Extending $\mathbb Q$ with different roots of $x^5-15$ results in different fields. Problem: Let $a,b$ be 2 different roots of $x^5-15$ over $\mathbb C$, show that $\mathbb Q(a) \neq \mathbb Q(b)$ (as subfields of $\mathbb C$).
My attempt: I don't know how to show this, obviously it suffices to show $a$ is the only root in $\mathbb Q(a)$. It's not a general truth that different roots of an irreducable polynomial result in different fields so it must be a special property of $x^5-15$, in particular the free coefficient (since the assertion is not true for $x^5-1$) . I also know the fields are ring-isomorphic but since I don't know all the isomorphisms from $\mathbb Q(a)$ to itself that does not help much either. Help would be appreciated.
 A: A stronger, perhaps simpler, assertion is true here, namely, that inside any algebraically closed field $K\supset\mathbb Q$, for distinct roots $\alpha,\beta$ of $x^5-15=0$, $\mathbb Q(\alpha)\not=\mathbb Q(\beta)$.
By Eisenstein's criterion and Gauss' lemma, for example, $x^5-15$ is irreducible in $\mathbb Q[x]$, so adjoining any single root (inside the extension $K$) generates a degree-five field.
Any two (distinct) roots $\alpha,\beta$ have ratio $\alpha/\beta$ that is a fifth root of unity (not $=1$). A separate discussion of roots of unity tells us that a fifth root of unity generates a degree-four extension of $\mathbb Q$. The multiplicativity of degrees in towers (and coprimality of $5$ and $4$) shows that $\mathbb Q(\alpha,\beta)$ is degree $20$ over $\mathbb Q$. Etc.
A: $x^5 - 15$ has 5 distinct roots since it is relatively prime with its derivative. It has exactly 1 real root, since $f(x) = x^5$ is a strictly increasing continuous function and $0^5 < 15 < 15^5$, so there is a root by the intermediate value theorem.
Let $a$ be the real root, and let $b$ be one of the four roots which is not real. Then $\mathbb{Q}(a) \subseteq \mathbb{R}$, and $b \notin \mathbb{R}$. Therefore, $b \notin \mathbb{Q}(a)$.
Therefore, we see that $\mathbb{Q}(a)$ contains only 1 root of $x^5 - 15$. Since $\mathbb{Q}(b) \simeq \mathbb{Q}(a)$ as rings for any other root $b$, this means that each $\mathbb{Q}(b)$ contains only 1 root of $x^5 - 15$. So for no two distinct roots $a, b$ is $a \in \mathbb{Q}(b)$. Thus, for any two distinct roots $a, b$, we have $\mathbb{Q}(a) \neq \mathbb{Q}(b)$.
