CHARACTERISATION OF PRIMES DIVIDING THE INDEX OF A CLASS OF POLYNOMIALS AND ITS APPLICATIONS

Abstract

Let ${\mathbb {Z}}_{K}$ denote the ring of algebraic integers of an algebraic number field $K = {\mathbb Q}(\theta )$, where $\theta $ is a root of a monic irreducible polynomial $f(x) = x^n + a(bx+c)^m \in {\mathbb {Z}}[x]$, $1\leq m<n$. We say $f(x)$ is monogenic if $\{1, \theta , \ldots , \theta ^{n-1}\}$ is a basis for ${\mathbb {Z}}_K$. We give necessary and sufficient conditions involving only $a, b, c, m, n$ for $f(x)$ to be monogenic. Moreover, we characterise all the primes dividing the index of the subgroup ${\mathbb {Z}}[\theta ]$ in ${\mathbb {Z}}_K$. As an application, we also provide a class of monogenic polynomials having non square-free discriminant and Galois group $S_n$, the symmetric group on n letters.

1 Introduction and statements of results

Let $K = {\mathbb Q}(\theta )$ be an algebraic number field with $\theta $ in the ring ${\mathbb {Z}}_K$ of algebraic integers of K and let $f(x)$ of degree n be the minimal polynomial of $\theta $ over the field ${\mathbb Q}$ of rational numbers. Let $d_K$ denote the discriminant of K and $D_f$ the discriminant of the polynomial $f(x)$ . It is well known that $d_K$ and $D_f$ are related by the formula

$$ \begin{align*}D_f = [{\mathbb{Z}}_K : {\mathbb{Z}}[\theta]]^2d_K.\end{align*} $$

We say that $f(x)$ is monogenic if ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ]$ , or equivalently, if $D_f = d_K$ . In this case, $\{1, \theta , \ldots , \theta ^{n-1}\}$ is an integral basis of K and K is a monogenic number field. A number field K is called monogenic if there exists some $\alpha \in {\mathbb {Z}}_K$ such that ${\mathbb {Z}}_K = {\mathbb {Z}}[\alpha ]$ .

The determination of monogenity of an algebraic number field is one of the classical and important problems in algebraic number theory. An arithmetic characterisation of monogenic number fields is a problem due to Hasse (see [6]). Gaál’s book [5] provides some classifications of monogenity in lower degree number fields. Using Dedekind’s Index Criterion, Jakhar et al. [8] gave necessary and sufficient conditions for ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ]$ when $\theta $ is a root of an irreducible trinomial $x^n+ax^m+b \in {\mathbb {Z}}[x]$ having degree n, providing infinitely many monogenic trinomials. Jones [9] computed the discriminant of the polynomial $f(x) = x^n+a(bx+c)^m \in {\mathbb {Z}}[x]$ with $1\leq m < n$ and proved that when $\gcd (n, mb) = 1$ , there exist infinitely many values of a such that ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ]$ where $K = {\mathbb Q}(\theta )$ and $\theta $ has minimal polynomial $f(x)$ . He also conjectured that if $\gcd (n,mb) = 1$ and a is a prime number, then the polynomial ${x^n+a(bx+c)^m \in {\mathbb {Z}}[x]}$ is monogenic if and only if $n^n + (-1)^{n+m}b^n(n-m)^{n-m}m^ma$ is square-free. Recently, Kaur and Kumar [12] proved that this conjecture is true. Jones [11] gave infinite families of number fields K generated by a root $\theta $ of an irreducible quadrinomial, quintinomial or sextinomial for which ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ].$ He also proved in [10] that if $\theta $ is a root of an irreducible polynomial of the type ${f(x) = x^p-2ptx^{p-1}+p^2t^2x^{p-2}+1 \in {\mathbb {Z}}[x]}$ and p is an odd prime with $p\nmid t$ , then ${\mathbb {Z}}_K\neq {\mathbb {Z}}[\theta ].$

Let $K = {\mathbb Q}(\theta )$ be an algebraic number field where $\theta $ has minimal polynomial $f(x) = x^n+a(bx+c)^m$ over ${\mathbb Q}$ with $1\leq m<n$ . We characterise all the primes dividing the index of ${\mathbb {Z}}[\theta ]$ in ${\mathbb {Z}}_K$ . As an application, we provide necessary and sufficient conditions for ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ].$ We also establish a more general result confirming [9, Conjecture 4.1]. Further, we give a class of monogenic polynomials of prime degree q having non square-free discriminant and Galois group isomorphic to the symmetric group $S_q$ . In some examples, we determine the index $[{\mathbb {Z}}_K:{\mathbb {Z}}[\theta ]]$ as well.

Throughout the paper, $D_f$ will stand for the discriminant of $f(x) = x^n +a(bx+c)^m$ with $1\leq m<n$ . Jones [9, Theorem 3.1] proved that the discriminant $D_f$ is given by

(1.1) $$ \begin{align} D_f = (-1)^{\binom{n}{2}} c^{n(m-1)}a^{n-1}[c^{n-m}n^{n} + (-1)^{m+n} a b^{n} m^{m}(n-m)^{n-m} ]. \end{align} $$

We prove the following result.

Theorem 1.1. Let $K= {\mathbb Q}(\theta )$ be an algebraic number field with $\theta $ in the ring ${\mathbb {Z}}_K$ of algebraic integers of K having minimal polynomial $f(x) = x^n+a(bx+c)^m$ , $1\leq m < n$ , over $\mathbb Q$ . A prime factor p of the discriminant $D_f$ of $f(x)$ does not divide $[{\mathbb {Z}}_{K} : {\mathbb {Z}}[\theta ]]$ if and only if p satisfies one of the following conditions:

1. (i) when $p \mid a$ , then $p^2 \nmid ac$ ;

2. (ii) when $p\nmid a$ , $p\mid b$ , $p\mid c$ , then $m=1$ and $p^2\nmid c$ ;

3. (iii) when $p \nmid ac$ and $p\mid b$ with $j \geq 1$ as the highest power of p dividing n, then either $p\mid b_1$ and $p\nmid c_2$ or p does not divide $b_1[(ac^m)b_{1}^{n}+(-c_2)^{n}]$ , where

   $$ \begin{align*}b_1 = \frac{mabc^{m-1}}{p}, \quad c_2 = \frac{1}{p}[ac^m+(-ac^m)^{p^j}];\end{align*} $$

4. (iv) when p does not divide $ab$ and $p\mid c$ , then $m=1$ and either $p \mid b_2 $ with $p\nmid c_1 $ or p does not divide $b_2[(ab ) b_2 ^{n -1} + (-c_1)^{n - 1}]$ , where

   $$ \begin{align*}b_2= \frac{1}{p}[ab+(-ab)^{p^{l}}], \quad c_1= \frac{ac}{p} \quad\text{and}\quad n-1 = p^ls', p\nmid s';\end{align*} $$

5. (v) when p does not divide $abc$ and $p\mid m$ with $n = s'p^k,~m = sp^k,~p \nmid \gcd (s', s)$ , then the polynomials

   $$ \begin{align*}x^{s'} + a(bx+c)^s \quad\text{and}\quad \frac{1}{p}\bigg[pt(bx+c)^m-\sum\limits_{j=1}^{p^k-1}\binom{p^k}{j}(x^{s'})^{p^k-j}(a(bx+c)^s)^{\kern1.2pt j}\bigg]\end{align*} $$
   are coprime modulo p, where $t\in {\mathbb {Z}}$ is an integer such that $a = a^{p^k} + pt$ ;

6. (vi) when $p\nmid abcm$ , then $p^2$ does not divide $D_f$ .

The following corollary is immediate. It extends the main results of [9].

Corollary 1.2. Let $K = {\mathbb Q}(\theta )$ and $f(x) = x^n + a(bx+c)^m$ be as in Theorem 1.1. Then ${\mathbb {Z}}_{K} = {\mathbb {Z}}[\theta ]$ if and only if each prime p dividing $D_f$ satisfies one of the conditions (i)–(vi) of Theorem 1.1.

If we take $\gcd (n,mb) = 1$ and $c=1$ , then conditions (ii)–(v) of Theorem 1.1 are not possible. So in the special case when $c=1$ and $\gcd (n,mb) = 1$ , the above corollary provides the main result of [12] stated below. This gives infinite families of monogenic polynomials and establishes a more general form of [9, Conjecture 4.1].

Corollary 1.3 [12].

Let $f(x) = x^n + a(bx+1)^m \in {\mathbb {Z}}[x]$ be a monic irreducible polynomial of degree n with $\gcd (n,mb) = 1$ . Then ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ]$ if and only if each prime p dividing $D_f$ satisfies either (i) $p\mid a$ and $p^2\nmid a$ or (ii) $p\nmid a$ and $p^2\nmid D_f$ .

The following proposition follows readily from the proof of Theorem 1.1(vi) and is of independent interest.

Proposition 1.4. Let $f(x) = x^q+a(bx+c)^m\in {\mathbb {Z}}[x]$ , $1\leq m < q$ , be an irreducible polynomial of prime degree. If there exists a prime p such that p divides $D_f$ and $p^2\nmid D_f$ with $p\nmid abcm$ , then the Galois group of $f(x)$ is $S_q$ .

The following result is an immediate consequence of Corollary 1.3 and Proposition 1.4. It provides a class of monogenic polynomials having non square-free discriminant and Galois group equal to a symmetric group.

Corollary 1.5. Let m be a positive odd integer and $f(x) = x^q+a(bx+1)^m\in {\mathbb {Z}}[x]$ be a polynomial having prime degree $q\geq 3$ with $q\nmid b$ . If $a \not \in \{0, \pm 1\}$ and $D_f/a^{q-1}$ are square-free numbers, then $f(x)$ is a monogenic polynomial having Galois group $S_q$ .

The following example is an application of Theorem 1.1, Corollary 1.3 and Proposition 1.4. In this example, $K={\mathbb Q}(\theta )$ with $\theta $ a root of $f(x)$ .

Example 1.6. Let p be a prime number. Consider $f(x) = x^p+p(x+1)^{p-1}$ . Note that $|D_f| = p^p(p^{p-1}-(p-1)^{p-1})$ . Using Proposition 1.4, it is easy to check that the Galois group of $f(x)$ is $S_p$ . By Corollary 1.3, ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ]$ if and only if $p^{p-1}-(p-1)^{p-1}$ is square-free. We now compute $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ for $p<20.$ For $p=2,3,7,11,17$ , it can be verified that the number $p^{p-1}-(p-1)^{p-1}$ is square-free; and hence ${\mathbb {Z}}_K = {\mathbb {Z}}[\theta ]$ . Next we calculate the exact value of $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ corresponding to $p = 5, 13$ and $19$ .

1. (i) For $p = 5$ , it can be easily checked that $D_{f} = 5^5\cdot 3^2\cdot 41.$ In view of Theorem 1.1(i), $5$ does not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ . Also, $3$ divides $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ and $41$ does not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ by Theorem 1.1(vi). Since $ D_{f} $ = $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]^2\cdot d_K$ , where $d_K$ is the discriminant of K, we see that $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ is $3$ when $p = 5$ .

2. (ii) Consider $p = 13$ . One can verify that $D_{f} = 13^{13}\cdot 5^2\cdot 7\cdot 67\cdot 109\cdot 157\cdot 229\cdot 313$ . By Theorem 1.1(i), $13$ does not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ . Also in view of Theorem 1.1(vi), $5$ divides $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ and the primes $7, 67, 109, 157, 229, 313$ do not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ . Since the exact power of $5$ dividing $D_{f}$ is $2$ , $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]=5$ .

3. (iii) When $p = 19$ , then one can check that the prime factorisation of $D_{f}$ is given by $19^{19}\cdot 7^3\cdot r$ with r a square-free number. Arguing as above, $19$ and each prime p dividing r do not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ and $7$ divides $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ . Therefore, $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]=7$ .

2 Proof of Theorem 1.1

In what follows, while dealing with a prime number p, for a polynomial $h(x)$ in $ {\mathbb {Z}}[x]$ , we shall denote by $\bar {h}(x)$ the polynomial over ${\mathbb {Z}}/p{\mathbb {Z}}$ obtained by interpreting each coefficient of $h(x)$ modulo p.

We first state the following well-known theorem. The equivalence of assertions (i) and (ii) of the theorem was proved by Dedekind (see [2, Theorem 6.1.4], [3]). A simple proof of the equivalence of assertions (ii) and (iii) is given in [7, Lemma 2.1].

Theorem 2.1. Let $f(x) \in {\mathbb {Z}}[x]$ be a monic irreducible polynomial having the factorisation $\bar {g}_1(x)^{e_1} \cdots \bar {g}_{t}(x)^{e_{t}}$ modulo a prime p 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 $K={\mathbb Q}(\theta )$ with $\theta $ a root of $f(x)$ . Then the following statements are equivalent:

1. (i) p does not divide $[{\mathbb {Z}}_K:{\mathbb {Z}}[\theta ]]$ ;

2. (ii) for each i, either $e_i =1 $ or $\overline g_i (x)$ does not divide $\overline M(x)$ where

   $$ \begin{align*}M(x) = \frac{1}{p}(f(x) - g_1 (x)^{e_1} \cdots g_{t} (x)^{e_{t}} );\end{align*} $$

3. (iii) $f(x)$ does not belong to the ideal $\langle p, g_i(x)\rangle ^2$ in ${\mathbb {Z}}[x]$ for any i, $1\leq i\leq t$ .

The next lemma (see [7, Corollary 2.3]) is easily proved using the binomial theorem.

Lemma 2.2. Let $k\geq 1$ be the highest power of a prime p dividing a number $n=p^{k}s'$ and c be an integer not divisible by p. If $\bar {g}_1(x)\cdots \bar {g}_r(x)$ is the factorisation of $x^{s'}-\bar {c}$ into a product of distinct irreducible polynomials over ${\mathbb {Z}}/p{\mathbb {Z}}$ with each $g_i(x)\in \mathbb Z[x] $ monic, then

$$ \begin{align*}x^n-c=(g_1(x)\cdots g_r(x)+pH(x))^{p^k}+pg_1(x)\cdots g_r(x)T(x)+p^2U(x)+c^{p^k}-c\end{align*} $$

for some polynomials $H(x), T(x), U(x)\in \mathbb Z[x]$ .

Proof of Theorem 1.1.

Let p be a prime dividing $D_f$ . In view of Theorem 2.1, p does not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ if and only if $f(x) \not \in \langle p, g(x) \rangle ^2$ for any monic polynomial $g(x) \in {\mathbb {Z}}[x]$ which is irreducible modulo p. Note that $f(x) \not \in \langle p,g(x)\rangle ^2 $ if $\bar g(x) $ is not a repeated factor of $\bar {f}(x) $ . We prove the theorem case by case.

Case (i): $p\mid a$ . In this case, $f(x) \equiv x^n \pmod p$ . Clearly, $f(x) \in \langle p, x\rangle ^2$ if and only if $p^2$ divides $ac^m$ ; consequently, $p\nmid [{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ if and only if $p^2\nmid ac$ .

Case (ii): $p \nmid a$ and p divides both b and c. In this situation, $f(x) \equiv x^n \pmod p$ and it is easy to see that $f(x) \in \langle p, x\rangle ^2$ if and only if $p^2$ divides $c^m$ . Therefore, $p\nmid [{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ if and only if $p^2\nmid c^m$ , that is, $m=1$ and $p^2\nmid c$ .

Case (iii): $p\nmid ac$ and $p\mid b$ . As $p\mid D_f$ , it is clear from (1.1) that $p\mid n$ . Write $n = p^j s', ~p\nmid s'$ . By the binomial theorem,

$$ \begin{align*} f(x) \equiv x^n + ac^m \equiv (x^{s'} + ac^m)^{p^{j}}\pmod p. \end{align*} $$

Let $\bar {g}_1(x) \cdots \bar {g}_{t}(x)$ be the factorisation of $h(x) = x^{s'} + ac^m$ over ${\mathbb {Z}}/p{\mathbb {Z}}$ , where $g_i(x) \in {\mathbb {Z}}[x]$ are monic polynomials which are distinct and irreducible modulo p. Write $h(x)$ as $g_1(x)\cdots g_t(x) + pH(x)$ for some polynomial $H(x) \in {\mathbb {Z}}[x]$ . Applying Lemma 2.2 to $h(x)$ and keeping in view that

$$ \begin{align*}f(x) = h(x^{p^{j}}) + a(bx)^m+\binom{m}{1}a(bx)^{m-1}c + \cdots + \binom{m}{m-1}a(bx)c^{m-1}\end{align*} $$

with $p\mid b$ , we see that

(2.1) $$ \begin{align}\hspace{-5pt} f(x)&=\bigg(\prod\limits _{i=1}^{t} g_i(x)+pH(x)\bigg)^{p^j}+pT(x)\prod\limits _{i=1}^{t} g_i(x)+p^2U(x) + ac^m +(-ac^m)^{p^j} + ma(bx)c^{m-1} \end{align} $$

for some polynomials $T(x), U(x) \in {\mathbb {Z}} [x].$ As $j\geq 1$ , the first three summands on the right-hand side of (2.1) belong to $\langle p, g_i(x) \rangle ^2$ for each i, $1\leq i\leq t$ . So $f(x) \in \langle p, g_i(x) \rangle ^2$ for some i, $1\leq i\leq t$ , if and only if $mabc^{m-1}x + ac^m +(-ac^m)^{p^j} = p(b_1 x +c_2)$ does so. Clearly, $p(b_1 x +c_2 )$ belongs to $\langle p, g_i(x) \rangle ^2$ for some i if and only if either p divides both $b_1 $ , $c_2 $ or $p\nmid b_1$ and the polynomials $\bar b_1 x +\bar c_2 ,~ x^n +\overline {ac^m} $ have a common root. One can easily check that the polynomials $\bar b_1 x + \bar c_2$ and $x^n + \overline {ac^m}$ have a common root if and only if $( -{\bar c_2}/{\bar b_1} ) ^{n }= -\overline {ac^m}$ , that is, if and only if $p\mid [(-ac^m) b_1 ^{n} -(-c_2)^{n}] $ . Hence, $f(x) \not \in \langle p, g_i(x) \rangle ^2$ for any i if and only if either $p\mid b_1 $ and $p\nmid c_2 $ or p does not divide $b_1 [(ac^m) b_1 ^{n} +(-c_2)^{n}]$ . This proves the theorem in case (iii) by virtue of Theorem 2.1.

Case (iv): $p\nmid ab$ and $p\mid c$ . In this case, $\overline {f}(x) = x^m(x^{n-m}+\overline {ab^m})$ . If $m\geq 2,$ then x is a repeated factor and it is easy to check that $f(x) \in \langle p, x\rangle ^2$ , that is, p always divides $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ by Theorem 2.1. So, assume now that $m =1.$ By (1.1), ${p\mid (n-1)}$ , say ${n-1 = p^l s'}$ with $p\nmid s'$ . Write $x^{s'} + ab = g_1(x)\cdots g_t(x) + pH(x)$ , where $g_1(x), \ldots , g_{t}(x)$ are monic polynomials which are distinct as well as irreducible modulo p and ${H(x) \in {\mathbb {Z}}[x]}$ . Applying Lemma 2.2 to $h(x)$ = $x^{s'} + ab$ , we can write ${f(x)= x (x^{n-1} +ab) +ac}$ as

(2.2) $$ \begin{align} f(x)=x\bigg[\bigg(\prod\limits_{i=1}^{t}g_i(x)+pH(x)\bigg)^{p^l}+pT(x)\prod\limits_{i=1}^{t}g_i(x)+ p^2U(x) + ab+(-ab)^{p^l}\bigg]+ac, \end{align} $$

where $T(x), U(x)$ belong to ${\mathbb {Z}} [x]$ . Note that $x,\bar {g}_1(x) , \ldots , \bar {g}_t(x)$ are distinct irreducible factors of $\bar f(x)$ . Since $l\geq 1$ , the first three summands inside the square bracket on the right-hand side of (2.2) belong to $\langle p, g_i(x) \rangle ^2$ for each i, $1\leq i\leq t$ . So $f(x) \in \langle p, g_i(x) \rangle ^2$ for some i, $1\leq i\leq t$ , if and only if $(ab +(-ab)^{p^l})x + ac = p(b_2 x +c_1)$ does so. Clearly, the polynomial $p(b_2 x +c_1 )$ belongs to $\langle p, g_i(x) \rangle ^2$ for some i if and only if either p divides both $b_2 $ , $c_1 $ or $p\nmid b_2 $ and the polynomials $\bar b_2 x +\bar c_1 ,~ x^{n-1} +\overline { ab} $ have a common root. The polynomials $\bar {b}_2 x + \bar {c}_{1}$ and $x^{n - 1} + \overline {ab}$ have a common root if and only if $( -{\bar c_1}/{\bar b_2} ) ^{n -1} = -\overline { ab} $ . Thus, $f(x) \in \langle p, g_i(x) \rangle ^2$ for some i if and only if either p divides both $b_2 $ , $c_1 $ or $p\nmid b_2 $ and $p\mid [(-ab ) b_2 ^{n -1} - (-c_1)^{n - 1}]$ . So we conclude that p does not divide $[{\mathbb {Z}}_K:{\mathbb {Z}}[\theta ]]$ if and only if $m=1$ and either $p \mid b_2 $ with $p\nmid c_1 $ or p does not divide $b_2[(ab ) b_2 ^{n -1} + (-c_1)^{n - 1}]$ . This proves the theorem in case (iv).

Case (v): $p \nmid abc$ and $p \mid m$ . As $p \mid D_f$ , p divides n in view of (1.1). Write $n = s'p^{k}$ , $m \kern1.3pt{=}\kern1.3pt sp^{k}$ with $p\kern1.3pt{\nmid}\kern1.3pt \gcd (s', s)$ so that $f(x) \kern1.3pt{=}\kern1.3pt (x^{s'})^{p^k} \kern1.3pt{+}\kern1.3pt a(bx+c)^{sp^k}$ . Set ${h(x) \kern1.3pt{=}\kern1.3pt x^{s'} \kern1.3pt{+}\kern1.3pt a(bx+c)^s}$ . Let $t\in {\mathbb {Z}}$ be an integer such that $a = a^{p^k} + pt$ . Then one can easily check that ${f(x) \equiv h(x)^{p^{k}}}$ (mod p). Let $h(x) \equiv g_1(x)^{d_1}\cdots g_{t}(x)^{d_{t}} \pmod p$ be the factorisation of $h(x)$ into a product of irreducible polynomials modulo p with $g_{i}(x) \in {\mathbb {Z}}[x]$ monic and $d_i> 0$ . Write

$$ \begin{align*} f(x) &= h(x)^{p^{k}} +pt(bx+c)^m-\sum\limits_{j=1}^{p^k-1}\binom{p^k}{j}(x^{s'})^{p^k-j}(a(bx+c)^s)^{\kern1.2pt j}. \end{align*} $$

Now $f(x) = (g_1(x)^{d_1} \cdots g_{t}(x)^{d_{t}})^{p^{k}} + pM(x)$ for some $M(x) \in {\mathbb {Z}}[x]$ . Since $k> 0$ , by Theorem 2.1, p does not divide $[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]$ if and only if $\overline {M}(x)$ is coprime to $\bar {h}(x)$ , which holds if and only if the polynomial

$$ \begin{align*}\frac{1}{p}\bigg[pt(bx+c)^m-\sum\limits_{j=1}^{p^k-1}\binom{p^k}{j}(x^{s'})^{p^k-j}(a(bx+c)^s)^{\kern1.2pt j}\bigg]\end{align*} $$

is coprime to $h(x)$ modulo p. This proves the theorem in case (v).

Case (vi): $p\nmid abcm$ . Since $p\mid D_f$ and $p\nmid abcm$ , it follows from (1.1) that $p\nmid n(n-m)$ . Let $\beta $ be a repeated root of $\bar {f}(x) = x^n+\bar {a}(\bar {b}x+\bar {c})^m$ in the algebraic closure of ${\mathbb {Z}}/p{\mathbb {Z}}$ . Then

(2.3) $$ \begin{align} \bar f(\beta)=\beta^n+\bar a (\bar b\beta+\bar c)^m= \bar 0; \quad \bar f'(\beta)=\bar n\beta^{n-1}+ \bar {m}\bar{a} \bar b (\bar b\beta + \bar{c})^{m-1}= \bar 0. \end{align} $$

On substituting $\bar {n}\beta ^{n-1} = -\bar {m}\bar {a} \bar b (\bar b\beta + \bar {c})^{m-1}$ in the first equation of (2.3), we see that

$$ \begin{align*} (b\beta + c)^{m-1}(ab(n-m)\beta + nac) \equiv 0 \pmod p. \end{align*} $$

Observe that $(b\beta + c) \not \equiv 0 \pmod p$ , otherwise $\beta = \bar {0}$ in view of the first equation of (2.3) which is not possible as $p\nmid ac$ . Therefore, keeping in mind that $p\nmid abcn(n-m)$ ,

(2.4) $$ \begin{align} \beta \equiv -\frac{nc}{b(n-m)} \pmod p \end{align} $$

is the unique repeated root of $\bar {f}(x)$ in ${\mathbb {Z}}/p{\mathbb {Z}}$ and it can be easily checked that $\beta $ has multiplicity $2$ . Assuming that $\beta $ is a positive integer satisfying (2.4), we can write

$$ \begin{align*} f(x) &= (x-\beta+\beta)^n + a(b(x-\beta + \beta)+c)^m,\\ &= \sum\limits_{k=0}^{n}\binom{n}{k}\hspace{1.5pt}\beta^{n-k}(x-\beta)^k + a\bigg(\sum\limits_{k=0}^{m}\binom{m}{k}(b\beta + c)^{m-k}b^k(x-\beta)^k\bigg),\\ &= (x-\beta)^2g(x) + f'(\hspace{2pt}\beta)(x-\beta)+f(\hspace{2pt}\beta), \end{align*} $$

where $f'(x)$ is the derivative of $f(x)$ and

$$ \begin{align*}g(x) = \sum\limits_{k=2}^{n}\binom{n}{k}\hspace{1.5pt}\beta^{n-k}(x-\beta)^{k-2} + a\bigg(\sum\limits_{k=2}^{m}\binom{m}{k}(b\beta + c)^{m-k}b^k(x-\beta)^{k-2}\bigg) \end{align*} $$

is in ${\mathbb {Z}}[x]$ . Then

(2.5) $$ \begin{align} \bar{f}(x) = (x-{\beta})^2\bar{g}(x), \end{align} $$

where $\bar {g}(x) \in ({\mathbb {Z}}/p{\mathbb {Z}})[x]$ is separable. Write $g(x) = g_1(x)\cdots g_t(x) + ph(x)$ , where $g_1(x), \ldots , g_t(x)$ are monic polynomials which are distinct as well as irreducible modulo p and $h(x) \in {\mathbb {Z}}[x]$ monic. Therefore, we can write

$$ \begin{align*}f(x) = (x-\beta)^2\bigg(\prod\limits_{i=1}^{t}g_i(x) + ph(x)\bigg) + f'(\beta)(x-\beta) + f(\beta).\end{align*} $$

So, by Theorem 2.1, p does not divide $[{\mathbb {Z}}_K:{\mathbb {Z}}[\theta ]]$ if and only if $\overline {M}(x)$ is coprime to $x-\beta $ , where

$$ \begin{align*}M(x) = \frac{1}{p}[p(x-\beta)^2h(x)+(x-\beta)f'(\beta)+f(\beta)],\end{align*} $$

that is, $f(\beta ) \not \equiv 0 \pmod p^2.$ By (2.4), since $p\nmid abcmn(n-m)$ , we see that $f(\hspace{2pt}\beta ) \not \equiv 0 \pmod p^2$ if and only if $(n^nc^{n-m}+(-1)^{n+m}b^n(n-m)^{n-m}m^ma) \not \equiv 0 \pmod p^2.$ This final case completes the proof of the theorem.

3 Proof of Proposition 1.4

The following two results on Galois groups will be used in the proof of Proposition 1.4.

Theorem 3.1 [1, Theorem 2.1].

Let $f(x)\in \mathbb {Z}[x]$ be a monic irreducible polynomial of degree n, having a root $\theta $ . Let p be a rational prime which is ramified in ${\mathbb Q}(\theta )$ . Suppose that $f(x)\equiv (x-c)^2\phi _2(x)\cdots \phi _r(x) \pmod p$ , where $(x-c), \phi _2(x), \ldots , \phi _r(x)$ are monic polynomials over ${\mathbb {Z}}$ which are distinct and irreducible modulo p. Then the Galois group of $f(x)$ over ${\mathbb Q}$ contains a nontrivial automorphism which keeps $n-2$ roots of $f(x)$ fixed.

Lemma 3.2 [4, Lemma 2].

Let $f(x)$ be an irreducible polynomial of degree $n\geq 2$ . If the Galois group of $f(x)$ over ${\mathbb Q}$ contains a transposition and a p-cycle for some prime $p>n/2$ , then the Galois group is $S_n.$

Proof of Proposition 1.4.

Let $\alpha $ be any root of $f(x)$ , so that $[{\mathbb Q}(\alpha ):{\mathbb Q}] = q.$ By the fundamental theorem of Galois theory, the Galois group of $f(x)$ , say $G_f$ , contains a subgroup whose index is q. By Lagrange’s theorem, q divides the order of $G_f$ . So, by Cauchy’s theorem, $G_f$ has an element of order q. Hence, $G_f$ contains a q-cycle. Now we show that $G_f$ contains a transposition. By hypothesis, there exists a prime p such that $p\mid D_f$ and $p\nmid abcm$ . As in (2.5) in the proof of Theorem 1.1(vi), $f(x) \equiv (x-\beta )^2g_1(x)\cdots g_t(x)$ (mod p), where $x-\beta , g_1(x), \ldots , g_t(x)$ are monic polynomials over ${\mathbb {Z}}$ which are distinct and irreducible modulo $p.$ Also, if $K ={\mathbb Q}(\theta )$ with $\theta $ a root of $f(x)$ , then keeping in mind the hypothesis $p^2\nmid D_f$ and the relation $D_f = [{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]^2d_K$ , we see that $p \mid d_K$ . Hence, p is ramified in K. Therefore, by Theorem 3.1, the Galois group of $f(x)$ contains a transposition. Hence, by Lemma 3.2, the Galois group is $S_q.$