ON A CONJECTURE OF LENNY JONES ABOUT CERTAIN MONOGENIC POLYNOMIALS

Abstract

Let $K={\mathbb {Q}}(\theta )$ be an algebraic number field with $\theta $ satisfying a monic irreducible polynomial $f(x)$ of degree n 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. Deciding whether or not a monic irreducible polynomial is monogenic is an important problem in algebraic number theory. In an attempt to answer this problem for a certain family of polynomials, Jones [‘A brief note on some infinite families of monogenic polynomials’, Bull. Aust. Math. Soc. 100 (2019), 239–244] conjectured that if $n\ge 3$, $1\le m\le n-1$, $\gcd (n,mB)=1$ and A is a prime number, then the polynomial $x^n+A (Bx+1)^m\in {\mathbb {Z}}[x]$ is monogenic if and only if $n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA$ is square-free. We prove that this conjecture is true.

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

(1.1) $$ \begin{align} \Delta_f=[{\mathbb{Z}}_K:{\mathbb{Z}}[\theta]]^2d_K. \end{align} $$

Clearly if $\Delta_f$ is square-free, then $f(x)$ is monogenic but the converse need not be true. Jones [4, 6] constructed infinite families of monogenic polynomials having non square-free discriminant. In [7], 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 [5], 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 [4, Theorem 3.1],

(1.2) $$ \begin{align} \Delta_f=(-1)^{{n(n-1)}/{2}}A^{n-1}[n^n+(-1)^{n+m}B^n(n-m)^{n-m}m^mA]. \end{align} $$

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 [4, 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, [2]; see also [1, 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 [3]). 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 [8, 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

(3.1) $$ \begin{align} \alpha\equiv -\frac{nA}{AB(n-m)}\,\mod p \end{align} $$

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.

Proof of Corollary 1.3.

It is easy to verify that $f(x)$ satisfies Eisenstein’s criterion with respect to the prime $p.$ So $f(x)$ is an irreducible polynomial. Hence, the result follows from Theorem 1.1.