Brian Bi
\[ \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal} \]
Return to table of contents for Brian's unofficial solutions to Artin's Algebra

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 \[ \alpha_{2,3} = -\frac{\alpha_1}{2}\left(1 \pm \frac{\delta}{2p\alpha_1 + 3q} \right) \]