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 [Reference Hasse6]). Gaál’s book [Reference Gaál5] provides some classifications of monogenity in lower degree number fields. Using Dedekind’s Index Criterion, Jakhar et al. [Reference Jakhar, Khanduja and Sangwan8] 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 [Reference Jones9] 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 [Reference Kaur and Kumar12] proved that this conjecture is true. Jones [Reference Jones11] 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 [Reference Jones10] 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 [Reference Jones9, 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 [Reference Jones9, Theorem 3.1] proved that the discriminant
$D_f$
is given by
$$ \begin{align} D_f = (-1)^{\binom{n}{2}} c^{n(m-1)}a^{n-1}[c^{n-m}n^{n} + (-1)^{m+n} a b^{n} m^{m}(n-m)^{n-m} ]. \end{align} $$
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:
-
(i) when
$p \mid a$
, then
$p^2 \nmid ac$
; -
(ii) when
$p\nmid a$
,
$p\mid b$
,
$p\mid c$
, then
$m=1$
and
$p^2\nmid c$
; -
(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
$$ \begin{align*}b_1 = \frac{mabc^{m-1}}{p}, \quad c_2 = \frac{1}{p}[ac^m+(-ac^m)^{p^j}];\end{align*} $$
-
(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
$$ \begin{align*}b_2= \frac{1}{p}[ab+(-ab)^{p^{l}}], \quad c_1= \frac{ac}{p} \quad\text{and}\quad n-1 = p^ls', p\nmid s';\end{align*} $$
-
(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 are coprime modulo p, where
$$ \begin{align*}x^{s'} + a(bx+c)^s \quad\text{and}\quad \frac{1}{p}\bigg[pt(bx+c)^m-\sum\limits_{j=1}^{p^k-1}\binom{p^k}{j}(x^{s'})^{p^k-j}(a(bx+c)^s)^{\kern1.2pt j}\bigg]\end{align*} $$
$t\in {\mathbb {Z}}$
is an integer such that
$a = a^{p^k} + pt$
;
-
(vi) when
$p\nmid abcm$
, then
$p^2$
does not divide
$D_f$
.
The following corollary is immediate. It extends the main results of [Reference Jones9].
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 [Reference Kaur and Kumar12] stated below. This gives infinite families of monogenic polynomials and establishes a more general form of [Reference Jones9, Conjecture 4.1].
Corollary 1.3 [Reference Kaur and Kumar12].
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
$19$
.
-
(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
$p = 5$
. -
(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
$2$
,
$[{\mathbb {Z}}_K : {\mathbb {Z}}[\theta ]]=5$
. -
(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 [Reference Cohen2, Theorem 6.1.4], [Reference Dedekind3]). A simple proof of the equivalence of assertions (ii) and (iii) is given in [Reference Jakhar, Khanduja and Sangwan7, 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:
-
(i) p does not divide
$[{\mathbb {Z}}_K:{\mathbb {Z}}[\theta ]]$
; -
(ii) for each i, either
$e_i =1 $
or
$\overline g_i (x)$
does not divide
$\overline M(x)$
where
$$ \begin{align*}M(x) = \frac{1}{p}(f(x) - g_1 (x)^{e_1} \cdots g_{t} (x)^{e_{t}} );\end{align*} $$
-
(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 [Reference Jakhar, Khanduja and Sangwan7, 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
$$ \begin{align}\hspace{-5pt} f(x)&=\bigg(\prod\limits _{i=1}^{t} g_i(x)+pH(x)\bigg)^{p^j}+pT(x)\prod\limits _{i=1}^{t} g_i(x)+p^2U(x) + ac^m +(-ac^m)^{p^j} + ma(bx)c^{m-1} \end{align} $$
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
$$ \begin{align} f(x)=x\bigg[\bigg(\prod\limits_{i=1}^{t}g_i(x)+pH(x)\bigg)^{p^l}+pT(x)\prod\limits_{i=1}^{t}g_i(x)+ p^2U(x) + ab+(-ab)^{p^l}\bigg]+ac, \end{align} $$
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
$$ \begin{align} \bar f(\beta)=\beta^n+\bar a (\bar b\beta+\bar c)^m= \bar 0; \quad \bar f'(\beta)=\bar n\beta^{n-1}+ \bar {m}\bar{a} \bar b (\bar b\beta + \bar{c})^{m-1}= \bar 0. \end{align} $$
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
$p\nmid abcn(n-m)$
,
$$ \begin{align} \beta \equiv -\frac{nc}{b(n-m)} \pmod p \end{align} $$
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
$$ \begin{align} \bar{f}(x) = (x-{\beta})^2\bar{g}(x), \end{align} $$
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 [Reference Bishnoi and Khanduja1, 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 [Reference Filaseta and Moy4, 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
$S_q.$
