Minimum polynomials of $\alpha^{37}$ over $\mathbb{Q}$ 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?
 A: Regarding the first part of your question, it's not true in general that if $f\in\mathbb{Q}[x]$ has no rational root then $f$ is irreducible. For example, consider $f(x)=(x^2-2)(x^2+1)$. However, if you are familiar with Eisenstein's criterion for irreducibility, you'll see that it can be used to show that $x^n-2$ is irreducible over $\mathbb{Q}$ for every $n>0$. So in fact this part doesn't depend on those two exponents being prime. However, the fact that they are prime will come into play in the second part of the problem.
A: $\mathbb{Q}(\alpha,\beta)$ is the composite field of $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$, which have relatively prime degrees 37 and 17. Hence, by a standard theorem on extensions, the degree of the composite is 37*17. You are right that the maximum possible degree of a composite is the product of the individual degrees. The minimum pssible degree is their least common multiple.
In regard to Andre's comment, the polynomial $x^{4}+4$ has no rational roots,but factors over the rationals as $(x^{2}+2x+2)(x^{2}-2x+2)$. 
As for part 2 of your question (b), consider a polynomial in $\mathbb{Q}[x]$ and evaluate it at $\alpha\beta$. Can you relate this element to any of the extensions in your final paragraph?
