let $a = e^{2πi/5}$. What is $[Q(2^{1/5},a) : Q]$?

Question

Where $Q$ is the field of rational numbers.

Attempt: I know that the minimal polynomial of $a$ over $Q$ is $x^4 + x^3 + x^2 + x + 1$. What is the minimal polynomial of $2^{1/5}$ over $Q$? What are the degrees of $Q(a)/Q$ and $Q(2^{1/5})/Q$?

Answer 1

Recall that for two finite field extensions $A \subseteq B \subseteq C$ we have $[C:A] = [C:B]\cdot [B:A]$. This is used extensively in the argument below.

$[\Bbb Q(2^{1/5}):\Bbb Q] = 5$, with minimal polynomial $x^5 - 2$ (irreducible, for instance, via Eisenstein for the prime $2$), while $[\Bbb Q(a):\Bbb Q] = 4$ as you've already figured out.

Now, both $[\Bbb Q(2^{1/5}):\Bbb Q]$ and $[\Bbb Q(a):\Bbb Q]$ must divide $[\Bbb Q(a, 2^{1/5}):\Bbb Q]$, since $\Bbb Q(a)$ and $\Bbb Q(2^{1/5})$ are both intermediate fields. Thus $[\Bbb Q(a, 2^{1/5}):\Bbb Q]$ is divisible by $20$.

But since $2^{1/5}$ also is a root of $x^5 - 2$ in $\Bbb Q(a)$, $[\Bbb Q(a, 2^{1/5}):\Bbb Q(a)]$ is at most $5$ (this time it's not so easy to tell directly that the polynomial is minimal, but that doesn't matter). This means that $[\Bbb Q(a, 2^{1/5}): \Bbb Q]$ cannot be more than $20$.

Take these two together, and you get that the answer is $20$.