all algebraic numbers over $\mathbb{Q}(g(\pi))$ are the elements in $\mathbb{Q}$

Question

Let $g(x)$ be a polynomial in $\mathbb{Q}[x]$ such the $deg(g)>1$. I wish to show that $\mathbb{Q}$ are the only algebraic elements in $\mathbb{Q(g(\pi))}$.

I tried to show that $g(\pi)$ is not algebraic, yet even if I assume it, I don't know how to explain why the only elements in $\mathbb{Q}(\alpha)$ when $\alpha$ is not algebraic are $\mathbb{Q}$.

Answer 1

Hint: If $\alpha$ is a complex number that is transcendental over $\mathbb{Q}$ then $\mathbb{Q}(\alpha)$ is isomorphic to $\mathbb{Q}(x)$ i.e., to the field of rational functions over $\mathbb{Q}$.

Transcendental Extensions. $F(\alpha)$ isomorphic to $F(x)$