Brian Bi
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.