On the Monogenity of Quartic Number Fields Defined by x⁴ + ax² + b

Abstract

For any quartic number field K generated by a root $\alpha$ of an irreducible trinomial of type $x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$, we characterize when $\mathbb{Z}\left[\alpha \right]$ is integrally closed. Also for $p=2, 3$, we explicitly give the highest power of p dividing $i\left(K\right)$, the common index divisor of K. For a wide class of monogenic trinomials of this type, we prove that up to equivalence, there is only one generator of power integral bases in $K=\mathbb{Q}\left(\alpha \right)$. We illustrate our statements with a series of examples.

1. Introduction

1.1. Monogenity of Number Fields and Polynomials

Let K be a number field of degree n with ring of integers ${\mathbb{Z}}_K$, and absolute discriminant $d_K$. The number field K is called monogenic if it admits a power integral basis, that is an integral basis of type $(1,\alpha ,\dots ,{\alpha }^{n-1})$ for some $\alpha \in {\mathbb{Z}}_K$. Monogenity of number fields is a classical problem of algebraic number theory, going back to Dedekind, Hasse, and Hensel, cf, e.g., [1,2,3] for the present state of this area. It is called a problem of Hasse to give an arithmetic characterization of those number fields which have a power integral basis [1,2,4]. For any primitive element $\alpha$ of ${\mathbb{Z}}_K$ (that is $\alpha \in {\mathbb{Z}}_K$ with $K=\mathbb{Q}\left(\alpha \right)$), we denote by

$$\mathrm{ind}\left(\alpha \right)=({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$$

the index of α, that is the index of the $\mathbb{Z}$-module $\mathbb{Z}\left[\alpha \right]$ in the free $\mathbb{Z}$-module ${\mathbb{Z}}_K$ of rank n. As is known [3], we have

$$▵\left(\alpha \right)=\mathrm{ind}{\left(\alpha \right)}^2·d_K$$

where $▵\left(\alpha \right)$ is the discriminant of $\alpha$. If $F\left(x\right)$ is the minimal polynomial of $\alpha$, then we have $\Delta \left(F\right)=\Delta \left(\alpha \right)$ and we also use the index of F, $\mathrm{ind}\left(F\right)=\mathrm{ind}\left(\alpha \right)$. We also say that the polynomial $F\left(x\right)\in \mathbb{Z}\left[x\right]$ is monogenic, if $\mathrm{ind}\left(F\right)=1$, that is a root $\alpha$ of $F\left(x\right)$ generates a power integral basis in $K=\mathbb{Q}\left(\alpha \right)$. Obviously, if $F\left(x\right)$ is monogenic, then K is also monogenic, but the converse is not true: there may exist a power integral basis in K, even if $F\left(x\right)$ is not monogenic. Further, note that $F\left(x\right)$ is monogenic (that is $(1,\alpha ,\dots ,{\alpha }^{n-1})$ is an integral basis of K) if, and only if, $\mathbb{Z}\left[\alpha \right]$ is integrally closed.

The elements $\alpha$ and $\beta$ of ${\mathbb{Z}}_K$ are called equivalent if $\alpha \pm \beta \in \mathbb{Z}$. Obviously, equivalent elements have the same indices.

Let $(1,{\omega }_1,\dots ,{\omega }_{n-1})$ be an integral basis of K. The discriminant $▵\left(L(x_1,\dots ,x_{n-1})\right)$ of the linear form $L(x_1,\dots ,x_{n-1})={\omega }_1{x}_1+\dots +{\omega }_{n-1}{x}_{n-1}$ can be written (see [3]) as

$$▵\left(L(x_1,\dots ,x_{n-1})\right)={\left(\mathrm{ind}(x_1,\dots ,x_{n-1})\right)}^2·d_K,$$

where $\mathrm{ind}(x_1,\dots ,x_{n-1})$ is the index form corresponding to the integral basis $(1,{\omega }_1,\dots ,{\omega }_{n-1})$, having the property that for any $\alpha =x_0+{\omega }_1{x}_1+\dots +{\omega }_{n-1}{x}_{n-1}\in {\mathbb{Z}}_K$ (with $x_0,x_1,\dots ,x_{n-1}\in \mathbb{Z}$) we have $\mathrm{ind}\left(\alpha \right)=|\mathrm{ind}(x_1,\dots ,x_{n-1})|$.

Obviously, $\mathrm{ind}\left(\alpha \right)=1$ if, and only if, $(1,\alpha ,\dots ,{\alpha }^{n-1})$ is an integral basis of K. Therefore, $\alpha$ is a generator of a power integral base if, and only if, $(x_1,\dots ,x_{n-1})\in {\mathbb{Z}}^{n-1}$ is a solution of the index form equation

$$\mathrm{ind}(x_1,\dots ,x_{n-1})=\pm 1\mathrm{in}(x_1,\dots ,x_{n-1})\in {\mathbb{Z}}^{n-1}.$$

1.2. Results on the Monogenity of Binomilas, Trinomials

The problem of testing the monogenity of number fields and constructing power integral bases has been intensively studied during the last decades (see [3] and the references therein). An especially delicate and intensively studied problem is the monogenity of pure fields K generated by a root $\alpha$ of an irreducible polynomial $x^n-m$ (see [5,6,7]).

Recently, many authors have become interested in the monogenity of trinomials and number fields defined by a root $\alpha$ of a three-term irreducible polynomial $x^n+a{x}^m+b\in \mathbb{Z}\left[x\right]$. Jakhar, Khanduja and Sangwan [8], Jhorar and Khanduja [9], and Jakhar and Kumar [10] studied the integral closedness of $\mathbb{Z}\left[\alpha \right]$, $\alpha$ being a root of a trinomial. Their results were refined by Ibarra, Lembeck, Ozaslan, Smith, and Stange [11]. Recall that the results given in [8,9], can only decide if $\mathbb{Z}\left[\alpha \right]$ is integrally closed ($\alpha$ a root of the trinomial), but cannot decide whether the field $K=\mathbb{Q}\left(\alpha \right)$ is monogenic or not.

Jones [12,13], Jones, and Tristan [14] and Jones and White [15] also investigated the monogenity of some irreducible trinomials. Ben Yakkou and El Fadil [16] gave sufficient conditions on the coefficients of certain trinomials which guarantee the non-monogenity of the number field defined by a root of such a trinomial. Also, El Fadil [17,18,19] gave the necessary and sufficient conditions on the monogenity of number fields, generated by quartic trinomials of type $x^4+ax+b$, and also by certain quintic and sextic trinomials, in terms of the coefficients a and b of the trinomials.

1.3. The Index of a Number Field

The greatest common divisor of the indices of all integral primitive elements of K is called the index of K, denoted by $i\left(K\right)$. A rational prime p dividing $i\left(K\right)$ is called a prime common index divisor of K. It is clear that if ${\mathbb{Z}}_K$ has a power integral basis, then the index of K is trivial, namely, $i\left(K\right)=1$. Therefore a field having a prime index divisor is not monogenic.

The first number field with a non-trivial index was given by Dedekind in 1871, who exhibited examples in cubic and quartic number fields. For example, he considered the cubic field K generated by a root of $x^3-x^2-2x-8$ and showed that the prime 2 splits completely in ${\mathbb{Z}}_K$. So, if K were monogenic, then there would be a cubic polynomial, a root of which generating K, that splits completely into distinct polynomials of degree 1 in ${\mathbb{F}}_2\left[x\right]$. Since there are only two distinct polynomials of degree 1 in ${\mathbb{F}}_2\left[x\right]$, this is impossible.

Based on these ideas and using Kronecker’s theory of algebraic number fields, Hensel gave the necessary and sufficient conditions for any prime integer p to be a prime common index divisor [20]. Hensel [2] also showed that the prime divisors of $i\left(K\right)$ must be less than the degree of the field K.

We shall denote by ${\nu }_p\left(a\right)$, the highest power of the prime p dividing the integer a. For arbitrary number fields of degree $n\le 7$, Engstrom [21] characterized ${\nu }_p\left(i\left(K\right)\right)$, the highest power of p dividing $i\left(K\right)$, by the factorization of $\left(p\right)$ into powers of prime ideals of ${\mathbb{Z}}_K$ for every positive prime $p\le n$. Problem 22 of Narkiewicz [22] asks for an explicit formula for the highest power of a given prime integer p dividing $i\left(K\right)$.

In [23], Nakahara studied the index of non-cyclic but abelian biquadratic number fields. In [24] Gaál, Pethö, and Pohst characterized the field indices of biquadratic number fields having a Galois group $V_4$. El Fadil [18,19] gave the necessary and sufficient conditions for a prime integer p to divide $i\left(K\right)$ in terms of a, b in number fields K defined by trinomials $x^4+ax+b$ and $x^5+a{x}^2+b$, respectively.

1.4. The Purpose of the Present Paper

In this paper, we consider trinomials of type $x^4+a{x}^2+b$. This is a special type of trinomial, but exactly these special properties enable us to formulate results that far exceed some corresponding statements on general types of trinomials.

Our paper was motivated by some results concerning trinomials of type $x^4+ax+b$. Alaca and Williams [25,26] constructed p-integral bases and integral bases of quartic fields generated by a root of such a trinomial. Davis and Spearman [27] characterized the prime divisors of quartic number fields K generated by a root of such a trinomial.

In this paper, for any quartic number field K generated by a root $\alpha$ of an irreducible trinomial $x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$, we give the necessary and sufficient conditions for the integral closedness of $\mathbb{Z}\left[\alpha \right]$ in terms of the coefficients a and b. We also evaluate ${\nu }_p\left(i\left(K\right)\right)$ for every prime integer p. Based on Engstrom’s results given in [21], the unique prime candidates to divide $i\left(K\right)$ are 2 and 3, that is $i\left(K\right)=2^{{\nu }_2}{3}^{{\nu }_3}$, with $0\le {\nu }_2\le 2$ and $0\le {\nu }_3\le 1$. In [27], for a quartic number field K defined by a trinomial $x^4+ax+b\in \mathbb{Z}\left[x\right]$, and for every prime integer p, Davis and Spearman [27] gave the necessary and sufficient conditions on a and b so that p is a common index divisor of K. Their method is based on the calculation of the p-index form of K (that is, the index form mod p), derived from the p-integral bases of K (see [27]). Our results in Theorems 2 and 3 are analogous to that given in [27], but our method is totally different; the proofs of Theorems 2 and 3 are based on prime ideal factorization, which is performed for quartic number fields in the thesis of Montes (1999).

In the present paper, we consider three types of problems: For any quartic number field K generated by a root $\alpha$ of an irreducible trinomial $x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$,

–

we characterize when $\mathbb{Z}\left[\alpha \right]$ is integrally closed (Theorem 1),

–

we give the necessary and sufficient conditions for 2 or 3 to divide the index of K, in terms of a, b,

–

we determine ${\nu }_p\left(i\left(K\right)\right)$ for $p=2,3$,

–

we study the monogenity of K (see Section 4.2).

–

For a wide class of monogenic trinomials of type $x^4+a{x}^2+b$, we show that up to equivalence, the root of the trinomial is the only generator of power integral bases.

We also provide a series of examples that illustrate our results.

Our results provide a complete characterization of the monogenity properties of trinomials of type $x^4+a{x}^2+b$, extending the work of Davis and Sperman [27] also to the case of trinomials of type $x^4+a{x}^2+b$, with several new aspects. Since trinomials of type $x^4+ax+b$ were considered previously in [19], and $x^4+a{x}^3+b$ can be reduced to $x^4+ax+b$, our results complete the description of the monogenity properties of quartic trinomials.

2. Main Results

Throughout this section unless otherwise stated, K is a number field generated by a root $\alpha$ of an irreducible trinomial $F\left(x\right)=x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$ and we assume that for every prime p, ${\nu }_p\left(a\right)<2$ or ${\nu }_p\left(b\right)<4$.

Throughout this paper, for any integer $a\in \mathbb{Z}$ and a prime p, we set $a_p={\displaystyle \frac{a}{p^{{\nu }_p\left(a\right)}}}$.

2.1. Integral Closedness of $\mathbb{Z}\left[\alpha \right]$

Our first theorem characterizes the integral closedness of $\mathbb{Z}\left[\alpha \right]$:

Theorem 1.

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

1. 

${\nu }_p\left(a\right)\ge 1$ and ${\nu }_p\left(b\right)=1$.

2. 

$p=2$, $b\equiv 2(\mathrm{mod}4)$ and $a\equiv 3(\mathrm{mod}4)$.

3. 

$p=2$ does not divide b and $a\equiv 1-b(\mathrm{mod}4)$.

4. 

$p=2$ does not divide $ab$ and $a\equiv 3(\mathrm{mod}4)$.

5. 

$p=2$ does not divide $ab$ and $b\not\equiv a(\mathrm{mod}4)$.

6. 

$p\ge 3$, p does not divide b and either $p^2$ does not divide $a^2-4b$ or $-a{2}^{-1}$ is not a square in ${\mathbb{F}}_p$, where ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$ is the finite field with p elements.

In particular, if for every prime integer, p, p satisfies one of these conditions, then $i\left(K\right)$ is trivial, that is $i\left(K\right)=1$.

2.2. The Divisibility of the Field Index by 2 and 3

Next, for $p=2, 3$, we give the necessary and sufficient conditions for p to divide the index of K in terms of a, b. Furthermore, in every case, we explicitly give ${\nu }_p\left(i\left(K\right)\right)$.

Theorem 2.

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

$conditions$				${\nu }_2\left(i\left(K\right)\right)$
${\nu }_2\left(a\right)\ge 1$	${\nu }_2\left(a\right)=1$	${\nu }_2\left(b\right)=3$	$b\equiv 48-4a(\mathrm{mod}64)$	1
${\nu }_2\left(b\right)\ge 1$		${\nu }_2\left(b\right)=2k+1(k\ge 2)$	$b\equiv -2^{2k}a(\mathrm{mod}{2}^{2k+4})$	1
${\nu }_2\left(a\right)=0$	${\nu }_2\left(b\right)odd$	$a\equiv 7-b(\mathrm{mod}8)$		1
${\nu }_2\left(b\right)\ge 1$		$a\equiv 1(\mathrm{mod}4)$	$b\equiv -4a(\mathrm{mod}32)$	1
	${\nu }_2\left(b\right)=2$	$a\equiv 3(\mathrm{mod}8)$	$b\equiv 8-4a(\mathrm{mod}16)$	1
		$a\equiv 7(\mathrm{mod}8)$	$b\equiv 16-4a(\mathrm{mod}32)$	1
		$a\equiv 7(\mathrm{mod}8)$	$b\equiv -4a(\mathrm{mod}32)$	2
		$a\equiv 1(\mathrm{mod}4)$	$b\equiv -2^{2k}a(\mathrm{mod}{2}^{2k+3})$	1
	${\nu }_2\left(b\right)=2k(k\ge 2)$	$a\equiv 7(\mathrm{mod}8)$	$b\equiv 2^{2k}(2-a)(\mathrm{mod}{2}^{2k+2})$	1
		$a\equiv 3(\mathrm{mod}8)$	$b\equiv 2^{2k}(4-a)(\mathrm{mod}{2}^{2k+3})$	1
		$a\equiv 7(\mathrm{mod}8)$	$b\equiv -2^{2k}a(\mathrm{mod}{2}^{2k+3})$	2
${\nu }_2\left(a\right)\ge 1$	$a\equiv 0(\mathrm{mod}4)$	$b\equiv 15-a(\mathrm{mod}16)$		1
${\nu }_2\left(b\right)=0$	$a\equiv 6(\mathrm{mod}8)$			$gotoTableA$
${\nu }_2\left(a\right)=0$	$a\equiv 1(\mathrm{mod}8)$	$b\equiv 1(\mathrm{mod}8)$		1
${\nu }_2\left(b\right)=0$	$a\equiv 5(\mathrm{mod}8)$	$b\equiv 1(\mathrm{mod}8)$		1
Otherwise				0

$$TableA:a\equiv 6(\mathrm{mod}8) and b\equiv 1(\mathrm{mod}2)$$

$conditions$	${\nu }_2\left(i\left(K\right)\right)$
$a\equiv 6(\mathrm{mod}32)$ and $b\equiv 231-5a(\mathrm{mod}256)$	1
${\nu }_2(a-6)=2+k,{\nu }_2(b-3a+9)\ge 4+2kandk\ge 3$	1
${\nu }_2(a-6)=2+k,{\nu }_2(b-3a+9)\ge 4+2j,k\ge 3,j\le k-2$	1
and ${(b-3a+9)}_2\equiv 3(\mathrm{mod}4)$
${\nu }_2(a-22)=2+k,{\nu }_2(b-11a+121)\ge 4+2k$ and $k\ge 3$	1
${\nu }_2(a-22)=2+k,{\nu }_2(b-11a+121)\ge 4+2j,k\ge 3,j\le k-2$	1
and ${(b-11a+121)}_2\equiv 3(\mathrm{mod}4)$
$a\equiv 22(\mathrm{mod}32)$ and $b\equiv 247+3a(\mathrm{mod}256)$	1
$a\equiv 14(\mathrm{mod}16)$ and $b\equiv 55+3a(\mathrm{mod}64)$	1
Otherwise	0

In particular, when ${\nu }_2\left(i\left(K\right)\right)\ge 1$, then K is not monogenic.

Theorem 3.

The prime integer 3 divides $i\left(K\right)$ if, and only if, one of the following conditions holds:

1. 

$a\equiv -1(\mathrm{mod}3)$, ${\nu }_3\left(b\right)=2k$ for some positive integer k, and $b_3\equiv 1(\mathrm{mod}3)$.

2. 

$a\equiv b\equiv 1(\mathrm{mod}3)$, ${\nu }_3(a^2-4b)=2k$ for some positive integer k and ${(a^2-4b)}_3\equiv 1(\mathrm{mod}3)$.

In particular, if one of these conditions holds, then K is not monogenic.

In all the above cases, we have ${\nu }_3\left(i\left(K\right)\right)=1$.

Remark 1.

1. 

The field K can be non-monogenic even if the index $i\left(K\right)=1$. It suffices to consider the number field K generated by a root of the polynomial $F\left(x\right)=x^4-13$, which is irreducible over $\mathbb{Q}$, as it is 13-Eisenstein. Since $13=5+8k$ for $k=1$, we conclude by [7] (Section 4.3, p. 138) that K is not monogenic. But $(a,b)=(0,-13)$ satisfies neither the conditions of Theorem 2 nor the conditions of Theorem 3; therefore $i\left(K\right)=1$.

2. 

If the field index is $i\left(K\right)>1$, then the field is obviously non-monogenic. However, if the field index is 1, the field can still be monogenic or non-monogenic. A number field K is monogenic if, and only if, there exists some $\theta \in {\mathbb{Z}}_K$ such that $\mathrm{ind}\left(\theta \right)=1$, that is, the index form equation has a solution. So, the unique method which allows to test whether K is monogenic is to calculate the solutions of the index form equation of the field K (see [3] and Section 4.2).

3. 

It is well known that the index of a quartic field satisfies $i\left(K\right)\in \{1,2,3,4,6,12\}$ (see [21], p. 234). Thus, for every prime integer $p\ge 5$, ${\nu }_p\left(i\left(K\right)\right)=0$.

2.3. The Number of Inequivalent Generators of Power Integral Bases

Number fields usually only have a few inequivalent generators of power integral bases (see, e.g., the tables of [3]). Considering recently some types of monogenic binomials $x^n-m$, we had the experience that up to equivalence, the root of the binomial is the only generator of power integral bases in the number field generated by the root. Calculating “small” solutions (with coefficients $<{10}^{100}$ in absolute values in the integral bases), we found that this phenomenon occurs in pure sextic fields generated by a root of a monogenic binomial $x^6-m$ and in pure octic fields generated by a root of a monogenic binomial $x^8-m$, both for $0<m<-5000$.

Therefore, we found that it will also be interesting to consider the number of inequivalent generators of power integral bases of monogenic trinomials of type $x^4+a{x}^2+b$, utilizing the special properties of this trinomial. Note that using [28,29] in any specific quartic field, it is possible to calculate all generators of power integral bases, but we would like to formulate here a statement in a parametric form covering an infinite number of fields. We succeeded in covering the cases $a>1,b>1$:

Theorem 4.

Assume $a>1,b>1$ and $F\left(x\right)=x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$ is irreducible and monogenic. If $a,b$ are not of type

$$a={\displaystyle \frac{u\pm 1}{v}},b={\displaystyle \frac{u^2-1}{4v^2}}$$

(1)

for some $u,v\in \mathbb{Z},v\ne 0,u\ne \pm 1$, then up to equivalence, the root α of $F\left(x\right)$ is the only generator of power integral bases in $K=\mathbb{Q}\left(\alpha \right)$.

Note that if $a,b$ are of type (1), then there may be several inequivalent generators of power integral bases in the corresponding quartic fields.

We list a few examples of pairs of positive parameters $(a,b)$, represented in the form (1), such that the trinomial $F\left(x\right)=x^4+a{x}^2+b$ is irreducible and monogenic but the number field $K=\mathbb{Q}\left(\alpha \right)$, generated by a root $\alpha$ of $F\left(x\right)$ admits several inequivalent generators of power integral bases. As in such fields $(1,\alpha ,{\alpha }^2,{\alpha }^3)$ is an integral basis, we shall give the triples $(x,y,z)$ such that $\gamma =x\alpha +y{\alpha }^2+z{\alpha }^3$ generates a power integral basis in K:

Case 1.

$a={\displaystyle \frac{u-1}{v}},b={\displaystyle \frac{u^2-1}{4v^2}}$.

$(u,v,a,b)=(3,1,2,2),(x,y,z)=(1,-1,0),(1,0,0),(1,0,1),(1,1,0).$

$(u,v,a,b)=(5,1,4,6),(x,y,z)=(1,-1,0),(1,0,0),(1,1,0).$

$(u,v,a,b)=(7,2,3,3),(x,y,z)=(1,0,0),(1,0,1),(2,-1,1),(2,1,1).$

$(u,v,a,b)=(9,2,4,5),(x,y,z)=(1,0,0),(2,0,1).$

Case 2.

$a={\displaystyle \frac{u+1}{v}},b={\displaystyle \frac{u^2-1}{4v^2}}$.

$(u,v,a,b)=(3,1,4,2),(x,y,z)=(1,0,0),(3,0,1).$

3. A Short Introduction to Newton Polygons

We use Newton polygon techniques. This is a standard method which is rather technical but very efficient to apply. We have introduced the corresponding concepts in several former papers. Here, we only give a brief introduction, which makes our proofs understandable. For a detailed description, we refer to Guardia and Nart [30] and to Guardia, Montes, and Nart [31].

Let $F\left(x\right)\in \mathbb{Z}\left[x\right]$ be a monic irreducible polynomial with a root $\alpha$ and set $K=\mathbb{Q}\left(\alpha \right)$. We shall use Dedekind’s theorem [32] (Chapter I, Proposition 8.3) relating the prime ideal factorization $p{\mathbb{Z}}_K$ and the factorization of $F\left(x\right)$ modulo p (for primes p not dividing $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$). Also, we shall need Dedekind’s criterion [33] (Theorem 6.1.4) on the divisibility of $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$ by primes p.

For any prime integer p, let ${\nu }_p$ be the p-adic valuation of $\mathbb{Q}$, ${\mathbb{Q}}_p$ its p-adic completion, and ${\mathbb{Z}}_p$ the ring of p-adic integers. Let ${\nu }_p$ be the Gauss’s extension of ${\nu }_p$ to ${\mathbb{Q}}_p\left(x\right)$. For any polynomial $P={\sum }_{i=0}^n{a}_i{x}^i\in {\mathbb{Q}}_p\left[x\right]$, we set ${\nu }_p\left(P\right)=min({\nu }_p\left(a_i\right),i=0,\dots ,n)$. For nonzero polynomials $P,Q\in {\mathbb{Q}}_p\left[x\right]$, we extend this valuation to ${\nu }_p(P/Q)={\nu }_p\left(P\right)-{\nu }_p\left(Q\right)$.

Let $\varphi \in {\mathbb{Z}}_p\left[x\right]$ be a monic polynomial whose reduction is irreducible in ${\mathbb{F}}_p\left[x\right]$; let ${\mathbb{F}}_{\varphi }$ be the field ${\displaystyle \frac{{\mathbb{F}}_p\left[x\right]}{\left(\overline{\varphi }\right)}}$. For any monic polynomial $F\left(x\right)\in {\mathbb{Z}}_p\left[x\right]$, upon the Euclidean division by successive powers of $\varphi$, we expand $F\left(x\right)$ as $F\left(x\right)={\sum }_{i=0}^l{a}_i\left(x\right)\varphi {\left(x\right)}^i,$ called the $\varphi$-expansion of $F\left(x\right)$ (for every i, deg$\left(a_i\left(x\right)\right)<$ deg$\left(\varphi \right)$). The $\varphi$-Newton polygon $N_{\varphi }\left(F\right)$ of $F\left(x\right)$ with respect to p is the lower boundary convex envelope of the set of points $\{(i,{\nu }_p\left(a_i\left(x\right)\right)),a_i\left(x\right)\ne 0\}$. It is the process of joining the obtained edges $S_1,\dots ,S_r$ ordered by increasing slopes, which can be expressed as $N_{\varphi }\left(F\right)=S_1+\cdots +S_r$. For every side $S_i$ of $N_{\varphi }\left(F\right)$, the length $l\left(S_i\right)$ of $S_i$ is the length of its projection to the x-axis and its height $h\left(S_i\right)$ is the length of its projection to the y-axis. We call $d\left(S_i\right)=gcd(l\left(S_i\right),h\left(S_i\right))$, the degree of S. The principal $\varphi$-Newton polygon $N_{\varphi }^{-}\left(F\right)$ of F is the part of the polygon $N_{\varphi }\left(F\right)$, which is determined by joining all sides of negative slopes. To every side S of $N_{\varphi }^{-}\left(F\right)$, with initial point $(s,u_s)$ and length l, and to every $0\le i\le l$, we attach the following residue coefficient $c_i\in {\mathbb{F}}_{\varphi }$:

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

where $(p,\varphi (x\left)\right)$ is the maximal ideal of ${\mathbb{Z}}_p\left[x\right]$ generated by p and $\varphi \left(x\right)$. Let $-\lambda =-h/e$ be the slope of S, where h and e are two positive coprime integers. Then, $d=l/e$ is the degree of S. Notice that, the points with integer coordinates lying on S are exactly $(s,u_s),$$(s+e,u_s-h),\cdots ,(s+de,u_s-dh)$. Thus, if i is not a multiple of e, then $(s+i,u_{s+i})$ does not lie in S, and so $c_i=0$. The polynomial $R_{\lambda }\left(F\right)\left(y\right)=t_d{y}^d+t_{d-1}{y}^{d-1}+\cdots +t_{1}y+t_0\in {\mathbb{F}}_{\varphi }\left[y\right],$ is called the residual polynomial of $F\left(x\right)$ associated to the side S, where for every $i=0,\dots ,d$, $t_i=c_{ie}$.

Let $N_{\varphi }^{-}\left(F\right)=S_1+\cdots +S_r$ be the principal $\varphi$-Newton polygon of f with respect to p. We say that F is a $\varphi$-regular polynomial with respect to p, if $F_{S_i}\left(y\right)$ is square free in ${\mathbb{F}}_{\varphi }\left[y\right]$ for every $i=1,\dots ,r$. The polynomial $F\left(x\right)$ is said to be p-regular if $\overline{F\left(x\right)}={\prod }_{i=1}^r{\overline{{\varphi }_i}}^{l_i}$ for some monic polynomials ${\varphi }_1,\dots ,{\varphi }_t$ of $\mathbb{Z}\left[x\right]$ such that $\overline{{\varphi }_1},\dots ,\overline{{\varphi }_t}$ are irreducible coprime polynomials over ${\mathbb{F}}_p$ and $F\left(x\right)$ is a ${\varphi }_i$-regular polynomial with respect to p for every $i=1,\dots ,t$.

Let $\varphi \in {\mathbb{Z}}_p\left[x\right]$ be a monic polynomial, with $\overline{\varphi \left(x\right)}$ irreducible in ${\mathbb{F}}_p\left[x\right]$. As defined in [34] (Def. 1.3), the $\varphi$-index of $F\left(x\right)$, denoted by ${\mathrm{ind}}_{\varphi }\left(F\right)$, is deg$\left(\varphi \right)$ times the number of points with natural integer coordinates that lie below or on the polygon $N_{\varphi }^{-}\left(F\right)$, strictly above the horizontal axis, and strictly beyond the vertical axis.

Let $\overline{F\left(x\right)}={\prod }_{i=1}^r{\overline{{\varphi }_i}}^{l_i}$ be the factorization of $\overline{F\left(x\right)}$ in ${\mathbb{F}}_p\left[x\right]$, where every ${\varphi }_i\in \mathbb{Z}\left[x\right]$ is monic polynomial, with $\overline{{\varphi }_i\left(x\right)}$ is irreducible in ${\mathbb{F}}_p\left[x\right]$, $\overline{{\varphi }_i\left(x\right)}$ and $\overline{{\varphi }_j\left(x\right)}$ are coprime when $i\ne j$ and $i,j=1,\dots ,t$. For every $i=1,\dots ,t$, let $N_{{\varphi }_i}^{+}\left(F\right)=S_{i1}+\cdots +S_{i{r}_i}$ be the principal ${\varphi }_i$-Newton polygon of $F\left(x\right)$ with respect to p. For every $j=1,\dots ,r_i$, let $F_{S_{ij}}\left(y\right)={\prod }_{k=1}^{s_{ij}}{\psi }_{ijk}^{a_{ijk}}\left(y\right)$ be the factorization of $F_{S_{ij}}\left(y\right)$ in ${\mathbb{F}}_{{\varphi }_i}\left[y\right]$. We shall use the following index theorem of Ore (see, e.g., [34], Theorem 1.7 and Theorem 1.9):

Theorem 5.

1. 

Using the above notation, we have

$${\nu }_p\left(\mathit{ind}\left(F\right)\right)\ge \sum _{i=1}^r{\mathit{ind}}_{{\varphi }_i}\left(F\right).$$

The equality holds if $F\left(x\right)$ is p-regular.

2. 

If $F\left(x\right)$ is p-regular, then

$$p{\mathbb{Z}}_K=\prod _{i=1}^r\prod _{j=1}^{r_i}\prod _{k=1}^{s_{ij}}{\mathfrak{p}}_{ijk}^{e_{ij}},$$

is the factorization of $p{\mathbb{Z}}_K$ into powers of prime ideals of ${\mathbb{Z}}_K$ lying above p, where $e_{ij}=l_{ij}/d_{ij}$, $l_{ij}$ is the length of $S_{ij}$, $d_{ij}$ is the degree of $S_{ij}$, and $f_{ijk}=deg\left({\varphi }_i\right)\times deg\left({\psi }_{ijk}\right)$ is the residue degree of the prime ideal ${\mathfrak{p}}_{ijk}$ over p.

Remark 2.

By Dedekind’s criterion, ${\nu }_p\left(\mathit{ind}\left(F\right)\right)=0$ if, and only if, ${\mathit{ind}}_{{\varphi }_i}\left(F\right)=0$ for every $i=1,\dots ,r$, which means $N_{{\varphi }_i}^{-}\left(F\right)$ has a single side of height 1 for every $i=1,\dots ,r$.

Let us illustrate this theorem with this example: Let $F\left(x\right)={\varphi }^8+(72x+90){\varphi }^5+(72x+216){\varphi }^3+(12x+48)\varphi +96$, with $\varphi =x-1$. $F\left(x\right)$ is irreducible over $\mathbb{Q}$. For $p=2$, the $\varphi$-Newton polygon of $F\left(x\right)$ is given in the following Figure 1.

In this example, ${\mathrm{ind}}_{\varphi }\left(F\right)=9$ × deg$\left(\varphi \right)=9$. Let K be the number field generated by $\alpha$ a root of $F\left(x\right)$. Since $\overline{F\left(x\right)}={\overline{\varphi }}^8$ by Theorem 5, we obtain $ind\left(\alpha \right)\ge 9$. Moreover, since $N_{\varphi }\left(F\right)=S_1+S_2+S_3$ has three sides, with respective degrees $1,2$, and 1 and $R_{{\lambda }_2}\left(F\right)\left(y\right)=y^2+y+1$ is irreducible over ${\mathbb{F}}_p$, we obtain $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3$ is a product of three prime ideals of ${\mathbb{Z}}_K$ with respective residue degrees $1,2$, and 1.

When the program of Ore fails, that is $F\left(x\right)$ is not p-regular, then in order to complete the factorization of $F\left(x\right)$, Guardia, Montes, and Nart introduced the notion of a high-order Newton polygon. They showed, thanks to a theorem on the index [31] (Theorem 4.18), that after a finite number of iterations, this process yields all monic irreducible factors of $F\left(x\right)$, all prime ideals of ${\mathbb{Z}}_K$ lying above a prime p, the index $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$, and the absolute discriminant of K. For more details, we refer to [31].

4. Examples

In this section, we give several examples that illustrate our main results.

4.1. Example 1: Some Monogenic Number Fields Defined by Non-Monogenic Trinomials

In the following statement, we give an infinite family of monogenic number fields generated by roots of non-monogenic trinomials:

Proposition 1.

Let K be the number field generated by a root α of an irreducible trinomial

$$F\left(x\right)=x^4+4a{x}^2+8b\in \mathbb{Z}\left[x\right],$$

assuming 2 does not divide b, b square-free, and for every odd prime p, if p does not divide b, then $p^2$ does not divide $a^2-2b$ or $-2^{-1}a\notin {\mathbb{F}}_p^2$.

Then, $F\left(x\right)$ is a non-monogenic polynomial, but $K=\mathbb{Q}\left(\alpha \right)$ is a monogenic number field and $\theta ={\displaystyle \frac{{\alpha }^3}{4}}$ is a generator of a power integral basis of K.

For the proof of this Proposition, see Section 5.

4.2. Example 2: Monogenic and Non-Monogenic Fields, Index Forms

Now, let K be a number field generated by a root $\alpha$ of $x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$ such that b is square-free $a\equiv 0(\mathrm{mod}2)$, $b\equiv 1(\mathrm{mod}2)$, and for every odd prime p either $p^2$ does not divide $a^2-4b$ or $-2^{-1}a\notin {\mathbb{F}}_p^2$. Then,

(1)

If ${\nu }_2(b+1+a)=1$, then by Theorem 1, $\mathbb{Z}\left[\alpha \right]$ is the ring of integers of K.

(2)

If ${\nu }_2(b+1+a)\ge 3$ and $a\equiv 0(\mathrm{mod}4)$, then by Theorem 2 (9), 2 divides $i\left(K\right)$, and so K is not mongenic.

(3)

If ${\nu }_2(b+1+a)=2$ and $a\equiv 0(\mathrm{mod}4)$, then $\overline{F\left(x\right)}={\varphi }^4$, where $\varphi =x-1$. The principal $\varphi$-Newton polygon of F with respect to 2 is $N_{\varphi }\left(F\right)=S$ having a single side joining $(0,2)$, $(2,1)$, and $(4,0)$. Thus, S is of degree 2, slope $\lambda =-1/2$, and $R_{1/2}\left(F\right)\left(y\right)=y^2+y+1$ is irreducible over ${\mathbb{F}}_{\varphi }$. Thus, $\left(1,\alpha ,{\displaystyle \frac{{\alpha }^2+1}{2}},{\displaystyle \frac{{\alpha }^3+\alpha }{2}}\right)$ is a $\mathbb{Z}$-basis of ${\mathbb{Z}}_K$.

The conditions ${\nu }_2(b+1+a)=2$ and $a\equiv 0(\mathrm{mod}4)$ imply $a=4k,b=4l-4k-1$ with integer parameters $k,l$. The index form corresponding to the above integral basis is $F_1·F_2$, where

$$\begin{gathered} F_1(x_1,x_2,x_3)=-x_1^2-x_3^{2}l+2x_3^{2}k-x_1{x}_3+2x_1{x}_{3}k, \\ {F}_2(x_1,x_2,x_3)=4x_2^4{k}^2-4x_3^{4}l-4x_2^{4}l+4x_2^{4}k+64x_3^4{k}^4-32x_3^4{k}^3+4x_3^4{l}^2+8x_1^3{x}_3+8x_1^2{x}_3^2+4x_1{x}_3^3-2x_3^2{x}_2^2 \\ -4x_3^{4}k+20x_3^4{k}^2+x_2^4+x_3^4+4x_1^4-32x_1^3{x}_{3}k-128x_1{x}_3^3{k}^3+8x_1^2{x}_2^{2}k-24x_1{x}_3^{3}k+8x_2^2{x}_3^2{k}^2 \\ +64x_1{x}_3^3{k}^2-40x_1^2{x}_3^{2}k-8x_1{x}_3^{3}l+32x_2^2{x}_3^2{k}^3+8x_3^{4}kl+8x_3^2{x}_2^{2}l-32x_3^4{k}^{2}l-8x_1^2{x}_3^{2}l+96x_1^2{k}^2{x}_3^2 \\ -4x_1{x}_2^2{x}_3+32x_1{x}_3^{3}kl-24x_2^2{x}_3^{2}kl-8x_1{x}_2^2{x}_{3}k+16x_3{x}_2^2{x}_{1}l-32x_1{x}_2^2{x}_3{k}^2. \end{gathered}$$

These number fields can be either monogenic or not.

(3.1) For $k=l=1$, we obtain the number field $K=\mathbb{Q}\left(\alpha \right)$ generated by a root $\alpha$ of $f\left(x\right)=x^4+4x^2-1$. The field K is a mixed quartic field with Galois group $D_4$ and discriminant $d_K=-400$. Up to equivalence, all generators of power integral bases are given by the solutions $(x_1,x_2,x_3)=(1,0,1),(2,0,1),(3,-1,2),(3,1,2)$ of the index form equation. (Here, and in the following, we used the algorithm described in Gaál, Pethő, and Pohst [29] (see also [3]) to solve the index form equations in order to find all generators of power integral bases of the corresponding quartic fields.)

(3.2) For $k=2,l=7$, we obtain the number field $K=\mathbb{Q}\left(\alpha \right)$ generated by a root $\alpha$ of $f\left(x\right)=x^4+8x^2+19$. The field K is a totally complex quartic field with Galois group $D_4$ and discriminant $d_K=2736$, which is not monogenic, since the index form equation has no solutions.

(4)

If ${\nu }_2(b+1+a)=3$ and $a\equiv 2(\mathrm{mod}4)$, then $\overline{F\left(x\right)}={\varphi }^4$, where $\varphi =x-1$, The principal $\varphi$-Newton polygon of f with respect to 2 is $N_{\varphi }\left(F\right)=S$ has a single side joining $(0,3)$ and $(4,0)$. Thus, S is of degree 1, and so the residual polynomial $R_{\lambda }\left(F\right)\left(y\right)$ is irreducible over ${\mathbb{F}}_{\varphi }$. Thus, $\left(1,\alpha ,{\displaystyle \frac{{\alpha }^2+1}{2}},{\displaystyle \frac{{\alpha }^3+{\alpha }^2+3\alpha +3}{4}}\right)$ is a $\mathbb{Z}$-basis of ${\mathbb{Z}}_K$.

The conditions ${\nu }_2(b+1+a)=3$ and $a\equiv 2(\mathrm{mod}4)$ imply $a=4k+2,b=8l-4k-3$ with integer parameters $k,l$. The index form corresponding to the above integral basis is $F_1·F_2$ where

$$\begin{gathered} F_1(x_1,x_2,x_3)=-2x_1^2-x_3^{2}l+2x_3^{2}k-2x_1{x}_3+2x_1{x}_{3}k, \\ {F}_2(x_1,x_2,x_3)=4x_1^4+4x_1^3{x}_3+4x_1^2{x}_3^2+16x_3{x}_2^2{x}_{1}l-12x_3{x}_2^2{x}_{1}k+16x_3^2{x}_2{x}_{1}l-12x_3^2{x}_2{x}_{1}k+8x_1{x}_3^{3}kl \\ -12x_3^2{x}_2^{2}kl-12x_3^3{x}_{2}kl-16x_3{x}_2^2{x}_1{k}^2-16x_3^2{x}_2{x}_1{k}^2+8x_3{x}_2{x}_1^{2}k+4x_1{x}_3^3{k}^2+8x_3{x}_2^3{k}^2 \\ +24x_1^2{x}_3^2{k}^2-2x_3^{4}kl-4x_1^2{x}_3^{2}l-4x_3^4{k}^{2}l-16x_1^3{x}_{3}k+4x_3^3{x}_{2}k+8x_3^2{x}_2^2{k}^3+2x_1{x}_3^{3}l \\ -8x_1^2{x}_3^{2}k+12x_3^2{x}_2^{2}k+2x_3^3{x}_{2}l-6x_3^2{x}_2^{2}l+16x_3{x}_2^{3}k+8x_3^3{x}_2{k}^2+8x_3^3{x}_2{k}^3-8x_1{x}_3^{3}k \\ +8x_2^2{x}_1^{2}k-16x_1{x}_3^3{k}^3-16x_3{x}_2^{3}l+12x_3^2{x}_2^2{k}^2+4x_3^4{k}^2+4x_3^4{k}^4+4x_2^4{k}^2+8x_2^{4}k+x_3^4{l}^2 \\ -8x_2^{4}l+4x_2^4+4x_3^2{x}_2^2+8x_3{x}_2^3+4x_2^2{x}_1^2+4x_3{x}_2{x}_1^2-4x_3{x}_2^2{x}_1-4x_3^2{x}_2{x}_1, \end{gathered}$$

These number fields can be either monogenic or not.

(4.1) For $k=l=1$, we obtain the number field $K=\mathbb{Q}\left(\alpha \right)$ generated by a root $\alpha$ of $f\left(x\right)=x^4+6x^2+1$. The field K is a totally complex quartic field with Galois group $V_4$ and discriminant $d_K=256$. Up to equivalence, all generators of power integral bases are given by the solutions $(x_1,x_2,x_3)=(1,-1,1),(1,0,1)$ of the index form equation.

(4.2) For $k=1,l=3$, we obtain the number field $K=\mathbb{Q}\left(\alpha \right)$ generated by a root $\alpha$ of $f\left(x\right)=x^4+6x^2+17$. The field K is a totally complex quartic field with Galois group $D_4$ and discriminant $d_K=4352$, which is not monogenic, since the index form equation has no solutions.

4.3. Examples 3: Applying Engstrom’s Results

Let K be a number field generated by a root of the irreducible polynomial $F\left(x\right)=x^4+a{x}^2+b\in \mathbb{Z}\left[x\right]$. In the following examples, we show that $i\left(K\right)\ne 1$, which implies that K is not monogenic. In all these examples, we shall use the result of Engstrom [21] (p. 234), stating that the exponents of 2 and 3 in the field index $i\left(K\right)$ of K only depend on the type of factorizations of $2{\mathbb{Z}}_K$ and $3{\mathbb{Z}}_K$ into prime ideals of ${\mathbb{Z}}_K$.

• If $a=-55$ and $b=-99$, then $F\left(x\right)$ is 11-Eisenstein, and so $F\left(x\right)$ is irreducible over $\mathbb{Q}$. Since $a\equiv 1(\mathrm{mod}8)$ and $b\equiv 5(\mathrm{mod}8)$, by Theorem 2 (1) and its proof, 2 divides $i\left(K\right)$ and $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each prime factor. By Engstrom’s theorem, we conclude that ${\nu }_2\left(i\left(K\right)\right)=1$. Again, by Theorem 3 (1) and its proof, 3 divides $i\left(K\right)$ and $3{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. By Engstrom’s theorem, we conclude that ${\nu }_3\left(i\left(K\right)\right)=1$. Hence, $i\left(K\right)=6$.
• Similarly, if $a=17$ and $b=765$, then $F\left(x\right)$ is irreducible over $\mathbb{Q}$. Since $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each prime factor and $3{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor, we conclude by Engstrom’s theorem that ${\nu }_2\left(i\left(K\right)\right)=1$, ${\nu }_3\left(i\left(K\right)\right)=1$. Hence, $i\left(K\right)=6$.
• If $a=3$ and $b=4$, then $F\left(x\right)$ is irreducible over $\mathbb{Q}$. Since ${\nu }_2\left(b\right)=2$ and $a+b_2\equiv 4(\mathrm{mod}8)$, we conclude that $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. By Engstrom’s theorem, it implies ${\nu }_2\left(i\left(K\right)\right)=2$. For $p=3$, since 3 divides a and does not divide b, hence, 3 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$. Thus, by ${\nu }_3\left(i\left(K\right)\right)=0$, ${\nu }_2\left(i\left(K\right)\right)=2$, we have $i\left(K\right)=4$.
• If $a=-1$ and $b=304$, then $F\left(x\right)$ is irreducible over $\mathbb{Q}$. Since ${\nu }_2\left(b\right)=4$ and $1+a+b\equiv 0$$(\mathrm{mod}8)$, we conclude that $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. For $p=3$, since 3 divides $a\equiv -1(\mathrm{mod}3)$, ${\nu }_3\left(b\right)=2$, and $b_3\equiv 1(\mathrm{mod}3)$, we have $3{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. By Engstrom’s theorem, we have ${\nu }_3\left(i\left(K\right)\right)=1$, ${\nu }_2\left(i\left(K\right)\right)=2$; hence, $i\left(K\right)=12$.
• If $a=4$ and $b=-86$, then $F\left(x\right)$ is 2-Eisenstein, and so $F\left(x\right)$ is irreducible over $\mathbb{Q}$ and 2 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$. Since $a\equiv 1(\mathrm{mod}3)$ and $b+1+a\equiv 0(\mathrm{mod}27)$, we conclude that $3{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. By Engstrom’s theorem, it implies ${\nu }_3\left(i\left(K\right)\right)=1$. By ${\nu }_2\left(i\left(K\right)\right)=0$, ${\nu }_3\left(i\left(K\right)\right)=1$, we have $i\left(K\right)=3$.
• If $a=-1$ and $b=304$, then $F\left(x\right)$ is irreducible over $\mathbb{Q}$. Since ${\nu }_2\left(b\right)=4$ and $1+a+b\equiv 0$$(\mathrm{mod}8)$, we conclude that $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. For $p=3$, since 3 divides $a\equiv -1(\mathrm{mod}3)$, ${\nu }_3\left(b\right)=2$, and $b_3\equiv 1(\mathrm{mod}3)$, we have $3{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime factor. By Engstrom’s theorem, we have ${\nu }_3\left(i\left(K\right)\right)=1$, ${\nu }_2\left(i\left(K\right)\right)=2$; hence, $i\left(K\right)=12$.

5. Proofs of Our Main Results

Proof of Theorem 1.

Let p be a prime integer candidate to divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$. Since $▵=16b{(a^2-4b)}^2$ is the discriminant of $F\left(x\right)$, we conclude that $p=2$, or p divides b or p divides $a^2-4b$.

• If p divides both of a and b, then $\overline{F\left(x\right)}={\varphi }^4$, where $\varphi =x$. If ${\nu }_p\left(b\right)=1$, then by Remark 2, ${\nu }_p\left(({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])\right)=0$ if, and only if, ${\nu }_p\left(b\right)=1$.
• For $p=2$,

  (a)

  If 2 divides a and b, then by the first point, 2 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$ if, and only if, ${\nu }_2\left(b\right)=1$.

  (b)

  If 2 divides b and does not a, then $\overline{F\left(x\right)}=x^2{(x+1)}^2$. Let $\varphi =x-1$. Then, $F\left(x\right)={\varphi }^4+4{\varphi }^3+(6+a){\varphi }^2+(4+2a)\varphi +(1+b+a)$. Thus, by Remark 2, we conclude that 2 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$ if, and only if, ind_x$\left(F\right)=0$ and ind_ϕ$\left(F\right)=0$, which means that ${\nu }_2\left(b\right)=1$ and ${\nu }_2(b+a+1)=1$. That is, $b\equiv 2(\mathrm{mod}4)$ and $a\equiv 2-(1+b)\equiv 3(\mathrm{mod}4)$.

  (c)

  If 2 divides a and does not divide b, then $\overline{F\left(x\right)}={(x+1)}^4$. Let $\varphi =x-1$. Then, $F\left(x\right)={\varphi }^4+4{\varphi }^3+(6+a){\varphi }^2+(4+2a)\varphi +(1+b+a)$. By Remark 2, we conclude that 2 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$ if, and only if, ${\nu }_2(b+a+1)=1$, which means $b\equiv 1-a(\mathrm{mod}4)$.

  (d)

  If 2 does not divide $ab$, then $\overline{F\left(x\right)}={(x^2+x+1)}^2$. Let $\varphi =x^2+x+1$ and $F\left(x\right)={\varphi }^2+(a-1-2x)\varphi +(1-a)x+b-a$. Thus, by Remark 2, 2 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$ if, and only if, ${\nu }_2(a-1)=1$ or ${\nu }_2(a-b)=1$, which means $a\equiv 3(\mathrm{mod}4)$ or $b\equiv 2+a(\mathrm{mod}4)$.

• If p is odd, p divides b and does not divide a, then $\overline{F\left(x\right)}=x^2(x^2+a)$. Since $x^2+a$ is square free in ${\mathbb{F}}_p\left[x\right]$, then by Remark 2, p does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$ if, and only if, ind_x$\left(F\right)=0$. That is, ${\nu }_p\left(b\right)=1$.
• Now assume that p is odd and p does not divide $ab$. If p does not divide $a^2-4b$, then $p^2$ does not divide $▵\left(F\right)$, and so p does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$. Also since $F^{\prime }\left(x\right)=2x(2x^2+a)$, if $-2^{-1}a\notin {\mathbb{F}}_p^2$, then $\overline{F\left(x\right)}$ is square free in ${\mathbb{F}}_p\left[x\right]$. Thus, by Dedekind’s criterion, we conclude that p does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$. If p divides $a^2-4b$ and $-2^{-1}a\in {\mathbb{F}}_p^2$, then by Hensel’s lemma, let $u\in {\mathbb{Z}}_p$ such that $2u^2+a\equiv 0(\mathrm{mod}{p}^r)$ with $r\ge 2$ a positive integer. Then, $2^{2}F\left(u\right)={(-a)}^2+2a·(-a)+4b\equiv -(a^2-4b)(\mathrm{mod}{p}^r)$. Let $\varphi =x-u$ and $F\left(x\right)={\varphi }^4+a_3{\varphi }^3+a_2{\varphi }^2+a_1\varphi +a_0$ be the $\varphi$-expansion of $F\left(x\right)$ with $a_1=F^{\prime }\left(u\right)$ and $a_0=F\left(u\right)$. Thus, $a_1\equiv F^{\prime }\left(u\right)\equiv 0(\mathrm{mod}{p}^2)$ and $a_0\equiv 0(\mathrm{mod}p)$. By Remark 2, $\mathrm{ind}\left(F\right)=0$ if, and only if, ${\nu }_p\left(F\left(u\right)\right)=1$, which means that $p^2$ does not divide $a^2-4b$.

□

For the proof of Theorems 2 and 3, based on Engstrom’s results, we need to factorize $2{\mathbb{Z}}_K$ and $3{\mathbb{Z}}_K$ into the product of powers of prime ideals of ${\mathbb{Z}}_K$. Recall the following lemma, which characterizes the prime index divisors of K. Its proof is an immediate consequence of Dedekind’s theorem:

Lemma 1.

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

Proof of Theorem 2.

By virtue of Engstrom’s results [21], we need to factorize $2{\mathbb{Z}}_K$ into powers of prime ideals of ${\mathbb{Z}}_K$. Moreover, 2 divides $i\left(K\right)$ if, and only if, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime ideal or $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each prime ideal or also $2{\mathbb{Z}}_K={\mathfrak{p}}_1^2{\mathfrak{p}}_2{\mathfrak{p}}_3$ with residue degree 1 each prime ideal. If $\mathbb{Z}\left[\alpha \right]$ is integrally closed, then 2 does not divide the index $i\left(K\right)$. Thus, according to Theorem 1, we deal only with the following cases:

• If ${\nu }_2\left(b\right)\ge 2$ and ${\nu }_2\left(a\right)\ge 1$, then $\overline{F\left(x\right)}={\varphi }^4$ in ${\mathbb{F}}_2\left[x\right]$, where $\varphi =x$. If $N_{\varphi }\left(F\right)=S$ has a single side, then let d be the degree of S. Since ${\nu }_2\left(a\right)\le 1$ or ${\nu }_2\left(b\right)\le 3$, we conclude that $d\in \{1,2\}$. If $d=1$, then there is a unique prime ideal of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1. If $d=2$ (${\nu }_2\left(a\right)\ge 1$ and ${\nu }_2\left(b\right)=2$), then there is a unique prime ideal of ${\mathbb{Z}}_K$ lying above 2 with residue degree 2 and ramification index 2 or two prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1 and ramification index 2 each. If $N_{\varphi }\left(F\right)=S_1+S_2$ has two sides, that is ${\nu }_2\left(a\right)=1$ and ${\nu }_2\left(b\right)\ge 3$, then $S_2$ is of degree 1 and it yields a unique prime ideal of ${\mathbb{Z}}_K$ lying above 2 and the degree of $S_1$ is 1 or 2. If ${\nu }_2\left(b\right)$ is even, then the degree of $S_1$ is 1 and there are two prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1 and ramification index 2 each. If ${\nu }_2\left(b\right)=2k+1$ for some positive integer k, then the degree of $S_1$ is 2. Let $\varphi =x+2^k$ and $F\left(x\right)={\varphi }^4+2^{k+2}{\varphi }^3+(a+3·2^{2k+2}){\varphi }^2+(2^{k+1}a+2^{3k+2})\varphi +(2^{4k}+2^{2k}a+b)$ be the $\varphi$-expansion of $F\left(x\right)$. Since ${\nu }_2(a+3·2^{2k+2})=1$, ${\nu }_2(2^{k+1}a+2^{3k+2})=k+2$ and ${\nu }_2(2^{4k}+2^{2k}a+b)\ge 2k+3$, we conclude that 2 divides $i\left(K\right)$ if, and only if, $N_{\varphi }\left(F\right)$ has three sides. That is, ${\nu }_2(2^{4k}+2^{2k}a+b)\ge 2k+4$, which means that $k\ge 2$ and ${\nu }_2(2^{2k}a+b)\ge 2k+4$ or $k=1$ and ${\nu }_2(2^4+4a+b)=6$ ($b\equiv 48-4a(\mathrm{mod}64)$). In these cases, there are three prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1 each. Thus, ${\nu }_2\left(i\left(K\right)\right)=1$.
• If ${\nu }_2\left(b\right)\ge 1$ and $a\equiv 1(\mathrm{mod}2)$, then $\overline{F\left(x\right)}={(x·\varphi )}^2$ in ${\mathbb{F}}_2\left[x\right]$, where $\varphi =x-1$. We have the following cases:

  (a)

  If ${\nu }_2\left(b\right)=1+2k$ for some positive integer k, then x provides a unique prime ideal of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1. In this case, 2 is a common index divisor of K if, and only if, $\varphi$ provides two prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1 each. For this reason, let $F\left(x\right)={\varphi }^4+4{\varphi }^3+(a+6){\varphi }^2+(2a+4)\varphi +(b+a+1)$. Since ${\nu }_2(2a+4)=1$, we conclude that 2 is a common index divisor of K if, and only if, ${\nu }_2(b+a+1)\ge 3$. That is, ${\nu }_2\left(b\right)=1+2k$ and $a\equiv 7-b(\mathrm{mod}8)$. In this case, ${\nu }_2\left(i\left(K\right)\right)=1$.

  (b)

  If ${\nu }_2\left(b\right)=2k$ for some positive integer k, then $N_x^{-}\left(F\right)=S$ has a single side joining $(0,2k)$ and $(2,0)$. Let ${\varphi }_2=x+2^k$ and $F\left(x\right)={\varphi }_2^4-2^{k+2}{\varphi }_2^3+(a+3·2^{2k+1}){\varphi }_2^2-(2^{k+1}a+2^{3k+2}){\varphi }_2+(b+2^{2k}a+2^{4k})$. Since ${\nu }_2(2^{k+1}a+2^{3k+2})=k+1$ and ${\nu }_2(b+2^{2k}a+2^{4k})\ge 2k+1$, we conclude that 2 is a common index divisor of K if, and only if, ${\nu }_2(b+a+1)\ge 3$ and ${\nu }_2(b+2^{2k}a)=2k+1$ or ${\nu }_2(b+a+1)=1$ and ${\nu }_2(b+2^{2k}a)\ge 2k+3$ or ${\nu }_2(b+a+1)=2$ and ${\nu }_2(b+2^{2k}a)=2k+2$ or also ${\nu }_2(b+a+1)\ge 3$ and ${\nu }_2(b+2^{2k}a))\ge 2k+3$ (in this last case, ${\nu }_2\left(i\left(K\right)\right)=2$).

  Remark that if $k=1$, then

  ${\nu }_2(b+a+1)\ge 3\mathrm{and}{\nu }_2(b+2^{2}a)=3⟺$

  $a\equiv 3(\mathrm{mod}8)\mathrm{and}b\equiv 4(2-a\left)(\mathrm{mod}16\right),$

  ${\nu }_2(b+a+1)=1\mathrm{and}{\nu }_2(b+2^{2}a)\ge 5⟺$

  $a\equiv 1(\mathrm{mod}4)\mathrm{and}b\equiv -4a(\mathrm{mod}32),$

  ${\nu }_2(b+a+1)=2\mathrm{and}{\nu }_2(b+2^{2}a)=4⟺$

  $a\equiv 7(\mathrm{mod}8)\mathrm{and}b\equiv 4(4-a\left)(\mathrm{mod}32\right)$,

  ${\nu }_2(b+a+1)\ge 3\mathrm{and}{\nu }_2(b+2^{2}a)\ge 5⟺$

  $a\equiv 3(\mathrm{mod}8)\mathrm{and}b\equiv -4a(\mathrm{mod}32).$

  For $k\ge 2$, then

  ${\nu }_2(b+a+1)\ge 3\mathrm{and}{\nu }_2(b+2^{2}a)=2k+1⟺$

  $a\equiv 7(\mathrm{mod}8)\mathrm{and}b\equiv 2^{2k}(2-a)(\mathrm{mod}{2}^{2k+2}),$

  ${\nu }_2(b+a+1)=1\mathrm{and}{\nu }_2(b+2^{2}a)\ge 2k+3⟺$

  $a\equiv 1(\mathrm{mod}4)\mathrm{and}b\equiv -2^{2k}a(\mathrm{mod}{2}^{2k+3}),$

  ${\nu }_2(b+a+1)=2\mathrm{and}{\nu }_2(b+2^{2k}a)=2k+2⟺$

  $a\equiv 3(\mathrm{mod}8)\mathrm{and}b\equiv 2^{2k}(4-a)(\mathrm{mod}{2}^{2k+3}),$

  ${\nu }_2(b+a+1)\ge 3\mathrm{and}{\nu }_2(b+2^{2k}a)\ge 2k+3⟺$

  $a\equiv 7(\mathrm{mod}8)\mathrm{and}b\equiv -2^{2k}a(\mathrm{mod}{2}^{2k+3}).$

• If ${\nu }_2\left(b\right)=0$ and ${\nu }_2\left(a\right)\ge 1$, then $\overline{F\left(x\right)}={\varphi }^4$ in ${\mathbb{F}}_2\left[x\right]$, where $\varphi =x-1$. Let $F\left(x\right)={\varphi }^4+4{\varphi }^3+(a+6){\varphi }^2+(2a+4)\varphi +(b+a+1)$.

  (a)

  For ${\nu }_2\left(a\right)\ge 2$, we have ${\nu }_2(a+6)=1$ and ${\nu }_2(2a+4)=2$. It follows that 2 is a common divisor of K if, and only if, ${\nu }_2(b+a+1)\ge 4$. That is, $b\equiv 15-a(\mathrm{mod}16)$. In this case, ${\nu }_2\left(i\left(K\right)\right)=1$.

  (b)

  For $a\equiv 2(\mathrm{mod}8)$, let $C=(b+a+1)$, $B=(2a+4)$ and $A=(a+6)$. According to $a\equiv 2(\mathrm{mod}16)$ or $a\equiv 10(\mathrm{mod}16)$, we obtain ${\nu }_2\left(B\right)=3$ and ${\nu }_2\left(A\right)\ge 3$. It follows that if ${\nu }_2(a+b+1)\in \{1,3\}$, then $N_{\varphi }^{-}\left(F\right)=S$ has a single side of degree 1. Thus, there is a single prime ideal of ${\mathbb{Z}}_K$ lying above 2. If ${\nu }_2(a+b+1)=2$, then $N_{\varphi }^{-}\left(F\right)=S$ has a single side of degree 2 and $R_{\lambda }\left(F\right)\left(y\right)={(y+1)}^2$. Thus, we have to use second-order Newton’s polygon techniques. Since $-\lambda =-1/2$ is the slope of S, we conclude that 2 divides the ramification index $e\left(\mathfrak{p}\right)$ of $\mathfrak{p}$ over 2 for every prime ideal $\mathfrak{p}$ of ${\mathbb{Z}}_K$ lying above 2. Thus, the factorization of $2{\mathbb{Z}}_K$ has one of these forms: ${\mathfrak{p}}_1^4$, ${\mathfrak{p}}_1^2{\mathfrak{p}}_2^2$ or ${\mathfrak{p}}_3^2$ with $f\left({\mathfrak{p}}_1\right)=f\left({\mathfrak{p}}_2\right)=1$ and $f\left({\mathfrak{p}}_3\right)=2$ are the residue degrees. In all these cases, 2 is not a common divisor of K. If ${\nu }_2(a+b+1)=4$, then $N_{\varphi }^{-}\left(F\right)=S$ has a single side of degree 4 and $R_{\lambda }\left(F\right)\left(y\right)=y^4+y+1$ is irreducible over ${\mathbb{F}}_{\varphi }={\mathbb{F}}_2$. So, there is a single prime ideal of ${\mathbb{Z}}_K$ lying above 2. If ${\nu }_2(a+b+1)\ge 5$, then $N_{\varphi }^{-}\left(F\right)=S_1+S_2$ has two sides such that $S_1$ is of degree 1, $S_2$ is of degree 3 and $R_{{\lambda }_2}\left(F\right)\left(y\right)=y^3+1=(y+1)(y^2+y+1)$ in ${\mathbb{F}}_{\varphi }\left[y\right]$. Hence, there are exactly two prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1 each and a single prime ideal of ${\mathbb{Z}}_K$ lying above 2 with residue degree 2. So, ${\nu }_2\left(i\left(K\right)\right)=0$.

  (c)

  For $a\equiv 6(\mathrm{mod}8)$, we have $F\left(x\right)={\varphi }^4+4{\varphi }^3+(a+6){\varphi }^2+(2a+4)\varphi +(b+a+1)$ with $\varphi =x-1$. If ${\nu }_2(b+a+1)\in \{1,3\}$, then $2{\mathbb{Z}}_K={\mathfrak{p}}^4$ with $\mathfrak{p}$ the prime ideal of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1. In this case, 2 does not divide $i\left(K\right)$.

  If ${\nu }_2(b+a+1)=2$, then $N_{\varphi }\left(F\right)=S$ has a single side with $R_{\lambda }\left(F\right)\left(y\right)={(y+1)}^2$. Thus, there are two cases, which are as follows: Either there are two prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 1 and ramification index 2 each. Or there is a unique prime ideal of ${\mathbb{Z}}_K$ lying above 2 with residue degree 2 and ramification index 2. Again, in this case, 2 does not divide $i\left(K\right)$.

  (d)

  If ${\nu }_2(b+a+1)\ge 4$, then let $G\left(x\right)=x^4+2x^3+A{x}^2+Bx+C$ be the minimal polynomial of $\theta ={\displaystyle \frac{\alpha -1}{2}}$ over $\mathbb{Q}$, where $A={\displaystyle \frac{a+6}{4}}$, $B={\displaystyle \frac{2a+4}{8}}$, and $C={\displaystyle \frac{1+b+a}{16}}$. Since $a\equiv 6(\mathrm{mod}8)$ and ${\nu }_2(b+a+1)\ge 4$, $G\left(x\right)\in \mathbb{Z}\left[x\right]$, $\theta \in {\mathbb{Z}}_K$ is a primitive element of K, $A\equiv 1(\mathrm{mod}2)$ and we can replace $F\left(x\right)$ by $G\left(x\right)$. Since $a\equiv 6(\mathrm{mod}8)$, then ${\nu }_2\left(A\right)={\nu }_2(a+6)-2=0$ and ${\nu }_2(2a+4)\ge 4$. So, if ${\nu }_2(b+a+1)=4$, then $\overline{G\left(x\right)}={(x^2+x+1)}^2$. Let $\varphi =x^2+x+1$ and $G\left(x\right)={\varphi }^2+{\displaystyle \frac{a-6}{4}}\varphi +{\displaystyle \frac{b-3a+9}{16}}$. It follows that:

  • If ${\nu }_2(b-3a+9)=5$, then 2 does not divide ind$\left(F\right)$, and so ${\nu }_2\left(i\left(K\right)\right)=0$.
  • If ${\nu }_2(b-3a+9)=6$ and $a\equiv 6(\mathrm{mod}32)$, then for $\varphi =x^2+x-1$, $G\left(x\right)={\varphi }^2+{\displaystyle \frac{a+10}{4}}+{\displaystyle \frac{b+5a+25}{16}}$. Since $b+5a+25\equiv b-3a+9+8a+16\equiv 8(a+10)(\mathrm{mod}128)$, we conclude that ${\nu }_2\left({\displaystyle \frac{b+5a+25}{16}}\right)\ge 3$. As ${\nu }_2\left({\displaystyle \frac{a+10}{4}}\right)=2$, then ${\nu }_2\left(i\left(K\right)\right)=1$ if, and only if, ${\nu }_2\left(b+5a+25\right)\ge 8$, which means $b\equiv 231-5a(\mathrm{mod}256)$.
  • If ${\nu }_2(b-3a+9)=7$ and $a\equiv 6(\mathrm{mod}16)$, then for $\varphi =x^2+x+1$, $N_{\varphi }^{-}\left(G\right)$ has a single side of degree 1 and so, ${\nu }_2\left(i\left(K\right)\right)=0$.
  • If ${\nu }_2(b-3a+9)\ge 8$ and $a\equiv 6(\mathrm{mod}32)$, then let $k={\nu }_2(a-6)-2$ and $v={\nu }_2(b-3a+9)-4$. If $2k\le v$, then $N_{\varphi }^{-}\left(G\right)$ has two sides, and so $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each. Therefore, ${\nu }_2\left(i\left(K\right)\right)=1$.
    If $v<2k$ and v is odd, then $N_{\varphi }^{-}\left(G\right)$ has a single side of degree 1, $2{\mathbb{Z}}_K={\mathfrak{p}}_1^2$, and so ${\nu }_2\left(i\left(K\right)\right)=0$.
    If $v=2k$, then $N_{\varphi }^{-}\left(G\right)$ has a single side with $R_{\lambda }\left(G\right)\left(y\right)=y^2+y+1=(y-x)(y-x^2)$, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each, and so ${\nu }_2\left(i\left(K\right)\right)=1$.
    If $v=2j$ for some positive integer j and $v<2k$, then let $\varphi =x^2+x+1+2^j$ and $G\left(x\right)={\varphi }^2+({\displaystyle \frac{a-6}{4}}+2^{j+1})\varphi +{\displaystyle \frac{b-3a+9-(a-6)2^{j+2}+2^{2j+4}}{16}}$. So, If $j<k-1$, then ${\nu }_2({\displaystyle \frac{a-6}{4}}+2^{j+1})=j+1$ and
    ${\nu }_2\left({\displaystyle \frac{b-3a+9-(a-6)2^{j+2}+2^{2j+4}}{16}}\right)\ge 2j+1$. Analogously to the previous point, ${\nu }_2\left(i\left(K\right)\right)=1$ if, and only if,
    ${\nu }_2\left({\displaystyle \frac{b-3a+9-(a-6)2^{j+2}+2^{2j+4}}{16}}\right)\ge 2j+2$. Say ${(b-3a+9)}_2\equiv 3(\mathrm{mod}4)$.
    If $k=j+1$, then ${\nu }_2({\displaystyle \frac{a-6}{4}}+2^{j+1})\ge j+1$ and
    ${\nu }_2\left({\displaystyle \frac{b-3a+9-(a-6)2^{j+2}+2^{2j+4}}{16}}\right)=2j-1$ is odd. Thus, $N_{\varphi }^{-}\left(G\right)$ has a single side of degree 1, $2{\mathbb{Z}}_K={\mathfrak{p}}_1^2$, and so ${\nu }_2\left(i\left(K\right)\right)=0$.
  • If ${\nu }_2(b-3a+9)=6$ and $a\equiv 22(\mathrm{mod}32)$, then for $\varphi =x^2+x+3$, $G\left(x\right)={\varphi }^2+{\displaystyle \frac{a-22}{4}}+{\displaystyle \frac{b-11a+121}{16}}$. let $k={\nu }_2(a-22)-2$ and $v={\nu }_2(b-11a+121)-4$. If $2k\le v$, then $N_{\varphi }^{-}\left(G\right)$ has two sides, and so $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each. Therefore, ${\nu }_2\left(i\left(K\right)\right)=1$.
    If $v<2k$ and v is odd, then $N_{\varphi }^{-}\left(G\right)$ has a single side of degree 1, $2{\mathbb{Z}}_K={\mathfrak{p}}_1^2$, and so ${\nu }_2\left(i\left(K\right)\right)=0$.
    If $v=2k$, then $N_{\varphi }^{-}\left(G\right)$ has a single side with $R_{\lambda }\left(G\right)\left(y\right)=y^2+y+1=(y-x)(y-x^2)$, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each, and so ${\nu }_2\left(i\left(K\right)\right)=1$.
    If $v=2j$ for some positive integer j and $v<2k$, then let $\varphi =x^2+x+3+2^j$ and
    $G\left(x\right)={\varphi }^2+({\displaystyle \frac{a-22}{4}}+2^{j+1})\varphi +{\displaystyle \frac{b-11a+121-(a-22)2^{j+2}+2^{2j+4}}{16}}$. So, if $j<k-1$, then ${\nu }_2({\displaystyle \frac{a-22}{4}}+2^{j+1})=j+1$ and
    ${\nu }_2\left({\displaystyle \frac{b-11a+121-(a-22)2^{j+2}+2^{2j+4}}{16}}\right)\ge 2j+1$. Analogously to the previous point, ${\nu }_2\left(i\left(K\right)\right)=1$ if, and only if,
    ${\nu }_2\left({\displaystyle \frac{b-11a+121-(a-22)2^{j+2}+2^{2j+4}}{16}}\right)\ge 2j+2$, say ${(b-11a+121)}_2\equiv 3(\mathrm{mod}4)$.
    If $k=j+1$, then ${\nu }_2({\displaystyle \frac{a-22}{4}}+2^{j+1})\ge j+1$ and
    ${\nu }_2\left({\displaystyle \frac{b-3a+9-(a-22)2^{j+2}+2^{2j+4}}{16}}\right)=2j-1$ is odd. Thus, $N_{\varphi }^{-}\left(G\right)$ has a single side of degree 1, $2{\mathbb{Z}}_K={\mathfrak{p}}_1^2$, and so ${\nu }_2\left(i\left(K\right)\right)=0$.
  • If ${\nu }_2(b-3a+9)\ge 8$ and $a\equiv 22(\mathrm{mod}32)$, then for $\varphi =x^2+x+1$, $G\left(x\right)={\varphi }^2+{\displaystyle \frac{a-6}{4}}+{\displaystyle \frac{b-3a+9}{16}}$. Since ${\nu }_2(a-6)-2=2$ and $v={\nu }_2(b-3a+9)-4\ge 4$, then $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each, and so ${\nu }_2\left(i\left(K\right)\right)=1$.
  • If $a\in \{14,30\}(\mathrm{mod}32)$, then ${\nu }_2\left({\displaystyle \frac{a-6}{4}}\right)=1$. It follows that if ${\nu }_2(b-3a+9)=6$, then $N_{\varphi }^{-}\left(G\right)$ has a single side with $R_{\lambda }\left(G\right)\left(y\right)=y^2+y+1=(y-x)(y-x^2)$. Therefore, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each, and so ${\nu }_2\left(i\left(K\right)\right)=1$. Finally, if ${\nu }_2(b-3a+9)\ge 7$, then $N_{\varphi }^{-}\left(G\right)$ has two sides with degree 1 each. Thus, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$ with residue degree 2 each, and so ${\nu }_2\left(i\left(K\right)\right)=1$.

• If ${\nu }_2\left(ab\right)=0$, then $\overline{F\left(x\right)}={\varphi }^2$ in ${\mathbb{F}}_2\left[x\right]$, where $\varphi =x^2+x+1$. In this case, 2 is a divisor of K if, and only if, $\varphi$ provides two prime ideals of ${\mathbb{Z}}_K$ lying above 2 with residue degree 2 each. Let $F\left(x\right)={\varphi }^2+(-2x+a-1)\varphi +(1-a)x+b-a$ be the $\varphi$-expansion of $F\left(x\right)$. It follows that:

  (a)

  If $a\not\equiv b(\mathrm{mod}4)$ or $a\equiv 3(\mathrm{mod}4)$, then by Theorem 1, 2 does not divide $i\left(K\right)$.

  (b)

  If $b\equiv a\equiv 1(\mathrm{mod}8)$, then $N_{\varphi }\left(F\right)$ has two sides joining $(0,v)$, $(1,1)$, and $(2,0)$ with $v\ge 3$. Thus, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$, with ${\mathfrak{p}}_i$ a prime ideal of ${\mathbb{Z}}_K$ of residue degree 2 for each $i=1,2$.

  (c)

  If $b\equiv 1(\mathrm{mod}8)$ and $a\equiv 5(\mathrm{mod}8)$, then $N_{\varphi }\left(F\right)$ has a single side joining $(0,2)$ and $(2,0)$, with residue degree 2 and $R_{\lambda }\left(F\right)=y^2+xy+x+1=(y-1)(y-x^2)$ in ${\mathbb{F}}_{\varphi }\left[y\right]$. Thus, $2{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2$, with ${\mathfrak{p}}_i$ a prime of ${\mathbb{Z}}_K$ of residue degree 2 for each i. In these cases, ${\nu }_2\left(i\left(K\right)\right)=1$.

  (d)

  If $b\equiv 5(\mathrm{mod}8)$ and $a\equiv 1(\mathrm{mod}8)$, then $N_{\varphi }\left(F\right)$ has a single side joining $(0,2)$ and $(2,0)$, with residue degree 2 and $R_{\lambda }\left(F\right)=y^2+xy+1$, which is irreducible over ${\mathbb{F}}_{\varphi }$. Thus, by Theorem 5, $2{\mathbb{Z}}_K={\mathfrak{p}}^2$, with $\mathfrak{p}$ a prime ideal of ${\mathbb{Z}}_K$ of residue degree 4. Similarly, if $b\equiv a\equiv 5(\mathrm{mod}8)$, then $N_{\varphi }\left(F\right)$ has a single side joining $(0,2)$, $(1,1)$, and $(2,0)$, with residue degree 2 and $R_{\lambda }\left(F\right)=y^2+xy+x$, which is irreducible over ${\mathbb{F}}_{\varphi }$. Thus, $2{\mathbb{Z}}_K=\mathfrak{p}$, with $\mathfrak{p}$ a prime ideal of ${\mathbb{Z}}_K$ of residue degree 4. In these cases, ${\nu }_2\left(i\left(K\right)\right)=0$.

□

Proof of Theorem 3.

By virtue of Lemma 1, we need to factorize $3{\mathbb{Z}}_K$ into powers of prime ideals of ${\mathbb{Z}}_K$. More precisely, 3 divides $i\left(K\right)$ if, and only if, $3{\mathbb{Z}}_K={\mathfrak{p}}_1{\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_4$ with residue degree 1 each prime ideal. Moreover, if this holds, then $i\left(K\right)=1$. Since $▵=2^{4}b{(a^2-4b)}^2$ is the discriminant of $F\left(x\right)$, if 3 divides $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$, then 9 divides b or 3 divides $a^2-4b$. If 3 does not divide b, then $\overline{F\left(x\right)}$ has a square factor in ${\mathbb{F}}_3\left[x\right]$ if, and only if, $-a\equiv b\equiv 1(\mathrm{mod}3)$ or $a\equiv b\equiv 1(\mathrm{mod}3)$. Else, $\overline{F\left(x\right)}$ is square free, and so by Dedekind’s criterion, 3 does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])$. Therefore, 3 does not divide $i\left(K\right)$.

• If $-a\equiv b\equiv 1(\mathrm{mod}3)$, then $\overline{F\left(x\right)}={(x^2+1)}^2$ in ${\mathbb{F}}_3\left[x\right]$. Thus, there are at most two prime ideals of ${\mathbb{Z}}_K$ lying above 3.
• If $a\equiv b\equiv 1(\mathrm{mod}3)$, then ${\nu }_3(a^2-4b)\ge 1$. Since $F^{\prime }\left(x\right)=2x(2x^2+a)$ has three simple roots in ${\mathbb{F}}_3$, namely, 0, 1, and $-1$, then by Hensel’s lemma, let $s\in \mathbb{Z}$ such that $2s^2+a\equiv 0(\mathrm{mod}{3}^r)$ with $r={\nu }_3(▵)+1$. Then, $\overline{F\left(x\right)}={(x-s)}^2{(x+s)}^2$ in ${\mathbb{F}}_3\left[x\right]$. Let ${\varphi }_1=x-s$, ${\varphi }_2=x+s$, $F\left(x\right)={\varphi }_1^4+4s{\varphi }_1^3+(6s^2+a){\varphi }_1^2+(2as+4s^3){\varphi }_1+(b+a{s}^2+s^4)$ and $F\left(x\right)={\varphi }_2^4-4s{\varphi }_2^3+(6s^2+a){\varphi }_2^2-(2as+4s^3){\varphi }_2+(b+a{s}^2+s^4)$. Then, $4(b+a{s}^2+s^4)=2^{2}F\left(s\right)\equiv -(a^2-4b)(\mathrm{mod}{3}^r)$ and $F^{\prime }\left(s\right)\equiv 0(\mathrm{mod}{3}^r)$. So, $N_{{\varphi }_i}^{-}\left(F\right)=S_i$ has a single side joining $(0,{\nu }_3(a^2-4b)$ and $(2,0)$. It follows that if ${\nu }_3(a^2-4b)$ is odd, then the degree of $S_i$ is 1, and so ${\varphi }_i$ provides two prime ideals of ${\mathbb{Z}}_K$ lying above 3. If ${\nu }_3(a^2-4b)$ is even, then $R_{{\lambda }_i}\left(F\right)\left(y\right)=y^2-{(a^2-4b)}_3$ for every $i=1,2$. Therefore, each ${\varphi }_i$ provides two prime ideals of ${\mathbb{Z}}_K$ lying above 3 if, and only if, ${(a^2-4b)}_3\equiv 1(\mathrm{mod}3)$.
• If ${\nu }_3\left(b\right)\ge 1$ and $a\equiv 1(\mathrm{mod}3)$, then $\overline{F\left(x\right)}=x^2(x^2+1)$ in ${\mathbb{F}}_3\left[x\right]$. Since $x^2+1$ is irreducible in ${\mathbb{F}}_3\left[x\right]$, then there are at most three prime ideals of ${\mathbb{Z}}_K$ lying above 3.
• If ${\nu }_3\left(b\right)=2k+1$ is odd and $a\equiv -1(\mathrm{mod}3)$, then $\overline{F\left(x\right)}=x^2(x+1)(x-1)$ in ${\mathbb{F}}_3\left[x\right]$. Since ${\nu }_3\left(b\right)=1$, then the factor x provides a unique prime ideal of ${\mathbb{Z}}_K$ lying above 3. Thus, there are at most three prime ideals of ${\mathbb{Z}}_K$ lying above 3.
• If ${\nu }_3\left(b\right)=2k$ for an odd natural integer k and $a\equiv -1(\mathrm{mod}3)$, then $\overline{F\left(x\right)}=x^2(x+1)$$(x-1)$ in ${\mathbb{F}}_3\left[x\right]$. Since ${\nu }_3\left(b\right)=2k$, $N_x^{-}\left(F\right)$ has a single side with $R_{\lambda }\left(F\right)\left(y\right)=y^2+b_3$ the attached residual polynomial of $F\left(x\right)$. Thus, the factor x provides two prime ideals of ${\mathbb{Z}}_K$ lying above 3 if, and only if, $b_3\equiv -1(\mathrm{mod}3)$. In this case, there are four prime ideals of ${\mathbb{Z}}_K$ lying above 3, and so 3 divides $i\left(K\right)$.
• Now, assume that 9 divides b and ${\nu }_3\left(a\right)\ge 1$. In this case, $\overline{F\left(x\right)}=x^4$ in ${\mathbb{F}}_3\left[x\right]$. It follows that:
  If $2{\nu }_3\left(a\right)<{\nu }_3\left(b\right)$, then ${\nu }_3\left(a\right)=1$, ${\nu }_3\left(b\right)\ge 3$, and $N_{\varphi }^{-}\left(F\right)=S_1+S_2$ has two sides, where $\varphi =x$. Let $d_i$ be the degree of $S_i$ for every $i=1,2$. Then, $d_2=1$ and $d_1\in \{1,2\}$. Thus, there are at most three prime ideals of ${\mathbb{Z}}_K$ lying above 3.
  If $2{\nu }_3\left(a\right)>{\nu }_3\left(b\right)$, then ${\nu }_3\left(b\right)\in \{2,3\}$ and ${\nu }_3\left(a\right)\ge 2$, and $N_{\varphi }^{-}\left(F\right)=S_1$ has a single side, where $\varphi =x$. Let d be the degree of $S_1$. Then, $d\in \{1,2\}$, and so there are at most two prime ideals of ${\mathbb{Z}}_K$ lying above 3.

□

Proof of Theorem 4.

We shall apply the main result of [28] (see also [3]) allowing to reduce the index form equation in quartic number fields to a cubic form equation and to a pair of quadratic form equations.

Let $F\left(x\right)=x^4+a{x}^2+b$ an irreducible monogenic trinomial, then $(1,\alpha ,{\alpha }^2,{\alpha }^3)$ is an integral basis in $K=\mathbb{Q}\left(\alpha \right)$, $\alpha$ being a root of $F\left(x\right)$. We can represent any $\gamma \in {\mathbb{Z}}_K$ in the form

$$\gamma =a+x\alpha +y{\alpha }^2+z{\alpha }^3$$

(2)

with $a,x,y,z\in \mathbb{Z}$. We are going to determine all triples $(x,y,z)$, such that $\gamma$ generates a power integral basis in K (for distinct $a\in \mathbb{Z}$, we obtain equivalent generators). According to [28], if $\gamma$ generates a power integral basis in K, then there exist $u,v\in \mathbb{Z}$ such that

$$(u-av)(u^2-4b{v}^2)=\pm 1$$

(3)

and

$$\begin{array}{ccc}\hfill Q_1(x,y,z)& =& x^2+a{y}^2-2axz+(a^2+b)z^2=u,\hfill \\ \hfill Q_2(x,y,z)& =& y^2-xz+a{z}^2=v.\hfill \end{array}$$

(4)

By (3), we have

$$u-av=\epsilon ,u^2-4b{v}^2=\delta$$

with $\epsilon =\pm 1,\delta =\pm 1$. Then, substituting $u=\epsilon +av$ into $u^2-4b{v}^2=\delta$, we obtain

$$v^2(a^2-4b)+2\epsilon av=\delta -{\epsilon }^2=\delta -1.$$

(5)

I. If $\delta =-1$, then this implies that $v\left|\right(\delta -1)=-2$; hence, $v=\pm 1$ or $v=\pm 2$.

IA. If $v=\pm 1$, then by $u=av+\epsilon$, we have $u^2=a^2+2av\epsilon +1$. On the other hand, we have $u^2-4b{v}^2=\delta =-1$, $u^2=4b-1$, whence

$$a^2+2av\epsilon +1=4b-1,$$

$$2av\epsilon +2=4b-a^2,$$

which can be satisfied neither for even values of a modulo 4, nor for odd values of a, modulo 2.

IB. If $v=\pm 2$, then by $u=av+\epsilon$, we have $u^2=4a^2\pm 4a+1$. On the other hand, we have $u^2-4b{v}^2=\delta =-1$, $u^2=16b-1$, whence

$$4a^2\pm 4a+1=16b-1,$$

$$4a^2\pm 4a-16b=-2,$$

which can not be satisfied modulo 4.

II. If $\delta =1$, then (5) implies

$$v=0\mathrm{or}v={\displaystyle \frac{-2\epsilon a}{a^2-4b}}.$$

(6)

(Note that $a^2-4b\ne 0$, otherwise K would only be a quadratic field).

IIA. In case,

$$v={\displaystyle \frac{-2\epsilon a}{a^2-4b}}=(-\epsilon ){\displaystyle \frac{2a}{a^2-4b}}$$

we have,

$$u=av+\epsilon ={\displaystyle \frac{-2\epsilon a^2}{a^2-4b}}+\epsilon$$

whence

$$u=(-\epsilon ){\displaystyle \frac{a^2+4b}{a^2-4b}}.$$

The above $u,v$ satisfies both $u-av=\epsilon$ and $u^2-4b{v}^2=\delta =1$. Set

$$u_0={\displaystyle \frac{u}{-\epsilon }}={\displaystyle \frac{a^2+4b}{a^2-4b}},v_0={\displaystyle \frac{v}{-\epsilon }}={\displaystyle \frac{2a}{a^2-4b}}$$

($v\ne 0,v_0\ne 0$ since $a>1$). We have

$$a^2-4b={\displaystyle \frac{2a}{v_0}}$$

whence

$$u_0={\displaystyle \frac{a^2+4b}{a^2-4b}}=1+{\displaystyle \frac{8b}{a^2-4b}}=1+{\displaystyle \frac{4b{v}_0}{a}}$$

($u_0\ne 1$ because $b>1$) and

$$a={\displaystyle \frac{4b{v}_0}{u_0-1}},a^2={\displaystyle \frac{16b^2{v}_0^2}{{(u_0-1)}^2}}.$$

Further, by $u_0={\displaystyle \frac{a^2+4b}{a^2-4b}}$, we obtain

$$u_0(a^2-4b)=a^2+4b,$$

whence

$$a^2(u_0-1)=4b(u_0+1),$$

that is,

$$a^2=4b{\displaystyle \frac{u_0+1}{u_0-1}}.$$

Comparing the two expressions obtained for $a^2$, we confer

$${\displaystyle \frac{16b^2{v}_0^2}{{(u_0-1)}^2}}=4b{\displaystyle \frac{u_0+1}{u_0-1}},$$

whence

$$b={\displaystyle \frac{u_0^2-1}{4v_0^2}}$$

and

$$a={\displaystyle \frac{4b{v}_0}{u_0-1}}={\displaystyle \frac{u_0+1}{v_0}}.$$

All together, we obtain

$$a={\displaystyle \frac{u\pm 1}{v}},b={\displaystyle \frac{u^2-1}{4v^2}}.$$

Parameters of this type were excluded.

IIB. Finally, if $v=0$, then Equation (3) implies $u=\pm 1$. Observe that in our case,

$$Q_1(x,y,z)=x^2+a{y}^2-2xza+z^2(a^2+b)={(x-za)}^2+a{y}^2+b{z}^2.$$

Hence, $Q_1(x,y,z)=u$ (see (4)) can only be satisfied for $u=1$ and then in view of $a>1,b>1$, we have $x=\pm 1,y=0,z=0$; therefore, up to equivalence, $\alpha$ is the only generator of power integral bases of K. □

Proof of Proposition 1.

Let $\varphi =x$. Then, $\overline{F\left(x\right)}={\varphi }^4$ in ${\mathbb{F}}_2\left[x\right]$ and $N_{\varphi }\left(F\right)=S$ has a single side of degree $gcd(4,3)=1$. Thus, $F\left(x\right)$ is irreducible over ${\mathbb{Q}}_2$. Let K be the number field generated by a root $\alpha$ of $F\left(x\right)$.

Then, there is a unique valuation $\omega$ of K extending ${\nu }_2$. Since ${\nu }_2({\mathbb{Z}}_K:\mathbb{Z}\left[\alpha \right])=3$, we conclude that $F\left(x\right)$ is not a monogenic polynomial. Now, let $\theta ={\displaystyle \frac{{\alpha }^3}{4}}$. Then, $\theta \in K$. Since 3 and 4 are coprime, we conclude that $K=\mathbb{Q}\left(\theta \right)$. Let us show that ${\mathbb{Z}}_K=\mathbb{Z}\left[\theta \right]$, and so K is monogenic. By [35] (Corollary 3.1.4), in order to show that $\theta \in {\mathbb{Z}}_K$, we need to show that $\omega \left(\theta \right)\ge 0$, where $\omega$ is the unique valuation of K extending ${\nu }_2$. Since $N_{\varphi }\left(F\right)=S$ has a single side of slope $-3/4$, we conclude that $\omega \left(\alpha \right)=3/4$, and so $\omega \left(\theta \right)={\displaystyle \frac{9}{4}}-2={\displaystyle \frac{1}{4}}$. Let $g\left(x\right)$ be the minimal polynomial of $\theta$ over $\mathbb{Q}$. By the formula relating the roots and coefficients of a monic polynomial, we conclude that $g\left(x\right)=x^4+{\sum }_{i=1}^4{(-1)}^i{s}_i{x}^{4-i}$, where ${\displaystyle s_i=\sum _{k_1<\cdots <k_i}{\theta }_{k_1}\cdots {\theta }_{k_i}}$ and ${\theta }_1,\dots ,{\theta }_4$ are the ${\mathbb{Q}}_p$-conjugates of $\theta$. Since there is a unique valuation extending ${\nu }_2$ to any algebraic extension of ${\mathbb{Q}}_2$, we conclude that $\omega \left({\theta }_i\right)=1/4$ for every $i=1,\dots ,4$. Thus, ${\nu }_2\left(s_4\right)=\omega ({\theta }_1\cdots {\theta }_4)=4\times 1/4=1$ and ${\nu }_2\left(s_i\right)\ge i/4$ for every $i=1,\dots ,3$, which means that $g\left(x\right)$ is a 2-Eisenstein polynomial. Hence, 2 does not divide the index $({\mathbb{Z}}_K:\mathbb{Z}\left[\theta \right])$. As $▵\left(F\right)=2^{15}b{(a^2-2b)}^2$ and by the definition of $\theta$, 2 is the unique positive prime integer candidate to divide $\left(\mathbb{Z}\right[\alpha ]:\mathbb{Z}[\theta \left]\right)$, we conclude that for every prime integer, p, p does not divide $({\mathbb{Z}}_K:\mathbb{Z}\left[\theta \right])$, which means that ${\mathbb{Z}}_K=\mathbb{Z}\left[\theta \right]$. □

6. Conclusions

Trinomials of type $x^4+ax+b$ were previously considered in [19], and trinomials of type $x^4+a{x}^3+b$ can be reduced to trinomials of type $x^4+ax+b$. Therefore, the present paper completes the description of the monogenity properties of quartic trinomials.