Brian Bi
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## Miscellaneous problems for chapter 16

Problem 16.M.1 The Galois extensions $$K_2/F_2$$ and $$K_1/F_1$$ are both generated by the roots of $$f$$ (regarded as elements of $$K_2$$) and an element of either Galois group is completely determined by the permutation it induces on those roots. Thus, if $$\sigma \in \Gal(K_1/F_1)$$, there is at most one $$\sigma' \in \Gal(K_2/F_2)$$ such that $$\sigma'$$ and $$\sigma$$ agree on $$K_1$$. However, such $$\sigma'$$ does not necessary exist, since $$\sigma$$ fixes $$F_1$$ but might not fix the larger field $$F_2$$. On the other hand, if $$\tau \in \Gal(K_2/F_2)$$ then, since $$\tau$$ sends each root of $$f$$ to a root of $$f$$, $$\tau$$ sends each element of the field generated by the roots of $$f$$, namely $$K_1$$, to another element of $$K_1$$; and since $$\tau$$ fixes $$F_2$$, it also fixes the smaller field $$F_1$$. Therefore, all $$F_2$$-automorphisms of $$K_2$$ can be restricted to $$F_1$$-automorphisms of $$K_1$$, but not all $$F_1$$-automorphisms of $$K_1$$ can be extended to $$F_2$$-automorphisms of $$K_2$$. Thus $$\Gal(K_2/F_2)$$ is a subgroup of $$\Gal(K_1/F_1)$$.

Problem 16.M.2 No. $$\Q(\sqrt[4]{2})$$ is a Galois extension of $$\Q(\sqrt{2})$$ and $$\Q(\sqrt{2})$$ is a Galois extension of $$\Q$$, but $$\Q(\sqrt[4]{2})$$ is not a Galois extension of $$\Q$$.

Problem 16.M.3 Let $$F$$ be a field of characteristic 0.

Suppose $$P \in F[u_1, \ldots, u_n]$$ has the property that it vanishes whenever any two of the $$u$$'s are equal. Then $$\delta \divides P$$, where $$\delta = \prod_{i < j} (u_i - u_j)$$.

If $$n = 1$$ then $$\delta = 1$$, and there is nothing to prove. Otherwise, perform polynomial division of $$P$$ by $$u_1 - u_2$$ in the variable $$u_1$$, giving $$P = (u_1 - u_2)Q + R$$ where $$R$$ depends only on $$u_2, \ldots, u_n$$. For all $$(x_2, \ldots, x_n) \in F^{n-1}$$, evaluating both sides with $$u_1 = u_2 = x_2, u_3 = x_3, \ldots, u_n = x_n$$ yields $$0 = 0 + R$$; therefore $$R$$ vanishes for all possible $$(n-1)$$-tuples of arguments, and must be the zero polynomial. Thus, $$u_1 - u_2 \divides P$$. Similar reasoning with each possible pair of indices $$i < j$$ shows that $$u_i - u_j \divides P$$ for all such pairs. But $$F[u_1, \ldots, u_n]$$ is a UFD and the elements $$u_i - u_j$$ are irreducible, so the factorization of $$P$$ contains at least one copy of each $$u_i - u_j$$. Therefore $$\delta \divides P$$.

We now return to the problem. The Vandermonde determinant is a polynomial in the variables $$u_1, \ldots, u_n$$, and vanishes when any pair of the $$u$$'s are equal, therefore it is a multiple of $$\delta$$. However, $$\delta$$ is homogeneous with degree $$n(n-1)/2$$, and by the Leibniz determinant formula, so is the Vandermonde determinant. Therefore, the latter is a constant multiple of the former. To determine the constant, observe that the coefficient of $$u_1^{n-1} u_2^{n-2} \cdot \ldots \cdot u_{n-1}$$ is 1 in $$\delta$$, and in the Vandermonde determinant it is the sign of the permutation $$(n\ n-1 \ldots 2\ 1)$$, which contains $$n(n-1)/2$$ inversions. Therefore the required constant factor is $$(-1)^{n(n-1)/2}$$.

Problem 16.M.4

1. Let $$\sigma$$ be an automorphism of the reals. Clearly $$\sigma$$ fixes the rationals. Furthermore, $$\sigma$$ is sign-preserving: if $$x \ge 0$$, then $$\sigma(x) = \sigma((\sqrt{x})^2) = \sigma(\sqrt{x})^2$$, so $$\sigma(x) \ge 0$$ as well.

Suppose that $$\sigma(x) = y$$ where $$x \ne y$$. Let $$q$$ be a rational number between $$x$$ and $$y$$, so $$\sigma(q) = q$$. Then $$\sigma(x - q) = \sigma(x) - \sigma(q) = y - q$$, but $$x - q$$ and $$y - q$$ have opposite signs, so a contradiction has been reached.

2. Let $$\sigma$$ be a continuous automorphism on $$\C$$. Since $$\sigma(i)^2 = \sigma(i^2) = \sigma(-1) = -1$$, $$\sigma(i)$$ is either $$i$$ or $$-i$$. If $$\sigma(i) = i$$, then $$\sigma$$ is the identity on $$\Q(i)$$. Otherwise, $$\sigma$$ is complex conjugation on $$\Q(i)$$. Since $$\Q(i)$$ is dense in $$\C$$, it follows that $$\sigma$$ is either the identity or complex conjugation on all of $$\C$$.

Problem 16.M.5

1. Presumably the wording of the exercise is wrong, and we are meant to show that the Frobenius map is an $$F$$-automorphism of $$K$$.

The fact that $$\varphi$$ is the identity on $$F$$ follows from Fermat's little theorem. Also, $$\varphi(x) + \varphi(y) = x^p + y^p = (x + y)^p = \varphi(x + y)$$ in $$K$$ since $$K$$ has characterstic $$p$$, and $$\varphi(x)\varphi(y) = x^p y^p = (xy)^p = \varphi(xy)$$, and obviously $$\varphi(1) = 1$$, so $$\varphi$$ is a field endomorphism of $$K$$. Finally, since $$p$$th roots are unique in characteristic $$p$$, $$\varphi$$ is injective, and as $$K$$ is finite, must be surjective as well. Therefore $$\varphi$$ is an $$F$$-automorphism of $$K$$.

2. By Theorem 15.7.3(c), the multiplicative group $$K^\times$$ is a cyclic group of order $$p^r - 1$$. Let $$x$$ be an element of $$K$$ that has order $$p^r - 1$$. Then the $$n$$-fold iterate of $$\varphi$$ on $$x$$, $$\varphi^n(x)$$, is equal to $$x^{p^n}$$, which will be equal to $$x$$ if and only if $$p^n \equiv 1 \pmod{p^r - 1}$$. Clearly, the smallest positive $$n$$ for which this occurs is $$n = r$$. Therefore $$\varphi^0, \varphi^1, \varphi^2, \ldots, \varphi^{r - 1}$$ are all distinct elements of $$\Gal(K/F)$$. Finally, by Theorem 16.6.2, $$|\Gal(K/F)|$$ divides $$[K : F]$$, which is $$r$$, so $$|\Gal(K/F)|$$ is exactly $$r$$, and the iterates of $$\varphi$$ are all the elements of $$\Gal(K/F)$$.

3. Since the order of $$\Gal(K/F)$$ equals the degree of the extension $$K/F$$, the extension is Galois, so the conditions are satisfied for the fundamental theorem. However, Artin probably wants us to demonstrate explicitly that the correspondence holds in this case. Since $$\Gal(K/F) \cong C_r$$, it has $$d(r)$$ subgroups (where $$d(r)$$ is the number of positive integer divisors of $$r$$). By Theorem 15.7.3(e), there are also $$d(r)$$ intermediate fields including $$F$$ and $$K$$ themselves, namely $$\mathbb{F}_{p^i}$$ for each positive $$i$$ that divides $$r$$. Verification of the remaining properties of the correspondence are left as an exercise for the reader.

Problem 16.M.6

1. There are 4 automorphisms $$\sigma_1, \ldots, \sigma_4$$, where $$\sigma_i(\zeta^j) = \zeta^{ij}$$ for each $$i, j$$. $$\sigma_1$$ is the identity and leaves the pentagon invariant. $$\sigma_4$$ maps each vertex to its complex conjugate, so it yields the same pentagon but with opposite orientation. $$\sigma_2$$ maps the pentagon to the polygon obtained by tracing $$1, \zeta^2, \zeta^4, \zeta, \zeta^3, 1$$, which is a pentagram, and $$\sigma_3$$ produces the same pentagram with opposite orientation. So the orbit of the pentagon consists only of the pentagon and the inscribed pentagram.

2. The field $$\Q(\zeta)$$ is closed under complex conjugation. The square of the length of the line segment between two vertices of the pentagon, located at $$1$$ and $$\zeta$$, is the squared magnitude of their difference, that is, $$(\zeta - 1)\overline{\zeta - 1}$$. This is an element of $$\Q(\zeta)$$. Explicitly, it is given by the Law of Cosines as $$\alpha^2 = 2(1 - \cos 2\pi/5) = \frac{5 - \sqrt{5}}{2}$$, from which the relation $$(\alpha^2)^2 - 5\alpha^2 + 5 = 0$$ is easily obtained. This is irreducible by the Eisenstein criterion, so it is the minimal polynomial for $$\alpha$$.

Problem 16.M.7

1. It suffices to show that $$\delta$$ changes sign under a transposition $$\sigma = (a\ b)$$, where without loss of generality we take $$a < b$$. Under such a transposition, factors of the form $$u_i - u_j$$ where $$i, j \notin \{a, b\}$$ in $$\delta$$ remain unchanged. For $$i < a$$, the factors $$(u_i - u_a)$$ and $$(u_i - u_b)$$ are exchanged; this exchange does not change the overall product. Likewise for $$i > b$$ the factors $$(u_a - u_i)$$ and $$(u_b - u_i)$$ are exchanged. Finally, for $$i$$ between $$a$$ and $$b$$, the factors $$(u_a - u_i)$$ and $$(u_i - u_b)$$ become $$(u_b - u_i)$$ and $$(u_i - u_a)$$, but $$(u_b - u_i)(u_i - u_a) = (u_a - u_i)(u_i - u_b)$$; that is, the two factors are exchanged and both change sign. This leaves only a single unpaired sign change from $$(u_a - u_b)$$, which becomes $$(u_b - u_a)$$. So $$\delta$$ changes sign, as required.

2. We begin with a simple lemma that we will need at the end:

Let $$F$$ be a field of characteristic 0. Let $$P$$ be a polynomial in $$n$$ variables $$u_1, \ldots, u_n \in F$$ such that $$P(t_1, \ldots, t_n) = 0$$ whenever $$t_1, \ldots, t_n$$ are pairwise distinct. Then $$P$$ vanishes for all arguments.

We prove the claim by induction in the number of distinct values $$k$$ taken on by $$t_1, \ldots, t_n$$. The base case is $$k = n$$, that is, all values distinct; the claim then holds by assumption. For smaller values of $$k$$ we proceed by induction. Fix $$t_1, \ldots, t_n$$. Since $$k < n$$, there exists some $$i$$ with $$1 \le i \le n$$ such that $$t_i$$ is not unique among the $$t$$'s. Let $$Q$$ be a polynomial in a single variable, defined by $$Q(u) = F(t_1, \ldots, t_{i-1}, u, t_{i+1}, \ldots, t_n)$$. For all but finitely many values of $$u$$, there are $$k + 1$$ distinct values among the arguments to $$F$$, so by the induction hypothesis, $$Q$$ vanishes for those values of $$u$$. Since $$Q$$ vanishes for infinitely many values of $$u$$, $$Q$$ vanishes identically. Therefore $$Q(t_i) = F(t_1, \ldots, t_n) = 0$$. This proves the claim.

We now return to the problem. Let $$P$$ be a $$\frac{1}{2}$$-symmetric polynomial in $$u_1, \ldots, u_n$$ over the field $$F$$, which, as usual, we will assume to have characteristic 0. If $$n = 1$$ then take $$f = P$$ and $$g = 0$$, and we are done. Otherwise, define \begin{align*} f(u_1, \ldots, u_n) &= \frac{1}{n!}\sum_{\sigma \in S_n} P(u_{\sigma(1)}, u_{\sigma(2)}, \ldots, u_{\sigma(n)}) \\ G(u_1, \ldots, u_n) &= \frac{1}{n!}\sum_{\sigma \in S_n} (\operatorname{sgn}\sigma)P(u_{\sigma(1)}, u_{\sigma(2)}, \ldots, u_{\sigma(n)}) \end{align*} The function $$f$$ is the symmetrization of $$P$$. By construction, it is evident that $$f$$ is symmetric and $$G$$ is skew-symmetric, and both are polynomials; furthermore, \begin{align*} f(u_1, \ldots, g_n) + G(u_1, \ldots, u_n) &= \frac{1}{n!} \sum_{\sigma \in S_n}(1 + \operatorname{sgn}\sigma) P(u_{\sigma(1)}, \ldots, u_{\sigma(n)}) \\ &= \frac{1}{n!} \sum_{\sigma \in A_n} 2P(u_{\sigma(1)}, \ldots, u_{\sigma(n)}) \\ &= \frac{1}{n!} \sum_{\sigma \in A_n} 2P(u_1, \ldots, u_n) \\ &= P(u_1, \ldots, u_n) \end{align*} Since $$G$$ is skew-symmetric, it follows that it must vanish whenever two of its arguments are equal. By the Lemma to the solution of Problem 16.M.3 above, $$\delta \divides G$$. Write $$G = g\delta$$. Whenever $$u_1, \ldots, u_n$$ are distinct and $$\sigma \in S_n$$, \begin{align*} g(u_{\sigma(1)}, \ldots, u_{\sigma(n)}) &= \frac{G(u_{\sigma(1)}, \ldots, u_{\sigma(n)})} {\delta(u_{\sigma(1)}, \ldots, u_{\sigma(n)})} \\ &= \frac{\operatorname{sgn}(\sigma)} {\operatorname{sgn}(\sigma)} \frac{G(u_1, \ldots, u_n)}{\delta(u_1, \ldots, u_n)} \\ &= g(u_1, \ldots, u_n) \end{align*} Define $$\overline{g}$$ to be the symmetrization of $$g$$. Then $$g - \overline{g}$$ is a polynomial that vanishes whenever all its arguments are distinct. By the Lemma, $$g - \overline{g}$$ vanishes for all values of its arguments, so $$g$$ is symmetric. Thus, as required, we have the decomposition $$P = f + g\delta$$, where $$f$$ and $$g$$ are symmetric.

Problem 16.M.8 I'm not sure exactly what we're supposed to do for this problem, so I've skipped it.

Problem 16.M.9 I don't think the problem is correct as written. A counterexample is given by $f(t, x) = x^3 - 3t^2 x - (t^4 + t^2)$ Since $$f$$ is monic in $$x$$, if $$f$$ is reducible, it must be reducible as a polynomial in $$f$$ (with coefficients in $$\C[t]$$). We will see shortly that this is not the case, therefore $$f$$ is irreducible in $$\C[t, x]$$. The roots of $$f$$ are given by \begin{align*} x_1 &= t^{4/3} + t^{2/3} \\ x_2 &= \omega t^{4/3} + \omega^2 t^{2/3} \\ x_3 &= \omega^2 t^{4/3} + \omega t^{2/3} \end{align*} These roots obviously don't belong to $$\C(t)$$, so $$f$$ is irreducible as a polynomial over $$\C(t)$$, and hence also $$\C[t]$$. The discriminant of $$f$$, regarded as a cubic in $$x$$, is obtained from (16.2.8): $D = -27t^8 + 54t^6 - 27t^4 = (i\sqrt{27}(t^4 - t^2))^2 = \delta^2$ so $$D$$ is a square in $$\C(t)$$ and the splitting field of $$f$$ has degree 3. Explicitly, the roots $$x_2, x_3$$ can be written as a rational function of $$x_1$$ and $$\delta$$ using the result of Exercise 16.8.5. However, substituting $$t = 1$$ yields $$f(x) = x^3 - 3x - 2$$, which has the single root 2 and the double root 1. This contradicts the problem statement.

I don't know where the error in the problem is. If you know what Artin intended, feel free to write to me.

Problem 16.M.10 We will prove the desired result for algebraic extensions $$K$$, which implies the result in the finite case. It suffices to prove the desired result when $$K = \overline{F}$$, an algebraic closure of $$F$$; for $$K[x]$$ embeds into $$\overline{F}[x]$$ for any algebraic extension $$K/F$$, and if $$f(x)g(x) = h(x)$$ where $$f(x) \in K[x]$$, $$g(x) \in \overline{F}[x]$$, and $$h(x) \in F[x]$$, then polynomial division shows that $$g(x) \in K[x]$$ as required.

We will now proceed assuming that $$K$$ is algebraically closed. Write $$f$$ in fully factored form, $$f(x) = \prod_{i=1}^n (x - x_i)$$. For each $$x_i$$, let its Galois conjugates over $$K/F$$ be $$\{y_{ij}\}$$ for $$j = 1, \ldots, d_i$$, and $$y_{i1} = x_i$$. Then, we may take \begin{equation*} g(x) = \prod_{i=1}^n \prod_{j=2}^{d_i} (x - y_{ij}) \end{equation*} so that \begin{equation*} f(x) g(x) = \prod_{i=1}^n \prod_{j=1}^{d_i} (x - y_{ij}) \end{equation*} and for each $$i$$, the inner product expression is the minimal polynomial for $$x_i$$ over $$F$$, and thus lies in $$F[x]$$.

Problem 16.M.11

1. We are told that the resolvent cubic has one root $$\beta = \alpha_1 \alpha_2 + \alpha_3 \alpha_4$$ in $$F$$. It has either one or three roots in $$F$$ in total, but if it had three roots in $$F$$, then $$\sqrt{D}$$ would be in $$F$$ too; we are told this is not the case. Therefore $$\beta$$ is the only root that the resolvent cubic has in $$F$$. Permuting the four roots $$\alpha_1, \ldots, \alpha_4$$ causes the three roots of the resolvent cubic to be permuted; since the two roots other than $$\beta$$ are not in $$F$$, the only permutations that fix $$\beta$$ are those that fix the expression $$\alpha_1 \alpha_2 + \alpha_3 \alpha_4$$ formally. By the orbit-stabilizer theorem, there are 8 such permutations. Explicitly, they are the identity, $$(1\ 2)$$, $$(3\ 4)$$, $$(1\ 2)(3\ 4)$$, $$(1\ 3)(2\ 4)$$, $$(1\ 4)(2\ 3)$$, $$(1\ 3\ 2\ 4)$$, and $$(1\ 4\ 2\ 3)$$. We classified groups of order 8 in Exercise 7.3.4. Since our subgroup of $$S_4$$ contains exactly two elements of order 4, it must be isomorphic to $$D_4$$.

2. By inspection, $$\gamma^2 = \beta^2 - 4s_1$$, where $$s_1$$ is the elementary symmetric polynomial of degree 1 in $$\alpha_1, \ldots, \alpha_4$$ (i.e., their sum). Since $$\beta \in F$$, this establishes that $$\gamma^2 \in F$$.

Similarly, $$\epsilon^2 = s_1^2 - 4s_2 + 4\beta$$, and therefore $$\epsilon^2 \in F$$.

3. It appears that the text has an error. It should say that if $$\gamma \ne 0$$, then $$\delta\gamma$$ is in $$F$$ iff $$G \cong C_4$$, and likewise for $$\delta\epsilon$$. They will not necessarily be squares. For example, take the polynomial $$x^4 - 4x^2 + 2$$, given in Example 16.9.2(b) as an example of an irreducible quartic with Galois group $$C_4$$. It has roots $$\alpha_1 = \sqrt{2 + \sqrt{2}}$$, $$\alpha_2 = -\alpha_1$$, $$\alpha_3 = \sqrt{2 - \sqrt{2}}$$, $$\alpha_4 = -\alpha_3$$, with resolvent cubic $$x^3 + 4x^2 - 8x - 32$$, which has a single rational root $$\beta = \alpha_1 \alpha_2 + \alpha_3 \alpha_4 = -4$$. The discriminant is $$D = 2048$$, and $$\gamma = \alpha_1\alpha_2 - \alpha_3\alpha_4 = -2\sqrt{2}$$. Thus, $$\delta\gamma = -2\sqrt{2} \cdot 32\sqrt{2} = -128$$, certainly not a square in $$\Q$$.

Anyway, let's prove this. In part (a) we determined the group $$H$$ of permutations that fix $$\beta$$, which is isomorphic to $$D_4$$. If $$G \cong D_4$$, then every element of $$G$$ fixes $$\beta$$, so $$G = H$$. If $$G \cong C_4$$, then again every element of $$G$$ fixes $$\beta$$ but there are 4 other elements of $$H$$ that don't fix all of $$F$$. $$G$$ must be a cyclic subgroup of $$H$$. There is exactly one cyclic subgroup of order 4 in a dihedral group of order 8; explicitly, $$G$$ is generated by $$(1\ 3\ 2\ 4)$$.

The permutation $$(1\ 3\ 2\ 4)$$ sends $$\gamma = \alpha_1\alpha_2 - \alpha_3\alpha_4$$ to $$\alpha_3\alpha_4 - \alpha_2\alpha_1 = -\gamma$$ and $$\epsilon$$ to $$-\epsilon$$ likewise, and since it's odd, by Exercise 16.M.7(a), it sends $$\delta$$ to $$-\delta$$. Thus, $$\delta\gamma$$ and $$\delta\epsilon$$ are invariant under the $$C_4$$-subgroup of $$H$$. The group $$H$$ is generated by this $$C_4$$-subgroup together with the element $$(1\ 2)$$, which leaves $$\gamma$$ and $$\epsilon$$ invariant but changes the sign of $$\delta$$.

Since $$\delta \notin F$$, obviously $$\delta \ne 0$$. If $$\gamma \ne 0$$ as well, and $$G \cong D_4$$, then $$G$$ contains elements that send $$\delta\gamma \mapsto -\delta\gamma$$; but $$\delta\gamma \ne 0$$, so it is not fixed by $$G$$, so it cannot be in $$F$$. Conversely, if $$\delta\gamma \notin F$$, then by Theorem 16.6.4(b), there exists an element of $$G$$ that doesn't fix $$\delta\gamma$$, so $$G$$ must be strictly larger than $$C_4$$, and must be $$D_4$$. By the same reasoning, analogous statements hold for $$\delta\epsilon$$. This is equivalent to what we were required to prove, so we are done.

4. Suppose $$\gamma = \epsilon = 0$$. Then $$\alpha_1\alpha_2 = \alpha_3\alpha_4$$ and $$\alpha_1 + \alpha_2 = \alpha_3 + \alpha_4$$. This implies that $$\{\alpha_1, \alpha_2\} = \{\alpha_3, \alpha_4\}$$. By Proposition 15.6.8(b), $$f$$ is reducible, and a contradiction has been reached.

Problem 16.M.12 With $$K$$ denoting the splitting field of $$f$$, suppose $$G = \Gal(K/F)$$ is solvable, with the sequence $$G = H_0 \supseteq H_1 \supseteq \ldots \supseteq H_k = \{1\}$$ given. By the fundamental theorem, each $$H_i$$ has a corresponding intermediate field, namely the subfield of $$K$$ that it fixes; call this field $$F_i$$. (Thus, $$F = F_0$$, and $$K = F_k$$.) For each $$i = 1, \ldots, k$$, $$H_{i-1} = \Gal(K/F_{i-1})$$, $$H_i = \Gal(K/F_i)$$, and since $$H_{i-1}$$ is a normal subgroup of $$H_i$$, the extension $$F_i/F_{i-1}$$ is Galois, with Galois group $$H_{i-1}/H_i$$, which is known to be cyclic. Thus, the extensions $$F_1/F$$, $$F_2/F_1$$, …, $$K/F_{k-1}$$ are Galois with cyclic (and therefore abelian) Galois groups. By Lemma 16.12.8, each such extension $$F_i/F_{i-1}$$ can be constructed by successive Galois extensions of prime degree. Thus, the roots of $$f$$ satisfy the criterion 16.12.2(b), and are solvable over $$F$$.

Conversely, suppose $$f$$ has at least one solvable root $$\alpha$$ in the sense of 16.12.2(a): that is, there exists a tower of fields $$\label{radical-tower1} F = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_n$$ where $$\alpha \in F_n$$ and for each $$i = 1, \ldots, n$$, $$F_i = F_{i-1}(\beta_i)$$ for some $$\beta_i$$ such that a power of $$\beta_i$$ lies in $$F_{i-1}$$ (we call this a simple radical extension). As in the proof of Proposition 16.12.2, we can assume that $$[F_i : F_{i-1}]$$ is a prime number $$p_i$$ for each $$i$$. Let $$q = p_1 p_2 \ldots p_n$$, and let $$F' = F(\zeta_q)$$ where $$\zeta_q$$ is a primitive $$q$$th root of unity.

Let $$\gamma$$ be a primitive element for the extension $$F_n/F$$, and let $$g$$ be the minimal polynomial for $$\gamma$$ over $$F$$. Let $$K'$$ be the splitting field of $$g$$ over $$F$$, and let $$\gamma_1, \ldots, \gamma_r$$ be the roots of $$g$$ in $$K'$$, with $$\gamma_1 = \gamma$$. For each $$j = 1, \ldots, r$$, let $$\sigma_j$$ be an element of $$\Gal(K'/F)$$ that takes $$\gamma_1 \mapsto \gamma_j$$. For each such $$\sigma_j$$, the image of \eqref{radical-tower1}, \begin{equation*} F = \sigma_j(F) = \sigma_j(F_0) \subseteq \sigma_j(F_1) \subseteq \ldots \subseteq \sigma_j(F_n) \end{equation*} is a tower of simple radical extensions since $$\sigma_j(F_i) = \sigma_j(F_{i-1})(\sigma_j(\beta_i))$$, and $$\sigma_j(\beta_i)^{p_i} = \sigma_j(\beta_i^{p_i})$$, which is the image of some element of $$F_{i-1}$$ under $$\sigma_j$$. Note also that $$\gamma_j \in \sigma_j(F_n)$$ and $$\sigma_j(F_n) \subseteq K'$$.

Now let us build a tower of fields by first adjoining $$\zeta_q$$ to $$F$$, then successively adjoining $$\beta_1, \beta_2, \ldots, \beta_n$$, then $$\sigma_2(\beta_1), \ldots, \sigma_2(\beta_n)$$, …, $$\sigma_r(\beta_1), \ldots, \sigma_r(\beta_n)$$. Each time we adjoin $$\sigma_j(\beta_i)$$, by Proposition 16.11.2, this is either a trivial extension or a simple radical extension of degree $$p_i$$. Having adjoined all of these roots, we end up with a field that contains $$F'$$ and all the roots of $$g$$ and is contained within the field $$K'(\zeta_q)$$, and is therefore equal to $$K'(\zeta_q)$$. Thus $$K'(\zeta_q)$$ has been constructed from $$F$$ by a series of simple radical extensions: $F \subseteq F' \subseteq F'(\beta_1) \subseteq \ldots \subseteq K'(\zeta_q)$ Since $$K'(\zeta_q)$$ is the splitting field of $$g(x)(x^q - 1)$$ over $$F$$, the extension $$K'(\zeta_q)/F$$ is Galois. The extension $$F'/F$$ is Galois since $$F'$$ is the splitting field of $$x^q - 1$$ over $$F$$. Each of the subsequent simple radical extensions is of degree $$p_i$$ over a base field that contains the $$p_i$$th roots of unity, so it too is Galois, with cyclic Galois group. By Lemma 16.12.8, the extension $$F'/F$$ can be built up by successive Galois extensions of prime degree, which therefore also have cyclic Galois groups. Thus, we have a tower $F \subseteq F' \subseteq F'' \subseteq \ldots \subseteq K'(\zeta_q)$ where each field is a Galois extension of the previous field with cyclic Galois group. Since $$F'/F$$ is a Galois extension, the fundamental theorem implies that $$\Gal(K'(\zeta_q)/F')$$ is a normal subgroup of $$\Gal(K'(\zeta_q)/F)$$, and that the quotient group is $$\Gal(F'/F)$$, which is cyclic. Similarly, $$\Gal(K'(\zeta_q)/F'')$$ is a normal subgroup of $$\Gal(K'(\zeta_q)/F')$$ with quotient group $$\Gal(F''/F')$$, which is cyclic, and so on, which establishes that $$\Gal(K'(\zeta_q)/F)$$ is solvable over $$F$$. Since $$K'(\zeta_q)$$ contains $$F_n$$, it contains $$\alpha$$, and since it is a Galois extension of $$F$$, Theorem 16.3.2 implies that $$f$$ splits completely over $$K'(\zeta_q)$$, that is, $$K$$ is a subfield of $$K'(\zeta_q)$$. Again, by the fundamental theorem, $$\Gal(K/F) = \Gal(K'(\zeta_q)/F) / \Gal(K'(\zeta_q)/K)$$. It is a well-known fact that any quotient of a solvable group is solvable; therefore, $$\Gal(K/F)$$ is solvable.

Corollary: Since $$A_5$$ is a simple group, its only composition series is $$1 \triangleleft A_5$$, and it is not a solvable group. This gives an alternative proof of Theorem 16.12.4.

Problem 16.M.13 The desired result follows directly from the normal basis theorem, but I have not been able to find the easier solution using the roots of unity condition that Artin likely intended. Write to me if you have a solution.