I'm trying to solve an exercise as it follows:
$\alpha, \beta \in \mathbb{C}$ such that $a^{37}=2=\beta^{17}$. Note that both are prime.
a) Find the minimum polynomials of $\alpha$ and $\beta$ over $\mathbb{Q}$.
Well, I guess that $(x^{37}-2)$ and $(x^{17}-2)$ are irreducible over $\mathbb{Q}$, hence minimum polynomials. But does that follow from 37,17 being prime, or just from the fact that the polynomial has no zero in $\mathbb{Q}$? How do I write this down rigorously?
b) Compute $[\mathbb{Q}(\alpha, \beta):\mathbb{Q}]$, and compute $[\mathbb{Q}(\alpha\beta):\mathbb{Q}]$!
My guess here on the first one is that this is equal to $[\mathbb{Q}(\alpha, \beta):\mathbb{Q(\beta)}]*[\mathbb{Q}( \beta):\mathbb{Q}]$ and $[\mathbb{Q}(\alpha, \beta):\mathbb{Q(\alpha)}]*[\mathbb{Q}( \alpha):\mathbb{Q}]$, so the first product is $17x$, the second is $37y$, so the dimension of field extension is at least $17\times 37$, but this is also the maximum number (right?), so $17\times 37$ is the answer? and this would be the same answer as for the last part, $\mathbb{Q}(\alpha \beta)$? How do I write this down nicely?