0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Don't forget that $\Bbb Q$ is an extension of $\Bbb Q$. $\endgroup$ – Arthur Feb 20 at 15:02
  • $\begingroup$ 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). $\endgroup$ – Andrés E. Caicedo Feb 20 at 15:02
  • $\begingroup$ @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. $\endgroup$ – Utsav Dewan Feb 20 at 15:08
1
$\begingroup$

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 $G \subset F$.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Andrés E. Caicedo Feb 20 at 16:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.