# On power basis of a class of number fields

Anuj Jakhar  ,  Sumandeep Kaur  and  Surender Kumar  Department of
Mathematics, Indian Institute of Technology (IIT) Madras  Department of
Mathematics, Panjab University Chandigarh  Department of Mathematics, Indian
Institute of Technology (IIT) Bhilai<EMAIL_ADDRESS>
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $f(x)=x^{n}+ax^{2}+bx+c\in{\mathbf{Z}}[x]$ be an irreducible polynomial
with $b^{2}=4ac$ and let $K={\mathbf{Q}}(\theta)$ be an algebraic number field
defined by a complex root $\theta$ of $f(x)$. Let ${\mathbf{Z}}_{K}$ deonote
the ring of algebraic integers of $K$. The aim of this paper is to provide the
necessary and sufficient conditions involving only $a,c$ and $n$ for a given
prime $p$ to divide the index of the subgroup ${\mathbf{Z}}[\theta]$ in
${\mathbf{Z}}_{K}$. As a consequence, we provide families of monogenic
algebraic number fields. Further, when
${\mathbf{Z}}_{K}\neq{\mathbf{Z}}[\theta]$, we determine explicitly the index
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ in some cases.

###### Key words and phrases:

Ring of algebraic integers; Integral basis and discriminant; Monogenic number
fields.

###### 2010 Mathematics Subject Classification:

11R04; 11R29.

##  1\. Introduction and statements of results

An algebraic number field $K$ of degree $n$ is said to be monogenic if it
possess an integral basis of the form
$\\{1,\alpha,\alpha^{2},\cdots,\alpha^{n-1}\\}$ for some algebraic integer
$\alpha.$ Such an integral basis of $K$ is called a power basis of $K$. The
determination of monogenity of an algebraic number field is one of the
classical problem in algebraic number theory. Hasse put forth the problem of
characterising monogenic fields, and it is still an active field of research.
Despite the lack of a standard classification, there has been some progress in
this direction. Gaál and Grás have done extensive research on the idea of
monogenic fields. Results relevant to lower degree monogenic extensions can be
found in Gaál’s book [3]. Grás demonstrates in [4] that for $n$ relatively
prime to 6, there are only finitely many abelian monogenic fields of degree
$n$. The discriminant $d_{K}$ of an algebraic number field
$K={\mathbf{Q}}(\theta)$ and the index of ${\mathbf{Z}}[\theta]$ in ring of
integers ${\mathbf{Z}}_{K}$ are closely connected by the well-known formula
$D_{f}=[{\mathbf{Z}}_{K}:\mathbb{Z}[\theta]]^{2}d_{K};$ where $D_{f}$ is the
discriminant of the minimal polynomial $f(x)$ of $\theta$. In 1878, Dedekind
gave a criterion which provides necessary and sufficient conditions for a
prime $p$ to divide the index of ${\mathbf{Z}}[\theta]$ in ${\mathbf{Z}}_{K}$
(see Theorem 2.4).

Let $K={\mathbf{Q}}(\theta)$ be an algebraic number field where $\theta$
satisfies an irreducible quadrinomial
$f(x)=x^{n}+ax^{2}+bx+c\in{\mathbf{Z}}[x]$ with $b^{2}=4ac$. Then, in this
paper using Dedekind’s criterion, we characterize all the primes which divide
the group index $[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$. This will provide
necessary and sufficient conditions involving only $a,c,n$ for
$\\{1,\theta,\cdots,\theta^{n-1}\\}$ to be an integral basis of $K$. We also
prove that if a prime $p\nmid b$ and divides the discriminant $d_{K}$ of $K$,
then the exact power of $p$ dividing $d_{K}$ is one. As an application of this
result, in Corollary 1.6, we provide an explicit formula for $d_{K}$ defined
by a class of quadrinomials. In the end of the paper, we give examples which
generate families of monogenic algebraic number fields. In some examples, we
determine the index $[{\mathbf{Z}}_{K}:{\mathbf{Z}}[{\theta}]]$ as well.

Precisely, we prove:

######  Theorem 1.1.

Let $K=\mathbb{Q}(\theta)$ be an algebraic number field defiend by a root
$\theta$ of an irreducible quadrinomial
$f(x)=x^{n}+ax^{2}+bx+c\in{\mathbf{Z}}[x]$ with $b^{2}=4ac$. A prime $p$
dividing the discriminant $D_{f}$ of $f(x)$ does not divide
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ if and only if $p$ satisfies one of
the following conditions:

  * (1)

If $p\mid a$ and $p\mid c,$ then $p^{2}\nmid c$.

  * (2)

If $p\mid a$ and $p\nmid c$ with $r\geq 1$ as the highest power of $p$
dividing $n$, then either $p\mid b_{1}$ and $p\nmid c_{1}$ or $p\nmid
b_{1}[(-c_{1})^{n}+cb_{1}^{n}]$ where $c_{1}=\frac{{c+(-c)^{p^{r}}}}{p}$ and
$b_{1}=\frac{b}{p}.$

  * (3)

If $p\nmid a$ and $p\mid c$ with $l\geq 1$ as the highest power of $p$
dividing $n-2$, then either $p\mid a_{1}$ and $p\nmid b_{1}$ or
$p\nmid\bar{a}_{1}[(-\bar{a})^{2^{1-d}}{\bar{a}_{1}}^{\frac{n-2}{2^{d}}}-{(-\bar{b}_{1})}^{\frac{n-2}{2^{d}}}]$
where $a_{1}=\frac{{a+(-a)^{p^{l}}}}{p}$, $b_{1}=\frac{b}{p}$ and $d$ equals
$1$ or $0$ according as $2$ divides $n-2$ or not.

  * (4)

If $p=2$ and $p\nmid ac$, then either $a\equiv 1(\text{mod }4)$ or $c\equiv
1(\text{mod }4)$.

  * (5)

If $p\nmid b,$ then $p^{2}\nmid D_{f}.$

The following corollaries are immediate consequence of the above theorem.

######  Corollary 1.2.

Let $K={\mathbf{Q}}(\theta)$, $f(x)$ be as in the above theorem. Then
$\\{1,\theta,\cdots,\theta^{n-1}\\}$ is an integral basis of $K$ if and only
if each prime $p$ dividing $D_{f}$ satisfies one of the conditions $(1)-(5)$
of Theorem 1.1

The following result proved in [7, Theorem 1.3] can be quickly deduced from
Corollary 1.2.

######  Corollary 1.3.

Let $K={\mathbf{Q}}(\theta)$ be an algebraic number field with $\theta$ a root
of an irreducible polynomial $x^{n}-c\in{\mathbf{Z}}[x]$. Then
$\\{1,\theta,\cdots,\theta^{n-1}\\}$ is an integral basis of $K$ if and only
if $c$ is a square free integer and whenever $r\geq 1$ is the highest power of
a prime $p$ dividing $n$, $p$ not dividing $c$, then $p^{2}$ does not divide
$c^{p^{r}}-c.$

######  Corollary 1.4.

Let $f(x)=x^{n}+ax^{2}+bx+c\in{\mathbf{Z}}[x]$ be a monic polynomial of degree
$n$ where $c\neq\pm 1$ is squarefree. Assume that the sets of primes dividing
$a,b,c$ are the same. Let either $2|c$ or $a\equiv 1(\text{mod }4)$ or
$c\equiv 1(\text{mod }4)$. Let $K={\mathbf{Q}}(\theta)$ with $\theta$ a root
of $f(x)$. Then ${\mathbf{Z}}_{K}={\mathbf{Z}}[\theta]$ if and only if for
each prime $p$ dividing $D_{f}$ and not dividing $a$, $p^{2}$ does not divide
$D_{f}$.

We now state the following result which shall be proved after the proof of
Theorem 1.1. It is of independent interest as well.

######  Proposition 1.5.

Let $K=\mathbb{Q}(\theta)$ be an algebraic number field where $\theta$
satisfies an irreducible quadrinomial
$f(x)=x^{n}+ax^{2}+bx+c\in{\mathbf{Z}}[x]$ of degree $n\geq 3$ with
$b^{2}=4ac$. If $p$ is a prime number such that $p\nmid b$ and $p$ divides the
discriminant $d_{K}$ of $K$, then the exact power of $p$ dividing the
discriminant $d_{K}$ of $K$ is one.

Using the fact that $D_{f}=[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]^{2}d_{K}$,
the above proposition and Theorem 1.1 quickly yield the following explicit
formula for the absolute value of $d_{K}$.

######  Corollary 1.6.

Let $K=\mathbb{Q}(\theta)$, $f(x)$ be as in Theorem 1.1 and
$|D_{f}|=\prod\limits_{i=1}^{k}p_{i}^{r_{i}}\prod\limits_{j=1}^{l}q_{j}^{t_{j}}$
be the prime factorization of $|D_{f}|$ into distinct prime numbers such that
$p_{i}|b$ and $q_{j}\nmid b$. If all the primes $p_{i}$’s satisfy one of the
conditions $(1)-(5)$ of Theorem 1.1, then the absolute value of the
discriminant $d_{K}$ is given by

$|d_{K}|=\prod\limits_{i=1}^{k}p_{i}^{r_{i}}\prod\limits_{j=1}^{l}q_{j}^{\frac{1-(-1)^{t_{j}}}{2}}.$

##  2\. Preliminary Results.

For two polynomials $f(x),g(x)$; $R(f,g)$ will denote the resultant of $f(x)$
and $g(x)$. Now we state some already known properties of the resultant.

######  Lemma 2.1.

Let $f(x)=a_{m}x^{m}+a_{m-1}x^{m-1}+\cdots+a_{0}$ and
$g(x)=b_{n}x^{n}+b_{n-1}x^{n-1}+\cdots+b_{0}$ be two polynomials with integer
coefficients of degree $m$ and $n$, respectively. Then
(1) $R(f,g)=(-1)^{mn}R(g,f).$
(2) If $f(x)=g(x)q(x)+r(x),$ then

$R(f,g)=(b_{n})^{m-k}R(r,g)$

where $k=\deg r(x).$
(3) $R(f,g)=(b_{n})^{m}\prod\limits_{i=1}^{n}f(\gamma_{i})$
where $\gamma_{1},\gamma_{2},\cdots,\gamma_{n}$ are roots of $g(x)$.

Note that the discriminant of a monic polynomial $f(x)\in{\mathbf{Z}}[x]$ and
the resultant are related by the following formula

$D_{f}=(-1)^{\frac{n(n-1)}{2}}R(f^{\prime},f)$ (2.1)

where $f^{\prime}$ is the derivative of $f.$

######  Lemma 2.2.

Let $f(x)=x^{n}+ax^{2}+bx+c\in{\mathbf{Z}}[x]$ be a quadrinomial of degree
$n\geq 3$ with $b^{2}=4ac.$ Then the following hold:

  1. (1)

The discriminant of $f(x)$ is

$D_{f}=(-1)^{\frac{n(n-1)}{2}}\left[n^{n}(-c)^{n-1}-4(n-2)^{n-2}\left(\frac{b}{2}\right)^{n}\right].$
(2.2)

  2. (2)

If $p\nmid b(n-2)$, then $D_{f}\equiv 0~{}(\text{mod}~{}p^{2})$ if and only if
$f(-\frac{nb}{2a(n-2)})\equiv 0~{}(\text{mod}~{}p^{2}).$

###### Proof.

(1) We can take $a\neq 0$, because if $a=0$, then using $b^{2}=4ac$ we have
$b=0$. It is easy to check that the discriminant of $f(x)=x^{n}+c$ is
$D_{f}=(-1)^{\frac{n(n-1)}{2}}n^{n}(-c)^{n-1}.$ So assume that $a\neq 0$. In
view of Lemma 2.1(1) and Equation (2.1), we have
$D_{f}=(-1)^{\frac{n(n-1)}{2}}R(f,f^{\prime}).$ Using division algorithm, one
can write

$f(x)=f^{\prime}(x)q(x)+\frac{a(n-2)}{n}x^{2}+\frac{b(n-1)}{n}x+c.$

We denote $\frac{a(n-2)}{n}x^{2}+\frac{b(n-1)}{n}x+c$ by $r(x)$. Keeping in
mind $(1)$ and $(2)$ of Lemma 2.1, we have
$D_{f}=(-1)^{\frac{n(n-1)}{2}}n^{n-2}R(f^{\prime},r).$ Since the roots of
$r(x)$ are $\gamma_{1}=-\frac{b}{2a}$ and $\gamma_{2}=-\frac{nb}{2a(n-2)},$
therefore by $(3)$ of Lemma 2.1, we have

$D_{f}=(-1)^{\frac{n(n-1)}{2}}n^{n-2}\left(\frac{a(n-2)}{n}\right)^{n-1}f^{\prime}(\gamma_{1})f^{\prime}(\gamma_{2}).$

A simple calculation shows that

$D_{f}=(-1)^{\frac{n(n-1)}{2}}\left[n^{n}(-c)^{n-1}-4(n-2)^{n-2}\left(\frac{b}{2}\right)^{n}\right].$

(2) Using hypothesis $b^{2}=4ac$, we have $2$ divides $b.$ Since $p\nmid
b(n-2)$, it follows that $f\left(-\frac{nb}{2a(n-2)}\right)\equiv
0~{}(\text{mod}~{}p^{2})$ if and only if

$\left(-\frac{nb}{2a(n-2)}\right)^{n}+a\left(-\frac{nb}{2a(n-2)}\right)^{2}+b\left(-\frac{nb}{2a(n-2)}\right)+c\equiv
0~{}(\text{mod}~{}p^{2}).$

Using $b^{2}=4ac$, the last equation can be rewritten as

$(-nb)^{n}+n^{2}c(2a)^{n}(n-2)^{n-2}-2nc(2a)^{n}(n-2)^{n-1}+c(2a(n-2))^{n}\equiv
0~{}(\text{mod}~{}p^{2}).$

On dividing by $2^{n}a^{n}c$ and simplifying, the last equation becomes

$\frac{1}{c}\left(\frac{-nb}{2a}\right)^{n}+4(n-2)^{n-2}\equiv
0~{}(\text{mod}~{}p^{2}).$

Since $2a=\frac{b^{2}}{2c},$ we see that
$f\left(-\frac{nb}{2a(n-2)}\right)\equiv 0\mod p^{2}$ if and only if

$n^{n}(-c)^{n-1}-4(n-2)^{n-2}\left(\frac{b}{2}\right)^{n}\equiv
0~{}(\text{mod}~{}p^{2}),$

which completes the proof of the lemma in view of (1). ∎

The following result will be used in the proof of Theorem 1.1.

######  Lemma 2.3.

Let $a,b,c,n$ be integers such that $2\nmid ac$, $2|b$ and $2^{2}\nmid b$. Let
$n>2$ be an even integer and
$F(x)=(\frac{a+a^{2}}{2})x^{2}+({ac+\frac{b}{2}})x+(\frac{c+c^{2}}{2})$ be a
polynomial. Then the polynomial $F(x)$ is coprime to $x^{\frac{n}{2}}+ax+c$
modulo $2$ if and only if $a\equiv 1$ $($mod $4)$ or $c\equiv 1$ $($mod $4)$.

###### Proof.

Denote the polynomial $x^{\frac{n}{2}}+ax+c$ by $h(x)$. Since $2|n,2|b,2\nmid
ac,4\nmid b$, one can write $n=2m,a=2A+1,b=2B,c=2C+1$ with
$A,B,C,m\in{\mathbf{Z}}$ and $2\nmid B$. So

$h(x)\equiv x^{m}+x+1~{}(\mbox{mod}~{}2),$

and $F(x)$ can be written as

$F(x)=(2A^{2}+3A+1)x^{2}+((2A+1)(2C+1)+B)x+(2C^{2}+3C+1),$

which leads to

$F(x)\equiv(A+1)x^{2}+C+1~{}(\mbox{mod}~{}2).$

Let $\alpha=A$ $($mod $2)$ and $\gamma=C$ $($mod $2)$. Then the values of
$\alpha$ and $\gamma$ lie in $\\{0,1\\}$. Hence it remains to prove the lemma
in the following four cases.

Case 1. $\alpha=0,\gamma=0.$ In this case

$F(x)\equiv(x+1)^{2}~{}(\mbox{mod}~{}2),$

and one can easily check that $F(x)$ is coprime to $h(x)$ modulo $2$.

Case 2. $\alpha=0,\gamma=1.$ Now we have

$F(x)\equiv x^{2}~{}(\mbox{mod}~{}2).$

Therefore, we see that $F(x)$ is coprime to $h(x)$ modulo $2$.

Case 3. $\alpha=1,\gamma=1.$ In this case

${F}(x)\equiv 0~{}(\mbox{mod}~{}2).$

So $h(x)$ divides $F(x)$ modulo $2$, hence $F(x)$ can not be coprime to $h(x)$
modulo $2$ in this situation.

Case 4. $\alpha=1,\gamma=0.$ Here we have

$F(x)\equiv 1~{}(\mbox{mod}~{}2),$

which shows that $F(x)$ is coprime to $h(x)$ modulo $2$.

Therefore, in view of the above four cases, we see that $F(x)$ is not coprime
to $h(x)$ modulo $2$ if and only if we are in Case 3, i.e., $\alpha=1$ and
$\gamma=1.$ Returning to the definitions of $\alpha,\gamma$ we obtain that
this happens if and only if one has $a\equiv 3$ (mod $4$) and $c\equiv 3$ (mod
$4$), which completes the proof of the lemma. ∎

In 1878, Dedekind gave the following simple criterion known as Dedekind
Criterion (cf. [1, Theorem 6.1.4], [2]) which gave necessary and sufficient
condition to be satisfied by $f(x)$ so that $p$ does not divide
$[{\mathbf{Z}}_{K}:Z[\theta]]$.

######  Theorem 2.4.

(Dedekind Criterion) Let $K=\mathbb{Q}(\theta)$ be an algebraic number field
with $f(x)$ as the minimal polynomial of the algebraic integer $\theta$ over
$\mathbb{Q}.$ Let $p$ be a prime and
$\overline{f}(x)=\overline{g}_{1}(x)^{e_{1}}\ldots\overline{g}_{t}(x)^{e_{t}}$
be the factorization of $\overline{f}(x)$ as a product of powers of distinct
irreducible polynomials over $\mathbb{Z}/p\mathbb{Z},$ with each
$g_{i}(x)\in\mathbb{Z}[x]$ monic. Let $M(x)$ denote the polynomial
$\frac{1}{p}(f(x)-g_{1}(x)^{e_{1}}\ldots g_{t}(x)^{e_{t}})$ with coefficients
from $\mathbb{Z}.$ Then $p$ does not divide
$[{\mathbf{Z}}_{K}:\mathbb{Z}[\theta]]$ if and only if for each $i,$ we have
either $e_{i}=1$ or $\bar{g}_{i}(x)$ does not divide $\overline{M}(x).$

We now state the following simple lemma proved in ([6, Lemma 2.2]), which will
be used in the sequel. It can be easily proved using the Binomial Theorem.

######  Lemma 2.5.

Let $h(x)=x^{m}+ax^{s}+b\in{\mathbf{Z}}[x]$ be a monic polynomial of degree
$m$. Let $p$ be a prime and $k\in{\mathbf{N}}$. Then
$h(x^{p^{k}})=h(x)^{p^{k}}+ph(x)T(x)+(ax^{sp^{k}}+b)+(-ax^{s}-b)^{p^{k}}$ for
some polynomial $T(x)\in{\mathbf{Z}}[x]$.

The following result will be used in the proof of Proposition 1.5 (cf. [8,
Theorem 4.24]). Its proof is omitted.

######  Theorem 2.6.

Let $p$ be a prime number and $K/{\mathbf{Q}}$ be an extension of degree $n$.
Let $p{\mathbf{Z}}_{K}=\wp_{1}^{e_{1}}\cdots\wp_{t}^{e_{t}}$, where
$\wp_{1},\cdots,\wp_{t}$ are distinct prime ideals of ${\mathbf{Z}}_{K}$ and
$N_{K/{\mathbf{Q}}}(\wp_{i})=p^{f_{i}}$. If $(p,e_{i})$ = 1 for $1\leq i\leq
t$, then the exact power of $p$ dividing $d_{K}$ is
${\sum\limits_{i=1}^{t}f_{i}(e_{i}-1)}$.

##  3\. Proof of Theorem 1.1.

###### Proof of Theorem 1.1..

Consider the first case, when $p\mid a$ and $p\mid c$. In this case
$f(x)\equiv x^{n}~{}(\text{mod}~{}p)$. As in Theorem 2.4, one can check that
$M(x)=\frac{a}{p}x^{2}+\frac{b}{p}x+\frac{c}{p},$ hence we see that
$p\nmid[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ if and only if $x$ does not
divide $\overline{M}(x)$. This is possible only when $p^{2}\nmid c.$

Consider now the case when $p\mid a$ and $p\nmid c$. Since $b^{2}=4ac$, we
have $p|b$ and hence $f(x)\equiv x^{n}+c~{}(\text{mod}~{}p).$ Keeping in mind
the fact that $p|D_{f}$ and Equation $(\ref{df}),$ we see that $p\mid n.$
Write $n=p^{r}m,$ where $r\geq 1$ and $p\nmid m.$ In view of Binomial theorem,
we have

$x^{n}+c\equiv(x^{m}+c)^{p^{r}}~{}(\text{mod}~{}p).$ (3.1)

Let $\prod\limits_{i=1}^{t}\bar{g}_{i}(x)$ be the factorization of
$x^{m}+\bar{c}$ over ${\mathbf{Z}}/p{\mathbf{Z}}$, where
$g_{1}(x),g_{2}(x),\cdots,g_{t}(x)$ belonging to ${\mathbf{Z}}[x]$ are monic
polynomials which are distinct and irreducible modulo $p$. Write $x^{m}+c$ as
$\prod\limits_{i=1}^{t}\bar{g}_{i}(x)+pA_{1}(x)$ for some polynomial
$A_{1}(x)\in{\mathbf{Z}}[x]$. Then

$f(x)=((x^{p^{r}})^{m}+c)+ax^{2}+bx=h(x^{p^{r}})+ax^{2}+bx$

where $h(x)=x^{m}+c.$ We now split this case into two sub cases according as
$p$ is an odd prime or not.

First consider the sub case when $p\neq 2.$ Keeping in mind that $p^{2}\mid a$
and applying Lemma 2.5 to $h(x),$ we see that

$f(x)=\big{(}\prod\limits_{i=1}^{t}g_{i}(x)+pA_{1}(x)\big{)}^{p^{r}}+pA(x)\prod\limits_{i=1}^{t}g_{i}(x)+p^{2}B(x)+c-c^{p^{r}}+bx$
(3.2)

for some polynomials $A(x),B(x)\in{\mathbf{Z}}[x]$. Denote
$\frac{{c-c^{p^{r}}}}{p}$ and $\frac{b}{p}$ by $c_{1}$ and $b_{1}$,
respectively. In view of $(\ref{gt})$,
$\bar{f}(x)=\big{(}\prod\limits_{i=1}^{t}\bar{g}_{i}(x)\big{)}^{p^{r}}.$ Write
$f(x)$ as $\prod\limits_{i=1}^{t}g_{i}(x)^{p^{r}}+pM(x)$,
$M(x)\in{\mathbf{Z}}[x]$, $M(x)\in{\mathbf{Z}}[x]$. Keeping in mind that
$r\geq 1$, it is immediate from (3.2) that

$\overline{M}(x)=\bar{A}(x)\prod\limits_{i=1}^{t}\bar{g}_{i}(x)+\overline{c_{1}}+\overline{b_{1}}x.$

Applying Theorem 2.4, $p\nmid[{\mathbf{Z}}_{K}:{\mathbf{Z}}_{\theta}]$ if and
only $\overline{M}(x)$ is coprime $\prod\limits_{i=1}^{t}\bar{g}_{i}(x)$ which
by virtue of the above equation holds if and only if
$(\overline{c_{1}}+\overline{b_{1}}x)$ is coprime to
$\prod\limits_{i=1}^{t}\bar{g}_{i}(x)^{p^{r}}=\bar{h}(x^{p^{r}})=x^{n}+\bar{c}$.
This happens if and only if either $p\mid b_{1}$ and $p\nmid c_{1}$ or $p\nmid
b_{1}((-c_{1})^{n}+cb_{1}^{n}).$
Now let $p=2.$ Since $b^{2}=4ac$, we see that $4\mid a$ and $4\mid b$.
Therefore

$f(x)=\big{(}\prod\limits_{i=1}^{t}g_{i}(x)+2A_{1}(x)\big{)}^{2^{r}}+2C(x)\prod\limits_{i=1}^{t}g_{i}(x)+4D(x)+c+c^{2^{r}}$

for some polynomials $C(x),D(x)\in{\mathbf{Z}}[x].$ Here $1\leq i\leq t,$
$\bar{g}_{i}(x)\nmid\overline{M}(x)$ if and only if
$\bar{g}_{i}(x)\nmid\overline{(\frac{{c+c^{2^{r}}}}{2})}.$ This happens only
when $4\nmid(c^{2^{r}}+c)$.

Consider now the case when $p\nmid a$ and $p\mid c.$ Note that $f(x)\equiv
x^{2}(x^{n-2}+a)~{}(\text{mod}~{}p).$ Let $l\geq 0$ denote the highest power
of $p$ dividing $n-2$. Consider first the possibility when $l=0$. If $p\neq
2,$ in view of $b^{2}=4ac,$ we have $p^{2}\mid c.$ Since $x$ is the only
repeated factor of $\bar{f}(x)$, we see that it divides
$\overline{M}(x)=\frac{\bar{b}}{p}x$. Hence by Theorem 2.4, $p$ divides
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$. If $p=2,$ then observe that
$p^{2}\mid b$ and $p^{2}\mid c$. Thus $\overline{M}(x)=\bar{0}$. Therefore, in
the case $l=0$, $p$ always divides $[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$.
Now assume that $l\geq 1$, say $n-2=p^{l}s^{\prime}$ and $p\nmid s^{\prime}$.
Write $x^{s^{\prime}}+a=g_{1}(x)\cdots g_{t}(x)+pH(x)$, where
$g_{1}(x),\cdots,g_{t}(x)$ are monic polynomials which are distinct as well as
irreducible modulo $p$ and $H(x)\in{\mathbf{Z}}[x]$. Keeping in mind that
$p^{2}\mid c$ and applying Lemma 2.5 to $h(x)$ = $x^{s^{\prime}}+a$, we can
write $f(x)=x^{2}(x^{n-2}+a)+bx+c$ as

$f(x)=x^{2}\left[(\prod\limits_{i=1}^{t}g_{i}(x)+pH(x))^{p^{l}}+pT(x)\prod\limits_{i=1}^{t}g_{i}(x)+p^{2}U(x)+a+(-a)^{p^{l}}\right]+bx+c,$
(3.3)

where $T(x),U(x)$ belong to ${\mathbf{Z}}[x]$. Note that
$x,\bar{g}_{1}(x),\cdots,\bar{g}_{t}(x)$ are distinct irreducible factors of
$\overline{F}(x)$. Denote $\frac{{a+(-a)^{p^{l}}}}{p}$ and $\frac{b}{p}$ by
$a_{1}$ and $b_{1}$, respectively. Since $l\geq 1$, by Theorem 2.4
$p\nmid[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ if and only if
$\bar{a}_{1}x^{2}+\bar{b}_{1}$ and $x^{n-2}+\bar{a}$ are coprime. Note that if
$p=2$, then $4|b$ and hence in this situation the polynomials
$\bar{a}_{1}x^{2}+\bar{b}_{1}$ and $x^{n-2}+\bar{a}$ will be coprime if and
only if $2\nmid\bar{a}\bar{a}_{1}.$ Now assume that $p\neq 2$. In this
situation, we see that $\bar{a}_{1}x^{2}+\bar{b}_{1}$ and $x^{n-2}+\bar{a}$
are coprime if and only if either $($I$)$ $p\mid a_{1}$ and $p\nmid b_{1}$ or
$($II$)$ $\bar{a}_{1}\neq\bar{0}$ and $x^{2}+\frac{\bar{b}_{1}}{\bar{a}_{1}}$,
$x^{n-2}+\bar{a}$ are coprime. Write $n-2=2^{d}s$ where $d$ equals $1$ or $0$
according as $2$ divides $n-2$ or not. Note that if $\xi$ is a common root of
the polynomials $x^{2}+\frac{\bar{b}_{1}}{\bar{a}_{1}}$, $x^{n-2}+\bar{a}$ in
the algebraic closure of ${\mathbf{Z}}/p{\mathbf{Z}}$, then we see that
$(\frac{-\bar{b}_{1}}{\bar{a}_{1}})^{\frac{n-2}{2^{d}}}=(-\bar{a})^{2^{1-d}}$.
Also a simple calculation shows that if
$\xi^{2^{d}}=(-\bar{a})^{-2^{d}}(-\frac{\bar{b}_{1}}{\bar{a}_{1}})^{\frac{n-1}{2^{1-d}}}$
with
$({\frac{{-\bar{b}_{1}}}{\bar{a}_{1}}})^{\frac{n-2}{2^{d}}}=(-\bar{a})^{2^{1-d}}$,
then $\xi$ is a root of the polynomials
$x^{2}+\frac{\bar{b}_{1}}{\bar{a}_{1}}$, $x^{n-2}+\bar{a}$. Hence one can
easily see that $($II$)$ holds if and only if
$p\nmid\bar{a}_{1}[(-\bar{a})^{2^{1-d}}{\bar{a}_{1}}^{\frac{n-2}{2^{d}}}-{(-\bar{b}_{1})}^{\frac{n-2}{2^{d}}}]$.

Consider now the case when $2\nmid ac.$ As $b^{2}=4ac,$ so $4\nmid b.$ Keeping
in mind that $2|D_{f}$ and Equation $(\ref{df}),$ we see that $2\mid n.$ Write
$n=2m$. Observe that $\bar{f}(x)=x^{n}+\bar{a}x^{2}+\bar{c}$. Denote
$x^{m}+ax+c$ by $h(x)$ so that $f(x)=h(x^{2})+bx.$ Let
$h(x)\equiv\prod\limits_{i=1}^{t}g_{i}(x)^{r_{i}}$ $($mod $2)$ be the
factorization of $h(x)$ into a product of irreducible polynomials modulo $2$
with $g_{i}(x)$ belonging to ${\mathbf{Z}}[x]$ monic and $r_{i}>0$. Using
Lemma 2.5, we see that

$f(x)=h(x^{2})+bx=(h(x))^{2}+2h(x)N(x)+ax^{2}+c+(-ax-c)^{2}+bx$

for some $N(x)\in{\mathbf{Z}}[x]$. Substituting
$h(x)=\prod\limits_{i=1}^{t}g_{i}(x)^{r_{i}}+2H(x)$ with
$H(x)\in{\mathbf{Z}}[x]$ in the above equation, we see that there exists
$N_{1}(x)\in{\mathbf{Z}}[x]$ such that

$f(x)=\big{(}\prod\limits_{i=1}^{t}g_{i}(x)^{r_{i}}\big{)}^{2}+4N_{1}(x)+2h(x)N(x)+ax^{2}+c+(-ax-c)^{2}+bx.$
(3.4)

Write $f(x)=\big{(}\prod\limits_{i=1}^{t}g_{i}(x)^{r_{i}}\big{)}^{2}+2M(x)$
for some $M(x)\in{\mathbf{Z}}[x]$. Applying Theorem 2.4, we see that
$2\nmid[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ if and only if
$\frac{1}{2}[ax^{2}+c+(-ax-c)^{2}+bx]$ is coprime to $h(x)$ modulo $2$, which
in view of Lemma 2.3 is equivalent to saying that either $a\equiv 1$ (mod $4$)
or $c\equiv 1$ (mod $4$).

Consider the last case when $p\nmid b.$ In view of $b^{2}=4ac$ and the fact
that $p|D_{f}$, we see that $p\neq 2$ and $p\nmid n(n-2).$ One can easily
check that $-(\overline{\frac{nb}{2(n-2)a}})$ is the only repeated root of
$\bar{f}(x)$ and has multiplicity two. Choose $d\in{\mathbf{Z}}$ such that

$nbd\equiv-2(n-2)a~{}~{}(\mbox{mod}~{}p^{2}).$ (3.5)

Therefore, we have

$\bar{f}(x)=(x-\bar{d})^{2}\bar{h}(x),$ (3.6)

where $\bar{h}(x)$ belonging to ${\mathbf{Z}}/p{\mathbf{Z}}[x]$ is a separable
polynomial. Write

$f(x)=(x-d)q(x)+f(d)$ (3.7)

for some monic polynomial $q(x)\in{\mathbf{Z}}[x]$. Observe that
$\bar{q}(x)=(x-\bar{d})\bar{h}(x)$ with $(x-\bar{d})$ does not divide
$\bar{h}(x)$. Let $\bar{h}(x)=\prod\limits_{i=1}^{t}\bar{g}_{i}(x)$ be the
factorization of $\bar{h}(x)$ into a product of distinct irreducible
polynomials over ${\mathbf{Z}}/p{\mathbf{Z}}$ with each
$g_{i}(x)\in{\mathbf{Z}}[x]$ monic. Write

$q(x)=(x-d)\prod\limits_{i=1}^{t}g_{i}(x)+ph_{1}(x)$

for some $h_{1}(x)$ belonging to ${\mathbf{Z}}[x]$. Substituting from the
above equation in $(\ref{2.2a})$, we see that

$f(x)=(x-d)^{2}\prod\limits_{i=1}^{t}g_{i}(x)+p(x-d)h_{1}(x)+f(d).$

Therefore applying Theorem 2.4, we see that
$p\nmid[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ if and only if $p^{2}$ does
not divide $f(d)$. Hence by Lemma 2.2, we have
$p\nmid[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ if and only if $p^{2}$ does
not divide $D_{f}.$ So the proof of the theorem is complete. ∎

###### Proof of Proposition 1.5.

Keeping in mind the formula
$D_{f}=[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]^{2}d_{K}$, we have $p|D_{f}$ as
$p$ divides $d_{K}$. Now recall the proof of the last case of Theorem 1.1, we
see that $p\neq 2$, $p\nmid n(n-2)$ and

$\bar{f}(x)=(x-\bar{d})^{2}\prod\limits_{i=1}^{t}\bar{g}_{i}(x),$

where $x-\bar{d},\bar{g}_{1}(x),\cdots,\bar{g}_{t}(x)$ are distinct and
irreducible polynomials modulo $p$ with $g_{i}(x)\in{\mathbf{Z}}[x]$ monic for
each $i$. Thus using Hensel’s Lemma, we see that $f(x)$ has a factorization in
${\mathbf{Q}}_{p}[x]$ such that

$f(x)=G(x)G_{1}(x)\cdots G_{t}(x),$

where $G_{1}(x),\cdots,G_{t}(x)$ are distinct irreducible polynomials. The
above factorization leads into the prime ideal factorization of
$p{\mathbf{Z}}_{K}$ as $p{\mathbf{Z}}_{K}=\mathfrak{p}\wp_{1}\cdots\wp_{t}$,
where $\wp_{i}$’s are distinct prime ideals of ${\mathbf{Z}}_{K}$ with
$N_{K/{\mathbf{Q}}}(\wp_{i})=p^{\deg G_{i}(x)}$, and as $p$ dividing $d_{K}$
is ramified in ${\mathbf{Z}}_{K}$, we have $\mathfrak{p}=\wp^{2}$ for some
prime ideal of ${\mathbf{Z}}_{K}$ with $N_{K/{\mathbf{Q}}}(\wp)=p$ and
$\gcd(\wp,\wp_{i})={\mathbf{Z}}_{K}$. Since $p\neq 2$, using Theorem 2.6, we
obtain that the exact power of $p$ dividing $d_{K}$ is one. ∎

##  4\. Examples.

We now provide some examples which illustrate our results. In these examples,
$K={\mathbf{Q}}(\theta)$ with $\theta$ a root of $f(x)$.

######  Example 4.1.

Suppose that $c\neq\pm 1$ is a squarefree integer. Consider the polynomial
$f(x)=x^{7}+c(x^{2}+2x+1)$. Then the discriminant $D_{f}$ of $f(x)$ is given
by $D_{f}=c^{6}[2^{2}\cdot 5^{5}c-7^{7}]$. By virtue of Theorem
$\ref{1.1}(i)$, any prime $p$ dividing $c$ will not divide
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$. Further in view of Corollary 1.2,
${\mathbf{Z}}_{K}={\mathbf{Z}}[\theta]$ if and only if for each prime $p$
dividing $D_{f}$ and not dividing $c$, $p^{2}$ does not divide $(2^{2}\cdot
5^{5}c-7^{7})$. Now we calculate the exact value of
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ corresponding to some values of $c$.

  1. (1)

For $c=2$, it can be easily seen that $|D_{f}|=2^{6}\cdot 3^{2}\cdot 83\cdot
1069.$ In view of Theorem $\ref{1.1},$ $2,83,1069$ do not divide
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ and $3$ divides
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$. Since
$D_{f}$=$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]^{2}\cdot d_{K}$, where
$d_{K}$ is the discriminant of $K$, we see that
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ is $3$ in this case.

  2. (2)

Consider $c=5$. One can check that $|D_{f}|=5^{6}\cdot 3\cdot 253681$. By
Theorem $\ref{1.1},$ the primes $5,3$ and $253681$ do not divide
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$. So
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]=1$, and hence
$\\{1,\theta,\cdots,\theta^{6}\\}$ is an integral basis of $K$ in this
situation.

  3. (3)

When $c=7$, then one can see that the prime factorization of $|D_{f}|$ is
given by $7^{7}\cdot 11^{3}\cdot 79$. Arguing as above, $7,79$ do not divide
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$ and $11$ divides
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]$. Therefore
$[{\mathbf{Z}}_{K}:{\mathbf{Z}}[\theta]]=11$.

######  Example 4.2.

Suppose that $c\equiv 1(\text{mod }4)$ and consider an irreducible111Suppose
there exists a prime factor $p$ of $c$ such that the highest power of $p$
dividing $c$ is $k$ and $\gcd(k,n)=1$, then
$f(x)=x^{n}+c(x^{2}+2x+1)\in{\mathbf{Z}}[x]$ will be irreducible over
${\mathbf{Q}}$ by Dumas irreduciblity criterion [5]. polynomial
$f(x)=x^{n}+c(x^{2}+2x+1)$. Then we have
$D_{f}=(-1)^{\frac{n(n-1)}{2}}c^{n-1}[(-1)^{n-1}n^{n}-4(n-2)^{n-2}c]$. By
virtue of Corollary 1.2, ${\mathbf{Z}}_{K}={\mathbf{Z}}[\theta]$ if and only
if for each prime $p$ dividing $D_{f}$, $p^{2}$ does not divide $c$ when
$p\mid c$ and if $p\nmid c$, then $p^{2}\nmid D_{f}$. For example, if we take
$n=5$ and $c\neq\pm 1$ is a squarefree integer, then
$\\{1,\theta,\theta^{2},\theta^{3},\theta^{4}\\}$ is an integral basis of $K$
if and only if $(3125-108c)$ is squarefree. It can be verified that
$(3125-108c)$ is squarefree when $c=-3,5,13,17,21$.

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