Fields and Galois theory

a. It is clear that $[F(\alpha) : F] \geq [F(\alpha, \beta) : F(\beta)]$ (in $F(\beta)$ we have more scalars than in $F$, so the dimension of the field generated by $\alpha$ can only decrease), and that $ [F(\alpha, \beta) : F(\beta)][F(\beta : F)] = [F(\alpha, \beta) : F]$, so

\[ [F(\alpha) : F][F(\beta : F)] \geq [F(\alpha, \beta) : F(\beta)][F(\beta : F)] = [F(\alpha, \beta) : F] \geq [F(\alpha\beta) : F] \] as $F(\alpha, \beta) \supseteq F(\alpha\beta)$.

To get a strict inequality, take for example $F = \mathbb{Q}$, $\alpha = \sqrt{2}$, and $\beta = \sqrt{3}$. We have \[ [ \mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = [ \mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2 \] so the RHS is $4$, but $ [ \mathbb{Q}(\sqrt{2}\sqrt{3}) : \mathbb{Q}] = [ \mathbb{Q}(\sqrt{6}) : \mathbb{Q}] = 2$ as $\sqrt{6}$ is a root of $x^2 - 6$ which has degree $2$. 

b.i. We have that $\mathbb{Q}(\sqrt{6}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$, so $[\mathbb{Q}(\sqrt{6}, \sqrt{3}) : \mathbb{Q}(\sqrt{3}) ] \leq 2$. But $\sqrt{2} \not \in \mathbb{Q}(\sqrt{3})$ (it is not a root of $x^2 - 3$), so the inequality cannot be strict, so \[ [\mathbb{Q}(\sqrt{6}, \sqrt{3}) : \mathbb{Q}(\sqrt{3}) ] = 2. \]

b.ii. We have that $2$ is a square modulo $7$ (as $3^2 = 9 = 2 \in \mathbb{F}_7$), so $[\mathbb{F}_7(\sqrt{2}) : \mathbb{F}_7] = 1$. On the other hand, $3$ is not a cube modulo $7$ (the only cubes are $1$ and $-1$), so since $3$ is prime, $[\mathbb{F}_7(\sqrt[3]{3}) : \mathbb{F}_7] = 3]$. In the end, \[ [\mathbb{F}_7(\sqrt{2}, \sqrt[3]{3}) : \mathbb{F}_7] = 3]. \]

b.iii Here, $t$ is a root of the polynomial $x^3 + x - (t^3+t) \in \mathbb{F}_2(t^3+t)[x]$. This polynomial factors as $(x-t)(x^2 + xt + t^2 + 1) \in \mathbb{F}_2(t)[x]$, so it is irreducible in $\mathbb{F}_2(t^3+t)[x]$ and so $[ \mathbb{F}_2(t) : \mathbb{F}_2(t^3+t) ] = 3$. Moreover, it is clear that $\mathbb{F}_4 \not \subseteq \mathbb{F}_2(t)$, so $[ \mathbb{F}_4(t) : \mathbb{F}_2(t) ] = 2$. Putting everything together, \[ [\mathbb{F}_4(t) : \mathbb{F}_2(t^3+t) ] = 6 \] 

c. Let $m=2$. We have that $\sqrt[3]{4\pi+1}$ is a root of $x^3 - 4\pi - 1 \in \mathbb{Q}(\pi)[x]$, so $\sqrt[3]{4\pi+1}$ must be algebraic over $\mathbb{Q}(\pi)$. Suppose by contradiction that it is algebraic over $\mathbb{Q}$. Since $\frac{1}{4}(x^3-1)$ is a polynomial in $\mathbb{Q}[x]$, then $\frac{1}{4}(\sqrt[3]{4\pi+1}^3 - 1) = \frac{1}{4}(4\pi +1 - 1) = \pi$ must also be algebraic over $\mathbb{Q}$, a contradiction.