The Galois group of $x^{2p}+bx^{p}+c^{p}$ over $\mathbb{Q}$

Akash Jim Princeton University
304 Washington Road
Princeton, NJ 08540 [email protected] and Thomas Hagedorn Department of Mathematics and Statistics
The College of New Jersey
2000 Pennington Road
Ewing, NJ 08618 [email protected]

Abstract.

We prove an irreducibility criterion for polynomials of the form $h(x)=x^{2m}+bx^{m}+c_{1}\in F[x]$ relating to the Dickson polynomials of the first kind $D_{p}$ . In the case when $F=\mathbb{Q}$ , $m$ is a prime $p>3$ , and $c_{1}=c^{p}$ , for $c\in\mathbb{Q}$ , we explicitly determine the Galois group of $d_{h}=D_{p}(x,c)+b$ , which is $\mathrm{Aff}(\mathbb{F}_{p})$ or $C_{p}\rtimes C_{(p-1)/2}\vartriangleleft\mathrm{Aff}(\mathbb{F}_{p})$ , and the Galois group of $h$ , which is $C_{2}\times\mathrm{Aff}(\mathbb{F}_{p}),\mathrm{Aff}(\mathbb{F}_{p})$ , or $C_{2}\times(C_{p}\rtimes C_{(p-1)/2})\vartriangleleft C_{2}\times\mathrm{Aff}(\mathbb{F}_{p})$ .

Key words and phrases:

Galois group, irreducible, trinomial, Dickson polynomial, power compositional

2020 Mathematics Subject Classification:

12F10, 12E05, 11R09

1. Introduction

Let $F$ be a field, $f(x)\in F[x]$ a polynomial, and let $K/F$ be the splitting field of $f(x)$ over $F$ . If $f(x)$ is a separable polynomial, the central result of Galois theory is that there is a bijective correspondence between the subgroups of $G=\mathrm{Gal}(K/F)$ , the automorphisms of $K$ that fix $F$ , and the subfields $L\subset K$ containing $F$ . The group $G$ contains much information about the field extension $K/F$ . In particular, Galois proved (see [1, VI, Thm. 7.2]) that the roots of $f(x)$ can be found via the usual arithmetic operations and $n$ th roots precisely when $G$ is a solvable group.

When $F=\mathbb{Q}$ , the Galois group of $f(x)$ has been determined for a number of classes of polynomials of small degree. For example, the Galois group of the trinomial $f(x)=x^{2k}+bx^{k}+c\in\mathbb{Q}[x]$ has been determined in the case of some small $k$ . The case $k=3$ was addressed in the work of [2] and [3], and the case $k=4$ was solved by [4], in the case when $c=1$ or a rational square. In [5], Jones solves the case when $k$ is a prime $p>3$ , $c=1$ , and $b$ is an integer with $|b|\geq 3$ . In this paper, we expand on these results and determine the Galois group of the trinomial polynomial $f(x)$ in the case when $k>3$ is prime and $c$ is a $p$ -th power in $\mathbb{Q}$ . In the process of determining this Galois group, we are also able to determine the Galois group of a related family of polynomials formed from the Dickson polynomials.

In [6], Dickson introduced a set of polynomials that often give automorphisms of the finite field $\mathbb{F}_{q^{r}}$ . The Dickson polynomials of the first kind are defined by

	$\displaystyle D_{1}(t,n)$	$\displaystyle=t$
	$\displaystyle D_{2}(t,n)$	$\displaystyle=t^{2}-2n$
	$\displaystyle D_{k}(t,n)$	$\displaystyle=tD_{k-1}(t,n)-nD_{k-2}(t,n)\text{ for }k>2.$

$D_{k}(t,n)$ is a degree $k$ polynomial and the next few polynomials are:

	$\displaystyle D_{3}(t,n)$	$\displaystyle=t^{3}-3nt$
	$\displaystyle D_{4}(t,n)$	$\displaystyle=t^{4}-4nt^{2}+2n^{2}$
	$\displaystyle D_{5}(t,n)$	$\displaystyle=t^{5}-5nt^{3}+5n^{2}t$
	$\displaystyle D_{6}(t,n)$	$\displaystyle=t^{6}-6nt^{4}+9n^{2}t^{2}-2n^{3}$
	$\displaystyle D_{7}(t,n)$	$\displaystyle=t^{7}-7nt^{5}+14n^{2}t^{3}-7n^{3}t$

A closed-form expression for $D_{k}(t,n)$ is given by:

D_{k}(t,n)=\sum_{i=0}^{\left\lfloor{\frac{k}{2}}\right\rfloor}{\frac{k}{k-i}}{\binom{k-i}{i}}(-n)^{i}t^{k-2i}\,.

We first establish a reducibility criterion that relates to the Dickson polynomials. It generalizes Theorem 1.1(1) of Jones [5], who considered the case when $h\in\mathbb{Z}[x]$ , $m$ is odd, $c=1$ , and $|b|\geq 3$ .

Theorem 1.1.

Let $F$ be a field and $m>1$ . The polynomial $h(x)=x^{2m}+bx^{m}+c\in F[x]$ is reducible if and only if one of the following conditions holds:

(1)

$f(x)=x^{2}+bx+c$ is reducible; or
(2)

For some prime $p$ divisor of $m$ , there exist $n,\,t\in F$ with $c=n^{p}$ and $b=-D_{p}(t,n)$ ; or
(3)

$4\mid m$ and there exist $n,t\in F$ with $c=16n^{4}$ and $b=4D_{4}(t,n)$ .

Remark 1.2.

If $F=\mathbb{Q}$ , Theorem 1.1 allows one to determine the reducibility of $x^{2m}+bx^{m}+c\in\mathbb{Q}[x]$ by using only the Rational Root Test to test for a rational root of the polynomials $d_{h}(x)=D_{p}(x,c)+b$ in (2) and $D_{4}(x,n)-b$ in (3).

When $p$ is a prime and $h(x)=x^{2p}+bx^{p}+c^{p}\in\mathbb{Q}[x]$ , we will see that the Galois group of $h$ is closely related to the Galois group of the polynomial

d_{h}(x)=d_{p,b,c}(x)=D_{p}(x,c)+b,

where $p$ is a prime and $b,c\in F$ . When $h$ is irreducible, we prove in Theorem 1.3 that $d_{h}(x)$ is irreducible and we explicitly classify the Galois groups of $h$ and $d_{h}$ . Theorem 1.3 generalizes Theorem 1.1(2) of Jones [5], who considered the case when $h\in\mathbb{Z}[x]$ , $c=1$ , and $|b|\geq 3$ .

Theorem 1.3.

Let $p>3$ be prime and assume $h(x):=x^{2p}+bx^{p}+c^{p}\in\mathbb{Q}[x]$ is irreducible. Then $d_{h}$ is irreducible and the Galois groups of $d_{h}$ and $h$ are determined as follows:

(1)

If $b^{2}-4c^{p}\notin(-1)^{p(p-1)/2}p\mathbb{Q}^{2}$ , the Galois group of $d_{h}$ is the affine group of $\mathbb{F}_{p}$ , $\mathrm{Aff}(\mathbb{F}_{p})\simeq C_{p}\rtimes C_{p-1}$ , and the Galois group of $h$ is $\mathrm{Aff}(\mathbb{F}_{p})\times C_{2}$ .
(2)

If $p\equiv 1\bmod 4$ and $b^{2}-4c^{p}\in p\mathbb{Q}^{2}$ , the Galois groups of $d_{h}$ and $h$ are identically $\mathrm{Aff}(\mathbb{F}_{p})$ .
(3)

If $p\equiv 3\bmod 4$ and $b^{2}-4c^{p}\in-p\mathbb{Q}^{2}$ , then the Galois group of $d_{h}$ is $C_{p}\rtimes C_{(p-1)/2}\vartriangleleft\mathrm{Aff}(\mathbb{F}_{p})$ , and the Galois group of $h$ is $(C_{p}\rtimes C_{(p-1)/2})\times C_{2}$ .

Remark 1.4.

In [5], the Galois group of $d_{h}$ in Theorem 1.3 (when $c=1$ ) is described using $\mathrm{Hol}(C_{n})$ , the Holomorph of $C_{n}$ . It is defined as the semi-direct product:

\mathrm{Hol}(C_{n}):=C_{n}\rtimes\mathrm{Aut}(C_{n})

Then the Galois group of $h$ is $\mathrm{Hol}(C_{2p})$ in part (1) and $\mathrm{Hol}(C_{p})$ in part (2). Part (3) does not appear as part of [5, Thm 1.1(2)] as $b^{2}-4>0$ by its assumption that $|b|\geq 3$ but the hypothesis of part (3) is that $b^{2}-4\leq 0$ .

Example 1.

Let $p=5$ , and consider $h(x)=x^{10}-3x^{5}+32$ . By Theorem 1.1, $h$ is irreducible. By Theorem 1.3, $h$ has Galois group $\mathrm{Aff}(\mathbb{F}_{5})\times C_{2}$ and

d_{h}(x)=x^{5}-10x^{3}+20x-3

has Galois group $\mathrm{Aff}(\mathbb{F}_{5})$ .

Remark 1.5.

The case of $p=2$ is a sub-case of the quartic, which has long been solved. Here, the Galois group of $h$ is $C_{4}$ , and the Galois group of $d_{h}$ is $C_{2}$ .

Remark 1.6.

The case of $p=3$ has been solved by Awtrey, Buerle, and Griesbach using resolvents as Example 4.1 of [3], over a general field. This generalizes Harrington and Jones’s [2]. In the case when $p=3$ , the statement of Theorem 1.3 is true, and can be derived from the work of [3]. (The Galois group must either be $C_{6}\simeq(C_{3}\rtimes C_{1})\times C_{2}$ or $D_{6}\simeq S_{3}\times C_{2}\simeq\mathrm{Aff}(\mathbb{F}_{3})\times C_{2}$ , which can easily be distinguished by the degree of the extension.) For technical reasons, our proof of Theorem 1.3 does not work when $p=3$ .

Remark 1.7.

The Dickson polynomials of the first kind appear here because of a special case of Waring’s identity, [7]:

\beta^{k}+\beta_{1}^{k}=D_{k}(\beta+{\beta_{1}},\beta{\beta_{1}})

The case of $n=1$ , with an arbitrary $k$ , is known as the $k$ th Vieta-Lucas polynomial of $t$ [8]. These polynomials are used in the work of Jones [5] in the case of polynomials $x^{2m}+Ax^{m}+1\in\mathbb{Z}[x]$ .

2. Some Preliminaries

Let $F$ denote an arbitrary field. $\mathrm{N}^{K}_{F}(\alpha)$ and $\mathrm{Tr}^{K}_{F}(\alpha)$ will denote the norm and trace of $\alpha\in K$ with respect to $F$ . Let $f(x)=f_{b,c}(x)$ denote the monic trinomial $x^{2}+bx+c\in F[x]$ , and denote its roots by $\alpha,\alpha_{1}$ and its determinant by $\Delta:=b^{2}-4c$ .

A key theorem in establishing irreducibility of power compositional polynomials (cf., e.g., [2], [4]) is Capelli’s Theorem (see Theorem 22 of [9] for proof):

Theorem 2.1 (Capelli’s Theorem).

Let $f(x),g(x)\in F[x]$ and assume $f(x)$ is irreducible with root $\alpha$ . Then $f(g(x))$ is reducible in $F[x]$ if and only if the polynomial $g(x)-\alpha$ is reducible in $F(\alpha)[x]$ . Moreover, if $g(x)-\alpha$ has the prime factorization $g(x)-\alpha=C\prod_{i=1}^{r}h_{i}(x)^{e_{i}}$ , with $C\in F(\alpha)$ and irreducible polynomials $h_{i}(x)\in F(\alpha)[x]$ , then

f(g(x))=\tilde{C}\prod_{i=1}^{r}\mathrm{N}_{F}^{F(\alpha)}(h_{i}(x))^{e_{i}},

where $\tilde{C}\in F$ and $\mathrm{N}_{F}^{F(\alpha)}(h_{i}(x))$ are irreducible.

Two well known consequences of this theorem are:

Theorem 2.2 (Capelli).

[1, VI, Thm. 9.1] Let $a\in F,n\in\mathbb{N}$ . $x^{n}-a$ is irreducible in $F[x]$ if and only if the following conditions hold:

(1)

$a\notin F^{p}$ for all primes $p\mid n$ ; and
(2)

If $4\mid n$ , $a\notin-4F^{4}$ .

Corollary 2.3.

For $p$ prime, $x^{p}-\alpha\in F[x]$ is reducible if and only if $\exists\beta\in F$ with $\alpha=\beta^{p}$ .

Thus, if $f(x^{p})$ is reducible, either $f(x)$ is reducible, or there exists $\beta\in F(\alpha)$ with $f(\beta^{p})=0$ and $\mathrm{N}^{F(\alpha)}_{F}(x-\beta)\mid h(x)$ .

Define $p(x):=h(x)/\mathrm{N}(x-\beta)$ . The polynomial $p(x)$ is symmetric in $\beta$ and its conjugate $\mathrm{N}(\beta)/\beta\in F(\alpha)$ , so it can be expressed in terms of the symmetric polynomials. By expanding and equating coefficients, we will obtain the reducibility criterion of 1.1.

3. A Reducibility Criterion for $x^{2m}+bx^{m}+c\in F[x]$

In this section, we prove Theorem 1.1, which gives a reducibility criteria for $x^{2m}+bx^{m}+c\in F[x]$ . Theorem 1.1 extends Jones’s reducibility criterion [5, Thm. 1.1(1)], which considered the case when $c=1$ . The proof of Theorem 1.1 is essentially the same as Jones’ proof in the case $c=1$ . It will follow as a corollary from Theorem 3.1, which gives a relationship of the coefficients of

f(x):=x^{2}+bx+c\in F[x].

when $f(x)$ is irreducible. Recall from the introduction that $D_{m}(t,n)$ are the Dickson polynomials.

Theorem 3.1.

Let $f(x):=x^{2}+bx+c\in F[x]$ be irreducible with a root $\alpha\in F(\sqrt{\Delta})\setminus F$ , where $\Delta:=b^{2}-4c$ . Then $\alpha\in F(\sqrt{\Delta})^{m}$ , for some positive integer $m$ if and only if there exists some $n,t\in F$ with $c=n^{m}$ and $b=-D_{m}(t,n)$ .

Before proving Theorem 3.1, we need to establish two theorems about polynomials of the form $f(x^{m})$ . Given variables $\beta,\beta_{1}$ , define:

t:=\beta+{\beta_{1}},\text{\qquad}n:=\beta{\beta_{1}}

For $m\in\mathbb{N}$ , define

\Psi_{m,\beta}(x):=\frac{x^{m}-\beta^{m}}{x-\beta}

and define the polynomial $p_{m}(x):=\Psi_{m,\beta}(x)\Psi_{m,\beta_{1}}(x)$ . For $m>0$ , let $a_{m}$ be the coefficient of $x^{m-1}$ in $p_{m}(x)$

Theorem 3.2.

The $a_{m}$ are given by $a_{1}=1$ , $a_{2}=t$ , and the recursive formula

a_{m+2}=ta_{m+1}-na_{m}

and $p_{m}(x)$ is expressible in terms of $\{a_{i}\}_{1\leq i\leq m}$ by:

p_{m}(x)=\sum_{i=1}^{m-1}a_{i}x^{2m-1-i}+a_{m}x^{m-1}+\sum_{i=1}^{m-1}a_{m-i}n^{i}x^{m-1-i}

(where the outer sums are defined to be 0 in the case of $m=1$ ).

Proof.

We first compute $p_{m}(x),a_{m}$ from their definition.

	$\displaystyle p_{m}(x)$	$\displaystyle=\Psi_{m,\beta}(x)\Psi_{m,\beta_{1}}(x)=(\sum_{i=0}^{m-1}\beta^{m-1-i}x^{i})(\sum_{j=0}^{m-1}{\beta_{1}}^{m-1-j}x^{j})$
		$\displaystyle=\sum_{l=0}^{2(m-1)}\Big{(}\sum_{\begin{subarray}{c}i+j=l\\ 0\leq i,j\leq m-1\end{subarray}}\beta^{m-1-i}{\beta_{1}}^{m-1-j}\Big{)}x^{l}$

The coefficient of the $x^{m-1}$ term is

\displaystyle{a_{m}=\sum_{\begin{subarray}{c}i+j=m-1\\ 0\leq i,j\leq m-1\end{subarray}}\beta^{m-1-i}{\beta_{1}}^{m-1-j}=\sum_{i=0}^{m-1}\beta^{m-1-i}{\beta_{1}}^{i}}.

We can then verify the base cases explicitly:

	$\displaystyle p_{1}(x)$	$\displaystyle=\Psi_{1,\beta}(x){\Psi_{1,\beta_{1}}(x)}=1\cdot 1=1=a_{1}$
	$\displaystyle p_{2}(x)$	$\displaystyle=\Psi_{2,\beta}(x){\Psi_{2,\beta_{1}}(x)}=(x+\beta)(x+\beta_{1})$
		$\displaystyle=x^{2}+tx+n$
		$\displaystyle=a_{1}x^{2}+a_{2}x^{1}+a_{1}n^{1}x^{0}$

With regard to the recursive relation of the $(a_{m})$ , we may compute:

\begin{split}ta_{m}-na_{m-1}&=(\beta+{\beta_{1}})\sum_{\begin{subarray}{c}i+j=m-1\\ 0\leq i,j\leq m-1\end{subarray}}\beta^{m-1-i}{\beta_{1}}^{m-1-j}\\ &\text{\hskip 36.135pt}-\beta{\beta_{1}}\sum_{\begin{subarray}{c}i+j=m-2\\ 0\leq i,j\leq m-2\end{subarray}}\beta^{m-2-i}{\beta_{1}}^{m-2-j}\\ &=\sum_{i=0}^{m-1}\beta^{m-i}{\beta_{1}}^{i}+\sum_{i=0}^{m-1}\beta^{m-1-i}{\beta_{1}}^{i+1}-\sum_{i=0}^{m-2}\beta^{m-1-i}{\beta_{1}}^{i+1}\\ &=\sum_{i=0}^{m-1}\beta^{m-i}{\beta_{1}}^{i}+\beta_{1}^{0}{\beta}^{m}=\sum_{i=0}^{m}\beta^{m-i}=a_{m+1}\qquad\end{split}

And with regard to the expression of $p_{m}$ , we assume for induction that the identity holds for $p_{m}$ and expand $p_{m+1}(x)$ :

	$\displaystyle p_{m+1}(x)$	$\displaystyle=\sum_{l=0}^{2m}\left(\sum_{\begin{subarray}{c}i+j=l\\ 0\leq i,j\leq m\end{subarray}}\beta^{m-i}{\beta_{1}}^{m-j}\right)x^{l}\allowdisplaybreaks[3]$
		$\displaystyle=\sum_{l=m+1}^{2m}\Big{(}\sum_{\begin{subarray}{c}i+j=l\\ 1\leq i,j\leq m\end{subarray}}\beta^{m-i}{\beta_{1}}^{m-j}\Big{)}x^{l}+\sum_{\begin{subarray}{c}i+j=m\\ 0\leq i,j\leq m\end{subarray}}\beta^{m-i}{\beta_{1}}^{m-j}x^{m}\allowdisplaybreaks[3]$
		$\displaystyle\text{\hskip 72.26999pt}+\sum_{l=0}^{m-1}\Big{(}\sum_{\begin{subarray}{c}i+j=l\\ 0\leq i,j\leq m-1\end{subarray}}\beta^{m-i}{\beta_{1}}^{m-j}\Big{)}x^{l}\allowdisplaybreaks[3]$

		$\displaystyle=x^{2}\sum_{l=m-1}^{2m-2}\Big{(}\sum_{\begin{subarray}{c}i+j=l\\ 0\leq i,j\leq m-1\end{subarray}}\beta^{m-1-i}{\beta_{1}}^{m-1-j}\Big{)}x^{l}+a_{m+1}x^{m}$
		$\displaystyle\text{\hskip 72.26999pt}+\beta{\beta_{1}}\sum_{l=0}^{m-1}\Big{(}\sum_{\begin{subarray}{c}i+j=l\\ 0\leq i,j\leq m-1\end{subarray}}\beta^{m-1-i}{\beta_{1}}^{m-1-j}\Big{)}x^{l}$

		$\displaystyle=n\Big{(}\sum_{i=1}^{m-1}a_{m-i}n^{i}x^{m-1-i}+a_{m}x^{m-1}\Big{)}+a_{m+1}x^{m}$
		$\displaystyle\text{\hskip 72.26999pt}+x^{2}\Big{(}a_{m}x^{m-1}+\sum_{i=1}^{m-1}a_{i}x^{2m-1-i}\Big{)}$

\displaystyle=\sum_{i=1}^{m}a_{m+1-i}n^{i}x^{m-i}+a_{m+1}x^{m}+\sum_{i=1}^{m}a_{i}x^{2m+1-i}

∎

The following theorem will be a tool to study polynomials of the form $f(x^{m})$ and prove Theorem 3.1.

Theorem 3.3.

Use the same notation as in Theorem 3.2, and define the polynomial $g(x):=(x-\beta^{m})(x-\beta_{1}^{m})$ . Then $g(x^{m})=x^{2m}-D_{m}(t,n)x^{m}+n^{m}$ , where $D_{m}$ is the $m$ th Dickson polynomial of the first kind.

Proof.

\begin{split}g(x^{m})&=(x^{m}-\beta^{m})(x-\beta_{1}^{m})=(x-\beta)\Psi_{m,\beta}(x)(x-\beta_{1})\Psi_{m,\beta_{1}}(x)\\ &=(x^{2}-tx+n)p_{m}(x)\\ &=(x^{2}-tx+n)\left(\sum_{i=1}^{m-1}a_{i}x^{2m-1-i}+a_{m}x^{m-1}+\sum_{i=1}^{m-1}a_{m-i}n^{i}x^{m-1-i}\right)\\ &=a_{1}x^{2m}+(na_{m-1}-ta_{m}+na_{m-1})x^{m}+a_{1}n^{m}\\ &=x^{2m}+(2na_{m-1}-ta_{m})x^{m}+n^{m}\\ \end{split}

We define

b_{m}=b_{m}(\beta,\beta_{1})=2na_{m-1}-ta_{m},

Then $g(x^{m})=x^{2m}+b_{m}x^{m}+n^{m}$ . We observe now that:

\begin{split}tb_{m+1}-nb_{m}&=t(2na_{m}-ta_{m+1})-n(2na_{m-1}-ta_{m})\\ &=2n(ta_{m}-na_{m-1})-t(ta_{m+1}-na_{m})\\ &=2na_{m+1}-ta_{m+2}\\ &=b_{m+2}\end{split}

Also, $b_{1}=-t$ and $b_{2}=2n-t^{2}$ . (Note: we define $a_{0}:=0$ so that $b_{1}=-t$ , in agreement with the direct expansion of $g(x^{m})$ .) So indeed $b_{m}=-D_{m}(t,n)$ , and as claimed, $g(x^{m})=x^{2m}-D_{m}(t,n)x^{m}+n^{m}$ . ∎

We now use Theorem 3.3 and Capelli’s Theorem to study polynomials of the form $f(x^{m})$ and prove Theorem 3.1.

Proof of Theorem 3.1.

If there exists $\beta\in F(\sqrt{\Delta})$ with $\beta^{m}=\alpha$ , $x^{m}-\alpha=(x-\beta)\Psi_{m}(x)$ is a valid factorization of $x^{m}-\alpha\in F(\sqrt{\Delta})[x]$ . We now identify the symbol $\beta_{1}$ from Theorems 3.2 and 3.3 with the conjugate of $\beta$ , namely $\mathrm{N}(\beta)/\beta\in F(\sqrt{\Delta})$ . From Capelli’s Theorem (and the multiplicativity of norms), this factorization induces the $F[x]$ -factorization

x^{2m}+bx^{m}+c=\mathrm{N}(x-\beta)\mathrm{N}(\Psi_{m,\beta}(x))=(x^{2}-tx+n)p_{m}(x)

Equating coefficients with the expression $f(x^{m})=x^{2m}-D_{m}(t,n)x^{m}+n^{m}$ from Theorem 3.3 gives the desired equality:

b=-D_{m}(t,n),c=n^{m}

with

t=\mathrm{Tr}(\beta),n=\mathrm{N}(\beta)\in F

Conversely, if $\exists n,t\in F$ with the given conditions, then by Theorem 3.3,

f(x^{m})=(x^{2}-tx+n)p_{m}(x)

By Capelli’s Theorem, $(x^{2}-tx+n)$ is the product of the norms of some irreducible polynomials in $F(\alpha)=F(\sqrt{\Delta})$ that divide $x^{m}-\alpha$ . $\alpha$ is quadratic, so $x^{2}-tx+n=\mathrm{N}(x-\beta)$ for some $(x-\beta)\mid(x^{m}-\alpha)$ , and $\alpha=\beta^{m}\in F(\sqrt{\Delta})^{m}$ . ∎

The criterion for reducibility, Theorem 1.1, follows as a corollary.

Proof of Theorem 1.1.

We use the notation common to Theorems 1.1 and 3.1. If $f$ is reducible, so is $h=f(x^{m})$ . If $f$ is irreducible, from Capelli’s Theorem, $h$ is reducible if and only if $x^{m}-\alpha$ is reducible in $F(\alpha)$ , where $\alpha$ is a root of $f$ . From Theorem 2.2, this occurs if and only if $\alpha\in F(\alpha)^{p}$ for some $p\mid m$ prime, or $\alpha\in-4F(\alpha)^{4}$ and $4\mid m$ . From Theorem 3.1, $\alpha\in F(\alpha)^{p}$ if and only if there exist $n,t\in F$ with $c=n^{p}$ and $b=-D_{p}(t,n)$ . If $\alpha=-4\beta^{4}$ for some $\beta\in F(\alpha)$ , denote the conjugate of $\beta$ over $F$ by $\beta_{1}$ , with $\alpha_{1}=-4\beta_{1}^{4}$ . Then

	$\displaystyle x^{4}+bx^{2}+c$	$\displaystyle=(x^{2}-\alpha)(x^{2}-\alpha_{1})$
		$\displaystyle=(x^{2}+4\beta^{4})(x^{2}+4\beta_{1}^{4})$
		$\displaystyle=x^{4}+4(\beta^{4}+\beta_{1}^{4})x^{2}+16\beta^{4}\beta_{1}^{4}$
		$\displaystyle=x^{4}+4D_{4}(\mathrm{Tr}(\beta),\mathrm{N}(\beta))+16(\mathrm{N}(\beta))^{4}$

By Waring’s identity (Remark 1.7). And indeed $t:=\mathrm{Tr}(\beta)\in F$ , $n:=\mathrm{N}(\beta)\in F$ . Conversely, if such $t,n$ exist, we observe that

\begin{split}x^{8}+bx^{4}+c&=x^{8}+4D_{4}(t,n)+16n^{4}\\ &=x^{8}+(4t^{4}-16nt^{2}+8n^{2})x^{4}+16n^{4}\\ &=(x^{4}+(2t^{2}-4n)x^{2}-4n^{2})(x^{4}-(2t^{2}-4n)x^{2}-4n^{2})\\ \end{split}

So $f(x^{4})$ and therefore $f(x^{m})$ are reducible. ∎

Remark 3.4.

We note that the criterion of Theorem 1.1 is very similar to parts (vi) and (vii) of Theorem 6 of Schinzel [10]. It is quite possible that Theorem 1.1 is encompassed by the results in [10] on the reducibility of trinomials. However, those results utilize elliptic curves and the proof of Theorem 1.1 does not.

4. Properties of $h=x^{2p}+bx^{p}+c^{p}\in\mathbb{Q}[x]$ and $d_{h}=D_{p}(x,c)+b$

We now restrict ourselves to the case when $F=\mathbb{Q}$ , $m=p>3$ is prime, and the constant term of $f$ is the form $c^{p}$ . With

	$\displaystyle f(x)$	$\displaystyle=x^{2}+bx+c^{p}$
	$\displaystyle h(x)$	$\displaystyle=f(x^{p})=x^{2p}+bx^{p}+c^{p}$

we assume in this section and the next that $h(x)$ is irreducible. By Theorem 1.1(1), $\Delta:=b^{2}-4c^{p}$ is not a rational square and the splitting field of $f$ is $\mathbb{Q}(\sqrt{\Delta})$ . Also, by Theorem 1.1(2), the irreducibility of $h$ implies that $b\neq-D_{p}(t,c)$ for any $t\in\mathbb{Q}$ . Hence the polynomial $d_{h}:=D_{p}(x,c)+b$ has no rational roots.

Let $\zeta_{n}$ be a primitive $n$ th root of unity and let

F_{n}=\mathbb{Q}(\zeta_{n})

be the $n$ th cyclotomic field. We will denote the splitting field of $h$ by $K$ and the splitting field of $d_{h}$ by $L$ . We denote the roots of $f$ by $\alpha,\alpha_{1}$ . We then choose $\beta$ so that $\beta^{p}=\alpha$ and $\beta_{1}:=\frac{c}{\beta}$ (and indeed $\beta_{1}^{p}=\frac{c^{p}}{\beta^{p}}=\frac{c^{p}}{\alpha}=\alpha_{1}$ ). $h$ has $2p$ roots, which are precisely

\{\zeta_{p}^{i}\beta\}_{i\in\mathbb{Z}_{p}}\cup\{\zeta_{p}^{i}\beta_{1}\}_{i\in\mathbb{Z}_{p}}

We claim now:

Lemma 4.1.

$\mathbb{Q}(\zeta_{p}^{i}\beta)=\mathbb{Q}(\zeta_{p}^{-i}\beta_{1})$ , and moreover, these are the only roots of $h(x)$ in this extension of $\mathbb{Q}$ .

Proof.

By construction $\zeta_{p}^{-i}\beta_{1}=\zeta_{p}^{-i}\frac{c}{\beta}=\frac{c}{\zeta_{p}^{i}\beta}$ , so $\mathbb{Q}(\zeta_{p}^{i}\beta)=\mathbb{Q}(\zeta_{p}^{-i}\beta_{1})$ . Suppose that some other root of $h$ lies in $\mathbb{Q}(\zeta_{p}^{i}\beta)$ , without loss of generality, $\zeta_{p}^{j}\beta$ . Then $\zeta_{p}^{i-j}\in\mathbb{Q}(\zeta_{p}^{i}\beta)$ . If $i\neq j$ , $F_{p}\subset\mathbb{Q}(\beta)$ . This is impossible because $[F_{p}:\mathbb{Q}]=p-1$ and $[\mathbb{Q}(\beta):\mathbb{Q}]=2p$ , but $p-1\nmid 2p$ given that $p>3$ . So $i=j$ , and $\zeta_{p}^{j}\beta=\zeta_{p}^{i}\beta$ , which is one of the given roots after all. ∎

As a corollary, there are no double roots: $\alpha\neq\alpha_{1}$ else $f$ is reducible, and the only pairs of roots which generate the same extension are of the form $\zeta_{p}^{i}\beta,\zeta_{p}^{-i}\beta_{1}$ , whose $p$ th powers are $\alpha,\alpha_{1}$ , respectively.

Now denote $B_{i}:=\mathbb{Q}(\zeta_{p}^{i}\beta)=\mathbb{Q}(\zeta_{p}^{-i}\beta_{1})$ and $\mathcal{B}:=\{B_{i}\}_{i\in\mathbb{Z}_{p}}$ , with $|\mathcal{B}|=p$ . Clearly, $[B_{i}:\mathbb{Q}]=2p$ . Also, $\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1}\in B_{i}$ . We show that it is a root of $d_{h}$ .

Lemma 4.2.

$d_{h}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1})=0$ .

Proof.

By factoring $f$ , we see that:

x^{2p}+bx^{p}+c^{p}=f(x^{p})=(x^{p}-(\zeta_{p}^{i}\beta)^{p})(x^{p}-(\zeta_{p}^{-i}\beta_{1})^{p})

And by Theorem 3.3:

f(x^{p})=x^{2p}-D_{p}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1},c)x^{p}+c^{p}

Equating coefficients, $-D_{p}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1},c)=b$ , i.e., $d_{h}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1})=0$ . ∎

We now define the fields $D_{i}:=\mathbb{Q}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1})$ , with $D_{i}\subset B_{i}$ , and define

\mathcal{D}:=\{D_{i}\}_{i\in\mathbb{Z}_{p}}

A priori, we do not know that the fields $D_{i}$ are distinct, but for now, it suffices that the distinct symbols $D_{i}$ are in bijective correspondence with $\mathbb{Z}_{p}$ and $\mathcal{B}$ . Before examining the fields $D_{i}$ , it will be useful to prove that:

Lemma 4.3.

$\mathrm{Gal}(K/\mathbb{Q})$ acts transitively and equivalently on $\mathcal{B}$ , on $\mathcal{D}$ , and on the roots of $d_{h}$ .

Proof.

Let $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ . $\sigma$ permutes the roots of $h$ , and from Lemma 4.1, it must be that $\sigma(\zeta_{p}^{i}\beta)\in B_{j}$ for some $j$ . Then $\sigma(B_{i})=B_{j}$ , and $\sigma$ acts on $\mathcal{B}$ . $\mathrm{Gal}(K/\mathbb{Q})$ is transitive on the roots of irreducible $h$ , so its action on $\mathcal{B}$ is also transitive. $\sigma$ maps the pair of roots in $B_{i}$ to the pair of roots in $B_{j}$ , so the action on $\mathcal{B}$ is equivalent to the action on the set of pairs of roots of $h$ . Then:

\sigma(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1})=\zeta_{p}^{j}\beta+\zeta_{p}^{-j}\beta_{1}

So $\sigma(D_{i})=D_{j}$ . Therefore, $\sigma$ acts equivalently on the roots of $d$ , on $\mathcal{D}$ , and on $\mathcal{B}$ , according to the correspondence $\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1}\in D_{i}\subset B_{i}$ , and these equivalent actions of $\mathrm{Gal}(K/\mathbb{Q})$ are transitive. ∎

We can now prove the following properties about the fields $D_{i}$ .

Lemma 4.4.

$d_{h}(x)=D_{p}(x,c)+b$ is irreducible, so $D_{i}\simeq\mathbb{Q}[x]/(d_{h}(x))$ . Also, $B_{i}=D_{i}(\sqrt{\Delta})$ , and $D_{i}=D_{j}$ if and only if $i=j$ .

Proof.

x^{2p}+bx^{p}+c^{p}=f(x^{p})=(x^{p}-\alpha)(x^{p}-\alpha_{1})=(x^{p}-(\zeta_{p}^{i}\beta)^{p})(x^{p}-(\zeta_{p}^{-i}\beta_{1})^{p})

By Theorem 3.3, $b=-D_{p}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1},c)$ , so $\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1}\in B_{i}$ is a root of $d_{h}$ . Thus, $[D_{i}:\mathbb{Q}]\leq p$ , and since $D_{i}\subset B_{i}$ , $[D_{i}:\mathbb{Q}]\mid 2p$ . Thus, $[D_{i}:\mathbb{Q}]$ could be $1$ , $2$ , or $p$ . $[D_{i}:\mathbb{Q}]\neq 1$ because this would give a rational root for $d_{h}$ , a contradiction.

Now assume temporarily that $[D_{i}:\mathbb{Q}]=2$ . Then $D_{i}(\sqrt{\Delta})\subset B_{i}$ is an extension of either degree 2 or degree 4. $4\nmid 2p$ given $p>3$ , so $[D_{i}(\sqrt{\Delta}):\mathbb{Q}]=2$ , and $D_{i}=\mathbb{Q}(\sqrt{\Delta})$ . By the transitive action of the Galois group, all $D_{i}$ are isomorphic and are thus quadratic extensions. So each root of $d_{h}$ is a root of a quadratic factor, and $d_{h}$ factors as a product of quadratics. But this would make $\deg(d_{h})=p$ even, a contradiction. So $[D_{i}:\mathbb{Q}]=p$ after all, and consequently, $d_{h}(x)$ is irreducible with $D_{i}\simeq\mathbb{Q}[x]/(d_{h}(x))$ , as claimed.

Then $\sqrt{\Delta}\notin D_{i}$ because $2\nmid p$ , so $[D_{i}(\sqrt{\Delta}):\mathbb{Q}]=2p=[B_{i}:\mathbb{Q}]$ and $D_{i}(\sqrt{\Delta})\subset B_{i}$ , so $D_{i}(\sqrt{\Delta})=B_{i}$ . If $i\neq j$ but $D_{i}=D_{j}$ , then $B_{i}=B_{j}$ , a contradiction of Lemma 4.1. So $D_{i}=D_{j}$ if and only if $i=j$ , as desired. ∎

As a technical lemma, we must now note that:

Lemma 4.5.

$K=F_{p}(\beta)$ . $[K:\mathbb{Q}]=2p(p-1)$ if $\sqrt{\Delta}\notin F_{p}$ and $[K:\mathbb{Q}]=p(p-1)$ if $\sqrt{\Delta}\in F_{p}$ .

Proof.

$F_{p}(\beta)=B_{0}(\zeta_{p})$ certainly contains all the roots of $h$ . Conversely, the roots of $h$ include $\beta$ and $\zeta_{p}\beta$ , so the splitting field is at least $\mathbb{Q}(\zeta_{p},\beta)=F_{p}(\beta)$ . Thus, $K=F_{p}(\beta)$ .

$[B_{i}:\mathbb{Q}]=2p$ and $[F_{p}:\mathbb{Q}]=p-1$ , and $\gcd(2p,p-1)=2$ because $p$ is odd. Thus, $p(p-1)\mid[K:\mathbb{Q}]\mid 2p(p-1)$ . If $\sqrt{\Delta}\in F_{p}$ , then $[F_{p}:\mathbb{Q}(\sqrt{\Delta})]=\frac{p-1}{2}$ is coprime to $[B:\mathbb{Q}(\sqrt{\Delta})]=p$ . Thus $[K:\mathbb{Q}]=2[K:\mathbb{Q}(\sqrt{\Delta})]=2\cdot\frac{p-1}{2}\cdot p=p(p-1)$ . If $\sqrt{\Delta}\notin F_{p}$ , then $[F_{p}:\mathbb{Q}(\sqrt{\Delta})]=p-1$ is again coprime to $[B:\mathbb{Q}(\sqrt{\Delta})]=p$ , and similarly $[K:\mathbb{Q}]=2[K:\mathbb{Q}(\sqrt{\Delta})]=2\cdot(p-1)\cdot p=2p(p-1)$ . ∎

Remark 4.6.

As $\mathbb{Q}(\sqrt{\Delta})/\mathbb{Q}$ and $F_{p}/\mathbb{Q}$ are Galois extensions, an automorphism $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ acts as an automorphism of $\mathbb{Q}(\sqrt{\Delta}),F_{p}$ .

Using the lemmas of this section, we are now able to specify the actions of $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ in terms of its action on $D_{0},\sqrt{\Delta}$ , and $\zeta_{p}$ . We define $\epsilon_{\sigma}=\sigma(\sqrt{\Delta})/\sqrt{\Delta}=\pm 1$ .

Theorem 4.7.

If $\sqrt{\Delta}\notin F_{p}$ , then there exists a bijection between $\mathrm{Gal}(K/\mathbb{Q})$ and $\mathcal{D}\times\{\pm 1\}\times\{\zeta_{p}^{i}\}_{i\in\mathbb{Z}_{p}^{\times}}$ given by $\sigma\mapsto(\sigma(D_{0}),\epsilon_{\sigma},\sigma(\zeta_{p}))$ . If $\sqrt{\Delta}\in F_{p}$ , then there exists a bijection between $\mathrm{Gal}(K/\mathbb{Q})$ and $\mathcal{D}\times\{\zeta_{p}^{i}\}_{i\in\mathbb{Z}_{p}^{\times}}$ given by $\sigma\mapsto(\sigma(D_{0}),\sigma(\zeta_{p}))$ .

Proof.

$\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ permutes the roots of $h$ . Because all roots of $h$ are expressible in terms of $\beta$ and $\zeta_{p}$ (as $\beta_{1}=c/\beta$ ), $\sigma$ is determined entirely by $\sigma(\beta)$ and $\sigma(\zeta_{p})$ .

If $\sqrt{\Delta}\notin F_{p}$ , there are $2p$ choices for $\sigma(\beta)$ and $p-1$ choices for $\sigma(\zeta_{p})$ , so $|\mathrm{Gal}(K/\mathbb{Q})|\leq 2p(p-1)$ . But $|\mathrm{Gal}(K/\mathbb{Q})|=2p(p-1)$ , so all choices must correspond to distinct elements of the Galois group. The action $\sigma(\beta)$ determines $(\sigma(B_{0}),\epsilon_{\sigma})$ , and vice-versa, and the actions on $B_{0},D_{0}$ are also equivalent. Therefore, $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ corresponds exactly to a choice of $(D_{i},\pm 1,\zeta^{j}_{p})$ . ∎

If $\sqrt{\Delta}\in F_{p}$ , $\sigma(\zeta_{p})$ determines $\sigma(\sqrt{\Delta})$ . Therefore, $(\sigma(\zeta_{p}),\sigma(D_{0}))$ determines $\sigma(\beta)$ and thus $\sigma$ . There are $p$ choices for $\sigma(D_{0})$ and $p-1$ choices for $\sigma(\zeta_{p})$ , so $|\mathrm{Gal}(K/\mathbb{Q})|\leq p(p-1)$ . But $|\mathrm{Gal}(K/\mathbb{Q})|=p(p-1)$ , so all choices must correspond to distinct elements of the Galois group. Therefore, $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ corresponds exactly to a choice of $(D_{i},\zeta_{p}^{j})$ . ∎

5. The Galois groups of $h$ and $d_{h}$

As in Section 4, we assume that

h(x)=f(x^{p})=x^{2p}+bx^{p}+c^{p},

that $h(x)$ is irreducible and let $d_{h}$ denote the polynomial $d_{h}=D_{p}(x,c)+b$ . The assumption on $h$ shows that $d_{h}$ is irreducible.

Theorem 5.1.

Let $L$ be the splitting field of $d_{h}$ over $K$ . Then $K=L(\sqrt{\Delta})$ and for all $i$ ,

D_{i}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))\subset L

Proof.

$L$ contains all of the roots of $d_{h}$ , which splits in $K$ , so $D_{i}\subset L\subset K$ for each $i$ . Since $\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1}\in D_{i}$ , each term is in $L$ . Then $L$ contains

\frac{(\zeta_{p}\beta+\zeta_{p}^{-1}\beta_{1})+(\zeta_{p}^{-1}\beta+\zeta_{p}\beta_{1})}{(\zeta_{p}^{0}\beta+\zeta_{p}^{-0}\beta_{1})}=\frac{(\beta+\beta_{1})(\zeta_{p}+\zeta_{p}^{-1})}{\beta+\beta_{1}}=\zeta_{p}+\zeta_{p}^{-1}

as well.

Now choose an arbitrary $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ which fixes $L$ . $\sigma$ acts trivially on $\mathcal{D}$ and $\mathcal{B}$ and acts on $\sqrt{\Delta}$ either trivially or by conjugation. If $\sigma(\sqrt{\Delta})=\sqrt{\Delta}$ , then $\sigma(\zeta_{p}^{i}\beta)=\zeta_{p}^{i}\beta$ , so $\sigma(\zeta_{p})=\zeta_{p}$ . Thus:

	$\displaystyle\sigma(\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$	$\displaystyle=\sigma(\sqrt{\Delta})(\sigma(\zeta_{p})-(\sigma(\zeta_{p}))^{-1})$
		$\displaystyle=\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1})$

Similarly, if $\sigma(\sqrt{\Delta})=-\sqrt{\Delta}$ , then $\sigma(\zeta_{p}^{i}\beta)=\zeta_{p}^{-i}\beta_{1}$ , so $\sigma(\zeta_{p})=\zeta_{p}^{-1}$ . Thus,

	$\displaystyle\sigma(\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$	$\displaystyle=\sigma(\sqrt{\Delta})(\sigma(\zeta_{p})-(\sigma(\zeta_{p}))^{-1})$
		$\displaystyle=-\sqrt{\Delta}(\zeta_{p}^{-1}-\zeta_{p})$
		$\displaystyle=\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1})$

Because $\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1})$ is fixed by the subgroup fixing $L$ , $\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1})\in L$ . Finally, $L(\sqrt{\Delta})\subset K$ contains each $B_{i}=D_{i}(\sqrt{\Delta})$ , so $K=L(\sqrt{\Delta})$ . ∎

From this point, we will need to construct automorphisms using Theorem 4.7, rather than considering arbitrary automorphisms as we have done in the previous section. We can prove:

Theorem 5.2.

The splitting field $L$ of $d_{h}$ is $D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$ .

To help to understand the proof of Theorem 5.2, Figures 1, 2, 3 provide diagrams of the relevant field inclusions in each case. All of the boxed fields are Galois over $\mathbb{Q}$ , and the fields marked $``p\times"$ are conjugates.

Proof.

$D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))\subset L$ from Theorem 5.1. Since

K=B_{0}(\zeta_{p})=D_{0}(\zeta_{p},\sqrt{\Delta})\subset D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}),\sqrt{\Delta})

and $\sqrt{\Delta}$ is of degree at most 2, we see that that

[K:L]\leq[K:D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))]\leq 2

Recall that $F_{p}=\mathbb{Q}(\zeta_{p})$ is the cyclotomic field. There are then 3 cases:

(1)

$\sqrt{\Delta}\notin F_{p}$ ; or
(2)

$\sqrt{\Delta}\in F_{p}$ and $p\equiv 1\bmod 4$ ; or
(3)

$\sqrt{\Delta}\in F_{p}$ and $p\equiv 3\bmod 4$ .

Case (1): Consider the case of $\sqrt{\Delta}\notin F_{p}$ . From Theorem 4.7, there exists an element of $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ defined by $(\sigma(D_{0}),\epsilon_{\sigma},\sigma(\zeta_{p}))=(D_{0},-1,\zeta_{p}^{-1})$ . Then

\sigma(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1})=\sigma(\zeta_{p})^{i}\sigma(\beta)+\sigma(\zeta_{p})^{-i}\sigma(\beta_{1})=\zeta_{p}^{-i}\beta_{1}+\zeta_{p}^{i}\beta

for all $i$ , so $\sigma$ fixes all roots of $d_{h}$ and therefore $L$ . The trivial element also fixes $L$ , so by the Galois correspondence, $[K:L]\geq 2$ . So $[K:L]=2$ , and $L=D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$ after all, and is of degree $\frac{1}{2}\cdot 2p(p-1)=p(p-1)$ , using Lemma 4.5.

Case (2): Now suppose that $\sqrt{\Delta}\in F_{p}$ and $p\equiv 1\bmod 4$ . Then $[\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1}):\mathbb{Q}]=\frac{p-1}{2}$ is even, so $\mathbb{Q}(\sqrt{\Delta})\subset\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})$ by the Galois correspondence. Thus

D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))=D_{0}(\zeta_{p}+\zeta_{p}^{-1},(\zeta_{p}-\zeta_{p}^{-1}),\sqrt{\Delta})\supset B_{0}(\zeta_{p})=K\supset L

and in fact $K=L=D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$ , which by Lemma 4.5 is of degree $p(p-1)$ .

Case (3): Finally, we suppose that $\sqrt{\Delta}\in F_{p}$ , but $p\equiv 3\bmod 4$ . Then $[\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1}):\mathbb{Q}]=\frac{p-1}{2}$ is odd, so $\sqrt{\Delta}\notin\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})$ by the Galois correspondence (else $\mathbb{Q}(\sqrt{\Delta})\subset\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})$ ). From Theorem 4.7, there exists an element $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ defined by $(\sigma(D_{0}),\sigma(\zeta_{p}))=(D_{0},\zeta_{p}^{-1})$ , and consequently $\sigma(\sqrt{\Delta})=-\sqrt{\Delta}$ because $\sqrt{\Delta}\notin\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})$ , the fixed field of conjugation in $F_{p}$ . Then as in case (1), $\sigma$ fixes all roots of $d_{h}$ and therefore $L$ , so $[K:L]\geq 2$ . Then $[K:L]=2$ , and $L=D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))=D_{0}(\zeta_{p}+\zeta_{p}^{-1})$ , after all, and is of degree $\frac{p(p-1)}{2}$ , using Lemma 4.5. ∎

Remark 5.3.

The choice of the automorphism $\sigma$ fixing $L$ follows Jones’ construction in [5, Section 3, p. 6].

Figure 1. The field diagram when

\sqrt{\Delta}\notin F_{p}

Figure 2. The field diagram when

\sqrt{\Delta}\in F_{p}

and

p\equiv 1\bmod 4

Figure 3. The field diagram when

\sqrt{\Delta}\in F_{p}

and

p\equiv 3\bmod 4

We are almost ready to prove Theorem 1.3, but first, we must recall the following fact about cyclotomic fields:

Lemma 5.4.

[1, VI, Thm 3.3] Let $p$ be an odd prime and let $(\frac{-1}{p})=(-1)^{p(p-1)/2}$ be the quadratic Legendre symbol. Let $K=\mathbb{Q}\left(\sqrt{(\frac{-1}{p})p}\,\right)$ . Then $K\subset\mathbb{Q}(\zeta_{p})$ and $K$ is the only quadratic extension of $\mathbb{Q}$ that is a subfield of $\mathbb{Q}(\zeta_{p})$ .

Corollary 5.5.

$\sqrt{\Delta}\in\mathbb{Q}(\zeta_{p})$ if and only if $\Delta\in(-1)^{p(p-1)/2}p\mathbb{Q}^{2}$ .

Finally, we can prove Theorem 1.3.

Proof of Theorem 1.3.

Recall that we are studying irreducible polynomials $h$ of the form $x^{2p}+bx^{p}+c^{p}$ . We note that for any irreducible $h$ of this form, $\Delta:=b^{2}-4c$ must fall into exactly one of the 3 cases given. And from Lemmas 4.4 and 5.2, $d_{h}$ is also irreducible with splitting field $L=\mathbb{Q}(\beta+\beta_{1},\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$ . We will use throughout the notation of $D_{i}:=\mathbb{Q}(\zeta_{p}^{i}\beta+\zeta_{p}^{-i}\beta_{1})$ . Note throughout that while we will be examining specific elements of $\mathrm{Gal}(K/\mathbb{Q})$ , we will be considering their actions on the roots of $d$ , which is equivalent to the canonical actions of their images in $\mathrm{Gal}(L/\mathbb{Q})$ under the division map.

Proof of Case 1. If $b^{2}-4c\notin(-1)^{p(p-1)/2}p\mathbb{Q}^{2}$ , then $\sqrt{\Delta}\notin F_{p}$ by Corollary 5.5. From above, this means that $L=D_{0}(\zeta_{p}+\zeta_{p}^{-1},\sqrt{\Delta}(\zeta_{p}-\zeta_{p}^{-1}))$ is an extension of degree $p(p-1)$ which does not contain $\sqrt{\Delta}$ .

Consider now the elements $\sigma,\tau\in\mathrm{Gal}(K/\mathbb{Q})$ defined using Theorem 4.7 by:

\begin{split}(\sigma(D_{0}),\epsilon_{\sigma},\sigma(\zeta_{p}))&=(D_{0},1,\zeta_{p}^{r})\\ (\tau(D_{0}),\epsilon_{\tau},\tau(\zeta_{p}))&=(D_{1},1,\zeta_{p})\\ \end{split}

Where $\zeta_{p}^{r}$ is a generator of the group of $p$ th roots of unity. By inspection, $\sigma(D_{i})=D_{ri}$ and $\tau(D_{i})=D_{i+1}$ . Let $S_{p}$ be the symmetric group on the set $\mathbb{Z}_{p}=\{0,\dots,p-1\}$ . The Galois group of $d_{h}$ therefore contains the elements

\begin{split}\tau_{1}=&\ (1\ r\ r^{2}\ \cdots\ r^{p-2})\in S_{p},\\ \tau_{2}=&\ (0\ 1\ 2\ \cdots\ p-1)\in S_{p}\end{split}

The elements $\tau_{1}$ , $\tau_{2}$ generate $\mathrm{Aff}(\mathbb{F}_{p})$ , whose size is $p(p-1)$ . As $|\mathrm{Gal}(L/\mathbb{Q})|=p(p-1)$ , $\mathrm{Gal}(L/\mathbb{Q})\simeq\mathrm{Aff}(\mathbb{F}_{p})$ . And $\mathrm{Gal}(\mathbb{Q}(\sqrt{\Delta})/\mathbb{Q})\simeq C_{2}$ , so the Galois group of $h$ is

\mathrm{Gal}(K/\mathbb{Q})\simeq\mathrm{Gal}(L/\mathbb{Q})\times\mathrm{Gal}(\mathbb{Q}(\sqrt{\Delta}/\mathbb{Q})\simeq\mathrm{Aff}(\mathbb{F}_{p})\times C_{2}

Proof of Case 2. If $p\equiv 1\bmod 4$ and $b^{2}-4c\in p\mathbb{Q}^{2}$ , then $\sqrt{\Delta}\in F_{p}$ by Corollary 5.5. From above, $L=K$ is the splitting field of $d_{h}$ and $h$ , and $|\mathrm{Gal}(K/\mathbb{Q})|=|\mathrm{Gal}(L/\mathbb{Q})|=p(p-1)$ . Now consider elements $\sigma,\tau\in\mathrm{Gal}(K/\mathbb{Q})$ defined using Theorem 4.7 by:

\begin{split}(\sigma(D_{0}),\sigma(\zeta_{p}))&=(D_{0},\zeta_{p}^{r})\\ (\tau(D_{0}),\tau(\zeta_{p}))&=(D_{1},\zeta_{p})\\ \end{split}

For $\zeta_{p}^{r}$ a generator. $\sigma$ conjugates $\sqrt{\Delta}\in F_{p}\setminus\mathbb{Q}$ , so $\sigma(\zeta_{p}^{i}\beta)=\zeta_{p}^{ri}\beta_{1}$ , and $\sigma(D_{i})=D_{-ri}$ . $\zeta_{p}^{-r}$ is a generator because

\left(\frac{-r}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{r}{p}\right)=(1)(-1)=-1

So $\sigma$ will be a $p-1$ -cycle. And again, $\tau(D_{i})=D_{i+1}$ . The Galois group of $d_{h}$ therefore contains the elements

$\tau_{1}$	$=$	$(1$	$-r$	$(-r)^{2}$	$\ldots$	$(-r)^{(p-2)}$	$)$
$\tau_{2}$	$=$	$(0$	$1$	$2$	$\ldots$	$p-1$	$)$

As in Case 1, the elements $\tau_{1}$ , $\tau_{2}$ generate $\mathrm{Aff}(\mathbb{F}_{p})$ , and $|\mathrm{Aff}(\mathbb{F}_{p})|=p(p-1)=|\mathrm{Gal}(K/\mathbb{Q})|=|\mathrm{Gal}(L/\mathbb{Q})|$ , so the Galois groups of $d_{h},h$ are identically

\mathrm{Gal}(K/\mathbb{Q})\simeq\mathrm{Gal}(L/\mathbb{Q})\simeq\mathrm{Aff}(\mathbb{F}_{p})

Proof of Case 3. If $p\equiv 3\bmod 4$ and $b^{2}-4c\in-p\mathbb{Q}^{2}$ , then $\sqrt{\Delta}\in F_{p}$ by Corollary 5.5. From above, $|\mathrm{Gal}(K/\mathbb{Q})|=p(p-1)$ , but $|\mathrm{Gal}(L/\mathbb{Q})|=\frac{p(p-1)}{2}$ . Again we consider elements $\sigma,\tau\in\mathrm{Gal}(K/\mathbb{Q})$ defined using Theorem 4.7 by:

\begin{split}(\sigma(D_{0}),\sigma(\zeta_{p}))&=(D_{0},\zeta_{p}^{r})\\ (\tau(D_{0}),\tau(\zeta_{p}))&=(D_{1},\zeta_{p})\\ \end{split}

For $\zeta_{p}^{r}$ a generator. $\sigma$ again conjugates $\sqrt{\Delta}\in F_{p}\setminus\mathbb{Q}$ , so $\sigma(\zeta_{p}^{i}\beta)=\zeta_{p}^{ri}\beta_{1}$ , and $\sigma(D_{i})=D_{-ri}$ . However, we see that $\zeta_{p}^{-r}$ is of order $\frac{p-1}{2}$ because

\left(\frac{-r}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{r}{p}\right)=(-1)^{p(p-1)/2}(-1)=1

So $\sigma$ will be two $\frac{p-1}{2}$ -cycles, the square of the $p-1$ -cycle generated by $\sqrt{-r}$ . And again, $\tau(D_{i})=D_{i+1}$ . The Galois group of $d_{h}$ therefore contains the elements $\tau=\tau_{1}\tau_{2},\tau_{3}\in S_{p}$ , where $\tau_{1},\tau_{2}$ are the $(p-1)/2$ cycles

$\tau_{1}$	$=$	$(1$	$-r$	$(-r)^{2}$	$\ldots$	$(-r)^{(p-3)/2}$	$)$
$\tau_{2}$	$=$	$(-1$	$r$	$-(-r)^{2}$	$\ldots$	$-(-r)^{(p-3)/2}$	$)$

and $\tau_{3}$ is the $p$ cycle

\tau_{3}

=

(0

1

2

\ldots

p-1

)

The elements $\tau$ , $\tau_{3}$ generate the normal subgroup $C_{p}\rtimes C_{(p-1)/2}\vartriangleleft\mathrm{Aff}(\mathbb{F}_{p})$ , whose size is $\frac{p(p-1)}{2}=|\mathrm{Gal}(L/\mathbb{Q})|$ . Thus, the Galois group of $d_{h}$ is

\mathrm{Gal}(L/\mathbb{Q})\simeq C_{p}\rtimes C_{(p-1)/2}\vartriangleleft\mathrm{Aff}(\mathbb{F}_{p})

Now $\sqrt{\Delta}\notin L$ and $L(\sqrt{\Delta})=K$ as in Case 1, so

\mathrm{Gal}(K/\mathbb{Q})\simeq\mathrm{Gal}(L/\mathbb{Q})\times C_{2}\simeq(C_{p}\rtimes C_{(p-1)/2})\times C_{2}

∎

Remark 5.6.

The automorphisms $\sigma,\tau$ we choose in the proof of Theorem 1.3 are again as in Jones’ [5, Section 3, p. 6].

Remark 5.7.

In the cases of $p=2$ and $p=3$ (the solutions to which we mention in Section 1), our methods of proof fail beginning at Lemma 4.1, where we use the fact $p-1\mid 2p$ . The issue is essentially that $\zeta_{2}=-1\in\mathbb{Q}$ for the case of $p=2$ and that if $\sqrt{\Delta}\in F_{3}$ , then $\mathbb{Q}(\sqrt{\Delta})=F_{3}$ in the case of $p=3$ . These issues both invalidate the distinctness of the $B_{i}$ and $D_{i}$ , which is key to Lemma 4.4, on which all of our subsequent work regarding $d_{h}$ and its Galois group is based.

Remark 5.8.

From Corollary 1.2 of [5], there exist an infinite number of polynomials of the form of case 1 of Theorem 1.3.

References

[1] Serge Lang “Algebra” Springer, 2002 DOI: 10.1007/978-1-4613-0041-0
[2] Joshua Harrington and Lenny Jones “The Irreducibility of Power Compositional Sextic Polynomials and Their Galois Groups” In Mathematica Scandinavica 102.2, 2017 DOI: https://doi.org/10.7146/math.scand.a-25850
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The Galois group of x2​p+b​xp+cpx^{2p}+bx^{p}+c^{p} over ℚ\mathbb{Q}

Abstract.

Key words and phrases:

2020 Mathematics Subject Classification:

1. Introduction

Theorem 1.1.

Remark 1.2.

Theorem 1.3.

Remark 1.4.

Example 1.

Remark 1.5.

Remark 1.6.

Remark 1.7.

2. Some Preliminaries

Theorem 2.1 (Capelli’s Theorem).

Theorem 2.2 (Capelli).

Corollary 2.3.

3. A Reducibility Criterion for x2​m+b​xm+c∈F​[x]x^{2m}+bx^{m}+c\in F[x]

Theorem 3.1.

Theorem 3.2.

Proof.

Theorem 3.3.

Proof.

Proof of Theorem 3.1.

Proof of Theorem 1.1.

Remark 3.4.

4. Properties of h=x2​p+b​xp+cp∈ℚ​[x]h=x^{2p}+bx^{p}+c^{p}\in\mathbb{Q}[x] and dh=Dp​(x,c)+bd_{h}=D_{p}(x,c)+b

Lemma 4.1.

Proof.

Lemma 4.2.

Proof.

Lemma 4.3.

Proof.

Lemma 4.4.

Proof.

Lemma 4.5.

Proof.

Remark 4.6.

Theorem 4.7.

Proof.

5. The Galois groups of hh and dhd_{h}

Theorem 5.1.

Proof.

Theorem 5.2.

Proof.

Remark 5.3.

Lemma 5.4.

Corollary 5.5.

Proof of Theorem 1.3.

Remark 5.6.

Remark 5.7.

Remark 5.8.

References

The Galois group of $x^{2p}+bx^{p}+c^{p}$ over $\mathbb{Q}$

3. A Reducibility Criterion for $x^{2m}+bx^{m}+c\in F[x]$

4. Properties of $h=x^{2p}+bx^{p}+c^{p}\in\mathbb{Q}[x]$ and $d_{h}=D_{p}(x,c)+b$

5. The Galois groups of $h$ and $d_{h}$