1 Introduction
Let F be a subfield of the complex numbers and
$f(x)\in F[x]$
an irreducible polynomial of degree n. Identifying the Galois group,
$\operatorname {\mathrm {Gal}}(f)$
, of
$f(x)$
as a transitive subgroup of
$S_n$
is a fundamental problem in computational algebra. In general, this is a difficult task; most modern approaches are based on [Reference Soicher and McKay9, Reference Stauduhar10]. However, when
$f(x)$
has a special form, the computation can be more straightforward.
For example, Galois groups of even quartic polynomials (
$x^4+ax^2+b$
), even sextic polynomials (
$x^6+ax^4+bx^2+c$
) and doubly even octic polynomials (
$x^8+ax^4+b$
) have elementary characterisations (see for example [Reference Altmann, Awtrey, Cryan, Shannon and Touchette1, Reference Awtrey and Jakes2]). In each case, the characterisation leverages information about the index-2 subfield of the field defined by the polynomial.
A natural extension of this technique is to irreducible reciprocal polynomials, which are polynomials satisfying
$f(x) = x^n\cdot f(1/x)$
, since the field extension defined by such a polynomial also has an index-2 subfield (see Theorem 2.1). Note that if
$f(x) = \sum _{i=0}^n f_i x^i$
is a reciprocal polynomial, then
$f_i = f_{n-i}$
; that is, the sequence of coefficients
$\{f_i\}$
forms a palindrome.
In this setting, some similar characterisations of
$\operatorname {\mathrm {Gal}}(f)$
are known. The following classical result of Dickson determines the Galois group of an irreducible reciprocal quartic polynomial by testing the squareness of two elements of F (see [Reference Dickson4]). In the theorem, as in the rest of the paper, we will use the following standard convention for describing groups:
$C_n$
denotes the cyclic group of order n,
$D_n$
the dihedral group of order
$2n$
, and
$A_n$
and
$S_n$
the alternating and symmetric groups on n letters, respectively. We also use
$\times $
to denote a direct product,
$\rtimes $
a semidirect product and
$\wr $
a wreath product.
Theorem 1.1 (Dickson)
Let
$x^4+ax^3+bx^2+ax+1 \in F[x]$
be irreducible. Then
$\operatorname {\mathrm {Gal}}(f)$
is isomorphic to:
-
•
$C_2\times C_2$
if and only if
$(b+2)^2-4a^2$
is a square in F; -
•
$C_4$
if and only if
$((b+2)^2-4a^2)(a^2-4b+8)$
is a square in F; -
•
$D_4$
if and only if neither
$(b+2)^2-4a^2$
nor
$((b+2)^2-4a^2)(a^2-4b+8)$
is a square in F.
Since reciprocal polynomials of odd degree are reducible (they have
$-1$
as a root), the next logical case to consider is Galois groups of irreducible reciprocal sextic polynomials; that is, those of the form
$x^6+ax^5+bx^4+cx^3+bx^2+ax+1 \in F[x]$
. The purpose of this paper is to provide a similar characterisation, reflecting the spirit of Theorem 1.1. In doing so, we will generalise the results in [Reference Jones6, Reference Jones7], which verify that certain families of irreducible reciprocal sextic polynomials with rational coefficients have Galois group isomorphic to either
$S_3$
,
$D_6$
or
$S_4\times C_2$
.
The remainder of the paper is organised as follows. In Section 2, we collect several results concerning field extensions defined by reciprocal polynomials and their Galois groups; these will be used to prove our main theorem in Section 3. Our main result, Theorem 3.5, gives a characterisation of the Galois group of
$x^6+ax^5+bx^4+cx^3+bx^2+ax+1$
that depends only on the squareness of three elements whose expressions involve a, b and c, along with whether or not a related quartic polynomial has a linear factor. As an application, we provide one-parameter families of reciprocal sextics for each possible Galois group (see Theorem 3.7).
2 Preliminary results
In this section, we let
$f(x) \in F[x]$
be a monic irreducible reciprocal polynomial of degree
$n=2m$
, and we let
$L = F(\alpha )$
where
$f(\alpha )=0$
. We collect results about L and
$\operatorname {\mathrm {Gal}}(f)$
that are used later in the paper.
The reader is referred to [Reference Lindstrom8] for an elementary overview of several standard facts about reciprocal polynomials. One such result shows there exists a polynomial
$g(x)$
of degree m such that
$f(x) = x^m\cdot g(x+1/x)$
(see [Reference Lindstrom8, Proposition 2.0.16]). It turns out that
$g(x)$
is the minimal polynomial of
$\alpha +1/\alpha $
. This result is straightforward to establish and shows L has an index-2 subfield. For convenience, we include a proof.
Theorem 2.1. The minimal polynomial,
$g(x)$
, of
$\beta = \alpha +1/\alpha $
has degree m. Thus, L has a subfield,
$K=F(\beta )$
, of index 2.
Proof. Let the roots of
$f(x)$
be
$\{\alpha = r_1, 1/r_1, \ldots , r_m, 1/r_m\}$
, and let
$g(x)$
be the polynomial whose roots are
$\{r_i+1/r_i\}$
for
$1\leq i\leq m$
. Thus, the degree of
$g(x)$
is m and
$g(\beta )=0$
. Let
$h(x) = x^m\cdot g(x+1/x)$
. It follows that
$f(x) = h(x)$
since both polynomials are monic, are of the same degree and have the same roots. If
$g(x)$
were reducible, say
$g(x) = k(x)\cdot l(x)$
, then this would imply
$f(x)$
is reducible since
$$ \begin{align*} f(x) = h(x) = x^{\text{deg}(k)} k(x+1/x) \cdot x^{\text{deg}(l)} l(x+1/x).\end{align*} $$
However, this contradicts the irreducibility of
$f(x)$
. Thus,
$g(x)$
is irreducible and is therefore the minimal polynomial of
$\beta $
.
Since subfields of L correspond to block systems of
$\operatorname {\mathrm {Gal}}(f)$
(see [Reference Dixon and Mortimer5, Section 1.6] for more information about block systems), it follows that
$\operatorname {\mathrm {Gal}}(f)$
can be embedded in a suitable wreath product. In particular, the following is an immediate consequence of Theorem 2.1 and [Reference Dixon and Mortimer5, Theorem 2.6 A].
Corollary 2.2. We have
$\operatorname {\mathrm {Gal}}(f)$
is a subgroup of
$C_2 \wr \operatorname {\mathrm {Gal}}(g) \simeq C_2^m \rtimes \operatorname {\mathrm {Gal}}(g)$
. Thus,
$2^m \cdot m!$
is an upper bound for
$|\kern-1.5pt\operatorname {\mathrm {Gal}}(f)|$
.
3 Galois groups of reciprocal sextics
For the rest of the paper, we let
$f(x) = x^6+ax^5+bx^4+cx^3+bx^2+ax+1 \in F[x]$
be an irreducible reciprocal polynomial,
$\alpha $
a root of
$f(x)$
and
$L = F(\alpha )$
. Let
$g(x)$
denote the minimal polynomial of
$\alpha +1/\alpha $
. Thus, we have the following result as a consequence of Theorem 2.1.
Corollary 3.1. The polynomial
$g(x) = x^3 + ax^2 + (b - 3)x-2a+c$
is irreducible and defines a cubic subfield K of L.
Since
$f(x)$
is irreducible of degree 6, it follows that
$\operatorname {\mathrm {Gal}}(f)$
is a transitive subgroup of
$S_6$
that is also a subgroup of
$C_2 \wr S_3 \simeq S_4 \times C_2$
, by Corollary 2.2. Among the 16 transitive subgroups of
$S_6$
, only 8 are subgroups of
$S_4\times C_2$
. Table 1 gives information about these 8 groups: their transitive numbers (as given in [Reference Butler and McKay3]), their orders, descriptive names and generators for (one representative of the conjugacy class of) each group.
Possible Galois groups of irreducible reciprocal sextic polynomials. Size gives the order of the group.
We point out that 6T7 and 6T8 are isomorphic copies of
$S_4$
that are distinguished by their parity; that is, 6T7 contains only even permutations while 6T8 does not; this is reflected in the table by the respective superscripts of each group’s ‘Name’. Note that each of these eight groups does appear as a Galois group over
$\mathbb {Q}$
. See Table 2, which gives one sample irreducible reciprocal sextic polynomial in
$\mathbb {Q}[x]$
for each possible Galois group.
One sample reciprocal sextic Polynomial in
$\mathbb {Q}[x]$
for each possible Galois group.
We can determine properties of
$\operatorname {\mathrm {Gal}}(f)$
from properties of
$\operatorname {\mathrm {Gal}}(g)$
, as our next general result shows.
Theorem 3.2. Let
$\phi (x)\in F[x]$
be irreducible of degree
$6$
and
$\rho $
be a root of
$\phi $
. Suppose that
$F(\rho )$
has a subfield M of degree
$3$
defined by
$\gamma (x)\in F[x]$
. Then:
-
(1)
$\operatorname {\mathrm {Disc}}(\phi )$
is a perfect square in F if and only if
$\operatorname {\mathrm {Gal}}(\phi )$
is either
$A_4$
or
$S_4^+$
; -
(2)
$\operatorname {\mathrm {Disc}}(\gamma )$
is a perfect square in F if and only if
$\operatorname {\mathrm {Gal}}(\phi )$
is either
$C_6$
,
$A_4$
or
$A_4\times C_2$
; -
(3)
$\operatorname {\mathrm {Disc}}(\phi )\cdot \operatorname {\mathrm {Disc}}(\gamma )$
is a perfect square in F if and only if
$\operatorname {\mathrm {Gal}}(\phi )$
is either
$S_3$
,
$A_4$
or
$S_4^-$
.
Proof. A standard result in Galois theory shows that
$\operatorname {\mathrm {Gal}}(\phi )$
is a subgroup of
$A_6$
if and only if
$\operatorname {\mathrm {Disc}}(\phi )$
is a square in F. Of the eight possibilities, only
$A_4$
and
$S_4^+$
are subgroups of
$A_6$
, proving item (1).
Since
$\gamma $
is a cubic polynomial, another standard result in Galois theory shows
$\operatorname {\mathrm {Gal}}(\gamma ) \simeq C_3$
if and only if
$\operatorname {\mathrm {Disc}}(\gamma )$
is a perfect square. Under the Galois correspondence, M corresponds to an index 3 subgroup H of
$\operatorname {\mathrm {Gal}}(\phi )$
containing the stabiliser of
$\rho $
. Let N be the normal core of H in
$\operatorname {\mathrm {Gal}}(\phi )$
; that is, the largest normal subgroup of
$\operatorname {\mathrm {Gal}}(\phi )$
contained in H. Therefore,
$\operatorname {\mathrm {Gal}}(\gamma )$
is isomorphic to
$\operatorname {\mathrm {Gal}}(\phi )/N$
. Using [11] to perform group computations, we see that each of the eight possibilities has a unique such subgroup H of index 3, up to conjugation. This means
$M/F$
is the unique cubic subfield of
$F(\rho )/F$
, up to isomorphism. Further group computations show that in the cases of
$C_6$
,
$A_4$
and
$A_4\times C_2$
,
$\operatorname {\mathrm {Gal}}(\phi )/N$
is isomorphic to
$C_3$
; in all other cases it is isomorphic to
$S_3$
, proving item (2).
If both
$\operatorname {\mathrm {Disc}}(\phi )$
and
$\operatorname {\mathrm {Disc}}(\gamma )$
are perfect squares, then
$\operatorname {\mathrm {Disc}}(\phi )\cdot \operatorname {\mathrm {Disc}}(\gamma )$
is a perfect square. According to the previous paragraphs, there is only one group among the eight where this occurs; namely,
$A_4$
. For the remainder of the proof, we suppose neither
$\operatorname {\mathrm {Disc}}(\phi )$
nor
$\operatorname {\mathrm {Disc}}(\gamma )$
is a perfect square. Thus, the polynomials
$x^2-\operatorname {\mathrm {Disc}}(\phi )$
and
$x^2-\operatorname {\mathrm {Disc}}(\gamma )$
define quadratic subfields of the splitting field of
$\phi (x)$
. By the Galois correspondence,
$F(\sqrt {\operatorname {\mathrm {Disc}}(\phi )})$
corresponds to
$H_1=A_6\cap \operatorname {\mathrm {Gal}}(\phi )$
. Similarly, if
$M'$
is the normal closure of
$\gamma (x)$
, then the subgroup fixing
$M'$
is N. Thus,
$F(\sqrt {\operatorname {\mathrm {Disc}}(\gamma )})$
corresponds to the unique subgroup
$H_2$
of
$\operatorname {\mathrm {Gal}}(\phi )$
of index 2 that contains N. It follows that
$\operatorname {\mathrm {Disc}}(\phi )\cdot \operatorname {\mathrm {Disc}}(\gamma )$
is a perfect square if and only if
$H_1=H_2$
. Among the four remaining possible Galois groups, direct computation shows
$S_3$
and
$S_4^-$
have
$H_1=H_2$
. The groups
$D_6$
and
$S_4\times C_2$
have
$H_1\neq H_2$
, proving item (3).
Our next result is an immediate consequence of Theorem 3.2 and the fact that
$\operatorname {\mathrm {Disc}}(f) = ((2a+c)^2-(2b+2)^2) \cdot \operatorname {\mathrm {Disc}}(g)^2$
.
Corollary 3.3. We have the following:
-
(1)
$(2a+c)^2-(2b+2)^2$
is a perfect square in F if and only if
$\operatorname {\mathrm {Gal}}(f)$
is either
$A_4$
or
$S_4^+$
; -
(2)
$\operatorname {\mathrm {Disc}}(g)$
is a perfect square in F if and only if
$\operatorname {\mathrm {Gal}}(f)$
is either
$C_6$
,
$A_4$
or
$A_4\times C_2$
; -
(3)
$((2a+c)^2-(2b+2)^2)\cdot \operatorname {\mathrm {Disc}}(g)$
is a perfect square in F if and only if
$\operatorname {\mathrm {Gal}}(f)$
is either
$S_3$
,
$A_4$
or
$S_4^-$
.
Next, we introduce a degree 4 resolvent polynomial that is helpful in determining
$\operatorname {\mathrm {Gal}}(f)$
.
Theorem 3.4. Let
$h(x) = x^4+Ax^3+Bx^2+Cx+D$
, where
-
•
$A = -4(a^2-2b-6)$
; -
•
$B = 2(3a^4 - 4a^2(3b+5) + 8(ac+b^2+4b+9))$
; -
•
$C = -4(a^4(a^2 - 6b-2) + 8a^2(ac+b^2 + 5) + 16(ac-2b^2 - b(ac+2) -4))$
; -
•
$D = (a^4-4a^2(b-1)+8(ac-2b))^2$
.
Then,
$h(x)$
is separable and has a linear factor if and only if
$\operatorname {\mathrm {Gal}}(f)$
is either
$C_6$
,
$S_3$
or
$D_6$
.
Proof. Let
$\{\alpha = r, 1/r, s, 1/s, t,1/t\}$
be the roots of
$f(x)$
. The roots of
$h(x)$
are
$\{(r-1/r) \pm (s-1/s) \pm (t-1/t)\}$
, which can be verified by using the theory of elementary symmetric functions to express the coefficients of
$h(x)$
in terms of a, b and c.
If
$h(x)$
were not separable, then two roots would be equal. There are six cases:
-
(1)
$(r-1/r) + (s-1/s) + (t-1/t) = (r-1/r) + (s-1/s) - (t-1/t)$
; -
(2)
$(r-1/r) + (s-1/s) + (t-1/t) = (r-1/r) - (s-1/s) + (t-1/t)$
; -
(3)
$(r-1/r) + (s-1/s) + (t-1/t) = (r-1/r) - (s-1/s) - (t-1/t)$
; -
(4)
$(r-1/r) + (s-1/s) - (t-1/t) = (r-1/r) - (s-1/s) + (t-1/t)$
; -
(5)
$(r-1/r) + (s-1/s) - (t-1/t) = (r-1/r) - (s-1/s) - (t-1/t)$
; -
(6)
$(r-1/r) - (s-1/s) + (t-1/t) = (r-1/r) - (s-1/s) - (t-1/t)$
.
We will show each case leads to a contradiction. Cases 2 and 5 imply
$s=1/s$
and Cases 1 and 6 imply
$t=1/t$
; these contradict the fact that
$f(x)$
is irreducible and thus separable. Case 4 implies
$st(s-t)=-(s-t)$
. Since
$f(x)$
is separable, this implies
$st=-1$
. Thus,
$s=-1/t$
and
$1/s=-t$
. Therefore,
$-a = r+1/r+s+1/s+t+1/t = r+1/r$
, which is rational. However, this contradicts Theorem 2.1, which shows
$r+1/r$
is not rational. Similarly, Case 3 implies
$st(s+t)=s+t$
. If
$s+t\neq 0$
, then
$st=1$
; which contradicts the separability of
$f(x)$
. If
$s=-t$
, then
$1/s=-1/t$
. We again reach the contradiction
$-a=r+1/r$
. Thus,
$h(x)$
is separable.
To prove the rest of the theorem, let
$G = S_4\times C_2$
and
$H = D_6$
be the subgroups of
$S_6$
as given in Table 1. Then, a complete set of a right coset representatives of
$G/H$
is
$\{\text {id}, (34), (56), (34)(56) \}$
. Further, the only block system of G is
$R = \{\{1,2\}, \{3,4\}, \{5,6\}\}$
. We identify r as root 1,
$1/r$
as root 2, s as root 3,
$1/s$
as root 4, t as root 5 and
$1/t$
as root 6.
A multivariable function stabilised by H is
$T(x_1,x_2,x_3,x_4,x_5,x_6) = x_1-x_2+x_3-x_4+x_5-x_6$
; the action on T is via subscripts. We form the resolvent polynomial corresponding to G, H and T (see [Reference Soicher and McKay9]); this produces the polynomial
$h(x)$
. By the theory of resolvent polynomials, the factorisation of
$h(x)$
corresponds to the orbits of
$\operatorname {\mathrm {Gal}}(f)$
acting on the cosets
$G/H$
. Direct computation on each possibility for
$\operatorname {\mathrm {Gal}}(f)$
shows that in the cases of
$C_6$
,
$S_3$
and
$D_6$
, there is an orbit of length 1 and an orbit of length 3. In the other five cases, there is a single orbit of length 4. This means
$h(x)$
factors as a linear times a cubic polynomial in those three cases and remains irreducible in the other five cases, proving the theorem.
We can now state our main result, which gives an elementary characterisation of
$\operatorname {\mathrm {Gal}}(f)$
. This is an immediate consequence of Corollary 3.3 and Theorem 3.4. For convenience, Table 3 summarises this characterisation.
Let
$f(x)=x^6+ax^5+bx^4+cx^3+bx^2+ax+1\in F[x]$
be irreducible,
$g(x) = x^3 + ax^2 + (b - 3)x-2a+c$
and
$h(x)$
as defined in Theorem 3.4. The table lists whether
$\operatorname {\mathrm {Disc}}(f)$
,
$\operatorname {\mathrm {Disc}}(g)$
and
$\operatorname {\mathrm {Disc}}(f)\cdot \operatorname {\mathrm {Disc}}(g)$
are perfect squares in F, and whether
$h(x)$
has a linear factor, according to
$\operatorname {\mathrm {Gal}}(f)$
.
Theorem 3.5. Let
$f(x)=x^6+ax^5+bx^4+cx^3+bx^2+ax+1\in F[x]$
be irreducible,
$g(x) = x^3 + ax^2 + (b - 3)x-2a+c$
and
$h(x)$
as defined in Theorem 3.4.
-
(1) If
$\operatorname {\mathrm {Disc}}(f)$
is a perfect square in F, then
$\operatorname {\mathrm {Gal}}(f)$
is
$A_4$
if
$\operatorname {\mathrm {Disc}}(g)$
is a square and is
$S_4^+$
otherwise. -
(2) If
$\operatorname {\mathrm {Disc}}(f)$
is not a square and
$\operatorname {\mathrm {Disc}}(g)$
is a square, then
$\operatorname {\mathrm {Gal}}(f)$
is
$C_6$
if
$h(x)$
has a linear factor and is
$A_4\times C_2$
otherwise. -
(3) If
$\operatorname {\mathrm {Disc}}(f)$
and
$\operatorname {\mathrm {Disc}}(g)$
are not squares and
$\operatorname {\mathrm {Disc}}(f)\cdot \operatorname {\mathrm {Disc}}(g)$
is a square, then
$\operatorname {\mathrm {Gal}}(f)$
is
$S_3$
if
$h(x)$
has a linear factor and is
$S_4^-$
otherwise. -
(4) If none of
$\operatorname {\mathrm {Disc}}(f)$
,
$\operatorname {\mathrm {Disc}}(g)$
and
$\operatorname {\mathrm {Disc}}(f)\cdot \operatorname {\mathrm {Disc}}(g)$
is a square, then
$\operatorname {\mathrm {Gal}}(f)$
is
$D_6$
if
$h(x)$
has a linear factor and is
$S_4\times C_2$
otherwise.
Example 3.6. As an example, we use Theorem 3.5 to compute the Galois group of a family of sextic reciprocal polynomials. Take
$t>-27/4$
and suppose the polynomial
$f(x) = x^6+3x^5+(t+6)x^4+(2t+7)x^3+(t+6)x^2+3x+1 \in \mathbb {Q}[x]$
is irreducible. Then,
$\operatorname {\mathrm {Disc}}(f) = -t^4(4t+27)^3$
, which is not a square. We also have
$g(x) = x^3 + 3x^2 + (t + 3)x + (2t + 1)$
. Then,
$\operatorname {\mathrm {Disc}}(g) = -t^2(4t+27)$
, which is also not a square. However,
$\operatorname {\mathrm {Disc}}(f)\cdot \operatorname {\mathrm {Disc}}(g)$
is a square. Further,
$h(x)$
has
$x+4t+27$
as a linear factor. By item (3) of Theorem 3.5,
$\operatorname {\mathrm {Gal}}(f)$
is
$S_3$
. Note, this also confirms item (3) of [Reference Jones6, Theorem 1].
As an application of Theorem 3.5, we give one-parameter families of reciprocal sextics defined over
$\mathbb {Q}$
for each possible Galois group.
Theorem 3.7. The polynomials in Table 4 have the indicated Galois group over
$\mathbb {Q}$
, except for values of t that result in reducible polynomials.
One-parameter families of reciprocal sextic polynomials with specified Galois group over
$\mathbf {Q}$
.
Verifying each family of polynomials in Table 4 has the indicated Galois group is a straightforward computation using Theorem 3.5 and a computer algebra system. For example, we can consider the polynomial in the first row of Table 4:
$f(x) = (2t^2{\kern-1pt}-2t+{\kern-1pt}13)x^6+(-4t+{\kern-1pt}2)x^5+(-2t^2{\kern-1pt}+{\kern-1pt}2t+19)x^4+(8t-4)x^3+(-2t^2+2t+19)x^2 +(-4t+2)x+(2t^2-2t+13)$
. Then,
$\operatorname {\mathrm {Disc}}(f) = -(t^2-t-1)^4(t^2-t+7)^4$
, which is not a square. We also have
$g(x) = (2t^2 - 2t + 13)x^3 + (-4t + 2)x^2 + (-8t^2 + 8t - 20)x + (16t - 8)$
. Then,
$\operatorname {\mathrm {Disc}}(g) = (t^2-t-1)^2(t^2-t+7)^2$
, which is a square. Furthermore,
$h(x)$
has
$(t^4 - 2t^3 + 14t^2 - 13t + 169/4)x + 4t^4 - 8t^3 + 36t^2 - 32t + 64$
as a linear factor. This proves
$\operatorname {\mathrm {Gal}}(f) = C_6$
, as claimed.
