Originally Posted by

**ThePerfectHacker** This is good.

This is good.

There is a problem here. (Surprised)

You need to verify that these are minimal polynomials by showing they are irreducible. It is easy to see that $\displaystyle t^5-3$ is irreducible over $\displaystyle \mathbb{Q}$ by Eisenstein. Now since $\displaystyle [\mathbb{Q}(\alpha):\mathbb{Q}]=5$ and $\displaystyle f(t) =t^4+t^3+t^2+t+1$ is irreducible over $\displaystyle \mathbb{Q}$ with $\displaystyle \gcd(4,5)=1$ it follows that $\displaystyle f(t)$ is irreducible over $\displaystyle \mathbb{Q}(\alpha)$. Therefore, $\displaystyle t^4+t^3+t^2+t+1$ is minimal polynomial over $\displaystyle \mathbb{Q}(\alpha)$. We have proven that $\displaystyle \alpha$ has degree $\displaystyle 5$ over $\displaystyle \mathbb{Q}$ and $\displaystyle \varepsilon$ has degree $\displaystyle 4$ over $\displaystyle \mathbb{Q}(\alpha)$. Therefore, a basis for $\displaystyle L/\mathbb{Q}(\alpha)$ is $\displaystyle \{1,\varepsilon,\varepsilon^2,\varepsilon^3\}$ and a basis for $\displaystyle \mathbb{Q}(\alpha)/\mathbb{Q}$ is $\displaystyle \{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$. Thus, $\displaystyle L/\mathbb{Q}$ has a basis consisting of all these possible products between these bases and so there are $\displaystyle 4\cdot 5=20$ elements which means $\displaystyle [L:\mathbb{Q}]=20$.

Just in case you are interested the Galois group is the Frobenius group $\displaystyle F_{20}$.