For $K=\mathbb Q(\alpha,\beta)$ with $\beta^2=\alpha^3$ if $\beta \in \mathbb Q(\alpha)$, then $|K:\mathbb Q|$ is finite.

Question

Let $K/\mathbb Q$ be a field extension and there exists $\alpha,\beta \in K$ with $K=\mathbb Q(\alpha,\beta)$ and $\beta^2=\alpha^3$. Then we have to show that:

If $\beta \in \mathbb Q(\alpha)$, then $|K:\mathbb Q|$ is finite.

Clearly $\beta \in \mathbb Q(\alpha)$ implies that $K=\mathbb Q(\alpha)$. I can't find a way to proceed.Contrapositively it will be enough to show that $\alpha$ is transcendental over $\mathbb Q$ will imply that $\beta \notin \mathbb Q(\alpha)$. Since $\{\alpha,\beta\}$ is algebraically dependent we must have $tr.deg.(K/\mathbb Q)=1$. How can I proceed in this way? Some other approach is also equally welcome. Thanks.

Answer 1

Hint: For any $f\in \mathbf Q(x)$, evaluating at any rational number (apart from the finitely many roots of the denominator) makes sense and yields a rational. This is not true for $x^{\frac{3}{2}}$.