1 Introduction and statements of results
Let
$K={\mathbb {Q}}(\theta )$
be an algebraic number field and let
$f(x)$
of degree n be the minimal polynomial of
$\theta $
over
${\mathbb {Q}}$
. The polynomial
$f(x)$
is said to be monogenic if
$\{1,\theta ,\ldots ,\theta ^{n-1}\}$
is an integral basis of K.
Denote the ring of algebraic integers of K by
${\mathbb {Z}}_K$
. The field K is said to be monogenic if there exists
$\alpha \in {\mathbb {Z}}_K$
such that
${\mathbb {Z}}_K={\mathbb {Z}}[\alpha ].$
It is well known that if
$f(x)$
is monogenic, then the number field K is monogenic but the converse is not always true (for example,
${\mathbb {Q}}(\sqrt {d}),$
where
$d\ne 1$
is a square-free integer congruent to
$1$
modulo
$4$
).
The discriminant of a monic polynomial over a field
$\mathbb {F}$
of degree n having roots
$\theta _1, \ldots , \theta _n$
in the algebraic closure of
$\mathbb {F}$
is
$\Delta _f=\prod _{1\leq i<j\leq n}^{}(\theta _i-\theta _j)^2.$
It is a classical result in algebraic number theory that if
$f(x)$
is the minimal polynomial of an algebraic integer
$\theta $
over
${\mathbb {Q}}$
, then the discriminant
$ \Delta _f$
of
$f(x)$
and the discriminant
$d_K$
of
$K={\mathbb {Q}}(\theta )$
are related by the formula
$$ \begin{align} \Delta_f=[{\mathbb{Z}}_K:{\mathbb{Z}}[\theta]]^2d_K. \end{align} $$
Clearly if
$\Delta_f$
is square-free, then
$f(x)$
is monogenic but the converse need not be true. Jones [Reference Jones4, Reference Jones6] constructed infinite families of monogenic polynomials having non square-free discriminant. In [Reference Jones7], Jones showed that there exist infinitely many primes
$p \kern1.3pt{\geq }\kern1.3pt 3$
and integers
$t \kern1.4pt{\geq}\kern1.4pt 1$
coprime to p, such that
${f(x) \kern1.3pt{=}\kern1.3pt x^p \kern1.3pt{-}\kern1.3pt 2ptx^{p-1} \kern1.3pt{+}\kern1.3pt p^2t^2x^{p-2} \kern1.3pt{+}\kern1.3pt 1}$
is nonmonogenic and, in [Reference Jones5], he gave infinite families of monogenic polynomials using a new discriminant formula.
Throughout the paper,
$f(x)=x^n+A(Bx+1)^m\in {\mathbb {Z}}[x]$
is an irreducible polynomial with
$n\ge 3$
and
$1\le m\le n-1$
,
$\theta $
is a root of f,
$K={\mathbb {Q}}(\theta )$
is the corresponding algebraic number field,
$\Delta _f$
denotes the discriminant of
$f(x)$
and
$\mathrm {Ind}_K(\theta )$
denotes the group index
$ [{\mathbb {Z}}_K: {\mathbb {Z}}[\theta ]]$
. From [Reference Jones4, Theorem 3.1],
$$ \begin{align} \Delta_f=(-1)^{{n(n-1)}/{2}}A^{n-1}[n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA]. \end{align} $$
Theorem 1.1. Let
$A,B,n,m$
be integers with
$1\le m\le n-1, n> 2$
and
$B\ne 0.$
Assume that
$\gcd (n,mB)=1.$
Then an irreducible polynomial of the type
$f(x)=x^n+A (Bx+1)^m$
is monogenic if and only if both A and
$n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA$
are square-free.
Remark 1.2. In Theorem 1.1, the assumption that
$\gcd (n,mB)=1$
cannot be dropped. For example, consider the polynomial
$f(x){\kern-1pt}={\kern-1pt}x^3{\kern-1pt}-{\kern-1pt}6(3x+{\kern-1pt}1)$
. Here
$n{\kern-1pt}={\kern-1pt}3, m{\kern-1pt}={\kern-1pt}1, A{\kern-1pt}={\kern-1pt}-6$
and
$B=3.$
Note that
$f(x)$
is irreducible over
${\mathbb {Q}}.$
The polynomial
$f(x)$
is monogenic and has discriminant
$\Delta _f=23.2^2.3^5.$
However,
$n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA=23.3^3$
is not square-free.
The following corollary is an immediate consequence of Theorem 1.1. It is conjectured by Jones in [Reference Jones4, Conjecture 4.1].
Corollary 1.3. Let p be a prime number, and
$n, m$
and B be positive integers with
$1\le m\le n-1,~n>2$
and
$\gcd (n,mB)=1.$
Then
$f(x)=x^n+p(Bx+1)^m$
is monogenic if and only if
$n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mp$
is square-free.
Example 1.4. Let p be an odd prime number and let
$a,b$
be positive integers with
$n>2$
. Consider the polynomial
$f(x)=x^n+ax^2+bx+p$
with
$b^2=4ap.$
Note that
$f(x)$
satisfies Eisenstein’s criterion with respect to p, so it is irreducible over
${\mathbb {Q}}$
. The polynomial
$x^n+ax^2+bx+p$
with
$b^2=4ap$
can be reduced to the form
$x^n+p(Bx+1)^2$
with
$B={b}/{2p}.$
If
$\gcd (n,2B)=1,$
that is,
$\gcd (n,{b}/{p})=1,$
then Corollary 1.3 implies that
$f(x)$
is monogenic if and only if
$n^n+(-1)^n4({b}/{2p})^n({n-2})^{n-2}p$
is square-free.
Example 1.5. Let B be an integer not divisible by
$3$
with
$|B|\ge 4$
and let
$A\neq \pm 1$
be a nonzero square-free integer. Then the polynomial
$f(x)=x^3+A(Bx+1)^2$
is irreducible by Perron’s criterion, which states that if
${f(x)=x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0\in {\mathbb {Z}}[x]}$
with
$a_0\ne 0 \hspace {0.1in} \text {and}~ \hspace {0.1in} |a_{n-1}|>1+|a_{n-2}|+\cdots +|a_0|,$
then the polynomial
$f(x)$
is irreducible over
$\mathbb {Q}$
. In view of Theorem 1.1, the polynomial
$x^3+A(Bx+1)^2$
is monogenic if and only if
$4AB^3-27$
is square-free.
2 Preliminary results
In what follows, for a prime number p and a given polynomial
$g(x) \in {\mathbb {Z}}[x],~ \overline{g} (x)$
will denote the polynomial obtained by reducing each coefficient of
$ g(x)$
modulo
$p $
.
Let
$f(x)\in {\mathbb {Z}}[x]$
be a monic irreducible polynomial having a root
$\theta $
and let
$L={\mathbb {Q}}(\theta )$
be an algebraic number field. In 1878, Dedekind proved the following criterion which gives a necessary and sufficient condition to be satisfied by
$f(x)$
so that p does not divide
$\mathrm {Ind}_L(\theta )$
.
Theorem 2.1 (Dedekind’s criterion, [Reference Dedekind2]; see also [Reference Cohen1, Theorem 6.1.4]).
Let
$L=\mathbb {Q}(\theta )$
be an algebraic number field and
$f(x)$
the minimal polynomial of the algebraic integer
$\theta $
over
$\mathbb {Q}.$
Let p be a prime and
$\overline {f}(x) = \overline {g}_{1}(x)^{e_{1}}\cdots \overline {g}_{t}(x)^{e_{t}}$
be the factorisation 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
$M(x) = (f(x)-g_{1}(x)^{e_{1}}\cdots g_{t}(x)^{e_{t}})/p \in \mathbb {Z}[x]$
. Then p does not divide
$\mathrm {Ind}_L(\theta )$
if and only if, for each i, either
$e_{i}=1$
or
$ \overline{g}_{i}(x)$
does not divide
$\overline {M}(x).$
With the notation as in Theorem 2.1, one can easily check that if
$f(x)$
is monogenic, then for each prime p dividing
$\Delta _f$
, either
$e_i=1$
or
$ \overline{g}_i(x)$
does not divide
$\overline {M}(x)$
for each i.
Definition 2.2. A polynomial
$a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$
in
${\mathbb {Z}}[x]$
with
$a_n\ne 0$
is called an Eisenstein polynomial with respect to a prime p if
$p\nmid a_n,$
$p\mid a_i$
for
$0\le i\le n-1$
and
$p^2\nmid a_0.$
The following result is known as Eisenstein’s criterion (see [Reference Ireland and Rosen3]). It will be used in the proof of Corollary 1.3.
Theorem 2.3. Let
$g(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0\in {\mathbb {Z}}[x]$
with
$n\ge 1.$
If there is a prime number p such that
$ p\nmid a_n, p\mid a_{n-1},\ldots , p\mid a_0$
and
$p^2\nmid a_0,$
then
$g(x)$
is irreducible over
${\mathbb {Q}}$
.
The following lemma will be used in the proof of Theorem 1.1.
Lemma 2.4 [Reference Narkiewicz8, Lemma 2.17].
Let
$\alpha $
be an algebraic integer and let
$L={\mathbb {Q}}(\alpha )$
. If the minimal polynomial of
$\alpha $
over
${\mathbb {Q}}$
is an Eisenstein polynomial with respect to the prime
$p,$
then
$\mathrm {Ind}_L(\alpha )$
is not divisible by
$p.$
3 Proof of Theorem 1.1 and Corollary 1.3
Proof of Theorem 1.1.
Clearly
$A\ne 0.$
Suppose that
$\theta $
is a root of
$f(x)$
and
$K={\mathbb {Q}}(\theta )$
. From (1.2),
$$ \begin{align*}\Delta_f=(-1)^{{n(n-1)}/{2}}A^{n-1}[n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA].\end{align*} $$
First suppose that the polynomial
$f(x)$
is monogenic. Then
$\mathrm {Ind}_K(\theta )=1.$
Let p be a prime dividing
$\Delta _f$
. The following cases arise.
Case 1:
$p\mid A$
. Then
$f(x)\equiv x^n\,\mod p$
and
$M(x)=A(Bx+1)^m/p$
. As
$n\ge 3$
, by Dedekind’s criterion, we see that
$\overline{x}$
should not divide
$\overline {M}(x).$
This implies that
$p^2\nmid A.$
Thus, A is square-free. Suppose that
$p^2$
divides
$(n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA).$
Then the hypothesis
$p\mid A$
implies that
$p\mid n.$
Since
$n\ge 3$
and A is square-free, we have
$p\mid B^n(n-m)^{n-m}m^m,$
that is,
$p\mid m(n-m)B,$
which is not true because
$\gcd (n,mB)= 1.$
It follows that
$p^2$
cannot divide
$(n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA)$
and so
$ (n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA)$
is square-free.
Case 2:
$p\nmid A$
. Then p will divide
$(n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA).$
Keeping in mind the hypothesis
$\gcd (n,mB)=1,$
it is easy to see that
$p\nmid n$
and so
$p\nmid Bm(n-m)$
. Let
$\alpha $
be a repeated root of
$\overline{f}(x)=x^n+\overline{A}(\overline{B}x+1)^m$
in the algebraic closure of
${\mathbb {Z}}/p{\mathbb {Z}}.$
Then
$$ \begin{align*} \alpha^n+A(B\alpha+1)^m\equiv 0\,\mod p \end{align*} $$
and
$$ \begin{align*} n\alpha^{n-1}+mAB(B\alpha+1)^{m-1}\equiv 0\,\mod p. \end{align*} $$
So
$n\alpha ^{n-1}\equiv -mAB(B\alpha +1)^{m-1}\,\mod p.$
By substitution,
$$ \begin{align*}-\alpha mAB(B\alpha+1)^{m-1}+n A(B\alpha+1)^{m}\equiv0\,\mod p,\end{align*} $$
that is,
$$ \begin{align*}(B\alpha+1)^{m-1}(\alpha AB(n-m)+nA)\equiv 0\,\mod p.\end{align*} $$
If
$B\alpha +1\equiv 0\,\mod p,$
then
$\alpha \equiv -{1}/{B}\,\mod p,$
which yields the contradiction
${(-1)^n}/{\overline {B}^n}=\overline {f}({-1}//{\overline {B}})=\overline {f}(\overline {\alpha })=0.$
Thus,
$\alpha AB(n-m)+nA\equiv 0\,\mod p,$
so that
$$ \begin{align} \alpha\equiv -\frac{nA}{AB(n-m)}\,\mod p \end{align} $$
is the unique repeated root of
$\overline {f}(x)$
in
${\mathbb {Z}}/p{\mathbb {Z}}$
and it is easy to show that
$\alpha $
has multiplicity
$2.$
So, assuming that
$\alpha $
is a positive integer satisfying (3.1), we can write
$$ \begin{align*} f(x) & =(x-\alpha+\alpha)^n+A(B(x-\alpha+\alpha)+1)^m,\\ & =\displaystyle\sum_{k=0}^{n}\binom{n}{k}\alpha^{n-k}(x-\alpha)^k+A\bigg(\displaystyle\sum_{k=0}^{m}\binom{m}{k}(B\alpha+1)^{m-k}B^k(x-\alpha)^k\bigg),\\ & =(x-\alpha)^2h(x)+f'(\alpha)(x-\alpha)+f(\alpha), \end{align*} $$
where
$f'(x)$
is the derivative of
$f(x)$
and
$$ \begin{align*}h(x)=\displaystyle\sum_{k=2}^{n}\binom{n}{k}\alpha^{n-k}(x-\alpha)^{k-2}+A\bigg(\displaystyle\sum_{k=2}^{m}\binom{m}{k}(B\alpha+1)^{m-k}B^k(x-\alpha)^{k-2}\bigg)\end{align*} $$
is in
${\mathbb {Z}}[x].$
Then
$\overline {f}(x)=(x-\overline {\alpha })^2\overline {h}(x),$
where
$\overline {h}(x)\in {\mathbb {Z}}[x]$
is separable. Let
$\prod _{i=1}^{t} \overline{h}_i(x)$
be the factorisation of
$\overline {h}(x)$
into a product of distinct irreducible polynomials
$\overline {h}_i(x)\in {\mathbb {Z}}/p{\mathbb {Z}}[x]$
with each
$h_i(x)\in {\mathbb {Z}}[x]$
monic. Then we can write
$$ \begin{align*} f(x)=(x-\alpha)^2\bigg(\displaystyle\prod_{i=1}^{t}h_i(x)+p g(x)\bigg)+f'(\alpha)(x-\alpha)+f(\alpha), \end{align*} $$
for some polynomial
$g(x)\in {\mathbb {Z}}[x]$
. This implies that
$$ \begin{align*}M(x)=\frac{1}{p}[p (x-\alpha)^2g(x)+(x-\alpha)f'(\alpha)+f(\alpha)].\end{align*} $$
In view of Dedekind’s criterion and the hypothesis that
$f(x)$
is monogenic, we see that
$f(\alpha )\not \equiv 0\,\mod p^2.$
Equivalently,
$$ \begin{align*}(n^n+(-1)^{n+m}(n-m)^{n-m}m^mB^nA)\not\equiv 0\,\mod p^2.\end{align*} $$
Hence,
$(n^n+(-1)^{n+m}(n-m)^{n-m}m^mB^nA)$
is square-free.
Conversely, suppose A and
$(n^n+(-1)^{n+m}(n-m)^{n-m}m^mB^nA)$
are square-free. If
$A=\pm 1,$
then using (1.1), we see that
$\mathrm {Ind}_K(\theta )=1,$
that is,
$f(x)$
is monogenic. If p be a prime divisor of
$A,$
then
$f(x)$
is an Eisenstein polynomial with respect to the prime p. Therefore, by Lemma 2.4,
$p\nmid \mathrm {Ind}_K(\theta ).$
Hence, by (1.1),
$f(x)$
is monogenic. This completes the proof of the theorem.
