2 Discriminants of power compositional polynomials

In this section, we give a formula for the discriminant of a polynomial of the form $g(x^n)$ that we will use later. We note that our result is a special case of [Reference Harrington and Jones10, Theorem 2.7], but our method of proof is different.

For complete generality, we let K be a field, $\overline {K}$ an algebraic closure of K, $f(x)\in K[x]$ a monic polynomial of degree n, $R_f$ the set of roots of $f(x)$ in $\overline {K}$ and $f'(x)$ the derivative of $f(x)$ . Recall that the discriminant of $f(x)$ , which we denote by $ {\mathrm{Disc}}(f)$ , can be computed as follows (see for example [Reference Cohen6, Section 3.3]):

$$ \begin{align*} {\mathrm{Disc}}(f) = (-1)^{n(n-1)/2} \prod_{\rho\in R_f} f'(\rho). \end{align*} $$

Lemma 2.1. Let K be a field and $f(x)\in K[x]$ a monic polynomial, where $f(x) = g(x^n)$ and $g(x)\in K[x]$ is monic of degree m. Let $c = f(0)$ . Then,

$$ \begin{align*} {\mathrm{Disc}}(f) = (-1)^{nm(n-1)(m+2)/2} \cdot n^{nm} \cdot c^{n-1} \cdot {\mathrm{Disc}}(g)^n. \end{align*} $$

Proof. Let $R_f$ and $R_g$ denote the roots of $f(x)$ and $g(x)$ in an algebraic closure $\overline {K}$ , respectively. Let $\zeta \in \overline {K}$ be a primitive nth root of unity. Thus, there exist $\rho _1,\ldots , \rho _m \in \overline {K}$ such that $R_f = \{ \rho _i\zeta ^j : 1\leq i\leq m, 0\leq j\leq n-1\}$ and $R_g = \{\rho _1^n, \ldots , \rho _m^n\}$ . We note that

$$ \begin{align*} c = f(0) = (-1)^{nm} \prod_{\rho\in R_f} \rho.\end{align*} $$

We define $d_f$ and $d_g$ by

$$ \begin{align*} d_f = \prod_{\rho\in R_f} g'(\rho^n), \quad d_g = \prod_{\rho\in R_g} g'(\rho). \end{align*} $$

Since the map $R_f$ to $R_g$ defined by $x\mapsto x^n$ is n-to-one, we have $d_f = d_g^n$ . Further,

$$ \begin{align*} {\mathrm{Disc}}(g) = (-1)^{m(m-1)/2} \prod_{\rho\in R_g} g'(\rho) = (-1)^{m(m-1)/2} d_g. \end{align*} $$

We can therefore conclude that

$$ \begin{align*} (-1)^{nm(1-m)/2} {\mathrm{Disc}}(g)^n = d_g^n. \end{align*} $$

Using the chain rule, we see that $f'(x) = nx^{n-1}g'(x^n)$ . Therefore, the discriminant of $f(x)$ is

$$ \begin{align*} {\mathrm{Disc}}(f) &= (-1)^{nm(nm-1)/2} \prod_{\rho\in R_f} f'(\rho) = (-1)^{nm(nm-1)/2} \prod_{\rho\in R_f} (n\cdot \rho^{n-1}\cdot g'(\rho^n)) \\ &= (-1)^{nm(nm-1)/2} \bigg(\prod_{\rho\in R_f} n \bigg) \bigg(\prod_{\rho\in R_f} \rho^{n-1} \bigg) \bigg(\prod_{\rho\in R_f} g'(\rho^n) \bigg) \\ &= (-1)^{nm(nm-1)/2} \cdot n^{nm} \cdot (-1)^{nm(n-1)}\cdot c^{n-1} \cdot d_f \\ &= (-1)^{nm(nm-2n-3)/2} \cdot n^{nm} \cdot c^{n-1} \cdot d_g^n \\ &= (-1)^{nm(n-1)(m+2)/2} \cdot n^{nm} \cdot c^{n-1} \cdot {\mathrm{Disc}}(g)^n.\\[-2.7pc] \end{align*} $$

3 Notation and preliminary results

For the rest of this paper, we fix the following notation:

• • $f(x) = x^9+ax^6+bx^3+c \in \mathbb {Q}[x]$ , irreducible;

• • $g(x) = x^3+ax^2+bx+c$ ;

• • $\zeta $ a primitive cube root of unity (that is, a root of $x^2+x+1$ );

• • $S_n$ the symmetric group of degree n;

• • $A_n$ the alternating group of degree n;

• • $ {\mathrm{Gal}}(h)$ the Galois group of a polynomial $h(x)$ , where the base field will be clear from context.

It follows that the complex roots of $f(x)$ are

$$ \begin{align*}\{\alpha, \alpha\zeta, \alpha\zeta^2, \beta, \beta\zeta, \beta\zeta^2, \gamma, \gamma\zeta, \gamma\zeta^2\},\end{align*} $$

where $\alpha ^3, \beta ^3, \gamma ^3$ are the roots of $g(x)$ . From Lemma 2.1,

$$ \begin{align*} {\mathrm{Disc}}(f) = -3^9c^2{\mathrm{Disc}}(g)^3.\end{align*} $$

This leads to the following results.

We recall two facts about discriminants and Galois groups (see for example [Reference Dummit and Foote8, Section 14.6]).

1. (1) For any irreducible polynomial $\tilde {g}(x) \in \mathbb {Q}[x]$ of degree n, $ {\mathrm{Gal}}(\tilde {g})$ is isomorphic to a subgroup of $A_n$ if and only if $ {\mathrm{Disc}}(\tilde {g}) \in \mathbb {Q}^2$ .

2. (2) If F is any field of characteristic 0 and $\tilde {g}(x) \in F[x]$ is irreducible of degree 3, then $ {\mathrm{Gal}}(\tilde {g})$ over F is isomorphic to $A_3$ (cyclic of order 3) if $ {\mathrm{Disc}}({\tilde {g}}) \in F^2$ and is isomorphic to $S_3$ if $ {\mathrm{Disc}}(\tilde {g}) \notin F^2$ .

Combining item (2) above with Corollary 3.1 yields the following result.

Let L denote the splitting field of $f(x)$ over $\mathbb {Q}$ . Thus, $L = \mathbb {Q}(\alpha ,\beta ,\gamma ,\zeta )$ . Since ${[\mathbb {Q}(\alpha ) : \mathbb {Q}] = 9}$ and $[\mathbb {Q}(\zeta ):\mathbb {Q}]=2$ , we see that $[L:\mathbb {Q}]\geq 18$ . Let K denote the splitting field of $g(x)$ . Thus, $K = \mathbb {Q}(\alpha ^3,\sqrt { {\mathrm{Disc}}(g)})$ , and we have $\beta ^3,\gamma ^3\in K$ as well. We note further that $[K:\mathbb {Q}]\leq 6$ and, therefore, $[K(\zeta ):\mathbb {Q}] \leq 12$ . It follows that for each $\rho \in \{\alpha ,\beta ,\gamma \}$ , we have $[K(\rho ,\zeta ): K(\zeta )] \leq 3$ . We have therefore established the following result.

We turn our attention to the relative extension $\mathbb {Q}(\alpha )/\mathbb {Q}(\alpha ^3)$ . Factoring $f(x)$ over $\mathbb {Q}(\alpha ^3)$ , we obtain the factorisation

$$ \begin{align*} f(x) = (x^3-\alpha^3)(x^6+(a+\alpha^3)x^3+(\alpha^6+a\alpha^3+b),\end{align*} $$

which can be verified by expanding the factored expression and using the fact that $f(\alpha )=0$ so that $c = -\alpha ^9-a\alpha ^6-b\alpha ^3$ . It follows that $x^3-\alpha ^3$ is irreducible, for if it were not, then this would contradict the fact that $[\mathbb {Q}(\alpha ):\mathbb {Q}] = 9$ , thereby contradicting the irreducibility of $f(x)$ . It therefore also follows that $x^3-\alpha ^3$ defines the relative extension $\mathbb {Q}(\alpha )/\mathbb {Q}(\alpha ^3)$ . The following result will be useful in the determination of $ {\mathrm{Gal}}(f)$ .

4 Possible Galois groups

Since $f(x)$ is irreducible of degree 9, $\text {Gal}(f)$ can be realised as a transitive subgroup of $S_9$ , the symmetric group of degree 9; it is well defined up to conjugation as different orderings of the roots correspond to conjugate groups.

There are 34 such transitive subgroups, and they can be accessed with GAP [16] or at the L-functions and Modular Forms Database [17]. We use the standard ‘T-number’ notation to identify transitive groups as given in [Reference Butler and McKay5]. For example, 9T1 represents cyclic groups of order 9 and 9T34 represents $S_9$ . See Table 1, which gives several pieces of information about representatives of conjugacy classes of transitive subgroups of $S_9$ .

Table 1 Transitive subgroups of $S_9$ by T-number, order, parity and subfield information, as defined in Section 4.

Specifically, let G be a representative from one of these conjugacy classes. Then, the table gives the following information:

• • the T-number of G;

• • the order of G;

• • the parity of G; the parity is $+1$ if $G \leq A_9$ and $-1$ otherwise;

• • the subfields of G.

The subfields of G are identified as follows. Let $G_1$ denote the stabiliser of 1 in G. For each subgroup H of G of index 3 containing $G_1$ up to conjugacy, we compute the action of G on the cosets $G/H$ and the action of H on the cosets $H/G_1$ , and we identify each action as a transitive subgroup of $S_3$ . We are justified in calling this list the ‘subfields’ of G for the following reason: if $ {\mathrm{Gal}}(f)$ is isomorphic to G and $f(\alpha )=0$ , then $G_1$ corresponds to $\mathbb {Q}(\alpha )$ under the Galois correspondence and the nontrivial proper subfields $\mathbb {Q}(\alpha )$ correspond to the proper subgroups H properly containing $G_1$ ; conjugate subgroups correspond to isomorphic subfields. For each such subgroup H, let K denote its fixed field. Let the irreducible cubic polynomials $g(x)$ and $\tilde {g}(x)$ define $K/\mathbb {Q}$ and $\mathbb {Q}(\alpha )/K$ , respectively. Then, by the Galois correspondence, $ {\mathrm{Gal}}(g)$ and $ {\mathrm{Gal}}(\tilde {g})$ are isomorphic to the actions of G on $G/H$ and H on $H/G_1$ , respectively. Each entry in this column of the table is of the form $[i,j]$ , where $ {\mathrm{Gal}}(g)$ is isomorphic to 3Ti and $ {\mathrm{Gal}}(\tilde {g})$ is isomorphic to 3Tj; there is an entry for each such subgroup H (up to conjugacy). This is slightly more general than what is listed at [17], which only includes the transitive number for $ {\mathrm{Gal}}(g)$ for each subfield $K/\mathbb {Q}$ .

We note that if $ {\mathrm{Gal}}(f)$ is isomorphic to 9T4, then it is not clear immediately if $ {\mathrm{Gal}}(g)$ is isomorphic to 3T1 or 3T2. However, it follows from Lemma 3.4 that $ {\mathrm{Gal}}(g)$ must be 3T1. We formalise this in the following corollary.

We also note that each of the 11 groups appearing in Table 1 does indeed occur as a Galois group of some irreducible power compositional polynomial of degree 9 over $\mathbb {Q}$ ; see Table 2 for one such polynomial per group. Also in the table, we give standard descriptive names, such as $C_n$ for the cyclic group of order n, $D_n$ for the dihedral group of order $2n$ and $S_n$ for the symmetric group of degree n. We use $\times $ for direct products and : for semidirect products (that are not direct products).

Table 2 Sample irreducible power compositional nonic polynomials with specified Galois group over  $\mathbb {Q}$ .

5 Determining $\text {Gal}(f)$

In this section, we fix a generic ordering of the roots of $f(x)$ as defined in Table 3.

As the roots of $g(x)$ are $\{\alpha ^3,\beta ^3,\gamma ^3\}$ , we see that $\mathcal {B} = \{B_1,B_2,B_3\}$ forms a complete block system for $ {\mathrm{Gal}}(f)$ , where $B_1 = \{1,4,7\}$ , $B_2 = \{2,5,8\}$ and $B_3 = \{3,6,9\}$ . In other words, for each $\sigma \in {\mathrm{Gal}}(f)$ , we have $\sigma (B_i) \in \mathcal {B}$ .

By reordering the roots within each block if necessary, it follows that $ {\mathrm{Gal}}(f)$ is a subgroup of the permutation group $G \simeq \text {9T24}$ , where

(5.1) $$ \begin{align} G = \langle (1,5,9)(2,3,7,8,6,4), (2,6,5,9,8,3)\rangle. \end{align} $$

Consider the subgroup $H \simeq \text {9T8}$ of G, where

(5.2) $$ \begin{align} H = \langle (1,3,2)(4,9,5,7,6,8), (1,4,7)(2,6,8,3,5,9)\rangle. \end{align} $$

Consider also the multivariable function $T(x_1, \ldots , x_9) = (x_1+x_2+x_3)^3$ . Letting each $\sigma \in G$ act on T via subscripts, we see that the stabiliser of T inside G is H; this straightforward computation can be carried out with [16], for example.

Let $r(x)$ be the resolvent polynomial corresponding to G, H and T, according to [Reference Cohen6, Definition 6.3.2]. More concretely, we can specify the roots of $r(x)$ in terms of the roots of $f(x)$ as follows. A group computation shows that a complete set of right coset representatives for $G/H$ is: $\text {Id}$ , $(3,6,9)$ , $(3,9,6)$ , $(2,3,5,6,8,9)$ , $(2,5,8)(3,6,9)$ , $(2,5,8)(3,9,6)$ , $(2,3,8,9,5,6)$ , $(2,6)(3,8)(5,9)$ and $(2,8,5)(3,9,6)$ . Letting each of these coset representatives act on T via subscripts and then evaluating each image of T at the roots of $f(x)$ as specified in Table 3, we see that the roots of $r(x)$ are of the form $(\alpha +\beta \zeta ^i+\gamma \zeta ^j)^3$ for $0 \leq i,j \leq 2$ . We can expand $r(x)$ and express its coefficients as elementary symmetric polynomials in the roots of $f(x)$ . Doing so leads us to the following definition.

Definition 5.1. Let $f(x) = x^9 +ax^6+bx^3+c \in \mathbb {Q}[x]$ be irreducible and let ${r(x) = x^9 + \sum _{i=0}^8 a_ix^i}$ , where

$$ \begin{align*} a_8 &= -9a;\\ a_7 &= 36a^2 - 81b;\\ a_6 &= -84a^3 + 486ab - 4293c;\\ a_5 &= 9(14a^4 - 135a^2b + 243b^2 + 189ac);\\ a_4 &= -9(14a^5 - 180a^3b + 729ab^2 - 1836a^2c + 4860bc);\\ a_3 &= -2673c(19a^3 - 81ab)(28a^6 - 405a^4b + 2187a^2b^2 - 6561b^3) + 61236c^2;\\ a_2 &= -9(4a^7 - 54a^5b + 243a^3b^2 + 108a^4c + 1458a^2bc - 6561b^2c + 13851ac^2);\\ a_1 &= 9(a^8 - 9a^6b + 432a^5c - 1701a^3bc + 7290a^2c^2 - 6561bc^2);\\ a_0 &= -(a^3 - 27c)^3. \end{align*} $$

We have the following result about $r(x)$ .

Proof. Item (2) follows from item (1) and a group computation, since the degrees of the irreducible factors of $r(x)$ correspond to the orbit lengths of the action of $ {\mathrm{Gal}}(f)$ on the cosets $G/H$ ; which in turn follows from a general result about irreducible factors of resolvent polynomials (see for example [Reference Cohen6, Theorem 6.3.3]).

Table 3 Generic ordering of the roots of $f(x) = x^9+ax^6+bx^3+c$ .

Table 4 Possible Galois groups of $f(x) = g(x^3)$ , where $g(x)=x^3+ax^2+bx+c$ . For each possible $G = {\mathrm{Gal}}(f)$ , we include the T-number of G, the parity of G, the T-number of $ {\mathrm{Gal}}(g)$ , the degrees of the irreducible factors of $r(x)$ and the degrees of the irreducible factors of $s(x)$ when $ {\mathrm{Gal}}(f)$ is either 9T10 or 9T21. The polynomials $r(x)$ and $s(x)$ are given in Definitions 5.1 and 5.3, respectively.

To prove item (1), we verify that $(\alpha +\beta \zeta ^i+\gamma \zeta ^j)^3$ is not equal to $(\alpha +\beta \zeta ^k+\gamma \zeta ^l)^3$ except when $(i,j)=(k,l)$ . This is equivalent to showing $\beta \zeta ^i+\gamma \zeta ^j$ is not equal to ${\beta \zeta ^k+\gamma \zeta ^l}$ for $(i,j)\neq (k,l)$ and $j,k,l\in \{0,1,2\}$ .

If $i=k$ (and $j\neq l)$ , then $\gamma =0$ ; this contradicts the irreducibility of $f(x)$ . We reach a similar contradiction if $j=l$ (and $i\neq k$ ). By dividing both expressions by $\zeta ^i$ , we may also assume $i=0$ . This leaves 12 cases to analyse; namely $j\in \{0,1,2\}$ , $k\in \{1,2\}$ and $l\in \{0,1,2\}\setminus \{j\}$ .

Suppose $(j,k,l) = (0,1,1)$ so that $\beta +\gamma = \beta \zeta + \gamma \zeta $ . In this case, $\beta = -\gamma $ , which implies that $\beta ^3 = -\gamma ^3$ . This in turn implies $-\alpha ^3 = a$ is a rational root of $g(x)$ , indicating that $g(x)$ is reducible, contradicting the irreducibility of $f(x)$ . Similar reasoning also applies to the cases $(j,k,l) \in \{(0,2,2), (1,1,2), (1,2,0), (2,1,0), (2,2,1)\}$ .

Suppose $(j,k,l) = (0,1,2)$ so that $\beta +\gamma = \beta \zeta + \gamma \zeta ^2$ . Thus, $\beta (1-\zeta ) = \gamma (\zeta ^2-1)$ . This implies $\beta = \gamma \zeta ^2$ , and thus $f(x)$ is not separable and therefore reducible. Similar reasoning also applies to the cases $(j,k,l) \in \{(0,2,1), (1,1,0), (1,2,2), (2,1,1), (2,2,0)\}$ .

An inspection of Table 4 shows that $ {\mathrm{Gal}}(f)$ is uniquely determined in all cases except for 9T10 versus 9T21 by considering: (1) whether $ {\mathrm{Disc}}(f)$ is a square in $\mathbb {Q}$ ; (2) whether $ {\mathrm{Disc}}(g)$ is a square in $\mathbb {Q}$ ; and (3) the degrees of the irreducible factors of $r(x)$ .

To determine whether $ {\mathrm{Gal}}(f)$ is 9T10 or 9T21, we use another resolvent polynomial, which we define next.

Definition 5.3. Let $f(x) = x^9 +ax^6+bx^3+c \in \mathbb {Q}[x]$ be irreducible and $r(x)$ be as in Definition 5.1. Let $s(x)$ be defined by

$$ \begin{align*} s(x)^2=\frac{\text{Resultant}_y(r(y),r(x-y))}{2^9\cdot r(x/2)}. \end{align*} $$

Thus, $s(x)$ is the polynomial whose roots are sums of the form $\rho _i+\rho _j$ for $i<j$ , where $\rho _1, \ldots , \rho _9$ are the roots of $r(x)$ .

We note that resultants can be computed via [Reference Cohen6, Algorithm 3.3.7]. The following result completes our classification.

Using Table 4, we can summarise the characterisation of $ {\mathrm{Gal}}(f)$ as follows.

6 Examples

In this section, we give several examples that apply Theorem 5.5 to compute Galois groups of power compositional nonic polynomials. Our first example recovers [Reference Jones and Phillips11, Theorem 1.1].

Proof. Factoring $r(x)$ gives three irreducible factors of degree 3:

$$ \begin{align*} &x^3 + 27mx^2 + 7047m^2x - 107811m^3, \\ &x^3 - 135mx^2 + 2187m^2x - 10125m^3, \\ &x^3 + 27mx^2 + 243m^2x + 81m^3. \end{align*} $$

By Theorem 5.5, this shows $ {\mathrm{Gal}}(f) = \text {9T3} \simeq D_9$ .

Table 5 gives one-parameter families of polynomials for each of the 11 possible Galois groups of $f(x)$ , assuming the resulting polynomial is irreducible when specialised at a particular rational value of the parameter. Here are two additional examples that illustrate the correctness of Table 5.

Table 5 One parameter families of polynomials of the form $f(x)=x^9+ax^6+bx^3+c \in \mathbb {Q}[x]$ with prescribed Galois group when $f(x)$ is irreducible.

Proof. Factoring $r(x)$ gives

$$ \begin{align*} r(x) = x(x^2 - 27tx + 189t^2)(x^6 + 459t^2x^4 + 7290t^4x^2 + (27t^2)^3). \end{align*} $$

Thus, $R = [1,2,6]$ . It follows from Theorem 5.5 that $ {\mathrm{Gal}}(f)$ is either 9T4 or 9T8. We have $g(x) = x^3+3tx^2-4t^2x+t^3$ and $ {\mathrm{Disc}}(g) = (7t^3)^2$ which is a square. Thus, $ {\mathrm{Gal}}(f) = \text {9T4} \simeq C_3\times S_3$ .

Proof. Using Mathematica, we see that $r(x)$ is irreducible over $\mathbb {Q}(t)$ . Thus, we assume that $t\in \mathbb {Q}$ is chosen so that $r(x)$ is irreducible. We also see that

$$ \begin{align*} {\mathrm{Disc}}(f) = (3^9t^4(t-1)^3(t^3 - 3t^2 - 24t - 1)^3)^2, \end{align*} $$

which is a perfect square. By Theorem 5.5, we see that $ {\mathrm{Gal}}(f)$ is either 9T10 or 9T21. Forming $s(x)$ as given in Definition 5.3 and factoring it, we see it has three factors:

$$ \begin{align*} &x^9 - 162tx^8 + (8019t^2 + 729t)x^7 + \cdots; \\ &x^9 - 162tx^8 + (10206t^2 + 729t)x^7 + \cdots; \\ &x^{18} + 324tx^{17} + \cdots. \end{align*} $$

Assuming t is chosen so that $s(x)$ is separable, we see that $S = [9,9,18]$ . Thus, $ {\mathrm{Gal}}(f) = \text {9T10} \simeq C_9:C_6$ .

7 Field extensions defined by polynomials of the form $g(x^3)$

In this section, we are interested in determining when a field extension $L/\mathbb {Q}$ of degree n can be defined by an irreducible polynomial $f(x)=g(x^3) \in \mathbb {Q}[x]$ . We make use of the following previous results.

Theorem 7.1 [Reference Awtrey, Beuerle and Griesbach1, Proposition 1.2].

Suppose $L/\mathbb {Q}$ is an extension of degree $mk$ . Then, L can be defined by an irreducible polynomial $g(x^k)\in \mathbb {Q}[x]$ if and only if L has a subfield K of degree m such that $L/K$ is defined by an irreducible polynomial $x^k-a\in K[x]$ .

Moreover, if $\tilde {g}(x)\in \mathbb {Q}[x]$ is irreducible of degree m defining K and $h(x)=x^k-a\in K[x]$ is irreducible defining $L/K$ , then $f(x)=g(x^k)$ defines $L/\mathbb {Q}$ , where

$$ \begin{align*} f(x) = \text{ Resultant}_y(x^k-a,\tilde{g}(y)).\end{align*} $$

We can guarantee that $f(x)$ is irreducible by replacing a by $ab^k$ for some $b\in K$ , if necessary (see for example [Reference Dixon7, Appendix A]).

As a consequence of Theorems 7.1 and 7.2, the following algorithm is guaranteed to produce an irreducible power compositional polynomial $g(x^3)$ defining $L/\mathbb {Q}$ , when it exists.

We have implemented [18, Algorithm 7.3] and created a web-interface [Reference Awtrey, Patane and Toone4], where a user can enter a polynomial $f(x)\in \mathbb {Q}[x]$ . The website returns a power compositional $g(x^3)$ that defines an extension isomorphic to the extension defined by $f(x)$ , when such a power compositional polynomial exists.

Example 7.4. To illustrate Algorithm 7.3, consider the irreducible polynomial in $\mathbb {Q}[x]$ :

$$ \begin{align*} f(x) = x^{18} &- 6x^{17} + 19x^{16} - 35x^{15} + 41x^{14} \\ &- 29x^{13} + 9x^{12} + 14x^{11} - 35x^{10} + 34x^9 - 9x^8 \\ & - 6x^7 + 4x^6 - 15x^5 + 12x^4 + 8x^3 - 6x^2 - x + 1. \end{align*} $$

Let $f(\alpha )=0$ and $L = \mathbb {Q}(\alpha )$ .

1. (1) Using [18], we see that the irreducible polynomial $g(x) = x^6 + 7x^5 + 18x^4 + 18x^3 + 2x^2 - 4x + 1$ defines a subfield K of L of degree 6.

2. (2) Let $g(\beta )=0$ . Factoring $f(x)$ over K and extracting an irreducible cubic factor, we obtain the polynomial $h(x) = x^3+h_2x^2+h_1x-\beta $ , where

   $$ \begin{align*} h_2 = \beta^5 + 6\beta^4 + 12\beta^3 + 8\beta^2 + \beta; \quad h_1 = -\beta^5 - 5\beta^4 - 7\beta^3 + 3\beta - 1. \end{align*} $$

3. (3) Using Theorem 7.2, we obtain

   $$ \begin{align*} \tilde{h}(x) = x^3 + (-9\beta^5 - 42\beta^4 - 60\beta^3 - 12\beta^2 - 9\beta - 3)\in K[x], \end{align*} $$

   which is irreducible over K and also defines $L/K$ .

4. (4) Computing $\text {Resultant}_\beta (\tilde {h}(x),g(\beta ))$ , we obtain

   $$ \begin{align*}x^{18} + 108x^{15} + 3834x^{12} + 43011x^9 - 107892x^6 - 2755620x^3 + 24111675, \end{align*} $$

   which is a power compositional polynomial and irreducible over $\mathbb {Q}.$