Prove that $\mathbb{Q}(\alpha) = \mathbb{Q}(\alpha^{2})$ 
Let $f = X^{4}+X^{3}-X+2 \in \mathbb{Q}[X]$ and suppose that $f(\alpha)=0$ with $\alpha \in \mathbb{C}$.  Prove that $\mathbb{Q}(\alpha) = \mathbb{Q}(\alpha^{2})$.

So far i've tried to look at the structure of the groups:
$\mathbb{Q}(\alpha) = \{b+c\alpha\}$
and (assuming that $\alpha^{2}$ is algebraic over $\mathbb{Q}$) 
$\mathbb{Q}(\alpha^{2}) = \{b+c\alpha^{2}\}$ 
it is easy to see that: $\mathbb{Q}(\alpha^{2})\subset\mathbb{Q}(\alpha)$ but it's the other side:
$\mathbb{Q}(\alpha)\subset\mathbb{Q}(\alpha^{2})$
 which bothers me.
any tips/tricks?
Thanks!

$\newcommand{\Q}{\mathbb{Q}}\newcommand{\a}{\alpha}$
As you say, it's clear that $\Q(\a^2) \subset \Q(\a)$, so you need to prove $\Q(\a) \subset \Q(\a^2)$. This is equivalent to $\a \in \Q(\a^2)$.
But you know that $\a^4 + \a^3 - \a + 2 = 0$. Equivalently, $(\a^2)^2 + \a (\a^2 - 1) + 2 = 0$. Besides neither $1$ nor $-1$ are roots of $f$, so $\a^2 \neq 1$. Therefore $$\a = \frac{2 + (\a^2)^2}{1-\a^2} \in \Q(\a^2)$$
Note that your description of $\Q(\a)$ and $\Q(\a^2)$ are a bit far off. They are respectively the minimal subfields of $\mathbb{C}$ containing $\a$, resp $\a^2$, so you need to consider all rational fractions (the ones that make sense that is, don't divide by 0). The description $\Q(\beta) = \{ b + c \beta \}$ is only valid in some cases, for example $\beta = \sqrt{2}$ (or more generally when it's a quadratic extension).
