Determine all algebraic extensions of $Bbb Q$ contained in $Bbb Q(sqrt{2},pi)$

Determine all algebraic extensions of $\Bbb Q$ contained in $\Bbb Q(\sqrt{2},\pi)$

Assuming $\pi$ to be transcendental over $\Bbb Q$ , it seems to me that the answer must be only $\Bbb Q(\sqrt{2})$ .

$\pi$ transcendental $\implies {\pi}^n $ transcendental $\forall n \in \Bbb N$ , any linear combination of $\pi$ with $\sqrt{2}$ and in fact, ${\sqrt{2}}^k {\pi}^l$ transcendental $\forall k,l \in \Bbb N$, $\sqrt{\pi}$ is transcendental but do not know how to show (or whether it is at all true!) that ${\pi}^{\frac{1}{n}}$ transcendental $\forall n \in \Bbb N$.

So my intuitive idea is that $\pi$ should not come into the picture and hence, $\Bbb Q(\sqrt 2)$ should be only such extension, but how make my argument rigorous (if it's at all true!)

Thanks in advance for help!

Answer

Let $F=\mathbb{Q}(\sqrt{2})$.

If $\pi$ was algebraic over $F$, $[F(\pi):F]$ would be finite. Since $[F: \mathbb{Q}]=2$, $[F(\pi):\mathbb{Q}]$ would be finite, thus $\pi$ would be algebraic. A contradiction.

So $\mathbb{Q}(\sqrt{2},\pi) =F(\pi) \cong F(X)$.

It is well-known that $F(X)$ contains no nontrivial algebraic extension of $F$: thus neither does $F(\pi)$.

So if $G$ is any algebraic extension of $\mathbb{Q}$ in $F(\pi)$, $G(\sqrt{2})$ is an algebraic extension of $F$ contained in $F(\pi)$: it is $F$. Thus $G \subset F$.