1 Introduction
Throughout this article, we let
We say
$f(x)$
is monogenic if
$f(x)$
is irreducible over
${\mathbb Q}$
and
$\{1,\theta ,\theta ^2,\ldots ,\theta ^{2p-1}\}$
is a basis for the ring of integers
${\mathbb Z}_K$
of
$K={\mathbb Q}(\theta )$
, where
$f(\theta )=0$
. It is well known (see [Reference Cohen2]) that for
$f(x)$
irreducible over
${\mathbb Q}$
,
where
$[{\mathbb Z}_K:{\mathbb Z}[\theta ]]$
is called the index of
$f(x)$
, while
$\Delta (f)$
and
$\Delta (K)$
denote the respective discriminants over
${\mathbb Q}$
of
$f(x)$
and the number field K. From (1.2),
$f(x)$
is monogenic if and only if
$\Delta (f)=\Delta (K)$
, or equivalently,
${\mathbb Z}_K={\mathbb Z}[\theta ]$
.
With
$f(x)$
irreducible over
${\mathbb Q}$
, the authors determined
${{\mathrm {{Gal}}}}(f)$
for
$p=3$
in [Reference Harrington and Jones3, Theorem 4.2]. The second author extended this result in [Reference Jones9] to the situation when
$p\ge 5$
with
$b=1$
. Recently, Jim and Hagedorn have determined
${{\mathrm {{Gal}}}}(f)$
for
$p\ge 5$
without the restriction that
$b=1$
[Reference Jim and Hagedorn8]. Combining all these results yields the following result.
Theorem 1.1. Suppose that
$f(x)$
, as defined in (1.1), is irreducible over
${\mathbb Q}$
and let
$C_N$
denote the cyclic group of order N. Then,
If
$p\nmid \delta $
, we see from Theorem 1.1 that
${{\mathrm {{Gal}}}}(f)\simeq (C_p\rtimes C_{p-1})\times C_2$
. Since, either all or none of the trinomials in
$\{\pm f(\pm x),\pm \tilde {f}(\pm x)\}$
are monogenic, where
$\tilde {f}$
denotes the reciprocal of f, the characterisation of the monogenic trinomials
$f(x)$
with
${{\mathrm {{Gal}}}}(f)\simeq (C_p\rtimes C_{p-1})\times C_2$
follows from [Reference Jones10, Reference Jones11]. Consequently, in this article, we focus on the situation when
$p\mid \delta $
and we provide a complete characterisation of the monogenic trinomials
$f(x)$
, according to their Galois groups in Theorem 1.1, in this situation. A rather striking consequence of this investigation is highlighted in Corollary 1.3, which links the monogenicity of a polynomial with the existence of primes of a certain form. This work extends prior investigations of the authors [Reference Harrington and Jones4–Reference Harrington and Jones6, Reference Jones9–Reference Jones11]. Our main results are as follows.
Theorem 1.2. Let p,
$f(x)$
and
$\delta $
be as given in (1.1). Suppose that
$f(x)$
is irreducible over
${\mathbb Q}$
and that
$p\mid \delta $
.
-
(1) If ${{\mathrm {{Gal}}}}(f)\simeq C_p\rtimes C_{p-1}$
, then $$ \begin{align*}f(x) \ \text{is monogenic} \ \Longleftrightarrow \ (p,a,b)\in \{(5,\pm 3,1), (a^2+4,a,-1)\}.\end{align*} $$
-
(2) If ${{\mathrm {{Gal}}}}(f)\simeq (C_p\rtimes C_{(p-1)/2})\times C_2$
, then $$ \begin{align*}f(x) \ \text{is monogenic} \ \Longleftrightarrow \ (p,a,b)\in \{(3,\pm 1,1)\}.\end{align*} $$
-
(3) Let ${{\mathrm {{Gal}}}}(f)\simeq (C_p\rtimes C_{p-1})\times C_2$
.-
(a) If $\delta =(-1)^{p(p+1)/2}py^2$
for some y, then $$ \begin{align*}f(x) \ \text{is monogenic} \ \Longleftrightarrow \ (p,a,b)\in \{(3,\pm4,1)\}.\end{align*} $$
-
(b) If $\delta \ne (-1)^{p(p+1)/2}py^2$
for any y, then $$ \begin{align*}f(x) \ \text{is monogenic} \ \Longleftrightarrow \ \begin{cases} b=1 & \text{with } a-2 \text{ and } a+2 \text{ squarefree}\\ \text{or}\\ b=-1 & \text{with } 4\nmid a \text{ and } {\mathcal W} \text{ squarefree.} \end{cases} \end{align*} $$Moreover, for any prime $p\ge 3$
, there exist infinitely many such trinomials
$f(x)$
for each case of item (3b).
-
The following interesting corollary of Theorem 1.2 emphasises the connection between the existence of infinitely many primes of a particular quadratic form and the existence of infinitely many monogenic trinomials
$f(x)$
of a certain form with
${{\mathrm {{Gal}}}}(f)\simeq C_p\rtimes C_{p-1}$
.
Corollary 1.3. Let a and z denote positive integers. For any integer
$N>2$
, there exists a prime
$p>N$
such that the polynomial
$f(x)=x^{2p}+ax^p-1$
is monogenic with
${{\mathrm {{Gal}}}}(f)\simeq C_p\rtimes C_{p-1}$
for some a (and only one such a exists) if and only if there exist infinitely many primes of the form
$z^2+4$
.
2 Preliminaries
The following theorem is due to Kaur et al. [Reference Kaur, Kumar and Remete12].
Theorem 2.1. Let
$g(x)$
be a monic polynomial with integer coefficients. Let
$k\ge 2$
be an integer such that
${\mathfrak f}(x):=g(x^k)$
is irreducible over
${\mathbb Q}$
. Then,
${\mathfrak f}(x)$
is monogenic if and only if all of the following conditions are satisfied:
-
(1) $g(0)$
is squarefree; -
(2) q does not divide the index of ${\mathfrak f}(x)$
for all primes
$q\mid k$
; -
(3) $g(x)$
is monogenic.
The next theorem, due to Jakhar et al. [Reference Jakhar, Khanduja and Sangwan7], gives necessary and sufficient conditions to determine when a prime divisor of
$\Delta ({\mathfrak F})$
does not divide the index of
${\mathfrak F}$
, where
${\mathfrak F}$
is an arbitrary irreducible monic trinomial.
Theorem 2.2. Let
$n,m\in {\mathbb Z}$
with
$n>m\ge 1$
. Let
$K={\mathbb Q}(\theta )$
be an algebraic number field with
$\theta \in {\mathbb Z}_K$
, the ring of integers of K, having minimal polynomial
${\mathfrak F}(x)=x^{n}+Ax^m+B$
over
${\mathbb Q}$
, where
$\gcd (m,n)=d_0$
,
$m=m_1d_0$
and
$n=n_1d_0$
. A prime factor q of
$\Delta ({\mathfrak F})$
does not divide
$[{\mathbb Z}_K:{\mathbb Z}[\theta ]]$
if and only if q satisfies one of the following conditions:
-
(i) when $q\mid A$
and
$q\mid B$
, then
$q^2\nmid B$
; -
(ii) when $q\mid A$
and
$q\nmid B$
, then $$ \begin{align*}\text{either } \; q\mid a_2 \text{ and } q\nmid b_1 \quad \text{ or } \quad q\nmid a_2((-B)^{m_1}a_2^{n_1}+(-b_1)^{n_1}),\end{align*} $$where $a_2=A/q$
and
$b_1={(B+(-B)^{q^j})}/{q}$
, such that
$q^j\mid \mid n$
with
$j\ge 1$
;
-
(iii) when $q\nmid A$
and
$q\mid B$
, then $$ \begin{align*}\text{either}\; q\mid a_1 \text{ and } q\nmid b_2 \quad\text{or}\quad q\nmid a_1b_2^{m-1}((-A)^{m_1}a_1^{n_1-m_1}-(-b_2)^{n_1-m_1}),\end{align*} $$where $a_1={(A+(-A)^{q^l})}/{q}$
, such that
$q^l\mid \mid (n-m)$
with
$l\ge 0$
, and
$b_2=B/q$
;
-
(iv) when $q\nmid AB$
and
$q\mid m$
with
$n=s^{\prime }q^r$
,
$m=sq^r$
,
$q\nmid \gcd (s^{\prime },s)$
, then the polynomials $$ \begin{align*} H_1(x):=x^{s^{\prime}}+Ax^s+B \quad \text{and}\quad H_2(x):=\dfrac{Ax^{sq^r}+B+(-Ax^s-B)^{q^r}}{q} \end{align*} $$are coprime modulo q;
-
(v) when $q\nmid ABm$
, then
$q^2\nmid (B^{n_1-m_1}n_1^{n_1}-(-1)^{m_1}A^{n_1}m_1^{m_1}(m_1-n_1)^{n_1-m_1})$
.
In light of Theorem 2.1, we use Theorem 2.2 to prove the following useful lemma.
Lemma 2.3. Let
$g(x)=x^2+ax+b$
, where a is a nonzero integer and
$b\in \{\pm 1\}$
. Suppose that
$g(x)$
is irreducible over
${\mathbb Q}$
. Then,
Proof. Observe that
$\Delta (g)=\delta $
. We examine Theorem 2.2 with
$A=a$
,
$B=b$
and q a prime divisor of
$\delta $
.
Suppose first that
$b=1$
. If
$q\mid a$
, then
$q=2$
, so that
$b_1=1$
in condition (ii) of Theorem 2.2. It follows that
$2\nmid [{\mathbb Z}_K:{\mathbb Z}[\theta ]]$
if and only if
$4\mid a$
. If
$q\nmid a$
, then we have
$q\nmid [{\mathbb Z}_K:{\mathbb Z}[\theta ]]$
if and only if
$q^2\nmid {\mathcal W}$
is squarefree by condition (v) of Theorem 2.2, which completes the proof when
$b=1$
.
Suppose next that
$b=-1$
. If
$q\mid a$
, then
$q=2$
, so
$b_1=0$
in condition (ii) of Theorem 2.2. Thus,
$2\nmid [{\mathbb Z}_K:{\mathbb Z}[\theta ]]$
if and only if
$4\nmid a$
. Suppose that
$q\nmid a$
. Then, by condition (v) of Theorem 2.2, we see that
$q\nmid [{\mathbb Z}_K:{\mathbb Z}[\theta ]]$
if and only if
${\mathcal W}$
is squarefree.
3 The proof of Theorem 1.2
We first examine the three conditions of Theorem 2.1 to determine criteria for the monogenicity of
$f(x)$
. In the context of Theorem 2.1, with
${\mathfrak f}(x)$
equal to our trinomial
$f(x)$
, we have
$f(x)=g(x^p)$
, where
$g(x)=x^{2}+ax+b^p$
. Note that
$p\mid \Delta (f)$
since
$\Delta (f)=b^{p(p-1)}p^{2p}(a^2-4b^p)^p$
, by a theorem due to Swan [Reference Swan14].
Condition (1) of Theorem 2.1 requires that
$g(0)$
is squarefree, which implies that
$b=\pm 1$
. For a prime
$p\ge 3$
, with
$j\in \{1,2\}$
and
$b=\pm 1$
, let
Condition (2) of Theorem 2.1 indicates that we only have to check the single prime
$q=p$
in Theorem 2.2 to determine conditions under which the index of
$f(x)$
is not divisible by a prime divisor q of
$\Delta (f)$
. Since
$p\mid \delta $
, where p is odd, and
$b=\pm 1$
, we deduce that we only have to check condition (iv) of Theorem 2.2.
Condition (3) of Theorem 2.1 requires that
$g(x)$
is monogenic for the monogenicity of
$f(x)$
. For the four cases
$(\Gamma _j,b)$
described in (3.1), it follows from Lemma 2.3 that
where
$4\mid a$
if
$2\mid a$
when
$b=1$
, and
$4\nmid a$
when
$b=-1$
. Clearly, of the four cases
$(\Gamma _j,b)$
, the case
$(\Gamma _2,-1)$
is not possible. Consequently, we examine the three cases
$(\Gamma _j,b)\in \{(\Gamma _1,1), (\Gamma _2,1), (\Gamma _1,-1)\}$
, and the two cases
$(\Omega ,\pm 1)$
.
The Case
${(\Gamma _1,1)}$
.
From (3.1) and (3.2), we have
$a^2-4\in \{p,4p\}$
.
If
$a^2-4=p$
, then either
$a-2=1$
and
$a+2=p$
or
$a-2=-p$
and
$a+2=-1$
. From the first possibility, we get
$(p,a)=(5,3)$
, while the second possibility produces the pair
$(p,a)=(5,-3)$
. These pairs correspond to the trinomials
Examining condition (iv) of Theorem 2.2 with
$q=5$
, we see that
so that
$H_1(x)$
and
$H_2(x)$
are coprime in
${\mathbb F}_5[x]$
. Thus, the two trinomials in (3.3) are monogenic and we deduce from Theorem 1.1 that
${{\mathrm {{Gal}}}}(f)\simeq C_5\rtimes C_4$
for both, since
$p\equiv 1 \pmod {4}$
.
If
$a^2-4=4p$
, then
$4\mid a$
so that
$(a/2)^2-1=p$
. Arguing as before, we see that
$(p,a)\in \{(3,\pm 4)\}$
. These pairs yield the monogenic trinomials
$f(x)=x^6\pm 4x^3+1$
, with
${{\mathrm {{Gal}}}}(f)\simeq (C_3\rtimes C_2)\times C_2$
from Theorem 1.1, since
$p\equiv 3 \pmod {4}$
.
Of course, all of the above conclusions can be easily verified using a computer algebra system.
The Case
${(\Gamma _2,1)}$
.
From (3.1) and (3.2), we have
$a^2-4\in \{-p,-4p\}$
. The only viable situation here is
$a^2-4=-p$
, which implies that
$p=3$
and
$a=\pm 1$
. Hence,
$f(x)=x^6\pm x^3+1$
. From condition (iv) of Theorem 2.2 with
$q=p$
,
which verifies that
$H_1(x)$
and
$H_2(x)$
are coprime in
${\mathbb F}_3[x]$
. Thus, the trinomials
$f(x)$
are monogenic and
${{\mathrm {{Gal}}}}(f)\simeq (C_3\rtimes C_{1})\times C_2\simeq C_6$
by Theorem 1.1 since
$p\equiv 3 \pmod {4}$
.
The Case
${(\Gamma _1,-1)}$
.
From (3.1) and (3.2), we have
$a^2+4\in \{p,4p\}$
.
If
$a^2+4=4p$
, then
$p=(a/2)^2+1$
, which implies that
$p=2$
since
$4\nmid a$
, and contradicts the fact that
$p\ge 3$
. Thus,
$a^2+4=p$
and
$p\equiv 1 \pmod {4}$
. For monogenicity, we examine condition (iv) of Theorem 2.2 with
$q=p$
. Then,
and
Hence,
$H_1(x)$
and
$H_2(x)$
are coprime in
${\mathbb F}_p[x]$
. Therefore,
$f(x)$
is monogenic with
${\mathrm {{Gal}}}(f)\simeq C_p\rtimes C_{p-1}$
from Theorem 1.1.
The Case
${(\Omega ,1)}$
.
From Lemma 2.3, we require that
for the monogenicity of
$f(x)$
. It is easy to see that the squarefree conditions in (3.4) are equivalent to each of the factors of
${\mathcal W}$
being squarefree, regardless of the parity of a. Furthermore, since
$p\ge 3$
and
$p\mid \delta $
, we see that
$p\mid (a-2)$
or
$p\mid (a+2)$
, whether or not
$2\mid a$
.
We give details only for
$p\mid (a-2)$
since
$p\mid (a+2)$
is similar. Writing
$a=pk+2$
for some
$k\in {\mathbb Z}$
and examining condition (iv) of Theorem 2.2, we have
and
since
$a-2=pk$
is squarefree. Hence,
$H_1(x)$
and
$H_2(x)$
are coprime in
${\mathbb F}_p[x]$
, so that
$f(x)$
is monogenic. We see from Theorem 1.1 that
${{\mathrm {{Gal}}}}(f)\simeq (C_p\rtimes C_{p-1})\times C_2$
. Since there exist infinitely many integers a such that
$(a-2)(a+2)$
is squarefree when
$2\nmid a$
, and infinitely many integers a such that
$(a/2-1)(a/2+1)$
is squarefree when
$2\mid a$
[Reference Booker and Browning1], it follows that there exist infinitely many
$a\in {\mathbb Z}$
such that
$f(x)$
is monogenic with
${{\mathrm {{Gal}}}}(f)\simeq (C_p\rtimes C_{p-1})\times C_2$
.
The Case
${(\Omega ,-1)}$
.
Since
$p\mid \delta $
and
$f(-x)$
is monogenic if and only if
$f(x)$
is monogenic, this case follows from [Reference Jones11, Theorem 1.1]. There exist infinitely many such trinomials since there exist infinitely many integers a such that
$(a^2+4)/\gcd (a,2)^2$
is squarefree [Reference Nagel13].
4 The proof of Corollary 1.3
Suppose that p is a prime with
$p>N$
such that, for some integer
$a>0$
,
$f(x)=x^{2p}+ax^p-1$
is monogenic with
${{\mathrm {{Gal}}}}(f)\simeq C_p\rtimes C_{p-1}$
. Then,
$p=a^2+4$
from Theorem 1.2 and, clearly, a is unique. Conversely, suppose that
$z^2+4>N$
is prime for some positive integer z, and let
$f(x)=x^{2(z^2+4)}+zx^{z^2+4}-1=g(x^{z^2+4})$
, where
$g(x)=x^2\, +zx-1$
. Since
$g(x)$
is irreducible over
${\mathbb Q}$
and
$f(x)$
is irreducible over
${\mathbb Q}$
if and only if
$f(-x)$
is irreducible over
${\mathbb Q}$
, it follows from [Reference Jones10, Lemma 3.1] that
$f(x)$
is irreducible over
${\mathbb Q}$
. Then, from Theorem 1.2,
$f(x)$
is monogenic with
${{\mathrm {{Gal}}}}(f)\simeq C_{z^2+4}\rtimes C_{z^2+3}$
, since
$z^2+4\equiv 1 \pmod {4}$
.
