(x\right)}$ in order ${\displaystyle r}$. For more details, we refer to [26].

4. Proofs of Our Main Results

Proof of Theorem 1. 

1.

If ${\displaystyle p}$ divides ${\displaystyle a}$ and ${\displaystyle b}$, then, by Theorem 6, ${\displaystyle p}$ does not divide ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$ if and only if ${\displaystyle {\nu }_p\left(b\right)=1}$.

2.

For ${\displaystyle p=2}$, ${\displaystyle 2}$ divides ${\displaystyle b}$ and does not divide ${\displaystyle a}$, we have ${\displaystyle \overline{F\left(x\right)}=x{(x-1)}^2{(x^2+x+1)}^2}$. Let ${\displaystyle {\varphi }_1=x-1}$ and ${\displaystyle {\varphi }_2=x^2+x+1}$. Since ${\displaystyle F\left(x\right)=\cdots -(4x-2){\varphi }_2+(a+1)x+b}$ and ${\displaystyle F\left(x\right)=\cdots +(a+7){\varphi }_1+(a+1+b)}$, by Theorem 6, ${\displaystyle 2}$ does not divide ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$ if and only if ${\displaystyle {\nu }_2(b+a+1)=1}$ and ${\displaystyle {\nu }_2((a+1)x+b)=1}$ , which means ${\displaystyle b\equiv 1-a(\mathrm{mod}4)}$ and ${\displaystyle a\equiv 1(\mathrm{mod}4)}$ or ${\displaystyle b\equiv 2(\mathrm{mod}4)}$. That is, ${\displaystyle (a,b)\in \left\{\right(1,0),(3,2\left)\right\}(\mathrm{mod}4)}$.

3.

For ${\displaystyle p=3}$, ${\displaystyle 3}$ divides ${\displaystyle b}$ and ${\displaystyle a\equiv 1(\mathrm{mod}3)}$, we have ${\displaystyle \overline{F\left(x\right)}=x{(x^2+1)}^3}$. Let ${\displaystyle \varphi =x^2+1}$. Since ${\displaystyle F\left(x\right)=x{\varphi }^3-3x{\varphi }^2+3x\varphi +(a-1)x+b}$, by Theorem 6, ${\displaystyle 3}$ does not divide ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$ if and only if ${\displaystyle {\nu }_3((a-1)x+b)=1}$, which means that ${\displaystyle a\not\equiv 1(\mathrm{mod}9)}$ or ${\displaystyle b\not\equiv 0(\mathrm{mod}9)}$. That is, ${\displaystyle (a,b)\not\equiv (1,0)(\mathrm{mod}9)}$.

4.

For ${\displaystyle p=3}$, ${\displaystyle 3}$ divides ${\displaystyle b}$ and ${\displaystyle a\equiv -1(\mathrm{mod}3)}$, we have ${\displaystyle \overline{F\left(x\right)}=x{(x-1)}^3{(x+1)}^2}$. Let ${\displaystyle {\varphi }_1=x-1}$ and ${\displaystyle {\varphi }_2=x+1}$. Since ${\displaystyle F\left(x\right)={\varphi }_1^7+7{\varphi }_1^6+21{\varphi }_1^5+35{\varphi }_1^4+35{\varphi }_1^3+21{\varphi }_1^2+(a+7){\varphi }_1+(a+1+b)}$ and ${\displaystyle F\left(x\right)={\varphi }_1^7-7{\varphi }_2^6+21{\varphi }_2^5-35{\varphi }_2^4+35{\varphi }_2^3-21{\varphi }_2^2+(a+7){\varphi }_2+(b-a-1)}$, by Theorem 6, ${\displaystyle 3}$ does not divide ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$ if and only if ${\displaystyle {\nu }_2(a+1+b)=1}$ and ${\displaystyle {\nu }_3(b-a-1)=1}$. That is, ${\displaystyle (a,b)\in \left\{\right(2,0),(8,3),(8,6\left)\right\}(\mathrm{mod}9)}$.

5.

For ${\displaystyle p=7}$, if ${\displaystyle 7}$ divides ${\displaystyle a}$ and ${\displaystyle 7}$ does not divide ${\displaystyle b}$, then ${\displaystyle \overline{F\left(x\right)}={(x+b)}^7}$. Let ${\displaystyle \varphi =x+b}$. Then ${\displaystyle F\left(x\right)={\varphi }^7-7b{\varphi }^6+21b^2{\varphi }^5-35b^3{\varphi }^4+35b^4{\varphi }^3-21b^5{\varphi }^2+(a+7b^6)\varphi +(b-ab-b^7)}$, by Theorem 6, ${\displaystyle 7}$ does not divide ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$ if and only if ${\displaystyle {\nu }_7(1-a-b^6)=1}$.

6.

For ${\displaystyle p\notin \{2,3,7\}}$ such that ${\displaystyle p}$ does not divide both ${\displaystyle a}$ and ${\displaystyle b}$, if ${\displaystyle p^2}$ does not divide ${\displaystyle 6^6{a}^7+7^7{b}^6}$, then, by the formula ${\displaystyle ▵={({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}^2{d}_K}$, ${\displaystyle p}$ does not divide ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$. If ${\displaystyle p^2}$ divides ${\displaystyle 6^6{a}^7+7^7{b}^6}$, then let ${\displaystyle t}$ be an integer such that ${\displaystyle 6at\equiv -7b(\mathrm{mod}{p}^2)}$. Then ${\displaystyle {\left(6a\right)}^6{F}^{\prime }\left(t\right)=7{(-7b)}^6+6^6{a}^7\equiv 0(\mathrm{mod}{p}^2)}$ and ${\displaystyle {\left(6a\right)}^{7}F\left(t\right)\equiv 0(\mathrm{mod}{p}^2)}$. Thus ${\displaystyle {(x-t)}^2}$ divides ${\displaystyle \overline{F\left(x\right)}}$ in ${\displaystyle {\mathbb{F}}_p\left[x\right]}$. As ${\displaystyle F\left(t\right)}$ is the remainder of the Euclidean division of ${\displaystyle F\left(x\right)}$ by ${\displaystyle x-t}$, by Theorem 6, ${\displaystyle p}$ divides the index ${\displaystyle ({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}$.

The proofs of Theorems 2 and 3 are based on Engstrom’s results [28]. ${\displaystyle \square }$

Proof of Theorem 2. 

By virtue of Engstrom’s results [28], the proof is achieved if we provide the factorization of ${\displaystyle 2{\mathbb{Z}}_K}$ into powers of prime ideals of ${\displaystyle {\mathbb{Z}}_K}$. Based on Theorem 1, we deal with the cases: ${\displaystyle 2|a}$ and ${\displaystyle 4|b}$ or ${\displaystyle (a,b)\in \left\{\right(1,2),(3,0\left)\right\}(\mathrm{mod}4)}$.

1.

If ${\displaystyle 2}$ divides ${\displaystyle a}$ and ${\displaystyle 4}$ divides ${\displaystyle b}$, then, for ${\displaystyle \varphi =x}$, we have ${\displaystyle \overline{F\left(x\right)}={\varphi }^7}$ in ${\displaystyle {\mathbb{F}}_2\left[x\right]}$.

(a)

If ${\displaystyle N_{\varphi }\left(F\right)=S}$ has a single side, that is, ${\displaystyle {\nu }_2\left(a\right)\ge {\nu }_2\left(b\right)}$, then the side, ${\displaystyle S}$, is of degree ${\displaystyle 1}$. Thus there is a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$.

(b)

If ${\displaystyle N_{\varphi }\left(F\right)=S_1+S_2}$ has two sides joining ${\displaystyle (0,{\nu }_2\left(b\right))}$, ${\displaystyle (1,{\nu }_2\left(a\right))}$, and ${\displaystyle (7,0)}$, that is, ${\displaystyle {\nu }_2\left(a\right)+1\le {\nu }_2\left(b\right)}$, 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 degree of ${\displaystyle S_2}$.

i.

If ${\displaystyle {\nu }_2\left(a\right)\notin \{2,3,4\}}$, 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.

ii.

If ${\displaystyle {\nu }_2\left(a\right)=2}$, then the slope of ${\displaystyle S_2}$ is ${\displaystyle \frac{-1}{3}}$ and ${\displaystyle R_1\left(F\right)\left(y\right)={(y+1)}^2}$ is the residual polynomial of ${\displaystyle F\left(x\right)}$ attached to ${\displaystyle S_2}$. Thus, we have to use second order Newton polygon techniques. Let ${\displaystyle {\omega }_2}$ be the valuation of a second order Newton polygon; defined by ${\displaystyle {\omega }_2\left(P\left(x\right)\right)=\min \{3{\nu }_2\left(p_i\right)+ih,ß=0,\dots ,n\}}$ for every non-zero polynomial ${\displaystyle P=\sum _{i=0}^n{p}_i{x}^i}$. Let ${\displaystyle {\varphi }_2}$ be the key polynomial of ${\displaystyle {\omega }_2}$ and let ${\displaystyle N_2\left(F\right)}$ be the ${\displaystyle {\varphi }_2}$-Newton polygon of ${\displaystyle F\left(x\right)}$ with respect to the valuation ${\displaystyle {\omega }_2}$. It follows that:

if ${\displaystyle {\nu }_2\left(b\right)=3}$, then for ${\displaystyle {\varphi }_2=x^3+2x+2}$, we have ${\displaystyle F\left(x\right)=x{\varphi }_2^2+(4-4x-4x^2){\varphi }_2+8x^2+(a-4)x+b-8}$. It follows that, if ${\displaystyle {\nu }_2(a-4)=3}$, then ${\displaystyle N_2\left(F\right)=T}$ 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 ${\displaystyle {\nu }_2(a-4)\ge 4}$ and ${\displaystyle {\nu }_2(b-8)\ge 4}$, then ${\displaystyle N_2\left(F\right)=T}$ has a single side joining ${\displaystyle (0,11)}$, ${\displaystyle (1,9)}$, and ${\displaystyle (2,7)}$, with ${\displaystyle R_2\left(F\right)\left(y\right)=y^2+y+1}$, 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 ${\displaystyle {\nu }_2\left(b\right)\ge 4}$ and ${\displaystyle {\nu }_2(a+4)=3}$, then, for ${\displaystyle {\varphi }_2=x^3+2}$, we have ${\displaystyle F\left(x\right)=x{\varphi }_2^2-4x{\varphi }_2+(a+4)x+b}$ is the ${\displaystyle {\varphi }_2}$-expansion of ${\displaystyle F\left(x\right)}$, and so ${\displaystyle N_2\left(F\right)=T}$ 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 ${\displaystyle {\nu }_2\left(b\right)=4}$ and ${\displaystyle {\nu }_2(a+4)\ge 4}$, then for ${\displaystyle {\varphi }_2=x^3+2}$, ${\displaystyle N_2\left(F\right)=T}$ has a single side joining ${\displaystyle (0,12)}$ 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 ${\displaystyle {\nu }_2\left(b\right)\ge 5}$ and ${\displaystyle {\nu }_2(a+4)=4}$, then, for ${\displaystyle {\varphi }_2=x^3+2}$, we have ${\displaystyle F\left(x\right)=x{\varphi }_2^2-4x{\varphi }_2+(a+4)x+b}$ as the ${\displaystyle {\varphi }_2}$-expansion of ${\displaystyle F\left(x\right)}$, and ${\displaystyle N_2\left(F\right)=T}$ 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 ${\displaystyle R_2\left(F\right)=y^2+y+1}$ 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 ${\displaystyle {\nu }_2\left(b\right)\ge 5}$ and ${\displaystyle {\nu }_2(a+4)\ge 5}$, then, for ${\displaystyle {\varphi }_2=x^3+2}$, ${\displaystyle N_2\left(F\right)=T_1+T_2}$ has two sides joining ${\displaystyle (0,v)}$ , ${\displaystyle (1,10)}$, and ${\displaystyle (2,7)}$ with ${\displaystyle v\ge 15}$. So, each ${\displaystyle T_i}$ has degree ${\displaystyle 1}$, and so ${\displaystyle S_2}$ 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 prime ideal factor. Based on Engstrom’s result, we conclude that ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=1}$.

iii.

For ${\displaystyle {\nu }_2\left(a\right)=3}$, we have ${\displaystyle R_1\left(F\right)\left(y\right)=y^3+1=(y^2+y+1)(y+1)}$ as the residual polynomial of ${\displaystyle F\left(x\right)}$ attached to ${\displaystyle T_1}$. Thus, ${\displaystyle T_1}$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$, with residue degree ${\displaystyle 1}$ and a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 2}$. Thus, ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=0}$.

iv.

The case ${\displaystyle {\nu }_2\left(a\right)=4}$ is similar to the case ${\displaystyle {\nu }_2\left(a\right)=2}$. In this case, ${\displaystyle {\nu }_2\left(i\left(K\right)\right)\ge 1}$ if and only if ${\displaystyle {\nu }_2\left(b\right)\ge 7}$ and ${\displaystyle {\nu }_2(a+16)\ge 7}$. 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 ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=1}$.

2.

${\displaystyle (a,b)\in \left\{\right(1,2),(3,0\left)\right\}(\mathrm{mod}4)}$. In this case, ${\displaystyle \overline{F\left(x\right)}=x{(x-1)}^2{(x^2+x+1)}^2}$ modulo ${\displaystyle 2}$. Let ${\displaystyle \varphi =x-1}$, ${\displaystyle g\left(x\right)=x^2+x+1}$, ${\displaystyle F\left(x\right)=\cdots -21{\varphi }^2+(7+a)\varphi +(b+a+1)}$, and ${\displaystyle F\left(x\right)=(x-3)g^3+(5+3x)g^2-(4x+2)g+(a+1)x+b}$. 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 ${\displaystyle \varphi }$ provides two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ of degree ${\displaystyle 1}$ each or ${\displaystyle \varphi }$ 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:

(a)

If ${\displaystyle a\equiv 1(\mathrm{mod}4)}$ and ${\displaystyle b\equiv 2(\mathrm{mod}4)}$, then ${\displaystyle {\nu }_2(▵)\ge 7}$ and ${\displaystyle N_g^{+}\left(F\right)}$ 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 ${\displaystyle N_{\varphi }^{+}\left(F\right)}$, let ${\displaystyle u=\frac{-7b_2}{3a}}$. Then ${\displaystyle u\in {\mathbb{Z}}_2}$. Let ${\displaystyle F(x+u)=x^7+\cdots +21u^5{x}^2+Ax+B}$, where ${\displaystyle A=7u^6+a=\frac{-▵}{6^6{a}^6}}$ and ${\displaystyle B=u^7+au+b=\frac{b▵}{6^7{a}^7}}$. It follows that ${\displaystyle {\nu }_2\left(A\right)={\nu }_2\left(B\right)={\nu }_2(▵)-6}$, and so ${\displaystyle N_{\varphi }^{+}\left(F\right)=S_1}$ has a single side joining ${\displaystyle (0,{\nu }_2(▵)-6)}$ and ${\displaystyle (2,0)}$. Thus, if ${\displaystyle {\nu }_2(▵)}$ is odd, then ${\displaystyle \varphi }$ provides a unique prime ideal ${\displaystyle {\mathfrak{p}}_{211}}$ of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 1}$. If ${\displaystyle {\nu }_2(▵)=2(k+3)}$ for some positive integer, ${\displaystyle k}$, then let ${\displaystyle F(x+u+2^k)=x^7+\cdots +21{(u+2^k)}^5{x}^2+A_{1}x+B_1}$, where ${\displaystyle A_1=7u^6+a+3·2^{k+1}{u}^5+2^{2k}D=A+3·2^{k+1}{u}^5+2^{2k}D}$ and ${\displaystyle B_1=B+A·2^k+2^{2k}·21u^5+2^{3k}H=2^{2k}(\frac{b_2{▵}_2}{3^7{a}^7}+21u^5)+2^{3k}H}$ for some ${\displaystyle D\in {\mathbb{Z}}_2}$ and ${\displaystyle H\in {\mathbb{Z}}_2}$. Thus, ${\displaystyle B_1=2^{2k}(3·a·b_2{▵}_2+3·a·b_2)+2^{2k+3}L}$ for some ${\displaystyle L\in {\mathbb{Z}}_2}$. Hence if ${\displaystyle k\ge 2}$, then ${\displaystyle {\nu }_2\left(A_1\right)=k+1}$ and ${\displaystyle {\nu }_2\left(B_1\right)\ge 2k+1}$. More precisely, if ${\displaystyle {▵}_2\equiv 1(\mathrm{mod}4)}$, then ${\displaystyle {\nu }_2\left(B_1\right)=2k+1}$, and so ${\displaystyle \varphi }$ provides a unique prime ideal ${\displaystyle {\mathfrak{p}}_{211}}$ of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 1}$. If ${\displaystyle {▵}_2\equiv 3(\mathrm{mod}4)}$, then ${\displaystyle {\nu }_2\left(B_1\right)\ge 2k+2}$. It follows that, if ${\displaystyle {\nu }_2\left(B_1\right)=2k+2}$, and so ${\displaystyle \varphi }$ provides a unique prime ideal ${\displaystyle {\mathfrak{p}}_{211}}$ of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 2}$. If ${\displaystyle {\nu }_2\left(B_1\right)\ge 2k+3}$, then ${\displaystyle {\nu }_2\left(B_1\right)\ge 2k+3}$, and so ${\displaystyle \varphi }$ 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 ${\displaystyle i\left(K\right)}$ and ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=1}$.

For ${\displaystyle k=1}$, we have ${\displaystyle {\nu }_2(▵)=8}$ and ${\displaystyle a\equiv 5(\mathrm{mod}8)}$. In this case, ${\displaystyle \overline{F\left(x\right)}=x{(x-1)}^2{(x^2+x+1)}^2}$ modulo ${\displaystyle 2}$. Let ${\displaystyle \varphi =x-1}$, ${\displaystyle g\left(x\right)=x^2+x+1}$, ${\displaystyle F\left(x\right)=\cdots -21{\varphi }^2+(7+a)\varphi +(b+a+1)}$, and ${\displaystyle F\left(x\right)=(x-3)g^3+(5+3x)g^2-(4x+2)g+(a+1)x+b}$. 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 ${\displaystyle {\nu }_2\left(i\left(K\right)\right)\ge 1}$ if and only if ${\displaystyle \varphi }$ provides a unique prime ideal of of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 2}$ or ${\displaystyle \varphi }$ provides two distinct prime ideals of of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 1}$ each. If ${\displaystyle (a,b)\in \left\{\right(5,10),(13,2\left)\right\}(\mathrm{mod}16)}$, then ${\displaystyle \varphi }$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 2}$, and so ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=1}$. If ${\displaystyle (a,b)\in \left\{\right(5,10),(13,10\left)\right\}(\mathrm{mod}16)}$, then ${\displaystyle \varphi }$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 1}$, and so ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=0}$. For ${\displaystyle (a,b)\in \left\{\right(5,6),(5,14),(13,6),(13,14\left)\right\}(\mathrm{mod}16)}$, let us replace ${\displaystyle \varphi }$ by ${\displaystyle {\varphi }^{\prime }=x-3}$ and consider the ${\displaystyle {\varphi }^{\prime }}$-Newton polygon of ${\displaystyle F\left(x\right)}$ with respect to ${\displaystyle {\nu }_2}$. It follows that, if ${\displaystyle (a,b)\in \left\{\right(5,6),(13,14\left)\right\}(\mathrm{mod}16)}$, then ${\displaystyle {\varphi }^{\prime }}$ provides two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 1}$ each, and so ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=1}$. If ${\displaystyle (a,b)\in \left\{\right(5,14),(13,6\left)\right\}(\mathrm{mod}16)}$, then ${\displaystyle {\varphi }^{\prime }}$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 1}$, and so ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=0}$.

(b)

${\displaystyle a\equiv 3(\mathrm{mod}4)}$ and ${\displaystyle b\equiv -(a+1)(\mathrm{mod}8)}$ because ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ has two sides.

(c)

If ${\displaystyle a\equiv 3(\mathrm{mod}8)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}8)}$, then ${\displaystyle \varphi }$ 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 ${\displaystyle N_g^{+}\left(F\right)}$ 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 ${\displaystyle F\left(x\right)}$. In this case, ${\displaystyle 2{\mathbb{Z}}_K={\mathfrak{p}}_{111}{\mathfrak{p}}_{211}{\mathfrak{p}}_{311}{\mathfrak{p}}_{312}}$ with residue degrees ${\displaystyle f_{111}=1}$ and ${\displaystyle f_{211}=f_{311}=f_{312}=2}$, and so ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=3}$.

(d)

${\displaystyle a\equiv 7(\mathrm{mod}8)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}8)}$. In this case, ${\displaystyle \varphi }$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 2}$ with residue degree ${\displaystyle 2}$ and ${\displaystyle N_g^{+}\left(F\right)}$ has two sides. More precisely, ${\displaystyle 2{\mathbb{Z}}_K={\mathfrak{p}}_{111}{\mathfrak{p}}_{211}{\mathfrak{p}}_{311}{\mathfrak{p}}_{321}}$ with residue degrees ${\displaystyle f_{111}=1}$ and ${\displaystyle f_{211}=f_{311}=f_{312}=2}$, and so ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=3}$.

(e)

If ${\displaystyle a\equiv 5(\mathrm{mod}8)}$ and ${\displaystyle b\equiv -(a+1)(\mathrm{mod}16)}$ because if ${\displaystyle b\equiv -(a+1)(\mathrm{mod}32)}$, then ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ has two sides, and if ${\displaystyle b\equiv -(a+1)+16(\mathrm{mod}32)}$, then ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ 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 ${\displaystyle N_g^{+}\left(F\right)}$ 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.

(f)

If ${\displaystyle a\equiv 5(\mathrm{mod}8)}$ and ${\displaystyle {\nu }_2(b+(a+1))=2}$, then ${\displaystyle {\nu }_2(b-(a+1))\ge 3}$. If ${\displaystyle {\nu }_2(b-(a+1))=3}$, then, for ${\displaystyle \varphi =x+1}$, ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ has a single side of degree ${\displaystyle 1}$. Since ${\displaystyle {\nu }_2(a+1)=1}$, then ${\displaystyle N_g^{+}\left(F\right)}$ 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 ${\displaystyle \varphi =x+1}$, ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ has a single side of degree ${\displaystyle 2}$, and its attached residual polynomial of ${\displaystyle F}$ is ${\displaystyle R_1\left(F\right)\left(y\right)=y^2+y+1}$. Since ${\displaystyle b\equiv 6(\mathrm{mod}8)}$, we conclude that ${\displaystyle N_g^{+}\left(F\right)}$ has a single side of degree ${\displaystyle 1}$, and 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 ${\displaystyle i\left(K\right)}$. If ${\displaystyle {\nu }_2(b-(a+1))\ge 5}$, then, for ${\displaystyle \varphi =x+1}$, ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ 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 ${\displaystyle i\left(K\right)}$.

${\displaystyle \square }$

Proof of Theorem 3. 

By virtue of Engstrom’s results [28], the proof is achieved if we provide the factorization of ${\displaystyle 3{\mathbb{Z}}_K}$ into powers of prime ideals of ${\displaystyle {\mathbb{Z}}_K}$. Based on Theorem 1, we deal with the cases:

1.

${\displaystyle 3|a}$ and ${\displaystyle 9|b}$.

2.

${\displaystyle (a,b)\notin \left\{\right(2,0),(8,3),(8,6\left)\right\}(\mathrm{mod}9)}$.

3.

${\displaystyle (a,b)\equiv (1,0)(\mathrm{mod}9)}$.

1.

${\displaystyle 3|a}$ and ${\displaystyle 9|b}$, then for ${\displaystyle \varphi =x}$, ${\displaystyle \overline{F\left(x\right)}={\varphi }^7}$ in ${\displaystyle {\mathbb{F}}_3\left[x\right]}$. It follows that:

(a)

If ${\displaystyle {\nu }_3\left(a\right)\ge {\nu }_3\left(b\right)}$, then ${\displaystyle N_{\varphi }\left(F\right)}$ has a single side of degree ${\displaystyle 1}$, and so there is a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$.

(b)

If ${\displaystyle {\nu }_3\left(a\right)+1\le {\nu }_3\left(b\right)}$, then ${\displaystyle N_{\varphi }\left(F\right)=S_1+S_2}$ has two sides joining ${\displaystyle (0,{\nu }_3\left(b\right))}$, ${\displaystyle (1,{\nu }_3\left(a\right))}$, and ${\displaystyle (7,0)}$. Since ${\displaystyle S_1}$ is of degree ${\displaystyle 1}$, ${\displaystyle S_1}$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$. Thus, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)\ge 1}$ if and only if ${\displaystyle S_2}$ provides at least three prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$, with residue degree ${\displaystyle 1}$ each. If ${\displaystyle {\nu }_3\left(a\right)\notin \{2,3,4\}}$, 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 ${\displaystyle {\nu }_3\left(a\right)\in \{2,4\}}$, 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 ${\displaystyle {\nu }_3\left(a\right)=3}$, then ${\displaystyle S_2}$ is of degree ${\displaystyle 3}$ and its attached residual polynomial of ${\displaystyle F\left(x\right)}$ is ${\displaystyle R_1\left(F\right)\left(y\right)=y^3+a_3={(y+a_3)}^3}$. So, we have to use a second order Newton polygon. Let ${\displaystyle {\omega }_2}$ be the valuation of the second order Newton polygon. ${\displaystyle {\omega }_2}$ is defined by ${\displaystyle {\omega }_2\left(P\right)=\min \{2{\nu }_3\left(p_i\right)+i,i=0,\cdots ,n\}}$ for every non-zero polynomial ${\displaystyle p=\sum _{i=0}^n{p}_i{x}^i}$ of ${\displaystyle {\mathbb{Q}}_3\left[x\right]}$. Let ${\displaystyle {\varphi }_2=x^2+3a_3}$ be a key polynomial of ${\displaystyle {\omega }_2}$ and ${\displaystyle N_2\left(F\right)}$ the ${\displaystyle {\varphi }_2}$-Newton polygon of ${\displaystyle F\left(x\right)}$ with respect to ${\displaystyle {\omega }_2}$. It follows that: if ${\displaystyle a_3\equiv 1(\mathrm{mod}3)}$, then, for ${\displaystyle {\varphi }_2=x^2+3}$, we have ${\displaystyle F\left(x\right)=x{\varphi }_2^3-9x{\varphi }_2^2+27x{\varphi }_2+(a-27)x+b}$ as the ${\displaystyle {\varphi }_2}$-expansion of ${\displaystyle F\left(x\right)}$. We have the following cases:

i.

If ${\displaystyle {\nu }_3\left(b\right)=4}$, then ${\displaystyle N_2\left(F\right)=T}$ 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 ${\displaystyle 1}$.

ii.

If ${\displaystyle {\nu }_3\left(b\right)\ge 5}$ and ${\displaystyle {\nu }_3(a-27)=4}$, then ${\displaystyle N_2\left(F\right)=T}$ 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 ${\displaystyle 1}$.

iii.

If ${\displaystyle {\nu }_3\left(b\right)=5}$ and ${\displaystyle {\nu }_3(a-27)\ge 5}$, then ${\displaystyle N_2\left(F\right)=T}$ has a single side joining ${\displaystyle (0,10)}$ and ${\displaystyle (3,7)}$ and its attached residual polynomial of ${\displaystyle F}$ is ${\displaystyle R_2\left(F\right)\left(y\right)=x{y}^3+xy+b_3}$, which is irreducible over ${\displaystyle {\mathbb{F}}_2={\mathbb{F}}_{\varphi }}$ because ${\displaystyle \varphi }$ 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 ${\displaystyle 3}$.

iv.

If ${\displaystyle {\nu }_3\left(b\right)\ge 6}$ and ${\displaystyle {\nu }_3(a-27)\ge 5}$, then ${\displaystyle N_2\left(F\right)=T_1+T_2}$ has two sides joining ${\displaystyle (0,v)}$, ${\displaystyle (2,9)}$, and ${\displaystyle (3,7)}$ with ${\displaystyle v\ge 11}$. Thus, ${\displaystyle T_1}$ is of degree ${\displaystyle 1}$, ${\displaystyle T_2}$ of degree ${\displaystyle 2}$, and ${\displaystyle R_2\left(F\right)\left(y\right)=x{y}^2+x}$ is its attached residual polynomial of ${\displaystyle F\left(x\right)}$, which is irreducible over ${\displaystyle {\mathbb{F}}_2={\mathbb{F}}_{\varphi }}$. 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 ${\displaystyle a_3\equiv -1(\mathrm{mod}3)}$, let ${\displaystyle {\varphi }_2=x^2-3}$. Then, ${\displaystyle F\left(x\right)=x{\varphi }_2^3+9x{\varphi }_2^2+27x{\varphi }_2+(a+27)x+b}$ is the ${\displaystyle {\varphi }_2}$-expansion of ${\displaystyle F\left(x\right)}$. Analogous to the case ${\displaystyle a_3\equiv 1(\mathrm{mod}3)}$, in every case ${\displaystyle 3}$ does not divide ${\displaystyle i\left(K\right)}$. If ${\displaystyle a\equiv 1(\mathrm{mod}3)}$, then ${\displaystyle \overline{F\left(x\right)}=x{(x^2+1)}^3}$ in ${\displaystyle {\mathbb{F}}_3\left[x\right]}$. So, there is exactly a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$, and the other prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ are of residue degrees at least ${\displaystyle 2}$ each prime ideal factor. Hence, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$.

(c)

If ${\displaystyle a\equiv -1(\mathrm{mod}3)}$, then ${\displaystyle \overline{F\left(x\right)}=x{(x-1)}^3{(x+1)}^3}$ in ${\displaystyle {\mathbb{F}}_3\left[x\right]}$. Let ${\displaystyle {\varphi }_1=x-1}$, ${\displaystyle {\varphi }_2=x+1}$, ${\displaystyle F\left(x\right)={\varphi }_1^7+7{\varphi }_1^6+21{\varphi }_1^5+35{\varphi }_1^4+35{\varphi }_1^3+21{\varphi }_1^2+(7+a){\varphi }_1+(b+a+1)}$, and ${\displaystyle F\left(x\right)={\varphi }_2^7-7{\varphi }_2^6+21{\varphi }_2^5-35{\varphi }_2^4+35{\varphi }_2^3-21{\varphi }_2^2+(7+a){\varphi }_2+(b-(a+1))}$. It follows that:

i.

If ${\displaystyle a\equiv 8(\mathrm{mod}9)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}9)}$, then ${\displaystyle {\nu }_3(b+(1+a))\ge 2}$ and ${\displaystyle {\nu }_3(b-(1+a))\ge 2}$. 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 ${\displaystyle {\varphi }_i}$ provides two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$ each prime ideal factor. In these two cases, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=2}$.

ii.

If ${\displaystyle a\equiv 5(\mathrm{mod}9)}$ and ${\displaystyle b\equiv 3(\mathrm{mod}9)}$, then ${\displaystyle {\nu }_3(b+(1+a))\ge 2}$ and ${\displaystyle {\nu }_3(b-(1+a))=1}$. Thus, ${\displaystyle x}$ and ${\displaystyle {\varphi }_2}$ each provide a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$, and ${\displaystyle {\varphi }_1}$ provides two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$ each. Similarly, if ${\displaystyle a\equiv 5(\mathrm{mod}9)}$ and ${\displaystyle b\equiv 6(\mathrm{mod}9)}$, then ${\displaystyle {\nu }_3(b-(1+a))\ge 2}$ and ${\displaystyle {\nu }_3(b+(1+a))=1}$. Thus, ${\displaystyle x}$ and ${\displaystyle {\varphi }_1}$ each provide a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$, and ${\displaystyle {\varphi }_2}$ provides two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$ each. In these two cases, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=1}$.

iii.

If ${\displaystyle a\equiv 2(\mathrm{mod}9)}$ and ${\displaystyle b\equiv (1+a)\pm 9(\mathrm{mod}27)}$, then ${\displaystyle N_{{\varphi }_2}^{+}\left(F\right)}$ has a single side joining ${\displaystyle (0,2)}$ and ${\displaystyle (3,0)}$, and ${\displaystyle N_{{\varphi }_1}^{+}\left(F\right)}$ 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 ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$.

iv.

Similarly, if ${\displaystyle a\equiv 2(\mathrm{mod}9)}$ and ${\displaystyle b\equiv -(1+a)\pm 9(\mathrm{mod}27)}$, then there are ${\displaystyle 3}$ prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$ each, and so ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$.

v.

If ${\displaystyle a\equiv 2(\mathrm{mod}9)}$ and ${\displaystyle {\nu }_3\left(b\right)=1}$, then ${\displaystyle {\nu }_3(▵)\ge 8}$. Let ${\displaystyle u=\frac{-7b_3}{2a}}$. Then ${\displaystyle u\in {\mathbb{Z}}_3}$. Let ${\displaystyle \varphi =x-u}$ and ${\displaystyle F(x+u)=x^7+\cdots +35u^4{x}^3+21u^5{x}^2+Ax+B}$, where ${\displaystyle A=7u^6+a=\frac{-▵}{6^6{a}^6}}$ and ${\displaystyle B=u^7+au+b=\frac{b▵}{6^7{a}^7}}$. It follows that ${\displaystyle {\nu }_3\left(A\right)={\nu }_3\left(B\right)={\nu }_3(▵)-6}$, and so ${\displaystyle N_{\varphi }^{+}\left(F\right)=S_1}$ has a single side joining ${\displaystyle (0,{\nu }_3(▵)-6)}$ and ${\displaystyle (3,0)}$. Remark that, since ${\displaystyle {\nu }_3\left(b\right)=1}$ and ${\displaystyle {\nu }_3\left(B\right)\ge 2}$, ${\displaystyle {\nu }_3(-u^7-au+b)=1}$, 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, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)\ge 1}$ if and only if ${\displaystyle \varphi }$ provides at least two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$ each prime ideal factor.

A.

If ${\displaystyle {\nu }_3(▵)=8}$, then ${\displaystyle N_{\varphi }^{+}\left(F\right)}$ has a single side of degree ${\displaystyle 1}$, and so ${\displaystyle \varphi }$ provides a unique prime ideal of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$.

B.

If ${\displaystyle {\nu }_3(▵)=9}$, then ${\displaystyle N_{\varphi }^{+}\left(F\right)=S}$ has a single side joining ${\displaystyle (0,3)}$ and ${\displaystyle (3,0)}$, with ${\displaystyle R_1\left(F\right)\left(y\right)=-u^4{y}^3+u^5{y}^2+B_3}$ its attached residual polynomial of ${\displaystyle F\left(x\right)}$. Since ${\displaystyle a\equiv -1(\mathrm{mod}3)}$ and ${\displaystyle B=\frac{b▵}{6^7{a}^7}}$, we have ${\displaystyle u\equiv -b_3(\mathrm{mod}3)}$ and ${\displaystyle B_3\equiv b_3{▵}_3(\mathrm{mod}3)}$. Thus, ${\displaystyle R_1\left(F\right)\left(y\right)=-y^3-b_3{y}^2+b_3{▵}_3}$. Since ${\displaystyle R_1\left(F\right)\left(y\right)}$ is square free and ${\displaystyle R_1\left(F\right)\left(0\right)\ne 0}$, then ${\displaystyle R_1\left(F\right)\left(y\right)}$ has at most one root in ${\displaystyle {\mathbb{F}}_{\varphi }}$. 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, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$.

C.

If ${\displaystyle {\nu }_3(▵)\ge 10}$, then ${\displaystyle N_{\varphi }^{+}\left(F\right)=S_1+S_2}$ has two sides joining ${\displaystyle (0,v-6)}$ 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 ${\displaystyle {\nu }_3(▵)}$ is even, then ${\displaystyle S_1}$ is of degree ${\displaystyle 1}$, and so ${\displaystyle \varphi }$ provides two prime ideals of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 3}$ with residue degree ${\displaystyle 1}$ each. In this case, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=1}$. If ${\displaystyle {\nu }_3(▵)=2(k+3)+1}$, then ${\displaystyle S_1}$ is of degree ${\displaystyle 2}$, with residual polynomial ${\displaystyle R_1\left(F\right)\left(y\right)=u{y}^2+b_3{▵}_3}$. Since ${\displaystyle a\equiv -1(\mathrm{mod}3)}$, we have ${\displaystyle 2a\equiv 1(\mathrm{mod}3)}$ and ${\displaystyle u\equiv -b_3(\mathrm{mod}3)}$. Thus, ${\displaystyle R_1\left(F\right)\left(y\right)=-b_3(y^2-{▵}_3)}$. It follows that, if ${\displaystyle \left(\frac{{▵}_3}{3}\right)=1}$, then ${\displaystyle R_1\left(F\right)\left(y\right)}$ 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, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=2}$. But, if ${\displaystyle \left(\frac{{▵}_3}{3}\right)=-1}$, then ${\displaystyle R_1\left(F\right)\left(y\right)}$ is irreducible over ${\displaystyle {\mathbb{F}}_{\varphi }={\mathbb{F}}_3}$, 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 ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$.

${\displaystyle \square }$

Proof of Theorem 4. 

We start by showing that ${\displaystyle 5}$ does not divide ${\displaystyle i\left(K\right)}$ for every integer of ${\displaystyle a}$ and ${\displaystyle b}$, such that ${\displaystyle x^7+ax+b}$ is irreducible. By virtue of Engstrom’s results [28], the proof is achieved if we provide the factorization of ${\displaystyle 5{\mathbb{Z}}_K}$ into powers of prime ideals of ${\displaystyle {\mathbb{Z}}_K}$. Using the index formula ${\displaystyle ▵={({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])}^2{d}_K}$, if ${\displaystyle 5^2}$ does not divide ${\displaystyle ▵}$, then ${\displaystyle {\nu }_5\left(i\left(K\right)\right)=0}$. So, we assume that ${\displaystyle 5^2}$ divides ${\displaystyle ▵}$.

1.

So, ${\displaystyle 6^6{a}^7+7^7{b}^6\equiv 0(\mathrm{mod}5)}$. Since ${\displaystyle a^5\equiv a(\mathrm{mod}5)}$ and ${\displaystyle b^5\equiv b(\mathrm{mod}5)}$, then ${\displaystyle a^3\equiv 2b^2(\mathrm{mod}5)}$, which means ${\displaystyle (a,b)\in \left\{\right(0,0),(3,1),(2,2),(2,3),(3,4\left)\right\}(\mathrm{mod}5)}$. In order to show that ${\displaystyle {\nu }_5\left(i\left(K\right)\right)=0}$, 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 ${\displaystyle (a,b)\in \left\{\right(0,0),(2,1),(3,2),(3,3),(2,34\left\}(\mathrm{mod}5\right)}$, 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 ${\displaystyle x^7+ax+b}$ .

(a)

For ${\displaystyle (a,b)\equiv (0,0)(\mathrm{mod}5)}$, if ${\displaystyle {\nu }_5\left(a\right)\ge {\nu }_5\left(b\right)}$, then ${\displaystyle N_{\varphi }\left(F\right)=S}$ 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 ${\displaystyle {\nu }_5\left(a\right)+1\le {\nu }_5\left(b\right)}$, then ${\displaystyle N_{\varphi }\left(F\right)=S_1+S_2}$ 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.

(b)

For ${\displaystyle (a,b)\equiv (3,1)(\mathrm{mod}5)}$, since ${\displaystyle \overline{F\left(x\right)}={(x+4)}^2(x+3)(x^4+4x^3+x^2+x+2)}$ in ${\displaystyle {\mathbb{F}}_5\left[x\right]}$, there are at most three prime ideals, ${\displaystyle \mathfrak{p}}$, of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 5}$ with residue degree ${\displaystyle 1}$ each.

(c)

For ${\displaystyle (a,b)\equiv (2,2)(\mathrm{mod}5)}$, since ${\displaystyle \overline{F\left(x\right)}=(x^4+2x^3+4x^2+2x+2)(x+4){(x+2)}^2}$ in ${\displaystyle {\mathbb{F}}_5\left[x\right]}$, there are at most three prime ideals, ${\displaystyle \mathfrak{p}}$, of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 5}$ with residue degree ${\displaystyle 1}$ each.

(d)

For ${\displaystyle (a,b)\equiv (2,3)(\mathrm{mod}5)}$, since ${\displaystyle \overline{F\left(x\right)}=(x+1){(x+3)}^2(x^4+3x^3+4x^2+3x+2)}$ in ${\displaystyle {\mathbb{F}}_5\left[x\right]}$, there are at most three prime ideals ${\displaystyle \mathfrak{p}}$ of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 5}$ with residue degree ${\displaystyle 1}$ each.

(e)

For ${\displaystyle (a,b)\equiv (3,4)(\mathrm{mod}5)}$, since ${\displaystyle \overline{F\left(x\right)}=(x^4+x^3+x^2+4x+2){(x+1)}^2(x+2)}$ in ${\displaystyle {\mathbb{F}}_5\left[x\right]}$, there are at most three prime ideals, ${\displaystyle \mathfrak{p}}$, of ${\displaystyle {\mathbb{Z}}_K}$ lying above ${\displaystyle 5}$ with residue degree ${\displaystyle 1}$ each.

We conclude that in all cases ${\displaystyle {\nu }_5\left(i\left(K\right)\right)=0}$.

For ${\displaystyle p\ge 7}$, 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 ${\displaystyle p\ge 7}$ monic irreducible polynomials of degree ${\displaystyle f}$ in ${\displaystyle {\mathbb{F}}_p\left[x\right]}$ for every positive integer ${\displaystyle f\in \{1,2,3\}}$, we conclude that ${\displaystyle p}$ does not divide ${\displaystyle i\left(K\right)}$. ${\displaystyle \square }$

5. Examples

Let ${\displaystyle F=x^7+ax+b\in \mathbb{Z}\left[x\right]}$ be a monic irreducible polynomial and ${\displaystyle K}$ a number field generated by a root, ${\displaystyle \alpha }$, of ${\displaystyle F\left(x\right)}$. In the following examples, we calculate the index of the field ${\displaystyle K}$. First based on Theorem 4, ${\displaystyle {\nu }_p\left(i\left(K\right)\right)=0}$ for every prime integer ${\displaystyle p\ge 5}$. Thus, we need only to calculate ${\displaystyle {\nu }_p\left(i\left(K\right)\right)}$ for ${\displaystyle p=2,3}$.

1.

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

2.

For ${\displaystyle a=28}$ and ${\displaystyle b=32}$, since ${\displaystyle \overline{F\left(x\right)}}$ is irreducible over ${\displaystyle {\mathbb{F}}_5}$, ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$. By the first item of Theorem 2, we have ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=1}$. By Theorem 3, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$. Thus, ${\displaystyle i\left(K\right)=2}$.

3.

For ${\displaystyle a=3}$ and ${\displaystyle b=8}$, since ${\displaystyle \overline{F\left(x\right)}}$ is irreducible over ${\displaystyle {\mathbb{F}}_5}$, ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$. Again, since ${\displaystyle a\equiv 3(\mathrm{mod}4)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}8)}$, by Theorem 2, ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=3}$. By Theorem 3, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=0}$. Thus, ${\displaystyle i\left(K\right)=8}$.

4.

For ${\displaystyle a=-1}$ and ${\displaystyle b=9}$, since ${\displaystyle \overline{F\left(x\right)}}$ is irreducible over ${\displaystyle {\mathbb{F}}_2}$, ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$. Since ${\displaystyle 2{\mathbb{Z}}_K}$ is a prime ideal of ${\displaystyle {\mathbb{Z}}_K}$, ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=0}$. Also, since ${\displaystyle a\equiv 8(\mathrm{mod}9)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}9)}$, by Theorem 3, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=2}$. Thus, ${\displaystyle i\left(K\right)=9}$.

5.

For ${\displaystyle a=803}$ and ${\displaystyle b=2112}$, since ${\displaystyle \overline{F\left(x\right)}}$ is irreducible over ${\displaystyle {\mathbb{F}}_5}$, ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$. Since ${\displaystyle a\equiv 3(\mathrm{mod}4)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}8)}$, by Theorem 2, ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=3}$. Similarly, since ${\displaystyle a\equiv 5(\mathrm{mod}9)}$ and ${\displaystyle b\equiv 6(\mathrm{mod}9)}$, by Theorem 3, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=1}$. Thus, ${\displaystyle i\left(K\right)=24}$.

6.

For ${\displaystyle a=35}$ and ${\displaystyle b=72}$, since ${\displaystyle \overline{F\left(x\right)}}$ is irreducible over ${\displaystyle {\mathbb{F}}_{11}}$, ${\displaystyle F\left(x\right)}$ is irreducible over ${\displaystyle \mathbb{Q}}$. Since ${\displaystyle a\equiv 3(\mathrm{mod}4)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}8)}$, by Theorem 2, ${\displaystyle {\nu }_2\left(i\left(K\right)\right)=3}$. Similarly, since ${\displaystyle a\equiv 8(\mathrm{mod}9)}$ and ${\displaystyle b\equiv 0(\mathrm{mod}9)}$, by Theorem 3, ${\displaystyle {\nu }_3\left(i\left(K\right)\right)=2}$. Thus, ${\displaystyle i\left(K\right)=72}$.