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 15.8. Primitive Elements

Exercise 15.8.1 Suppose \(F\) is the finite field \(\mathbb{F}_{p^r}\). Let \(K\) be a finite extension of \(F\). Then \(K\) is also a finite field. According to Theorem 15.7.3(c), the group \(K^\times\) is generated by some element \(\alpha\), so \(F(\alpha) = K\).

Exercise 15.8.2 The field extension \(\Q(\sqrt{2}, \sqrt{3})\) has degree 4 over \(\Q\), so if \(\alpha \in \Q(\sqrt{2}, \sqrt{3}\), then \(d = [\Q(\alpha) : \Q]\) can only be 1, 2, or 4. Obviously \(d = 1\) exactly when \(\alpha \in \Q\). If \(d = 2\), then \(\alpha\) is the root of some quadratic \(x^2 + ax + b\) with \(a, b \in \Q\); this implies that \(\alpha = -\frac{a}{2} \pm \frac{\sqrt{a^2 - 4b}}{2}\), so \(\alpha\) must take the form \(e + f\sqrt{g}\) for some \(e, f, g \in \Q\), and we can choose \(g\) to be a squarefree nonnegative integer. We will use the fact that \(\{1, \sqrt{2}, \sqrt{3}, \sqrt{6}, \sqrt{g}\}\) will be linearly independent over \(\Q\) if \(g\) is any squarefree integer other than 1, 2, 3, or 6. This implies that \(g \in \{1, 2, 3, 6\}\). So if \(\alpha = p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}\) where \(p, q, r, s \in \Q\), then \(\alpha\) must be primitive (have degree 4) whenever at least two of \(q, r, s\) are nonzero, and it is easy to see that this condition is also necessary.