Brian Bi
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## Section 16.9. Quartic Equations

Exercise 16.9.1 By Theorem 15.8.1, degrees of elements in $$K$$ over $$F$$ correspond to degrees of intermediate fields over $$F$$. By the fundamental theorem, degrees of intermediate fields over $$F$$ correspond to indices of subgroups of $$\Gal(K/F)$$. $$S_4$$ has subgroups of all indices dividing its order: $$S_4$$ itself of index 1, $$A_4$$ of index 2, $$D_4$$ of index 3, $$S_3$$ of index 4, $$V_4$$ of index 6, $$C_3$$ of index 8, $$C_2$$ of index 12, and the trivial group of index 24. Therefore, all divisors of 24 occur as degrees of elements of $$K$$ over $$F$$.

Exercise 16.9.2 $$\alpha + \alpha' = \sqrt{4 + \sqrt{5}} + \sqrt{4 - \sqrt{5}}$$, so $$(\alpha + \alpha')^2 = (4 + \sqrt{5}) + (4 - \sqrt{5}) + 2\sqrt{(4 + \sqrt{5})(4-\sqrt{5})} = 8 + 2\sqrt{11}$$. Therefore $$\alpha + \alpha' = \sqrt{8 + 2\sqrt{11}}$$. It's clear that $$K$$ also contains some nested square roots with $$\sqrt{55}$$ under the radical, but $$K$$ can't contain an element of the form $$\sqrt{p + \sqrt{q}}$$ where $$p, q \in \Q, q \notin \Q(\sqrt{5}, \sqrt{11})$$ since that would imply that $$K$$ was generated by $$\sqrt{5}, \sqrt{11}, \sqrt{q}$$, having Galois group $$C_2^3$$.

Exercise 16.9.3 Suppose $$\sqrt{a} + \sqrt{b} = \sqrt{4 + \sqrt{7}}$$ with $$a, b \in \Q$$. Then $$a + b + 2\sqrt{ab} = (\sqrt{a} + \sqrt{b})^2 = 4 + \sqrt{7}$$. We must then have $$\sqrt{ab} \in \Q(\sqrt{7})$$, which is only possible if $$ab/7$$ is a square in $$\Q$$. Therefore, $$\sqrt{ab} = k\sqrt{7}$$ for some $$k \in \Q$$, and as $$1, \sqrt{7}$$ is a basis for $$\Q(\sqrt{7})$$, comparing coefficients gives $$a + b = 4$$ and $$2\sqrt{ab} = \sqrt{7}$$, therefore $$ab = 7/4$$. Solving, we obtain $$\sqrt{4 + \sqrt{7}} = \sqrt{1/2} + \sqrt{7/2}$$. We note here that $$\sqrt{7/2} - \sqrt{1/2} = \sqrt{4 - \sqrt{7}}$$ as well (we will use this in the next problem).

Exercise 16.9.4

1. We quickly check with the rational root theorem that $$x^4 - 8x^2 + 11$$ has no rational roots. If it splits over $$\Q$$, it must split into two quadratic factors. Over $$\C$$, the factorization is $$(x + \alpha)(x - \alpha)(x + \beta)(x - \beta)$$ where $$\alpha = \sqrt{4 + \sqrt{5}}$$ as in 16.9.2(a) (but we will not make use of this value in this paragraph) and $$\beta$$ is either $$\alpha'$$ or $$-\alpha'$$. Over $$\Q$$, either the quadratic factors are $$(x + \alpha)(x - \alpha)$$ and $$(x + \beta)(x - \beta)$$, or they are $$(x + \alpha)(x + \beta)$$ and $$(x - \alpha)(x - \beta)$$ (note that we can relabel $$\beta$$ as $$-\beta$$ if necessary). In the first case, the two factors would be $$x^2 - \alpha^2$$ and $$x^2 - \beta^2$$, but 11 cannot be written as the product of two squares in $$\Q$$, so $$\alpha^2$$ and $$\beta^2$$ can't both be rational. In the second case, the two factors would be $$x^2 \pm (\alpha + \beta)x + \alpha\beta$$, but 11 isn't a square in $$\Q$$, so $$\alpha\beta$$ can't be rational. Therefore $$x^4 - 8x^2 + 11$$ is irreducible over $$\Q$$.

Alternatively, we could use the expressions for the roots of $$x^4 - 8x^2 + 11$$ explicitly, namely $$\pm \sqrt{4 \pm \sqrt{5}}$$. None of the four roots is rational. If the quartic splits into two quadratics, one of them will have the product of two of the four roots as its constant term, and $$-1$$ times their sum as its linear coefficient; the other likewise with the other two roots. But none of the six possible products of a pair of roots is rational, since each is $$\pm 1$$ times either $$\sqrt{4 + \sqrt{5}} \sqrt{4 + \sqrt{5}} = 4 + \sqrt{5}$$ or $$\sqrt{4 + \sqrt{5}}\sqrt{4 - \sqrt{5}} = \sqrt{11}$$.

2. Performing the substitution $$x = y+1$$, we obtain the polynomial $$y^4 + 4y^3 - 2y^2 - 12y + 2$$, which is irreducible by the Eisenstein criterion, so $$x^4 - 8x^2 + 9$$ is also irreducible.

The roots are $$\pm \sqrt{4 \pm \sqrt{7}}$$, all four of which are irrational. In this case it's possible to pair them so that the products of the two pairs are rational, since $$\sqrt{4 + \sqrt{7}}\sqrt{4 - \sqrt{7}} = 3$$. However, the sums of the form $$\pm \sqrt{4 + \sqrt{7}} \pm \sqrt{4 - \sqrt{7}}$$ aren't rational, so again the factorization into two quadratics can't exist. To prove that these sums are irrational, we use the result of Exercise 16.9.3: $$\sqrt{4 + \sqrt{7}} \pm \sqrt{4 - \sqrt{7}} = (\sqrt{7/2} + \sqrt{1/2}) \pm (\sqrt{7/2} - \sqrt{1/2})$$, which works out to either $$\sqrt{2}$$ or $$\sqrt{14}$$, neither of which is rational.

3. We saw in Example 16.9.2(a) that $$\Gal(K/\Q) \cong D_4$$ and the Galois group is generated by $$\sigma$$ and $$\tau$$ (defined in (16.9.3)) where $$\sigma^4 = \tau^2 = 1$$ and $$\tau\sigma = \sigma^{-1}\tau$$. This group has 1 subgroup of order 1; 5 subgroups of order 2 (generated by the 5 elements of order 2, namely $$\sigma^k \tau$$ and $$\sigma^2$$); 3 subgroups of order 4, namely $$\langle \sigma \rangle$$, $$\langle \sigma^2, \tau\rangle$$, and $$\langle \sigma^2, \sigma\tau\rangle$$; and 1 subgroup of order 8. Thus, there are 3 intermediate fields of degree 2 over $$\Q$$, and 5 intermediate fields of degree 4.

By previous results, the 3 intermediate fields of degree 2 are obviously $$\Q(\sqrt{5})$$, $$\Q(\sqrt{11})$$, and $$\Q(\sqrt{55})$$.

One of the 5 intermediate fields of degree 4, namely the fixed field of $$\sigma^2$$, was previously described, $$L = \Q(\sqrt{5}, \sqrt{11})$$. Since $$\tau$$ fixes $$\alpha$$ and $$-\alpha$$, its fixed field contains $$\Q(\alpha)$$, but $$\Q(\alpha)$$ has degree 4, so it is the fixed field. Likewise $$\sigma^2 \tau$$ has fixed field $$\Q(\alpha')$$. The automorphism $$\sigma\tau$$ swaps $$\alpha$$ and $$\alpha'$$, so it fixes their product, $$\sqrt{11}$$, and their sum, $$\sqrt{8 + 2\sqrt{11}}$$, so the fixed field is $$\Q(\sqrt{8 + 2\sqrt{11}})$$. Finally, $$\sigma^3\tau$$ swaps $$\alpha$$ and $$-\alpha'$$, so it fixes $$\alpha - \alpha' = \sqrt{8 - 2\sqrt{11}}$$, and its fixed field is $$\Q(\sqrt{8 - 2\sqrt{11}})$$.

Exercise 16.9.5

1. We are asked to assume that $$\alpha$$ has degree 4, so we are looking for a quartic with coefficients in $$F$$ of which $$\alpha$$ is a root. Since $$\alpha^2 = r + \sqrt{t}$$, it follows that $$(\alpha^2 - r)^2 = t$$. This is the desired quartic. Explicitly, $$\alpha$$ is a root of $$x^4 - 2rx^2 + (r^2 - t)$$. The roots of $$\alpha$$ in $$K$$ are $$\pm \alpha$$ and $$\pm \alpha'$$. As discussed in Example 16.9.2, any $$F$$-automorphism sending $$\alpha$$ to some other root also sends $$-\alpha$$ to $$-1$$ times that root, implying that the Galois group is a subgroup of $$D_4$$. The subgroups of $$D_4$$ that act transitively on the four roots are isomorphic to $$D_4$$, $$C_4$$, or $$V_4$$, so the Galois group $$\Gal(K/F)$$ is one of these. Note also that the unique $$C_4$$-subgroup of $$D_4$$ and the two $$V_4$$-subgroups of $$D_4$$ all contain the element $$\rho = \sigma^2$$, so the latter is always in $$\Gal(K/F)$$.

2. Let $$\alpha' = \sqrt{r - \sqrt{t}}$$. The four roots of the minimal polynomial of $$\alpha$$ are $$\pm\alpha$$ and $$\pm\alpha'$$. Example 16.9.2 can be generalized to three cases:

• Case 1: $$r^2 - t$$ is a perfect square in $$F$$, say, $$q^2$$. Then we have $$\alpha\alpha' = q$$ (we may have to relabel $$-q$$ as $$q$$). Every element of $$\Gal(K/F)$$ must preserve this relation, so it follows that the automorphism is uniquely determined by the image of $$\alpha$$, and $$\Gal(K/F)$$ has order 4. We can see that if $$\alpha \rightsquigarrow \alpha'$$ then $$\alpha' \rightsquigarrow \alpha$$. Also, if $$\alpha \rightsquigarrow -\alpha$$ then clearly $$-\alpha \rightsquigarrow \alpha$$, and if $$\alpha \rightsquigarrow -\alpha'$$ then $$\alpha' \rightsquigarrow -\alpha$$ therefore $$-\alpha' \rightsquigarrow \alpha$$. So the three nontrivial elements of $$\Gal(K/F)$$ have order 2. This establishes that $$\Gal(K/F) \cong V_4$$.
• Case 2: $$r^2 - t$$ isn't a perfect square, but $$(r^2 - t)/t$$ is. (We can safely assume that $$t \ne 0$$ since if $$t = 0$$ then $$\alpha$$ cannot have degree 4.) Consequently we have $$\alpha' = \sqrt{r^2 - t}/\alpha = k\sqrt{t}/\alpha$$ where $$k \in F$$; but $$\sqrt{t} \in F(\alpha)$$, so $$\alpha' \in F(\alpha)$$. That is, $$F(\alpha)$$ is the splitting field for the minimal polynomial of $$\alpha$$, and $$\Gal(K/F)$$ has order 4. Let $$\sigma'$$ denote the element of $$\Gal(K/F)$$ that sends $$\alpha$$ to $$\alpha'$$. Then $$\sigma'(r + \sqrt{t}) = \sigma'(\alpha^2) = \alpha'^2 = r - \sqrt{t}$$, implying that $$\sigma'(\sqrt{t}) = -\sqrt{t}$$, and also $$\sigma'(\sqrt{r^2 - t}) = \sigma'(k\sqrt{t}) = -k\sqrt{t} = -\sqrt{r^2 - t}$$. But $$\sqrt{r^2 - t} = \alpha \alpha'$$, so $$\sigma'(\alpha\alpha') = -\alpha\alpha'$$. Since $$\sigma'(\alpha) = \alpha'$$, it follows that $$\sigma'(\alpha') = -\alpha$$, and $$\sigma'$$ has order 4. Therefore $$\Gal(K/F) \cong C_4$$.
• Case 3: Neither case 1 nor case 2 holds. Then, the intermediate field $$L = F(\sqrt{t}, \sqrt{r^2 - t})$$ has degree 4 over $$F$$. The relations $$\alpha^2 - r = \sqrt{t}$$ and $$\alpha\alpha' = \sqrt{r^2 - t}$$ are invariant under $$\rho$$, so $$\rho$$ fixes $$L$$. But $$[K : F]$$ is at most 8 (the order of $$D_4$$), the intermediate field fixed by $$\rho$$ has index 2 in $$K$$, and said fixed field contains $$L$$, which has degree 4 over $$F$$; thus, the only possibility is that $$L$$ is the fixed field, and $$[K : F] = 8$$. Therefore, $$\Gal(K/F) \cong D_4$$.

To summarize, the Galois group is $$V_4$$ when $$r^2 - t$$ is a square in $$F$$; $$C_4$$ when $$r^2 - t$$ is a nonsquare in $$F$$ with $$(r^2 - t)/t$$ a square in $$F$$; and $$D_4$$ otherwise.

3. We are in case 3. As in Exercise 16.9.4(c), we know there are 3 intermediate fields of degree 2 over $$F$$. One of these is clearly generated by $$\sqrt{t}$$. Since $$F(\sqrt{t})$$ doesn't contain $$\sqrt{r^2 - t}$$, but $$K$$ does, another intermediate field of degree 2 is generated by $$\sqrt{r^2 - t}$$, and a third by their product. Following the same reasoning used in Exercise 16.9.4(c) involving the explicit forms of the subgroups of degree 2, we see that the five intermediate fields of index 2 must be those generated by $$\alpha$$, $$\alpha'$$, $$\alpha + \alpha'$$, and $$\alpha - \alpha'$$, plus one more generated by $$\sqrt{t}$$ and $$\sqrt{r^2 - t}$$ together. We need to find a primitive element for this last extension. By the reasoning from Example 16.7.4 (applying the fundamental theorem to the Galois extension $$F(\sqrt{t},\sqrt{r^2 - t})/F$$), $$\sqrt{t} + \sqrt{r^2 - t}$$ is such an element.

Exercise 16.9.6 By Exercise 16.2.4(b), the discriminant is 256. We quickly check that $$x^4 + 1$$ is irreducible over $$\Q$$ using similar techniques as in Exercise 16.9.4. Thus, each of its roots has degree 4 over $$\Q$$. This quartic can be put into the form of Exercise 16.9.5(a) with $$r = 0, t = -1$$; indeed, $$\sqrt{0 + \sqrt{-1}}$$ is clearly one of the roots. Here $$r^2 - t = 1$$, which is a perfect square in $$\Q$$, so by Exercise 16.9.5(b), the Galois group is $$V_4$$.

Exercise 16.9.7 The cases where $$t$$ has a square root in $$F$$ are not interesting, so we will concentrate on the cases where this is not true. The field $$F(\sqrt{r + \sqrt{t}})$$ contains $$\sqrt{t}$$, so $$\sqrt{t} \in F(\sqrt{a}, \sqrt{b})$$, which implies that $$t = k^2 a$$ or $$t = k^2 b$$ or $$t = k^2 ab$$ for some $$k \in F$$. Without loss of generality we can simply exchange $$a$$ and $$b$$, so we can assume that $$t = k^2 a$$ or $$t = k^2 ab$$. Furthermore, $$F(\sqrt{a}, \sqrt{b}) = F(\sqrt{a}, \sqrt{ab})$$ assuming that the former is an extension of degree 4, so in fact there is also no loss of generality from assuming $$t = k^2 a$$.

We can now proceed by writing $$(m + n\sqrt{a} + p\sqrt{b} + q\sqrt{ab})^2 = r + k\sqrt{a}$$ with $$m, n, p, q \in F$$ (note that we can change the sign of $$k$$ if necessary). Expanding and equating coefficients of $$\sqrt{b}$$ and $$\sqrt{ab}$$, we obtain \begin{align*} 2mq + 2np &= 0 \\ 2mp + 2anq &= 0 \end{align*} Assuming that all four unknowns are nonzero, we solve the first equation to obtain $$q = -np/m$$; substituting this into the second equation then implies $$m^2 = an^2$$, which is possible only if $$m = n = 0$$. Analysis of cases where some variables vanish shows that in all cases we must either have $$m = n = 0$$ or $$p = q = 0$$ at least. Thus a square root of $$r + k\sqrt{a}$$ must always take the form $$m + n\sqrt{a}$$ or $$\sqrt{b}(m + n\sqrt{a})$$. These respectively square to $$(m^2 + an^2) + 2mn\sqrt{a}$$ and $$b(m^2 + an^2) + 2bmn\sqrt{a}$$, thus elements of the form $$\sqrt{r + \sqrt{t}}$$ can belong to $$F(\sqrt{a}, \sqrt{b})$$ when $$r = m^2 + an^2$$ and $$t = 4am^2 n^2$$ for some $$m, n \in F$$, or else when $$r = b(m^2 + an^2)$$ and $$t = 4ab^2m^2n^2$$. Similarly elements of the form $$m + n\sqrt{b}$$ and $$\sqrt{a}(m + n\sqrt{b})$$ can equal a nested radical of the form $$\sqrt{r + \sqrt{t}}$$ (we omit the explicit expressions for $$r$$ and $$t$$), as can those of the form $$m + n\sqrt{ab}$$; this enumeration is exhaustive for reasons argued above.

Exercise 16.9.8 We begin by considering the form that the unnested square root could take. Obviously it would contain at least one square root, and possibly two; could it ever require more than two? This is answered by the following Lemma.

Let $$F$$ be a field of characteristic other than 2, and let $$K = F(\sqrt{a_1}, \ldots, \sqrt{a_n})$$ be a field extension of degree $$2^n$$. Let $$r, t \in F$$ and suppose that $$x^2 = r + \sqrt{t}$$ where $$x \in K$$. Then $$x$$ can be written in the form $$\sqrt{b} + \sqrt{c}$$ for some $$b, c \in F$$.

The case where $$\sqrt{t} \in F$$ is obvious. The result is also immediate for $$n = 1$$, and the case $$n = 2$$ is worked out in Exercise 16.9.7. Consider now $$n \ge 3$$. Write $x = \sum_{\alpha \in \{0, 1\}^n} k_\alpha \sqrt{a^\alpha}$ where $$\alpha$$ is a multi-index. Since $$x^2 = r + \sqrt{t}$$, it follows that $$[F(x) : F]$$ is either 2 or 4. If it is 2, the desired result follows immediately. Otherwise, $$[F(x) : F] = 4$$. The subfield $$F(x)$$ is the fixed field of a subgroup $$H \subseteq \Gal(K/F)$$ with index 4. In particular, since $$\Gal(K/F) \cong C_2^n$$, it follows that $$H \cong C_2^{n-2}$$, and by the fundamental theorem, $$\Gal(F(x)/F) \cong C_2^n/C_2^{n-2} \cong V_4$$. Again by the fundamental theorem, $$F(x)/F$$ contains three intermediate subfields that are quadratic extensions of $$F$$. Proposition 15.3.3 implies that $$F(x)/F$$ can be generated by adjoining two square roots, $$F(x) = F(\sqrt{b'}, \sqrt{c'})$$ with $$b', c' \in F$$. We have thus reduced the problem to the case $$n = 2$$.

Thanks to the Lemma, we know that we are looking for an unnested expression of the form $$a + \sqrt{b}$$ or $$\sqrt{a} + \sqrt{b}$$. If $$(a + \sqrt{b})^2 = r + \sqrt{t}$$, then it's clear that $$(a - \sqrt{b})^2 = r - \sqrt{t}$$ and therefore $$(a^2 - b)^2 = r^2 - t$$; that is, $$r^2 - t$$ must be a square in $$\Q$$. If the square root is of the form $$\sqrt{a} + \sqrt{b}$$ then likewise $$(\sqrt{a} - \sqrt{b})^2 = r - \sqrt{t}$$ so $$(a - b)^2 = r^2 - t$$. We conclude that (a), (b), and (e), namely $$\sqrt{2 + \sqrt{11}}$$, $$\sqrt{10 + 5\sqrt{2}}$$, and $$\sqrt{11 + \sqrt{6}}$$, can't be denested. This also provides a means of finding the denested expression when it exists. The form $$\sqrt{a} + \sqrt{b}$$ obviously includes the form $$a + \sqrt{b}$$, so we will focus on this one. We have that $$(a - b)^2 = r^2 - t$$. In (c), this yields $$(a - b)^2 = 121 - 72 = 49$$, so $$a - b = \pm 7$$. Combining this with $$2\sqrt{ab} = 6\sqrt{2}$$ implying $$ab = \pm 18$$, we arrive at the solution $$\sqrt{9} + \sqrt{2}$$ or $$3 + \sqrt{2}$$; indeed $$(3 + \sqrt{2})^2 = 11 + 6\sqrt{2}$$. In (d), we have $$(a - b)^2 = 36 - 11 = 25$$, so $$a - b = \pm 5$$, and $$2\sqrt{ab} = \sqrt{11}$$ so $$ab = \pm 11/4$$; this can be solved to obtain $$a = 11/2, b = 1/2$$ so $$\sqrt{6 + \sqrt{11}} = \sqrt{11/2} + \sqrt{1/2}$$.

Exercise 16.9.9

1. The discriminant was computed in Exercise 16.2.4(b); it equals $$-27r^4 + 256s^3$$.

The resolvent cubic is $$x^3 - ax^2 + bx - c$$ where, as in (16.9.7), we define \begin{align} \begin{split} \beta_1 &= \alpha_1 \alpha_2 + \alpha_3 \alpha_4 \\ \beta_2 &= \alpha_1 \alpha_3 + \alpha_2 \alpha_4 \\ \beta_3 &= \alpha_1 \alpha_4 + \alpha_2 \alpha_3 \end{split} \label{eqn:bdef} \end{align} with $$\alpha_i$$ being the roots of $$x^4 + rx + s$$, and by Vieta's formulas, we know that \begin{align} \alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 &= 0 \label{eqn:a1v} \\ \alpha_1 \alpha_2 + \alpha_1 \alpha_3 + \alpha_1 \alpha_4 + \alpha_2 \alpha_3 + \alpha_2 \alpha_4 + \alpha_3 \alpha_4 &= 0 \label{eqn:a2v} \\ \alpha_1 \alpha_2 \alpha_3 + \alpha_2 \alpha_3 \alpha_4 + \alpha_3 \alpha_4 \alpha_1 + \alpha_4 \alpha_1 \alpha_2 &= -r \label{eqn:a3v} \\ \alpha_1 \alpha_2 \alpha_3 \alpha_4 &= s \label{eqn:a4v} \\ \beta_1 + \beta_2 + \beta_3 &= a \label{eqn:b1v} \\ \beta_1 \beta_2 + \beta_2 \beta_3 + \beta_3 \beta_1 &= b \label{eqn:b2v} \\ \beta_1 \beta_2 \beta_3 &= c \label{eqn:b3v} \end{align} Combining $$\eqref{eqn:b1v}$$ with $$\eqref{eqn:bdef}$$ and $$\eqref{eqn:a2v}$$, we find $$a = 0$$. The expression $$\eqref{eqn:b2v}$$ for $$b$$ expands to $b = \sum \alpha_1^2 \alpha_2 \alpha_3$ with the sum taken over the cyclic permutations of $$\alpha_1 \ldots \alpha_4$$. This equals $$\left(\sum \alpha_1 \alpha_2 \alpha_3 \right) \left(\sum \alpha_1 \right) - 4\alpha_1\alpha_2\alpha_3\alpha_4 = (-r)(0) - 4s = -4s$$, where we have used $$\eqref{eqn:a1v}, \eqref{eqn:a3v}, \eqref{eqn:a4v}$$. We could also use the expression $$\eqref{eqn:b3v}$$ for $$c$$ to compute $$c$$ in the same manner, but this is laborious, so instead we will use the fact that the discriminant of the quartic equals that of its resolvent cubic, and we know how to compute the latter since it is already known to have vanishing quadratic term. That is, $$-27r^4 + 256s^3 = -4b^3 - 27(-c)^2 = -4(-4s)^3 - 27c^2 = 256s^3 - 27c^2$$, therefore $$c$$ is either $$+r^2$$ or $$-r^2$$. The same choice of sign must apply in all cases, since the symmetric functions theorem implies that $$c$$, like $$a$$ and $$b$$, is a polynomial function of $$r$$ and $$s$$. To determine the sign, we merely need to consider the quartic $$x^4 - x$$, that is, with $$r = -1, s = 0$$. Its four roots are $$0, 1, \omega, \omega^2$$, with $$\omega$$ a primitive cube root of unity, so one of the $$\beta$$'s equals $$0 \cdot 1 + \omega \cdot \omega^2 = 1$$. This implies that 1 is a root of the resolvent cubic, which is either $$x^3 + 1$$ or $$x^3 - 1$$. Obviously this means the latter choice is correct, and the general form of the resolvent cubic is $x^3 - 4sx + r^2$

2. The first task is to verify that these polynomials are in fact irreducible. We can quickly check using the rational root test that they don't have a linear factor in $$\Q[x]$$. If a quartic splits into two quadratics, where, say, we label the roots of the first quadratic $$\alpha_1, \alpha_2$$ and those of the second $$\alpha_3, \alpha_4$$, then $$\beta_1 = \alpha_1 \alpha_2 + \alpha_3 \alpha_4$$ will be rational (in fact, Gauss's lemma implies that it will be an integer, but we won't need that result here). By the result of part (a), the quartics $$x^4 + 8x + 12$$ and $$x^4 + 8x - 12$$ respectively have resolvent cubics $$x^3 - 48x + 64$$ and $$x^3 + 48x + 64$$. Using the rational root test we find that neither has a rational root, so both quartics are irreducible.

The discriminant of $$x^4 + 8x + 12$$ is $$-27\cdot 8^4 + 256\cdot 12^3 = 331776 = 576^2$$. We have already determined that its resolvent cubic has no rational root, so it is irreducible, and by (16.9.9), its Galois group is $$A_4$$. The discriminant of $$x^4 + 8x - 12$$ is $$-27\cdot 8^4 + 256\cdot(-12)^3 = -552960$$, and its resolvent cubic is also irreducible, so its Galois group is $$S_4$$.

3. The resolvent cubic of $$x^4 + x - 5$$ is $$x^3 + 20x + 1$$ and the discriminant is $$-27\cdot 1^4 + 256(-5)^3 = -32027$$. The resolvent cubic is irreducible by the rational root test, and the quartic is irreducible by the argument we used in part (b). Therefore, by (16.9.9), the Galois group is $$S_4$$. If $$\alpha$$ is one of the roots of the quartic, this implies that $$\Gal(K/\Q(\alpha))$$ is a subgroup of $$S_4$$ of index 4, which must therefore be isomorphic to $$S_3$$. If $$\Q(\alpha)$$ has a subfield that is a quadratic extension of $$\Q$$, then its Galois group must have index 2 in $$S_4$$ and contain $$S_3$$ as a subgroup. But there is no such subgroup (as the only subgroup of index 2 in $$S_4$$, namely $$A_4$$, has no subgroup of index 2 of its own). Therefore $$\Q(\alpha)$$ can't be obtained by successive quadratic extensions, and is not constructible by ruler and compass. (Likewise, if an irreducible quartic has Galois group $$A_4$$, then the same argument shows that it can't have any subfield of degree 2 over $$\Q$$ at all, so its roots are also not constructible.)

Exercise 16.9.10 According to the result of Exercise 16.2.2(b), a quartic with four distinct roots has positive discriminant if it has zero or four real roots, and negative discriminant if it has two real roots. For irreducible quartics in characteristic zero, the roots are distinct, so (a) and (b) are equivalent. In both cases (16.9.9) tells us that the Galois group must be $$S_4$$, $$D_4$$, or $$C_4$$. However, since complex conjugation is a $$\R$$-automorphism of $$\C$$, it is also a $$\Q$$-automorphism of $$\C$$, so the Galois group of the quartic must include the restriction of complex conjugation to its splitting field, which acts by transposing the two complex roots. This implies that the Galois group cannot be $$C_4$$, so it must be $$D_4$$ or $$S_4$$, depending on whether the resolvent cubic is reducible.

Exercise 16.9.11

1. By the Eisenstein criterion, $$f$$ is irreducible. Write $$\alpha$$ in the form $$\alpha = \sqrt{r + \sqrt{t}}$$ where $$r = 0, t = 2$$. By Exercise 16.9.5(b), since $$r^2 - t = -2$$ which is not a square in $$\Q$$, and $$(r^2 - t)/t = -1$$ which is also not a square in $$\Q$$, the Galois group is $$D_4$$. This implies $$[K:\Q] = 8$$. But $$K = \Q(i, \alpha)$$, so $$\alpha$$ still has degree 4 over $$\Q(i)$$. The subgroup $$\Gal(K/\Q(i))$$ is therefore the Galois group of $$f$$ over $$\Q(i)$$, and contains some element $$\sigma$$ such that $$\sigma(\alpha) = i\alpha$$. But since $$\sigma$$ is a $$\Q(i)$$-automorphism, it follows that $$\sigma(i\alpha) = i\sigma(\alpha) = -\alpha$$, and $$\sigma$$ has degree 4. Therefore $$\sigma$$ belongs to the (unique) $$C_4$$-subgroup of $$\Gal(K/\Q)$$.

2. Given the above discussion in (a), this is trivial.

3. The group $$D_4$$ has five subgroups of order 2 and three subgroups of order 4, so there are five intermediate fields with degree 4 over $$\Q$$ and three with degree 2. The three intermediate fields of degree 2 are obviously $$\Q(\sqrt{2})$$, $$\Q(\sqrt{-2})$$, and $$\Q(i)$$. The five intermediate fields of degree 4 can be found using the result of Exercise 16.9.5(c); they are $$\Q(\alpha)$$, $$\Q(i\alpha)$$, $$\Q((1 + i)\alpha)$$, $$\Q((i - i)\alpha)$$, and $$\Q(\sqrt{2}, \sqrt{-2})$$, alternatively written as $$\Q(\sqrt{2}, i)$$.

Exercise 16.9.12

1. This polynomial is irreducible by the Eisenstein criterion, and has the form described in Exercise 16.9.5(a) with $$r = -2, t = 2$$. By the test we derived in 16.9.5(b), the Galois group is $$C_4$$.

2. The four roots of this polynomial are $$\pm \sqrt{-1 \pm \sqrt{-3}}$$. Obviously none of the four is rational. If this quartic splits over $$\Q$$, it must split into two quadratic factors. But the four roots may be written in denested form as $$\pm \sqrt{1/2} \pm i\sqrt{3/2}$$, so it is easy to see that no two of the four roots simultaneously have both their sum and product rational (or, indeed, real), implying that this quartic is in fact irreducible. That having been settled, we note that it has the form described in Exercise 16.9.5(a) with $$r = -1, t = -3$$. By the test we derived in 16.9.5(b), the Galois group is $$V_4$$.

3. In Exercise 16.9.6, we determined that Galois group of this polynomial is $$V_4$$.

4. By Exercise 16.9.9(a), the discriminant is 229 and the resolvent cubic is $$x^3 - 4x + 1$$. Since $$x^4 + x + 1$$ and $$x^3 - 4x + 1$$ can both be quickly verified to have no rational roots, by the reasoning in Exercise 16.9.9(b), they are both irreducible. By (16.9.9), the Galois group of the quartic is $$S_4$$.

5. We assume the reader knows that this quartic is irreducible and that its roots are $$\omega, \omega^2, \omega^3, \omega^4$$ where $$\omega$$ is one of the primitive fifth roots of unity. All of these lie in $$\Q(\omega)$$, so the splitting field is simply $$\Q(\omega)$$ and the Galois group is either $$C_4$$ or $$V_4$$. In fact, it's clear that the Galois group is just the automorphism group of the multiplicative group generated by $$\omega$$, which is $$C_4$$. (See Proposition 16.10.2(a).)

6. This quartic has the form described in Exercise 16.9.5(a) with $$r = -1/2, t = -3/4$$, and is shown to be irreducible by the same technique applied in (b). By the test we derived in 16.9.5(b), the Galois group is $$V_4$$.

Exercise 16.9.13 The roots of this polynomial are $$\pm \sqrt{1 \pm \sqrt{2}}$$. No root is rational, and two of the four are complex conjugates, so if this quartic splits, it must have a quadratic factor whose roots are the two complex conjugates. However, the product of those two roots is $$\sqrt{2} - 1$$, which is not rational, so we conclude that this quartic is irreducible. By Exercise 16.9.5(b), the Galois group is $$D_4$$. Let $$\alpha = \sqrt{1 + \sqrt{2}}, \alpha' = \sqrt{1 - \sqrt{2}}$$. We know that the Galois group contains an element $$\rho = x^2$$ that exchanges each element with its additive inverse; this preserves the value of $$\alpha^2$$, which is $$1 + \sqrt{2}$$, and the value of $$\alpha\alpha'$$, which is $$i$$, so the intermediate field $$\Q(\sqrt{2}, i)$$ corresponds to the subgroup $$\langle \rho \rangle$$. The other four subfields of degree 4 and their corresponding subgroups are as described in Exercise 16.9.5(c). There are three subgroups of order 4, and three intermediate fields that are quadratic extensions of $$\Q$$. One of these three subgroups is isomorphic to $$V_4$$ and contains the element that fixes $$\alpha$$ while exchanging $$\alpha'$$ and $$-\alpha'$$; this element fixes $$\alpha^2 = 1 + \sqrt{2}$$, so it corresponds to the intermediate field $$\Q(\sqrt{2})$$. The other $$V_4$$-subgroup contains the element that fixes none of the four roots, exchanging $$\alpha$$ with $$\alpha'$$; it fixes $$\alpha\alpha' = i$$, so it corresponds to the intermediate field $$\Q(i)$$. This leaves the last subgroup of index 2, which is isomorphic to $$C_4$$, to correspond to the intermediate field $$\Q(i\sqrt{2})$$.

Exercise 16.9.14

1. Let $$\alpha = \sqrt[3]{2 + \sqrt{2}}$$. It is easy to guess that the degree of $$\alpha$$ over $$\Q$$ should be 6, with the six roots given by choosing the three possible cube roots and two possible square roots, that is, the roots are $$\alpha, \omega\alpha, \omega^2\alpha$$ together with $$\alpha', \omega\alpha', \omega^2\alpha'$$ where $$\alpha' = \sqrt[3]{2 - \sqrt{2}}$$. The polynomial with these six roots is $$x^6 - 4x^3 + 2$$, which is irreducible at a glance by the Eisenstein criterion. In fact, since $$\Z[\omega]$$ is a UFD, and 2 is an Eisenstein prime, the generalized form of the Eisenstein criterion implies that this polynomial is also irreducible in $$\Q(\omega)$$.

Let $$\alpha_0, \ldots, \alpha_5$$ respectively equal $$\alpha$$, $$\alpha'$$, $$\omega\alpha$$, $$\omega\alpha'$$, $$\omega^2\alpha$$, and $$\omega^2\alpha'$$. Let $$K$$ denote the splitting field of $$\alpha$$. For each $$\sigma \in \Gal(K/\Q(\omega))$$, we must have that $$\sigma(\omega^j \alpha) = \omega^j \sigma(\alpha)$$ and $$\sigma(\omega^j \alpha') = \omega^j \sigma(\alpha')$$, or in other words, $$\sigma(\alpha_{j+2}) = \omega \sigma(\alpha_j)$$ for each $$j$$, modulo 6. This implies that the Galois group is a subgroup of the group of 18 possible permutations of the $$\alpha_i$$'s that satisfy this identity.

In Exercise 7.7.9(b) we classified the groups of order 18. In particular, the group of order 18 in question contains exactly 3 elements of order 2, and only one of the five groups of order 18, namely $$S_3 \times C_3$$, has this property. The degree $$[K : \Q(\omega)]$$ is a multiple of 6, so the Galois group has order either 6 or 18. Thus, $$\Gal(K/\Q(\omega))$$ is isomorphic to one of $$S_3 \times C_3$$, $$S_3$$, or $$C_6$$. It's not obvious that $$\alpha' \notin \Q(\omega, \alpha)$$; if we could establish this, it would immediately follow that $$[K : \Q(\omega)] = 18$$ and that the Galois group is $$S_3 \times C_3$$. Instead we will have to resort to Galois-theoretic arguments.

There is only a single subgroup of $$S_3 \times C_3$$ that is isomorphic to $$S_3$$, and it consists of the identity together with $$(0\ 1)(2\ 3)(4\ 5)$$, $$(0\ 3)(2\ 5)(4\ 1)$$, $$(0\ 5)(2\ 1)(4\ 3)$$, $$(0\ 2\ 4)(1\ 5\ 3)$$, and $$(0\ 4\ 2)(1\ 3\ 5)$$. There are three subgroups isomorphic to $$C_6$$, namely $$\langle(0\ 1\ 2\ 3\ 4\ 5)\rangle$$, $$\langle(0\ 3\ 2\ 5\ 4\ 1)\rangle$$, and $$\langle(0\ 5\ 2\ 1\ 4\ 3)\rangle$$. These four subgroups are all transitive.

Let us entertain the hypothesis that the Galois group might be the $$S_3$$-isomorphic group described above. All elements of this subgroup preserve the identity $$\alpha\alpha' = \sqrt[3]{2}$$, so their fixed field would contain $$\Q(\omega, \sqrt[3]{2})$$, which is larger than $$\Q(\omega)$$, a contradiction.

Conveniently, the three possible subgroups isomorphic to $$C_6$$ contain the common subgroup $$\langle(0\ 2\ 4)(1\ 3\ 5)\rangle$$, and this subgroup fixes the field $$\Q(\omega, \sqrt[3]{3 + 2\sqrt{2}})$$ since the permutation $$(0\ 2\ 4)(1\ 3\ 5)$$ fixes $$\alpha/\alpha' = \sqrt[3]{3 + 2\sqrt{2}}$$. Let us denote this latter quantity by $$\beta$$. $$\beta$$ is a root of the polynomial $$x^3 - (3 + 2\sqrt{2}) = 0$$ with coefficients in $$R = \Z[\sqrt{2}]$$. Observe that $$R$$ is norm-euclidean and that $$3 + 2\sqrt{2}$$ has norm 1. Any cube root of $$3 + 2\sqrt{2}$$ in $$R$$ would also have to have norm 1; this makes it fairly easy to determine that such a cube root doesn't exist. Again, since $$R$$ is norm-euclidean, it is also a UFD, and the irreducibility of $$x^3 - (3 + 2\sqrt{2})$$ over $$R[x]$$, by Gauss's lemma, implies its irreducibility over $$\Q(\sqrt{2})[x]$$; thus, $$\beta$$ has degree 3 over $$\Q(\sqrt{2})$$. Now $$\Q(\beta)$$ contains $$\sqrt{2}$$, so both 2 and 3 must divide $$[\Q(\beta) : \Q]$$, implying that the latter is equal to 6. Since $$\Q(\beta)$$ is a subfield of $$\R$$, we further have that $$\Q(\beta, \omega)$$ is strictly larger than $$\Q(\beta)$$, and thus has degree 12 over $$\Q$$. Then, since $$[\Q(\beta, \omega) : \Q] = [\Q(\beta, \omega) : \Q(\omega)] [\Q(\omega) : \Q]$$, it follows that $$\beta$$ has degree 6 over $$\Q(\omega)$$. But since $$\beta$$ is fixed by the subgroup of order 3 described above, this implies that $$[K : \Q(\omega)]$$ is at least 18, thus contradicting our hypothesis that the Galois group is isomorphic to $$C_6$$.

This leaves us with the conclusion that the only possible Galois group is the full $$S_3 \times C_3$$, and that $$[K : \Q(\omega)] = 18$$.

Note: As Artin points out in the text, we must be careful not to draw conclusions from permutations of the roots that do not correspond to any actual $$F$$-automorphism of $$K$$ when reasoning about the extension $$K/F$$. That's why the proofs above are by contradiction: attempting to determine the order of $$\Gal(K/\Q(\omega))$$ simply by exhibiting the fields $$\Q(\sqrt[3]{2}, \omega)$$ and $$\Q(\beta, \omega)$$ together with permutations of the roots that fix them, when we have not first proven that said permutations actually belong to the Galois group, would be circular reasoning.

2. Let $$\alpha = \sqrt{2 + \sqrt[3]{2}}$$, and write $$\alpha_0 = \alpha$$, $$\alpha_1 = \sqrt{2 + \omega\sqrt[3]{2}}$$, $$\alpha_2 = \sqrt{2 + \omega^2\sqrt[3]{2}}$$, $$\alpha_3 = -\alpha_0$$, $$\alpha_4 = -\alpha_1$$, $$\alpha_5 = -\alpha_2$$. As in part (a), it's easy to determine a polynomial with integer coefficients of which $$\alpha$$ is a root, namely $$\prod_{j=0}^5 (x - \alpha_j) = x^6 - 6x^4 + 12x^2 - 10$$, which, conveniently, can again be shown to be irreducible over $$\Q(\omega)$$ by the Eisenstein criterion.

Since $$\omega^2(\alpha_1^2 - 2) = \sqrt[3]{2} \in \Q(\alpha, \omega)$$, the degree of $$\alpha_1$$ over $$\Q(\alpha, \omega)$$ is at most 2; in particular, if $$\sigma \in \Gal(K/\Q(\omega))$$ (where $$K$$ is the splitting field) and $$\sigma(\alpha_0) = \alpha_j$$, then since $$(\alpha_0^2 - 2) = \omega^2(\alpha_1^2 - 2) = \omega(\alpha_2^2 - 2)$$, it follows that $$\sigma(\alpha_1) = \pm \alpha_{j + 1}$$ and $$\sigma(\alpha_2) = \pm \alpha_{j+2}$$. Let $$G \subseteq S_6$$ be the group of permutations of the labels $$0, \ldots, 5$$ that satify the above identities. Thus, the Galois group is a subgroup of $$G$$. The group $$G$$ has order 24. We observe that we can freely and independently exchange each of $$\alpha_0, \alpha_1, \alpha_2$$ with its additive inverse, so $$G$$ contains a subgroup isomorphic to $$E_8$$ generated by $$\sigma_0 = (0\ 3)$$, $$\sigma_1 = (1\ 4)$$, $$\sigma_2 = (2\ 5)$$. $$G$$ also contains some elements of order 3, such as $$\tau = (0\ 2\ 4)(1\ 3\ 5)$$. In fact, conjugation by $$\tau$$ acts on the $$E_8$$-subgroup as cyclic permutation of $$(0\ 3)$$, $$(2\ 5)$$, and $$(1\ 4)$$, therefore $$G$$ is the regular wreath product $$C_2 \wr C_3$$, which is known to be isomorphic to $$A_4 \times C_2$$. (Note to readers: it's also possible to verify this isomorphism directly without introducing the wreath product; it's simply more work.) The centre of this group consists only of $$C_2$$, so the element $$(0, 1) \in A_4 \times C_2$$ corresponds to the element $$\sigma_0 \sigma_1 \sigma_2$$, which is central in $$G$$. The group $$A_4 \times C_2$$ has a single subgroup isomorphic to $$E_8$$, namely $$V_4 \times C_2$$, so the subgroup of $$\langle \sigma_0, \sigma_1, \sigma_2\rangle$$ of order 4 that is normal in $$G$$ corresponds to the $$V_4$$-subgroup of the $$A_4$$ factor of $$G$$. Bearing in mind the action of conjugation by $$\tau$$ described above, this $$V_4$$-subgroup can only be $$\{1, \sigma_0 \sigma_1, \sigma_1 \sigma_2, \sigma_2 \sigma_0\}$$. Thus, the $$A_4$$ factor of $$G$$ is generated by $$\tau$$ and products of an even number of the $$\sigma$$'s.

To show that $$G$$ is the Galois group of $$K/\Q(\omega)$$, we can take an approach similar to that of part (a), in which we show that none of its proper subgroups can be the Galois group sought. The proper subgroups of $$G$$ that can possibly be the Galois group have order 6 or 12, since the order of the Galois group must be a multiple of 6. There is a single subgroup of $$A_4 \times C_2$$ of order 12, namely $$A_4$$ itself (since $$A_4$$ has no subgroup of order 6). There are four subgroups of $$A_4 \times C_2$$ of order 6, each generated by one of the four independent 3-cycles of $$A_4$$ and $$C_2$$ (again, since $$A_4$$ has no subgroup of order 6); thus, each of the subgroups of order 6 is isomorphic to $$C_6$$ and a possible set of generators is $$(0\ 1\ 2\ 3\ 4\ 5)$$, $$(0\ 1\ 5\ 3\ 4\ 2)$$, $$(0\ 4\ 2\ 3\ 1\ 5)$$, $$(0\ 4\ 5\ 3\ 1\ 2)$$.

As we found earlier, the $$A_4$$-subgroup of $$G$$ is generated by $$\sigma_0 \sigma_1$$, $$\sigma_1 \sigma_2$$, and $$\tau$$. Now $$\alpha_0 \alpha_1 \alpha_2 = \sqrt{10}$$ and applying an even number of $$\sigma$$'s changes two signs on the LHS, leaving the result $$\sqrt{10}$$ unchanged, while $$\tau(\alpha_0 \alpha_1 \alpha_2) = \alpha_2\alpha_3\alpha_4$$ and again two signs are flipped, leaving the result $$\sqrt{10}$$ unchanged. If the Galois group were assumed to be $$A_4$$, then this entire group would fix $$\Q(\omega, \sqrt{10})$$, a contradiction. Thus, $$A_4$$ can be ruled out as a possible Galois group.

Suppose now that the Galois group is isomorphic to $$C_6$$. This would imply that $$\alpha_1, \alpha_2 \in \Q(\alpha, \omega)$$, which appears to not be the case, so we will see whether we can show that a hypothetical Galois group of order 6 would actually fix an intermediate field of degree 4. The permutation $$(0\ 1\ 2\ 3\ 4\ 5)$$ of the roots would fix the expression $$\alpha_0\alpha_1 + \alpha_2\alpha_3 + \alpha_4\alpha_5$$. Likewise (we can check) the other three generators of $$C_6$$-subgroups fix similar expressions, each obtained by changing the signs of two of the terms. We suspect that these four subgroups together fix four roots of a quartic whose Galois group is $$A_4$$, since we are expecting their intersection, the $$C_2$$-subgroup generated by $$(0\ 3)(1\ 4)(2\ 5)$$, to fix an intermediate field of degree 12. Indeed, some algebra shows that \begin{align*} \phantom{=}& \, (x - \alpha_0\alpha_1 - \alpha_2\alpha_3 - \alpha_4\alpha_5) (x - \alpha_0\alpha_1 + \alpha_2\alpha_3 + \alpha_4\alpha_5) (x + \alpha_0\alpha_1 - \alpha_2\alpha_3 + \alpha_4\alpha_5) (x + \alpha_0\alpha_1 + \alpha_2\alpha_3 - \alpha_4\alpha_5) \\ =&\, ((x - \alpha_0\alpha_1)^2 - (\alpha_2\alpha_3 + \alpha_4\alpha_5)^2) ((x + \alpha_0\alpha_1)^2 - (\alpha_2\alpha_3 - \alpha_4\alpha_5)^2) \\ =&\, (x^2 - 2\alpha_0\alpha_1 x + \alpha_0^2\alpha_1^2 - \alpha_2^2\alpha_3^2 -\alpha_4^2\alpha_5^2 - 2\alpha_2\alpha_3\alpha_4\alpha_5) (x^2 + 2\alpha_0\alpha_1 x + \alpha_0^2\alpha_1^2 - \alpha_2^2\alpha_3^2 -\alpha_4^2\alpha_5^2 + 2\alpha_2\alpha_3\alpha_4\alpha_5) \\ =&\, (x^2 + \alpha_0^2\alpha_1^2-\alpha_2^2\alpha_3^2-\alpha_4^2\alpha_5^2)^2 - (2\alpha_0\alpha_1x + 2\alpha_2\alpha_3\alpha_4\alpha_5)^2 \\ =&\, x^4 + 2(\alpha_0^2 \alpha_1^2 - \alpha_2^2\alpha_3^2 - \alpha_4^2\alpha_5^2)x^2 - 4\alpha_0^2\alpha_1^2 x^2 - 8\alpha_0\alpha_1\alpha_2 \alpha_3\alpha_4\alpha_5 x + (\alpha_0^2\alpha_1^2 - \alpha_2^2\alpha_3^2 - \alpha_4^2\alpha_5^2)^2 - 4\alpha_2^2\alpha_3^2\alpha_4^2\alpha_5^2 \\ =&\, x^4 - 2(\alpha_0^2\alpha_1^2 + \alpha_2^2\alpha_3^2 + \alpha_4^2\alpha_5^2)x^2 + 8\alpha_0^2\alpha_1^2\alpha_2^2 x + \alpha_0^4\alpha_1^4 + \alpha_2^4\alpha_3^4 + \alpha_4^4\alpha_5^4 - 2(\alpha_0^2 + \alpha_1^2 + \alpha_2^2)\alpha_0^2\alpha_1^2\alpha_2^2 \\ =&\, x^4 - 24x^2 - 80x - 96 \end{align*} For $$x \ge 7$$, the quartic term will exceed the sum of the absolute values of the other three terms, so the four roots of this quartic must have absolute value less than 7. Using the rational root theorem, we rule out $$\pm 1, \pm 2\, \pm 3, \pm 4, \pm 6$$ and conclude that there are no rational roots. The resolvent cubic is $$x^3 + 24x^2 + 384x + 2816$$ (we will see in Exercise 16.9.16 how to calculate this); any real roots must be negative, and less than 44 in absolute value. The integer candidates $$-1, -2, -4, -8, -11, -16, -22, -32$$ are ruled out, and the resolvent cubic is irreducible, implying that the quartic doesn't split into two quadratics in $$\Q[x]$$. The element $$\alpha_0 \alpha_1 + \alpha_2 \alpha_3 + \alpha_4 \alpha_5$$ therefore has degree 4 over $$\Q$$, which implies that its degree over $$\Q(\omega)$$ is either 2 or 4. Thus, if $$\langle(0\ 1\ 2\ 3\ 4\ 5)\rangle$$ is the Galois group, then it fixes a nontrivial extension of $$\Q(\omega)$$, a contradiction. The irreducibility of this quartic implies that its other three roots also have degree at least 2 over $$\Q(\omega)$$, so the other three candidate $$C_6$$-subgroups are also ruled out, and the Galois group must be the full 24-element group $$A_4 \times C_2$$.

(Note from Brian: This solution contains a lot of brute force calculation, and I imagine a better solution exists. Feel free to write me to contribute one.)

Exercise 16.9.15 The idea is based on Ferrari's method of solving the depressed quartic. The quartic $$y^4 + py^2 + qy + r$$ is given. We would like to factor it by writing it as a difference of squares, $$y^4 + py^2 + qy + r = (y^2 + z/2)^2 - P(y)^2 \label{eqn:quart-diffsq}$$ where $$P$$ is a linear polynomial. Explicitly, the above implies that $$P(y)^2 = (z-p)y^2 - qy + (z^2/4 - r) \label{eqn:quart-linsq}$$ and in order for this to be a perfect square, we must choose $$z$$ so that the discriminant $$q^2 - 4(z-p)(z^2/4 - r) = -(z^3 - pz^2 - 4rz + (4pr - q^2))$$ vanishes. The cubic $$z^3 - pz^2 - 4rz + (4pr - q^2)$$ that appears on the right-hand side is the resolvent cubic (we will see this in Exercise 16.9.16). We assume that a solution $$z$$ is known. By \eqref{eqn:quart-diffsq} and \eqref{eqn:quart-linsq}, we have \begin{align*} y^4 + py^2 + qy + r &= (y^2 + z/2)^2 - \left(\sqrt{z-p}y - \frac{q}{2\sqrt{z-p}}\right)^2 \\ &= \left(y^2 + \sqrt{z-p}y + z/2 - \frac{q}{2\sqrt{z-p}}\right) \left(y^2 - \sqrt{z-p}y + z/2 + \frac{q}{2\sqrt{z-p}}\right) \label{eqn:depr_fact} \end{align*} so one solution is given by applying the quadratic formula to, say, the first factor, $$y = \frac{1}{2}\left(\sqrt{\frac{2q}{\sqrt{z-p}} - z - p} - \sqrt{z-p}\right) \label{eqn:depr_sol}$$

The exercise does not specify that the quartic is in depressed form, so we will now work with the general quartic $$x^4 + ax^3 + bx^2 + cx + d$$. We must first rewrite it in depressed form by making the substitution $$x = y - a/4$$, giving $$y^4 + px^2 + qx + r$$ where \begin{align} p &= b - \frac{3}{8}a^2 \label{eqn:depr_p} \\ q &= \frac{1}{8}a^3 - \frac{1}{2}ab + c \label{eqn:depr_q} \\ r &= d - \frac{1}{4}ac + \frac{1}{16}a^2b - \frac{3}{256}a^4 \label{eqn:depr_r} \end{align} The resolvent cubic of the original quartic has roots $$\alpha_1\alpha_2 + \alpha_3\alpha_4$$, $$\alpha_1\alpha_3 + \alpha_2\alpha_4$$, and $$\alpha_1\alpha_4 + \alpha_2\alpha_3$$, where $$\alpha_i$$ are the roots of the original quartic. Let $$\gamma_i = \alpha_i + a/4$$. The resolvent cubic of the depressed quartic has roots $$\gamma_1 \gamma_2 + \gamma_3 \gamma_4$$ and so on. Now \begin{align} \gamma_1 \gamma_2 + \gamma_3 \gamma_4 &= (\alpha_1 + a/4)(\alpha_2 + a/4) + (\alpha_3 + a/4)(\alpha_4 + a/4) \\ &= \alpha_1 \alpha_2 + \alpha_3\alpha_4 + (\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4)(a/4) + a^2/16 + a^2/16 \\ &= \alpha_1 \alpha_2 + \alpha_3\alpha_4 - \frac{1}{8}a^2 \label{eqn:resolvent-diff} \end{align} so if $$\beta$$ is a root of the resolvent cubic of the original quartic, then the corresponding root $$z$$ of the resolvent cubic of the depressed quartic is $$z = \beta - a^2/8$$. Using \eqref{eqn:depr_sol}, \eqref{eqn:depr_p}, \eqref{eqn:depr_q}, and \eqref{eqn:depr_r}, we find: \begin{align*} \alpha_1 &= y - a/4 \\ &= -\frac{a}{4} + \frac{1}{2}\left(\sqrt{\frac{2q}{\sqrt{z-p}} - z - p} - \sqrt{z-p}\right) \\ &= -\frac{a}{4} - \frac{1}{2}\sqrt{\beta - a^2/8 - b + 3a^2/8} + \frac{1}{2} \sqrt{-\beta + a^2/8 - b + 3a^2/8 + \frac{a^3/4 - ab + 2c}{\sqrt{\beta - a^2/8 - b + 3a^2/8}}} \\ &= -\frac{a}{4} - \frac{1}{2}\sqrt{\frac{a^2}{4} - b + \beta} + \frac{1}{2} \sqrt{\frac{a^2}{2} - b - \beta + \frac{1}{4}\frac{a^3 - 4ab + 8c} {\sqrt{\frac{a^2}{4} - b + \beta}}} \end{align*}

Exercise 16.9.16 The solution to the general depressed quartic by Ferrari's method was given in Exercise 16.9.15. To solve the equation $$y^4 + py^2 + qy + r = 0$$ we made use of an auxiliary cubic, $$z^3 - pz^2 - 4rz + (4pr - q^2)$$. We will refer to the monic cubic with roots $$\alpha_1\alpha_2 + \alpha_3\alpha_4$$ and its two cyclic permutations as the resolvent cubic. Our objective is to show that the auxiliary and resolvent cubics are the same. (The proper logical order is to read the description of Ferrari's solution in Exercise 16.9.15 first, then read the explanation here of why the auxiliary and resolvent cubics are identical, and then finally go back to Exercise 16.9.15 to express the roots of the quartic in terms of that cubic.)

A depressed quartic with roots $$\alpha_1, \alpha_2, \alpha_3, \alpha_4$$ will have three possible factorizations into two quadratic factors, namely \begin{align*} x^4 + px^2 + qx + r &= (x^2 - (\alpha_1 + \alpha_2)x + \alpha_1\alpha_2) (x^2 - (\alpha_3 + \alpha_4)x + \alpha_3\alpha_4) \\ &= (x^2 - (\alpha_1 + \alpha_2)x + \alpha_1\alpha_2) (x^2 + (\alpha_1 + \alpha_2)x + \alpha_3\alpha_4) \\ &= \left(x^2 + \frac{\alpha_1\alpha_2 + \alpha_3\alpha_4}{2} - \frac{\alpha_3\alpha_4 - \alpha_1\alpha_2}{2} - (\alpha_1+\alpha_2)x\right) \left(x^2 + \frac{\alpha_1\alpha_2 + \alpha_3\alpha_4}{2} + \frac{\alpha_3\alpha_4 - \alpha_1\alpha_2}{2} + (\alpha_1+\alpha_2)x\right) \\ &= \left(x^2 + \frac{\alpha_1\alpha_2 + \alpha_3\alpha_4}{2}\right)^2 - \left((\alpha_1+\alpha_2)x + \frac{\alpha_3\alpha_4 - \alpha_1\alpha_2}{2}\right)^2 \end{align*} and its two cyclic permutations. That is, each of the three factorizations of the depressed quartic into two quadratics takes the form \eqref{eqn:quart-diffsq}, implying that $$z = \alpha_1\alpha_2 + \alpha_3\alpha_4$$ is a root of the auxiliary cubic previously described.

As discussed in the text, the difference between two roots of the resolvent cubic can be written as a product of differences of roots of the original quartic.

If all four roots of the depressed quartic are distinct, then the three possible factorizations into two quadratic factors are distinct. Since each of the three distinct factorizations can be put in the form \eqref{eqn:quart-diffsq}, each of the three must be obtained from some root of the auxiliary cubic (which is needed in order to make the second term on the right-hand side of \eqref{eqn:quart-diffsq} a perfect square). Thus, the three roots of the auxiliary cubic are precisely the three cyclic permutations of $$z = \alpha_1\alpha_2 + \alpha_3\alpha_4$$, that is, the roots of the resolvent cubic. Thus, the auxiliary and resolvent cubics coincide, $$R(z) = z^3 - pz^2 - 4rz + (4pr - q^2) \label{eqn:depr_resolv}$$

If the depressed quartic has fewer than four distinct roots, this presents a complication since we need to establish that the auxiliary and resolvent cubic not only have the same two roots but also with the same multiplicity. A way to circumvent this difficulty is by using a continuity argument. This is left as an exercise for the reader.

We now return to the general cubic $$x^4 + ax^3 + bx^2 + cx + d$$. According to \eqref{eqn:resolvent-diff}, the roots of the resolvent cubic equal $$a^2/8$$ plus that of the resolvent cubic of the depressed quartic. Thus, the resolvent cubic is \begin{align*} R(z) &= (z - a^2/8)^3 - p(z - a^2/8)^2 - 4r(z - a^2/8) + (4pr - q^2) \\ &= z^3 + (-p - 3a^2/8)z^2 + (3a^4/64 + a^2p/4 - 4r)z + (-a^6/512 - a^4p/64 + a^2r/2 + 4pr - q^2) \\ &= z^3 - bz^2 + (3a^4/64 + a^2(b - 3a^2/8)/4 - 4r)z \\ &\quad + (-a^6/512 - a^4(b - 3a^2/8)/64 + a^2r/2 + 4(b - 3a^2/8)r - q^2) \\ &= z^3 - bz^2 + (-3a^4/64 + a^2b/4 - 4r)z + (a^6/256 - a^4b/64 + 4br - a^2r - q^2) \\ &= z^3 - bz^2 + (-3a^4/64 + a^2b/4 -4d + ac - a^2 b/4 + 3a^4/64)z \\ &\quad + a^6/256 - a^4b/64 + 4bd - abc + a^2b^2/4 - 3a^4b/64 - a^2d + a^3c/4 \\ &\quad - a^4b/16 + 3a^6/256 - a^6/64 - a^2b^2/4 - c^2 + a^4b/8 + abc - a^3c/4) \\ &= z^3 - bz^2 + (ac - 4d)z + (4bd - a^2d - c^2) \end{align*}

Exercise 16.9.17 Let $$K$$ be the splitting field of a quartic. By Theorem 15.5.10, if $$K$$ can be obtained by successive quadratic extensions to $$\Q$$, then every element of $$K$$ is constructible, and therefore the four roots are constructible.

By the fundamental theorem, successive quadratic extensions of $$\Q$$ correspond to a nested sequence of subgroups of index 2 starting from the full Galois group and ending in the trivial group. Such nested sequences exist for the groups $$C_4$$, $$V_4$$, and $$D_4$$, so it is clear that if $$\Gal(K/\Q)$$ is one of these three groups, then all four roots are constructible.

Conversely, suppose all four roots $$\alpha_1, \ldots, \alpha_4$$ are constructible. The field $$K$$ is $$\Q(\alpha_1, \ldots, \alpha_4)$$. By Theorem 15.5.6, each $$\Q(\alpha_i)$$ can be obtained by successive quadratic extensions, although the final field in the tower is not necessarily contained within $$K$$: $\Q = F_0 \subseteq F_1 \subseteq F_2 \subseteq \ldots \subseteq F_n$ where $$\alpha_1 \in F_n$$, and $$[F_j : F_{j-1}] = 2$$ for each $$j = 1, 2, \ldots, n$$. Let us successively adjoin the square roots needed to reach $$F_n$$, which contains $$\alpha_1$$, then further square roots needed to reach a field containing $$\alpha_2$$, then $$\alpha_3$$ and $$\alpha_4$$. We thus obtain a field extension of $$\Q$$ with degree equal to a power of two, which contains all four roots, and therefore has $$K$$ as an intermediate field. Therefore $$[K : \Q]$$ must also be a power of two. This implies that when the Galois group is $$A_4$$ (order 12) or $$S_4$$ (order 24), it's not possible for all four roots to be constructible.

We haven't yet addressed the question of whether it's possible for some of the roots to be constructible but not others. We now do so, based on an idea provided Math Stack Exchange. Suppose $$\alpha_1$$ is constructible. By Theorem 15.5.6, we have a sequence of quadratic extensions $\Q = F_0 \subseteq F_1 \subseteq \ldots \subseteq F_n$ with $$\alpha_1 \in F_n$$. Note that $$F_n$$ is not necessarily contained within $$K$$. By Theorem 16.6.6(c), there exists $$\sigma \in \Gal(K/\Q)$$ such that $$\sigma(\alpha_1) = \alpha_2$$. Let $$L$$ be a Galois extension of $$\Q$$ that contains the composite field $$KF_n$$. Since $$K$$ and $$L$$ are both Galois extensions of $$\Q$$, and $$K \subseteq L$$, Theorem 16.7.5 implies that the $$F$$-automorphism $$\sigma : K \to K$$ may be (possibly non-uniquely) extended to an $$F$$-automorphism $$\overline\sigma : L \to L$$ sending $$\alpha_1$$ to $$\alpha_2$$. Then $\Q = \overline\sigma(\Q) = \overline\sigma(F_0) \subseteq \overline\sigma(F_1) \subseteq \ldots \subseteq \overline\sigma(F_n)$ with $$\alpha_2 \in \overline\sigma(F_n)$$ and each extension in this tower quadratic, so by Theorem 15.5.10, $$\alpha_2$$ is also constructible. Thus, if one root is constructible, so is any other.

We conclude that if the Galois group of the quartic is $$C_4$$, $$V_4$$, or $$D_4$$, then all four roots are constructible, while if it is $$A_4$$ or $$S_4$$, then none of the four roots are constructible.

Exercise 16.9.18 Let $$K/F$$ be a Galois extension whose Galois group is $$D_4 = \langle x, y \mid x^4 = y^2 = 1, yx = x^{-1}y \rangle$$. (Although $$D_4$$ has nontrivial automorphisms, it doesn't matter which elements we choose to label as $$x$$ and $$y$$ as long as they satisfy the identities above.) The Klein four-subgroup $$\langle x^2, y \rangle$$ fixes a quadratic field extension $$L/F$$ which, by Proposition 15.3.3, is equal to $$F(\sqrt{d})$$ for some $$d \in F$$. The fixed field of the subgroup $$\langle y\rangle$$ of order 2 is itself a quadratic extension $$L'/L$$, which is, again, generated by adjoining some square root to $$L$$; that is, $$L' = L(\sqrt{p + q\sqrt{d}})$$ for some $$p, q \in F$$.

If $$q = 0$$, then $$L' = F(\sqrt{d}, \sqrt{p})$$ and $$L'$$ contains a subfield $$F(\sqrt{p})$$. The subgroup of $$D_4$$ corresponding to $$F(\sqrt{p})$$ must have order 4 and contain $$y$$. But the only subgroup of $$D_4$$ of order 4 containing $$y$$ is $$\langle x^2, y\rangle$$, which already corresponds to $$F(\sqrt{d})$$. Therefore $$L'$$ only has degree 2 over $$F$$, a contradiction. So in fact $$q \ne 0$$, and $$F(\sqrt{p + q\sqrt{d}})$$ contains $$\sqrt{d}$$, so $$L' = L(\sqrt{p + q\sqrt{d}}) = F(\sqrt{p + q\sqrt{d}})$$. This implies that $$\alpha = \sqrt{p + q\sqrt{d}}$$ has degree 4 over $$F$$. Its minimal polynomial is $$(z^2 - p)^2 - q^2d = 0$$.

We know that $$\alpha$$ is fixed by $$y$$ but not by $$x^2$$; however, $$\alpha^2 = p + q\sqrt{d}$$ is in $$F(\sqrt{d})$$, which is fixed by $$x^2$$. Therefore $$x^2(\alpha)$$ can only be the other square root, namely $$-\sqrt{p + q\sqrt{d}}$$.

Since $$F(\alpha)$$ is the fixed field of $$\langle y\rangle$$, which has order 2, the orbit of $$\alpha$$ under the full Galois group has cardinality 4, and therefore consists of the four roots of the minimal polynomial, which are $$\pm\alpha$$ and $$\pm\alpha'$$, where $$\alpha' = \sqrt{p - q\sqrt{d}}$$. We can arbitrarily choose which of the two square roots of $$p - q\sqrt{d}$$ to label as $$\alpha'$$ and which to label as $$-\alpha'$$; let's say $$x(\alpha) = \alpha'$$. The subgroup of $$D_4$$ fixing $$\alpha'$$ is then $$x\langle y\rangle x^{-1} = \langle x^2 y\rangle$$. Therefore $$F(\alpha)$$ and $$F(\alpha')$$ are different fields, and as $$F(\alpha)$$ already has degree 4 over $$F$$, the composite field $$F(\alpha, \alpha')$$ must be the entire field $$K$$. Therefore $$K$$ is the splitting field of the polynomial $$(z^2 - p)^2 - q^2d = 0$$, which has the desired form $$z^4 + bz^2 + c = 0$$ with $$b = -2p$$, $$c = p^2 - q^2d$$.