# 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!

• Don't forget that $\Bbb Q$ is an extension of $\Bbb Q$. – Arthur Feb 20 at 15:02
• Any element of $\mathbb Q(\sqrt2,\pi)$ is a rational expression (a quotient of polynomials) in $\sqrt2,\pi$. Use this to verify your intuition. Regarding your question about $\pi^{1/n}$, note that if $\alpha$ is algebraic, then so is $\alpha^n$ for any $n$. In fact, you may find useful in general to check that the algebraic numbers form a field (so, they are closed under addition, multiplication, and nonzero inverses). – Andrés E. Caicedo Feb 20 at 15:02
• @AndrésE.Caicedo Got it! so the answer should be $\Bbb Q$ and $\Bbb Q(\sqrt{2})$ right? Anyway my arguments look a bit unorganized. Can you give a short rigorous answer and so I could close the question by accepting an answer. – Utsav Dewan Feb 20 at 15:08

## 1 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 $$G \subset F$$.

• I am not sure this answer helps much. If we already know that $F(X)$ has no nontrivial algebraic extensions of $F$, there is not much to the question. I think the point is to show this from scratch. – Andrés E. Caicedo Feb 20 at 16:12