Brian Bi
$\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal}$

## Section 16.8. Cubic Equations

Exercise 16.8.1 (Why is this problem in this section?) The group $$V_4$$ has three subgroups of order 2. This implies that there are three intermediate fields $$L_1, L_2, L_3$$ with degree 2 over $$F$$. We are assuming characteristic zero, so by Proposition 15.3.3, these intermediate fields are generated by square roots: $$L_i = F(\sqrt{\delta_i})$$ with $$\delta_i \in F$$ for $$i = 1, 2, 3$$. The field $$F(\sqrt{\delta_1}, \sqrt{\delta_2})$$ contains both $$L_1$$ and $$L_2$$ and is contained within $$K$$, so the only possibility is that it is a quadratic extension of each of $$L_1$$ and $$L_2$$, and equal to $$K$$. Thus, we have constructed $$K$$ by adjoining to $$F$$ the square roots of two elements.

Each subgroup of $$V_4$$ of order 2 is the fixed field of one of the intermediate fields $$L_i$$. Obviously a subgroup of order 2 contains the identity and one of the three non-identity elements of $$V_4$$; for each of the intermediate fields $$L_i$$, there is therefore exactly one nontrivial element of $$G$$ that fixes $$\sqrt{\delta_i}$$, which is a primitive element for the extension $$L_i/F$$. Thus, the three nontrivial elements of $$G$$ can be labelled $$\sigma_1, \sigma_2, \sigma_3$$ where $$\sigma_i$$ fixes only $$\sqrt{\delta_i}$$. Its action on $$\sqrt{\delta_j}$$ where $$j \ne i$$ must be to take it to $$-\sqrt{\delta_j}$$, since an $$F$$-automorphism must take a root of $$x^2 - \delta_j$$ to another root.

Exercise 16.8.2

1. By (16.2.8), the discriminant is $$-108$$, which is not a square in $$\Q$$, so by Theorem 16.8.5, the Galois group is $$S_3$$.
2. The discriminant is $$-5400$$, which is not a square in $$\Q$$, so the Galois group is $$S_3$$.
3. Upon making the substitution $$x = y + 1$$, the cubic is transformed into the cubic $$y^3 - 3y - 1$$, which has the same discriminant. We can now explicitly evaluate the discriminant using (16.2.8); it is $$81$$, which is a square in $$\Q$$, so the Galois group is $$A_3$$.
4. The discriminant is $$35721$$, which is a square in $$\Q$$, so the Galois group is $$A_3$$.
5. We make the substitution $$x = y - 1/3$$, and the cubic becomes $$y^3 - 7y/3 - 7/27$$, which has discriminant 49, which is a square in $$\Q$$, so the Galois group is $$A_3$$.
6. Using the same substitution as in (e), we must end up with the cubic $$y^3 - 7y/3 + 47/27$$, which has discriminant $$-31$$, which is not a square in $$\Q$$, so the Galois group is $$S_3$$.

Exercise 16.8.3 The quadratic in (16.8.2) can be determined by performing polynomial long division of $$x^3 - a_1 x^2 + a_2 x - a_3$$ by $$x - \beta_1$$ (I changed the symbol for the roots from $$\alpha$$ to $$\beta$$ so it would be less confusing) and assuming that the remainder is zero since $$\beta_1$$ is assumed to be a root. The result is $$x^2 + (\beta_1 - a_1) x + (\beta_1^2 - a_1\beta_1 + a_2)$$. Another way to obtain this result is to solve for the coefficients of the quadratic using the expressions for $$a_1, a_2, a_3$$ in terms of the roots $$\beta_1, \beta_2, \beta_3$$ and the knowledge that the roots of the quadratic will be $$\beta_2, \beta_3$$.

Exercise 16.8.4 Let $$f(x) = x^3 + 2x + 1$$ and $$g(x) = x^3 + x + 1$$. By the rational root test, both $$f$$ and $$g$$ have no roots in $$\Q$$, so they are irreducible over $$\Q$$. Let $$\alpha_1, \alpha_2, \alpha_3$$ denote the roots of $$f$$ in its splitting field $$L \subseteq \C$$. The discriminant of $$f$$ is $$-59$$, so $$[L : \Q] = 6$$. Let $$\beta$$ be a root of $$g$$ in $$\C$$. Assume for the sake of contradiction that $$\beta \in L$$. Since $$L$$ is a splitting field over $$\Q$$, $$g$$ must split completely in $$L$$. Therefore $$L$$ must contain the square root of the discriminant of $$g$$, that is, $$\sqrt{-31}$$. But $$L$$ also contains $$\sqrt{-59}$$, so $$L$$ must contain a subfield with degree 4 over $$\Q$$, namely $$\Q(\sqrt{-31}, \sqrt{-59})$$. Since 4 doesn't divide 6, a contradiction has been reached. So our assumption that $$g$$ has a root in $$L$$ must have been false. Since $$g$$ doesn't have a root in $$L$$, it cannot have a root in the subfield $$\Q(\alpha)$$ either.

Exercise 16.8.5 To make the problem interesting, we have to assume that Artin intended for square roots and irrational constants to be disallowed: that is, we must show explicitly that $$\alpha_2, \alpha_3 \in \Q(\alpha_1, \delta)$$. The approach to do this is described here, although I believe the final result is wrong. Doing the algebra, we obtain $\alpha_{2,3} = -\frac{\alpha_1}{2}\left(1 \pm \frac{\delta}{2p\alpha_1 + 3q} \right)$