Odd degree element over $\mathbb{Q}$

Question

I am working on the following problem:

Let $\alpha$ be an element of odd degree over $\mathbb{Q}$. Prove that $\alpha$ is in the field $\mathbb{Q}(\beta)$ generated over $\mathbb{Q}$ by $\beta = \alpha + \frac{1}{\alpha}$.

I'm not quite sure how to approach this one. If $\alpha$ is an element of odd degree over $\mathbb{Q}$, I know that the corresponding minimal polynomial of $\alpha$ over $\mathbb{Q}$ has odd degree, call it $n$. Then, since $\alpha$ is algebraic of degree $n$ over $\mathbb{Q}$, my idea was to show that $\beta$ is algebraic over $\mathbb{Q}$ of degree greater than $n$. Then, this would mean that $\mathbb{Q}(\alpha)$ is actually a subfield of $\mathbb{Q}(\beta)$, and so it would follow that $\alpha$ is contained in the field $\mathbb{Q}(\beta)$.

How can I show that $\beta$ is algebraic of degree greater than $n$ over $\mathbb{Q}$ ? Does this amount to exploiting the equation $\beta = \alpha + \frac{1}{\alpha}$ ? I don't automatically know that, for example, $\alpha^n = 0$, so I wasn't sure how to do this to prove what I want.

Thanks!

Answer 1

Hint: What are the possible values for the degree $[\mathbb{Q}(\alpha):\mathbb{Q}(\beta)]$ given that $\beta \alpha= \alpha^2 + 1$ ?