I'm studying transcendental numbers and I have a question, $\mathbb{Q}(\pi) \subsetneq \mathbb{Q}(\sqrt{\pi})$??
I know that $\sqrt{\pi} \notin \mathbb{Q}(\pi),$ because if $\sqrt{\pi}$ is in $\mathbb{Q}(\pi)$ then there exists $p,q \in \mathbb{Q}[x]$ s.t $\sqrt \pi = \frac{p(\pi)}{q(\pi)} \Rightarrow \pi$ is root of $x(q(x))^2 - (p(x))^2$ (Actually is not clear why this polynomial is nonzero).
My question is: How can I prove that $\pi \in \mathbb{Q}(\sqrt{\pi})$?
$\pi\in\Bbb Q(\sqrt \pi)$ because $\pi=(\sqrt \pi)^2$.
Also, from $\sqrt\pi=\frac{p(\pi)}{q(\pi)}$ you obtain that $\sqrt \pi$ is a root of $xq(x^2)-p(x^2)$, which is non-zero because in $xq(x^2)$ has all monomials of odd degree (and it isn't the zero polynomial) and $p(x^2)$ has only of even degree. $\sqrt\pi$ can't be algebraic because then $\pi$ would be too.