Brian Bi

## Section 16.12. Quintic Equations

Exercise 16.12.1 Yes. Denote the extension by $$K/F$$. According to Exercise 7.8.4, $$\Gal(K/F)$$ is isomorphic to either $$C_{10}$$ or $$D_5$$. In both cases, there is a normal subgroup of order 5; by the fundamental theorem, this corresponds to an intermediate field $$L$$ such that $$[K : L] = 5$$, and the extension $$L/F$$ is a Galois extension of degree 2. The chain $$F \subseteq L \subseteq K$$ therefore satisfies condition 16.12.2(b).

Exercise 16.12.2 Let $$G$$ be a transitive subgroup of $$S_5$$. By the orbit-stabilizer theorem, $$5 \divides |G|$$. By the first Sylow theorem, $$G$$ contains an element of order 5, that is, a 5-cycle. Without loss of generality, say $$G$$ contains $$\sigma = (1\ 2\ 3\ 4\ 5)$$.

If $$G \subseteq A_5$$, then by Theorem 6.12.1, $$G$$ is cyclic or dihedral, or is isomorphic to one of $$A_4$$, $$S_4$$, or $$A_5$$. However, $$5 \not\divides |A_4|$$, and $$|S_4| \not\divides |A_5|$$, so we can eliminate two possibilities in that fashion. Again, since $$5 \divides |G|$$, and all elements of $$A_5$$ have order 1, 2, 3, or 5, the only cyclic possibility is $$C_5$$. And since $$D_n$$ always contains $$C_n$$ as a subgroup, the only dihedral possibility is $$D_5$$. Both may be realized as transitive subgroups of $$S_5$$ by their usual action on the vertices of a pentagon.

Otherwise, half of the elements of $$G$$ are even permutations and half are odd, and the even ones form a subgroup of $$A_5$$. Thus, $$G$$ can have order 10, 20, or 120. Obviously if $$|G| = 120$$ then $$G = S_5$$. If $$|G| = 10$$ then by Exercise 7.8.4, $$G$$ must be either cyclic or dihedral, but $$S_5$$ contains no elements of order 10, so $$G$$ must be dihedral, and this possibility was noted previously. If $$|G| = 20$$, by Exercise 7.7.9(c), $$G$$ belongs to one of five possible isomorphism classes, but all of these contain an element of order 10 (and thus cannot be embedded in $$S_5$$) except for $$GA(1, 5)$$, which can be generated in $$S_5$$ by $$\sigma$$ and $$\tau = (2\ 3\ 5\ 4)$$.

We thus conclude that there are five isomorphism classes of transitive subgroups of $$S_5$$, namely $$C_5, D_5, GA(1, 5), A_5, S_5$$.

Exercise 16.12.3 The desired result follows immediately from Exercise 16.12.2 and Lagrange's theorem.

Exercise 16.12.4

1. The field $$F(u)$$ is generated by $$u_1, \ldots, u_n$$, which are the roots of the polynomial $$(x - u_1) \cdot \ldots \cdot (x - u_n)$$, whose coefficients are $$1, s_1, s_2, \ldots, s_n$$ up to sign. Therefore $$F(u)$$ is the splitting field of a polynomial with coefficients in $$F(s_1, \ldots, s_n)$$, hence a Galois extension of the latter field.

2. Consider the tower of fields $$K_0 \subseteq K_1 \subseteq \ldots \subseteq K_n$$, where $$K_i = F(s_1, \ldots, s_n)(u_1, \ldots, u_i)$$. Thus, $$K_0 = F(s_1, \ldots, s_n)$$, and $$K_n = F(u)$$. Since $$K_n/K_0$$ is generated by $$u_1, \ldots, u_n$$, its Galois group is a subgroup of $$S_n$$. Now consider one of the extensions $$K_{i+1}/K_i$$ where $$i \in \{0, 1, \ldots, n-1\}$$. The minimal polynomial $$P(x)$$ for $$u_{i+1}$$ over $$K_i$$ must have the factor $$(x - u_{i+1})$$ in $$K_{i+1}$$. Writing $$P(x) = x^m + a_{m-1}x^{m-1} + \ldots + a_0$$, observe that since $$a_0, \ldots, a_{m-1} \in K_i$$, it follows that they are functions of $$s_1, \ldots, s_n, u_1, \ldots, u_i$$, and are therefore invariant under permutations of the $$u_{i+1}, \ldots, u_n$$. Therefore, writing $$u_{i+1}^m + a_{m-1} u_{i+1}^{m-1} + \ldots + a_0 = 0$$ and transposing $$u_{i+1}$$ with $$u_j$$ ($$j > i+1$$), we obtain $$u_j^m + a_{m-1} u_j^{m-1} + \ldots + a_0 = 0$$, so $$x - u_j$$ also divides $$P(x)$$. Thus, $$(x - u_{i+1})(x - u_{i+2}) \cdot \ldots \cdot (x - u_n) \divides P(x)$$, and the degree of $$u_{i+1}$$ over $$K_i$$ is at least $$n - i$$. Thus $$[K_n : K_0] \ge n \cdot (n-1) \cdot \ldots \cdot 2 = n!$$. But $$[K_n : K_0] \le n!$$ since $$\Gal(K_n/K_0) \subseteq S_n$$. Therefore $$\Gal(K_n/K_0) \cong S_n$$.

3. $$\Gal(F(u)/F(s,w))$$ is the subgroup of $$\Gal(F(u)/F(s))$$ fixing $$w$$, that is, the stabilizer of $$w$$ under the action of $$S_5$$ permuting the variables $$u_1, \ldots, u_5$$. It is clearly a transitive subgroup. The orbit of $$w$$ is the number of Hamiltonian cycles of the complete graph $$K_5$$, which is $$4 \cdot 3 \cdot 2 / 2$$ as we have 4 choices of neighbour for $$u_1$$, 3 choices of neighbour for $$u_2$$, and so on, and we traverse each possible cycle in both directions with this counting scheme. Thus the orbit has cardinality 12, and the stabilizer has cardinality $$5!/12 = 20$$. By Exercise 16.12.2, the Galois group is $$GA(1, 5)$$.

4. By Cayley's theorem, there exists $$n$$ such that $$G$$ can be embedded in $$S_n$$. Consider the Galois extension $$F(u_1, \ldots, u_n)/F(s_1, \ldots, s_n)$$ with Galois group $$S_n$$, as we found in part (a). Let $$L$$ be the subfield of $$F(u)$$ fixed by $$G$$, regarded as a subgroup of $$S_n$$. Then, by the fundamental theorem, $$F(u)/L$$ is a Galois extension with Galois group $$G$$.

Exercise 16.12.5 We will prove a more general claim: Suppose $$F$$ and $$K$$ are fields with $$\Q \subseteq F \subseteq K \subseteq \R$$, suppose all elements of $$F$$ are constructible, and suppose that $$[K : F] = 2^n$$ where $$n$$ is a nonnegative integer. Then, all elements of $$K$$ are constructible.

The proof is by induction on $$n$$. If $$n = 0$$, then $$K = F$$ and we are done. If $$n = 1$$, then $$[K : F] = 2$$ and Theorem 15.5.10 gives the desired result. Otherwise, $$n \ge 2$$. Let $$G = \Gal(K/F)$$, a 2-group. If $$G$$ is abelian, then let $$N$$ be a subgroup of order 2; $$N$$ is automatically normal since $$G$$ is abelian. Otherwise, let $$N = Z(G)$$, a normal subgroup. By Proposition 7.3.1, $$N$$ is nontrivial, but since $$G$$ isn't abelian, $$N$$ is a proper subgroup of $$G$$. Let $$L = K^N$$. By the fundamental theorem, $$\Gal(K/L) = N$$ and $$L/F$$ is also a Galois extension with Galois group $$G/N$$. Both of these groups $$N$$ and $$G/N$$ are 2-groups with order strictly less than $$2^n$$. Applying the induction hypothesis to $$L/F$$, we see that all elements of $$L$$ are constructible, and applying it again to $$K/L$$, we see that all elements of $$K$$ are constructible.

The result that the Exercise asks for is the special case of the above result, with $$F = \Q$$.

Exercise 16.12.6 Let us first assume that $$f$$ is irreducible. This removes an ambiguity in the statement of the exercise, as criterion 16.12.2(b) implies that either all roots of $$f$$ are solvable or no roots of $$f$$ are solvable.

The proof of Theorem 16.12.4 uses three properties of the group $$A_5$$: it is simple, it is not cyclic, and it contains an element of order 5. When $$A_5$$ is replaced by an arbitrary nonabelian simple group $$G$$, the first two requirements continue to hold. The third assumption is unnecessary in the current setting—the first two properties imply that any tower of the form 16.12.2(b) terminates in a field $$K$$ over which the Galois group of $$f$$ is still $$G$$, therefore $$f$$ does not split completely over $$K$$, therefore (Theorem 16.3.2) $$f$$ does not have any root in $$K$$, therefore no root of $$f$$ is solvable.

If the assumption that $$f$$ is irreducible is dropped, the statement must be weakened slightly: some roots of $$f$$ may be solvable (after all, we can take any irreducible $$f$$ and multiply it by $$x$$ without changing its Galois group) but at least one root is not solvable. To see this, let $$K$$ is the splitting field of $$f$$ and let $$\alpha \in \C$$ be a primitive element for the extension $$K/\Q$$. Let $$g$$ be the minimal polynomial for $$\alpha$$. Since $$g$$ is irreducible, we can apply the result we derived above for irreducible polynomials; $$g$$ has no solvable root, and in particular $$\alpha$$ is not solvable. Therefore, at least some elements of $$K$$ are not solvable, which implies that the roots of $$f$$ (which generate $$K$$ over $$\Q$$) cannot all be solvable.

Exercise 16.12.7 We can follow Example 16.12.7 here, using Exercise 16.12.8 (the generalization of Lemma 16.12.5, which we will prove later) and the resulting generalization of Corollary 16.12.6: if an irreducible polynomial of prime degree $$p$$ has exactly $$p - 2$$ real roots, then its Galois group is $$S_p$$. We may begin with the polynomial $$x(x^2 - 4)(x^2 + 4)(x^2 - 16) = x^7 - 16x^5 - 16x^3 + 256x$$, which has 5 real roots and 2 imaginary roots, and add a small constant term to obtain $$x^7 - 16x^5 - 16x^3 + 256x + 2$$, which is 2-Eisenstein, hence irreducible, and still has exactly 5 real roots.

Exercise 16.12.8 Without loss of generality, we may assume that the $$p$$-cycle takes the form $$\sigma = (1\ 2 \ldots p)$$. Let $$\tau$$ denote the transposition. We want to show that $$\langle \sigma, \tau\rangle = S_p$$. Repeated conjugation by $$\sigma$$ shifts the elements in $$\tau$$; thus, without loss of generality, we can assume $$\tau = (1\ x)$$, with $$x \ne 1$$. Since $$p$$ is prime, $$\sigma^{x-1}$$ is also a $$p$$-cycle, which begins $$(1\ x \ldots)$$. These two elements $$(1\ x)$$ and $$(1\ x\ldots)$$ both belong to $$\langle \sigma, \tau\rangle$$. Now let $$\sigma'$$ be a permutation that sends $$x \mapsto 2$$ and so on, so that $$\sigma'(1\ x)\sigma'^{-1} = (1\ 2)$$ and $$\sigma'(1\ x\ldots) \sigma'^{-1} = (1\ 2\ldots p)$$. The conjugate subgroup $$\sigma'\langle \sigma, \tau\rangle\sigma'^{-1}$$ thus contains $$(1\ 2)$$ and $$\sigma = (1\ 2 \ldots p)$$; it suffices to show that this conjugate subgroup is all of $$S_p$$. Repeated conjugation of $$(1\ 2)$$ by $$\sigma$$ can be used to generate all adjacent transpositions, $$(2\ 3), \ldots, (p-1\ p)$$. By Exercise 7.5.1(a), these adjacent transpositions generate $$S_p$$.