1 Introduction and statement of results
Let
$f(x) \in \mathbb {Z}[x]$
be a monic irreducible polynomial with a root
$\theta $
and
$K = \mathbb {Q}(\theta )$
. We denote by
$\mathbb {Z}_K$
the ring of integers of K. It is well known that
$\mathbb {Z}_K$
is a free
$\mathbb {Z}$
-module of rank
$\deg (f)$
and
$\mathbb {Z}[\theta ]$
is a submodule of
$\mathbb {Z}_K$
of the same rank, so that
$[\mathbb {Z}_K : \mathbb {Z}[\theta ]]$
is finite. The index of
$f(x)$
, denoted
$\operatorname {ind}(f)$
, is defined to be the index
$[ \mathbb {Z}_K : \mathbb {Z}[\theta ]]$
. The polynomial
$f(x)$
is called monogenic if the index of
$f(x)$
is one, that is,
$\mathbb {Z}_K = \mathbb {Z}[\theta ]$
. The number field K is said to be monogenic if there exists some
$\alpha \in \mathbb {Z}_K$
such that
$\mathbb {Z}_K = \mathbb {Z}[\alpha ]$
. Note that the monogenity of the polynomial
$f(x)$
implies the monogenity of the number field K. However, the converse is false in general. For example, let
$K = \mathbb {Q}(\theta )$
, where
$\theta $
is a root of
$f(x)=x^2-5$
. Then, K is monogenic, since
$\mathbb {Z}_K = \mathbb {Z}[{(1 + \sqrt {5})}/{2}]$
, whereas
$f(x)$
is not monogenic because
$[\mathbb {Z}_K: \mathbb {Z}[\sqrt {5}]] = 2$
.
We let
$\Delta (f)$
and
$\Delta (K)$
denote the discriminants of the polynomial
$f(x)$
and the number field K, respectively. It is well known that
$\Delta (f)$
and
$\Delta (K)$
are related by
$$ \begin{align} \Delta(f) = \operatorname{ind}(f)^2 \cdot \Delta(K). \end{align} $$
The determination of whether an algebraic number field is monogenic is a classical and fundamental problem in algebraic number theory. We easily see from (1.1) that if
$\Delta (f)$
is squarefree, then
$f(x)$
is monogenic. However, the converse does not hold in general. In this connection, several classes of monogenic polynomials with nonsquarefree discriminant have been investigated (see for example [Reference Barman, Narode and Wagh1, Reference Jones11, Reference Jones12]). When the discriminant of
$f(x)$
is not squarefree, it can be quite difficult to establish that
$[\mathbb {Z}_K : \mathbb {Z}[\theta ]]=1$
. One of the standard techniques is known as Dedekind’s index criterion [Reference Cohen2, Theorem 6.1.4]. This method is applied to determine whether or not a particular prime p is a divisor of
$[\mathbb {Z}_K : \mathbb {Z}[\theta ]]$
. Using this criterion, Jakhar et al. [Reference Jakhar, Khanduja and Sangwan8] obtained necessary and sufficient conditions for determining the primes dividing the index of a trinomial. In [Reference Jones10], Jones computed the discriminant of the polynomial
${g(x)=x^{n}+a(bx+c)^{m}\in \mathbb {Z}[x]}$
with
$1\leq m<n$
and proved that when
${\gcd (n,\,mb)=c=1}$
, there exist infinitely many values of a for which
$g(x)$
is irreducible and monogenic. In the same article, he conjectured that if
$\gcd (n,\,mb)=1$
and a is prime, then
$g(x)$
is monogenic if and only if
$n^{n}+(-1)^{\,n+m} b^{n} (n-m)^{\,n-m} m^{m}a$
is squarefree. This conjecture was subsequently proved by Kaur et al. in [Reference Kaur and Kumar13]. In [Reference Jakhar5], Jakhar established necessary and sufficient conditions for the primes dividing the index of
$g(x)$
, and further proved that, if
$\deg (g)=q$
is prime and if there exists a prime p such that
$p\mid \Delta (g)$
,
$p^{2}\nmid \Delta (g)$
and
$p\nmid abcm$
, then the Galois group of
$g(x)$
is isomorphic to the symmetric group
$S_{q}$
. Later, in [Reference Jakhar, Kalwaniya and Kotyada6], Jakhar et al. considered the more general polynomial
$g_1(x)=x^{n}+a(bx^{k}+c)^{n}$
, and studied both its discriminant and the primes dividing its index. In a similar vein, Jones [Reference Jones12] determined the discriminant of the family
$g_{a}(x)=x^{\,n-m}(x+k)^{m}+a$
and proved that there exist infinitely many primes p for which
$g_{p}(x)$
is monogenic. More recently, Jakhar et al. [Reference Jakhar, Laishram and Yadav9] characterised the primes dividing the index of
$g_{a}(x)$
.
We study the discriminant and index of the polynomial
$f(x) = (x^2 + 1)^n - ax^n$
, where
$a\in \mathbb {Z}$
. In the first theorem, we calculate the discriminant of
$f(x)$
.
Theorem 1.1. Let
$f(x) = (x^2 + 1)^n - a x^n \in \mathbb {Z}[x]$
, where
$n \ge 2$
. If
$f(x)$
is irreducible, then
$$ \begin{align} \Delta(f) = (-1)^{2n \choose 2} n^{2n} a^{2n-2}(2^n - a)(2^n - (-1)^n a). \end{align} $$
In the following theorem, we characterise all the primes dividing the index of
$f(x)$
.
Theorem 1.2. 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
$f(x) = (x^2 + 1)^n - ax^n$
over
$\mathbb {Q}$
. A prime factor p of the discriminant
$\Delta (f)$
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
$p^2 \nmid a$
; -
(ii) when
$p \nmid a$
and
$p \mid n$
, then
$p^2 \nmid (a^{p^j} - a)$
, where j is the highest power of p dividing n; -
(iii) when
$p \nmid an$
and n is odd, then
$p^2$
does not divide
$\Delta (f)$
; -
(iv) when
$p \nmid an$
and n is even, then
$p^2$
does not divide
$2^n - a$
.
The following corollary is an immediate consequence of Theorem 1.2.
Corollary 1.3. 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
$f(x) = (x^2 + 1)^n - ax^n$
over
$\mathbb {Q}$
. Then,
$\mathbb {Z}_K =~\mathbb {Z}[\theta ]$
if and only if each prime p dividing
$\Delta (f)$
satisfies one of the conditions (i)–(iv) of Theorem 1.2.
2 Proof of Theorems 1.1 and 1.2
Dedekind’s criterion [Reference Cohen2, Theorem 6.1.4] has been extensively studied and generalised in the literature (see, for example, [Reference Ershov3, Reference Khanduja and Jhorar14, Reference Khanduja and Kumar15]). In [Reference Khanduja and Jhorar14], Khanduja et al. studied equivalent versions of the generalised Dedekind’s criterion. We use the following equivalent version to prove Theorem 1.2 (see also [Reference Jakhar, Khanduja and Sangwan7, Lemma 2.1]).
Theorem 2.1 [Reference Khanduja and Jhorar14, Theorem 1.1]
Let
$f(x) \in \mathbb {Z}[x]$
be a monic irreducible polynomial having the factorisation
$\overline {g}_1(x)^{e_1} \cdots \overline {g}_t(x)^{e_t}$
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 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,
$1 \leq i \leq t.$
Next, we recall a proposition that will be used to calculate
$\Delta (f)$
in Theorem 1.1.
Proposition 2.2 [Reference Ireland and Rosen4, Proposition 12.1.4]
Let
$f(x) \in \mathbb {Z}[x]$
be a monic irreducible polynomial of degree n. Let
$\theta $
be a root of
$f(x)$
and
$K = \mathbb {Q}(\theta )$
. Then,
$$ \begin{align*} \Delta(f) = (-1)^{n \choose 2} \mathcal{N}(f'(\theta)),\end{align*} $$
where
$\mathcal {N} : = \mathcal {N}_{K / \mathbb {Q}}$
is the algebraic norm.
Now, we prove our main results.
Proof of Theorem 1.1
Suppose
$f(\theta ) = 0$
. Then,
$$ \begin{align} (\theta^2 + 1)^n = a \theta^n. \end{align} $$
Since
$f'(x) = 2nx(x^2 + 1)^{n-1} - anx^{n-1}$
, (2.1) yields
$$ \begin{align*} (\theta^2 + 1) f'(\theta) & = 2n\theta(\theta^2 + 1)^{n} - an\theta^{n-1}(\theta^2 + 1) \\ & = 2n \theta a \theta^n -na\theta^{n-1}(\theta^2 + 1) \\ & = na\theta^{n-1}(\theta -1 )(\theta + 1). \end{align*} $$
Hence,
$$ \begin{align} \mathcal{N}(f'(\theta)) = \frac{\mathcal{N}(a)\mathcal{N}(n)\mathcal{N}(\theta)^{n-1}\mathcal{N}(\theta-1)\mathcal{N}(\theta+1)}{\mathcal{N}(\theta^2 + 1)}. \end{align} $$
Note that if
$f(x)$
is the minimal polynomial of
$\theta $
, then
$f(x - 1)$
and
$f(x + 1)$
are the minimal polynomials of
$\theta + 1$
and
$\theta - 1$
, respectively. Thus,
$\mathcal {N}(\theta ) = 1$
,
$\mathcal {N}(\theta + 1) = (2^n - (-1)^na)$
and
$\mathcal {N}(\theta - 1) = (2^n - a)$
. Further, from (2.1), we have
$\mathcal {N}(\theta ^2 + 1)^n = \mathcal {N}(a)\mathcal {N}(\theta )^n = a^{2n} $
, so that
$\mathcal {N}(\theta ^2 + 1) = a^2$
. Now, (2.2) yields
$$ \begin{align*} \mathcal{N}(f'(\theta)) = \frac{a^{2n}n^{2n}(2^n -a )(2^n -(-1)^na)}{a^2} = a^{2n-2}n^{2n}(2^n -a )(2^n -(-1)^na). \end{align*} $$
Hence, employing Proposition 2.2,
$$ \begin{align*} \Delta(f) = (-1)^{2n \choose 2} \mathcal{N}(f'(\theta))= (-1)^{2n \choose 2} n^{2n} a^{2n-2}(2^n - a)(2^n - (-1)^n a). \end{align*} $$
This completes the proof of the theorem.
Proof of Theorem 1.2
Let p be a prime dividing
$\Delta (f)$
. In view of Theorem 2.1, p does not divide
$[\mathbb {Z}_K : \mathbb {Z}[\theta ]]$
if and only if
$f(x)\notin \langle p, g(x) \rangle ^2$
for any monic polynomial
$g(x)\in \mathbb {Z}[x]$
which is irreducible modulo p. We prove the theorem by considering four cases.
Case (i):
$p \mid a$
. Then,
$f(x) \equiv (x^2 + 1)^n \pmod p$
. If
$\overline {g}(x)$
is an irreducible factor of
$x^2 + 1$
over
$\mathbb {Z}/p\mathbb {Z}$
, then
$f(x) \not \in \langle p, g(x) \rangle ^2$
if and only if
$p^2 \nmid a$
. Thus, by Theorem 2.1,
${p \nmid [\mathbb {Z}_K : \mathbb {Z}[\theta ]]}$
if and only if
$p^2 $
does not divide a.
Case (ii):
$p \mid n $
and
$p \nmid a$
. Let
$n = sp^j$
with
$\gcd (s,p) = 1$
. By the binomial theorem,
$$ \begin{align*}f(x) \equiv ((x^2+1)^s - ax^s)^{p^j} \pmod p.\end{align*} $$
Let
$h(x) = (x^2+1)^s - ax^s$
and
$\prod _{i=1}^{t} \overline {g_i}(x)^{e_i}$
be the factorisation of
$h(x)$
over
$\mathbb {Z}/ p \mathbb {Z}$
, where
$g_i(x)\in \mathbb {Z}[x]$
are monic polynomials which are distinct and irreducible modulo p. Now, raising both sides of
$(x^2+ 1)^s = h(x) + ax^s$
to the
$p^j$
th power, we obtain
$$ \begin{align*}(x^2 + 1)^n = h(x)^{p^j} + a^{p^j} x^{s{p^j}} + p h(x) T(x)\end{align*} $$
for some polynomial
$T(x)\in \mathbb {Z}[x]$
, where
$n=sp^j$
. This yields
$$ \begin{align} f(x) = h(x)^{p^j} + p h(x) T(x) +(a^{p^j} -a) x^{n}. \end{align} $$
Therefore, by (2.3),
$f(x) \not \in \langle p , g_i(x) \rangle ^2$
if and only if
$p^2$
does not divide
$a^{p^j} -a$
. Thus, by Theorem 2.1,
$p \nmid [\mathbb {Z}_K : \mathbb {Z}[\theta ]]$
if and only if
$p^2$
does not divide
$a^{p^j} -a$
. This completes the proof in this case.
Case (iii):
$p \nmid an$
and n is odd. Since
$p \mid \Delta (f)$
, the discriminant of
$\overline {f}(x) $
is
$0$
in
$\mathbb {Z}/ p\mathbb {Z}$
so that
$\overline {f}(x)$
has a repeated root in the algebraic closure of
$\mathbb {Z}/p\mathbb {Z}$
. Let
$\alpha $
be a repeated root of
$\overline {f}(x)$
. Then,
$$ \begin{align} \overline{f}(\alpha) & = (\alpha^2 + 1)^n - \overline{a}\alpha^n = 0,\end{align} $$
$$ \begin{align} (\overline{f})'(\alpha) & = \overline{2n} \alpha (\alpha^2 + 1 )^{n-1} - \overline{n} \overline{a} \alpha^{n-1} = 0. \end{align} $$
Substituting
$(\alpha ^2 + 1)^n = \overline {a}\alpha ^n$
in (2.5) yields
$\overline {an} \alpha ^{n-1} (\alpha ^2 - 1) = 0$
. Since
$p \nmid an$
and
${\alpha \neq 0}$
, any repeated root must satisfy
$\alpha ^2 = 1$
in the algebraic closure of
$\mathbb {Z}/p\mathbb {Z}$
. If
$\alpha $
is an integer such that
$\alpha \equiv \pm 1 \pmod {p}$
, then by (2.4), we obtain
$a \equiv 2^n \alpha ^{-n} \pmod {p}$
. In particular,
$$ \begin{align} a\equiv \begin{cases} 2^n \pmod{p} & \text{if } \alpha\equiv 1 \pmod{p} \\ (-2)^n \pmod{p} & \text{if } \alpha\equiv -1\pmod{p}. \end{cases} \end{align} $$
Both the congruences in (2.6) can hold simultaneously only if
$p = 2$
. However, this would force
$2 \mid a$
, which contradicts the assumption that
$p \nmid a$
. Hence, either
$\alpha \equiv 1 \pmod {p}$
or
$\alpha \equiv -1 \pmod {p}$
is a root of
$\overline {f}(x)$
, but not both.
Next, we show that any multiple root of
$\overline {f}(x)$
has multiplicity two, that is,
${(\overline {f})"(\alpha ) \neq 0}$
. We have
$$ \begin{align} f"(x) = n (4x^2 (n-1)(x^2 + 1)^{n-2} + 2(x^2 +1 )^{n-1} -a (n-1)x^{n-2}). \end{align} $$
Substituting
$\alpha ^2 = 1$
and
$a\alpha ^n \equiv 2^n\pmod {p}$
in (2.7) yields
$(\overline {f})"(\alpha ) = \overline {n2^{n}}$
. As
$p \nmid n$
,
$f"( \alpha ) \neq 0$
in the algebraic closure of
$\mathbb {Z}/p\mathbb {Z}$
. Hence,
$\alpha $
has multiplicity two.
Now, let
$\beta $
be an integer satisfying
$\beta \equiv -1 \pmod {p^2}$
and
$\alpha \equiv \beta \pmod p$
. Then,
$$ \begin{align} f(x) \nonumber &= ( (x - \beta + \beta)^2 + 1)^n - a(x -\beta + \beta)^n \\\nonumber &= ((x -\beta)^2 + 2\beta(x -\beta) + \beta^2 + 1)^n - a( (x-\beta)^2 + \beta )^n \\\nonumber &= (x-\beta)^{2n} + {n\choose 1}(x -\beta)^{2n-1}(2\beta(x -\beta) + \beta^2 + 1) + \cdots \\\nonumber &\quad + {n \choose {n-1}} (x - \beta)^{2n-n+1} (2\beta(x -\beta) + \beta^2 + 1)^{n-1} + (2\beta(x -\beta) + \beta^2 + 1)^n \\\nonumber &\quad - a\bigg((x - \beta)^n + {n \choose 1}\ \beta (x -\beta)^{n-1} + \cdots + {n \choose 1}\ \beta^{n-1} (x - \beta) + \beta^n \bigg) \\\nonumber &= (x - \beta)^2 g(x) + (2n\beta(\beta^2 + 1)^{n-1} - na \beta^{n-1})(x - \beta) + ((\beta^2 +1)^n - a \beta^n) \\&= (x - \beta)^2 g(x) + f'(\beta) (x - \beta) + f(\beta), \end{align} $$
where
$$ \begin{align*} g(x) = \sum_{k = 1 }^{n} {n \choose k} u^{2k-2}(2 \beta u + t)^{n-k} - \sum_{k =2}^{n} a {n \choose k} u^{k-2} \beta^{n-k} \end{align*} $$
with
$u = x -\beta $
and
$t = \beta ^2 + 1$
. Alternatively, one can find the Taylor series expansion of
$f(x)$
at
$\beta $
as follows:
$$ \begin{align*} f(x) &= f(\beta) + (x - \beta) f'(\beta) + (x - \beta)^2 \frac{f"(\beta) }{2} + \cdots + (x - \beta)^{2n} \frac{f^{2n}(\beta)}{(2n)!} \\ & = f(\beta) + (x - \beta) f'(\beta) + (x - \beta)^2 \bigg( f"(\beta) + \cdots + (x - \beta)^{2n -2} \frac{f^{2n}(\beta)}{(2n)!} \bigg). \end{align*} $$
Then,
$$ \begin{align*} \overline{f}(x) = (x - \beta)^2 \overline{g}(x).\end{align*} $$
Note that
$\overline {g}(x) \in (\mathbb {Z}/p\mathbb {Z})[x]$
is a separable polynomial, since
$\beta $
is the only repeated root of
$\overline {f}(x)$
in the algebraic closure of
$\mathbb {Z}/p\mathbb {Z}$
. Hence,
$$ \begin{align*} g(x) = \prod_{i=1}^{t} g_i(x) + p h(x),\end{align*} $$
where
$g_1(x), g_2(x), \ldots , g_t(x)$
are monic polynomials over
$\mathbb {Z}$
which are distinct and irreducible modulo p, and
$h(x) \in \mathbb {Z}[x]$
. Therefore, (2.8) yields
$$ \begin{align*} f(x) = (x - \beta)^2 \bigg(\prod_{i=1}^{t} g_i(x) + p h(x)\bigg) + (x - \beta) f'(\beta) + f(\beta).\end{align*} $$
From Theorem 2.1, it follows that
$ p \nmid [\mathbb {Z}_K : \mathbb {Z}[\theta ]] $
if and only if
$\gcd ((x - \beta ), \overline {M}(x)) = 1$
, where
$$ \begin{align*} M(x) = \frac{1}{p}\bigg(f(x) - \prod_{i=1}^{t} g_i(x)\bigg) = \frac{1}{p}(p(x - \beta)^2 h(x) + (x - \beta) f'(\beta) + f(\beta)).\end{align*} $$
Moreover,
$\gcd ((x - \beta ), \overline {M}(x)) = 1$
if and only if
$f(\beta ) \not \equiv 0 \pmod {p^2}$
. Since
$\beta \equiv -1 \pmod {p}$
, we have
$p \mid (2^n+a)$
and
$p \nmid (2^n -a)$
. In addition,
$p \nmid an$
. Consequently,
$f(\beta ) \not \equiv 0 \pmod {p^2}$
if and only if
$p^2 \nmid ( 2^n + a)$
. As
$p \nmid (2^n -a)$
and
$p \nmid an$
, it follows from (1.2) that
$$ \begin{align*}p \nmid [\mathbb{Z}_K : \mathbb{Z}[\theta]] \quad \text{ if and only if } p^2 \nmid \Delta(f).\end{align*} $$
Proceeding similarly to the case
$ \alpha \equiv -1 \pmod p$
as shown previously, we can prove that, if
$ \alpha \equiv 1 \pmod p$
, then
$p \nmid [\mathbb {Z}_K : \mathbb {Z}[\theta ]]$
if and only if
$p^2 \nmid \Delta (f)$
. This completes the proof of Case (iii).
Case (iv):
$p \nmid an$
and n is even. Since n is even, from (2.6), we have
$p \mid 2^n -a$
if
$ \alpha \equiv \pm 1 \pmod p$
. The rest of the proof goes along similar lines as shown in Case (iii). This completes the proof of the theorem.
3 Examples and remarks
Example 3.1. Let p be an odd prime and consider the polynomial
$$ \begin{align*} f_p(x) = (x^{2}+1)^{p} - p x^{p}. \end{align*} $$
Assume that
$f_p(x)$
is irreducible over
$\mathbb {Q}$
. Using SageMath, we find that
$f_p(x)$
is irreducible over
$\mathbb {Q}$
for all primes
$p<200$
. By Theorem 1.1, we have
$\Delta (f_p) = p^{4p-2} H(p)$
, where
$H(p) = (2^{p} - p)(2^{p} + p)$
. Observe that for any odd prime p, we have
$p \nmid H(p)$
. Let r be a prime divisor of
$\Delta (f_p)$
. If
$r = p$
, then by Theorem 1.2(i),
$$ \begin{align*}r \nmid \operatorname{ind}(f_p).\end{align*} $$
If
$r \neq p$
, then
$r \mid H(p)$
, and by Theorem 1.2(iii),
$$ \begin{align*} \operatorname{ind}(f_p) = 1\quad \text{ if and only if } r^{2} \nmid H(p).\end{align*} $$
Consequently, if
$H(p)$
is squarefree, then
$\operatorname {ind}(f_p)=1$
and, hence, the polynomial
$f_p(x)$
is monogenic. We find that the set of primes up to
$100$
for which
$H(p)$
is squarefree is given by
$\{3, 11, 13, 17, 19, 29, 37,47,67,71,73,89\}$
.
We now compute the index of
$f_p(x)$
for the remaining primes. For
$p = 5$
, the discriminant is
$$ \begin{align*} \Delta(f_5) = 3^3\cdot 5^{10}\cdot 37.\end{align*} $$
By Theorem 1.2(i),
$5$
does not divide
$\operatorname {ind}(f_5)$
, and by Theorem 1.2(iii),
$37$
also does not divide
$\operatorname {ind}(f_5)$
. Moreover, applying Theorem 1.2(iii) again shows that
$3$
divides
$\operatorname {ind}(f_5)$
. Recalling the identity
$$ \begin{align*} \Delta(f_5) = \operatorname{ind}(f_5)^{2}\,\Delta(K) \end{align*} $$
from (1.1), we conclude that
$\operatorname {ind}(f_5)=3$
. Applying the same reasoning, one can compute
$\operatorname {ind}(f_p)$
for all the remaining primes
$p<100$
. In Table 1, we summarise these index values.
Values of p and
$\operatorname {ind}(f_p)$
.
Remark 3.2. Numerical experiments conducted in SageMath suggest that there are many primes p for which
$H(p)$
is squarefree. It would be interesting to establish, using analytic methods, the existence of infinitely many such primes.
Remark 3.3. Let
$q\neq p$
be a prime dividing
$\Delta (f_p)$
. In Example 3.1, we have
$\nu _q(\Delta (f_p)) \le 3$
. Consequently, Theorem 1.2, together with (2.1), allows one to compute
$\operatorname {ind}(f_p)$
. However, when
$\nu _q(\Delta (f_p)) \ge 4$
, Theorem 1.2 alone is insufficient to decide whether
$q^{2}$
divides
$\operatorname {ind}(f_p)$
.
