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I want to prove the following fact:

Let $m,n \in \mathbb{N}$ be coprime and $\alpha, \beta \in \mathbb{C}$ with $\alpha^m=2$ and $\beta^n=3$. Then $\alpha\beta$ is a primitive element of $\mathbb{Q}(\alpha,\beta)$.

My idea was to use the fact that $m,n$ are coprime to see that, for example, $(\alpha\beta)^m = \alpha^m\beta^m = 2 \beta^m \in \mathbb{Q}(\alpha\beta)$ implies that $\beta \in \mathbb{Q}(\alpha\beta)$. But there is obviously some argument missing and I don't know how to fill the gap. I tried to use the fact that $m,n$ are coprime to write something like $am+bn=1$ for some $a,b \in \mathbb{Z}$ but I don't think this helps.

How can I prove this fact?

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  • $\begingroup$ I couldn't find a similar question using search, so I hope this is no duplicate! $\endgroup$ – namsap Feb 17 '15 at 15:23
  • $\begingroup$ Are you asking about the step $2\beta^m\in\Bbb{Q}(\alpha\beta)\implies \beta\in\Bbb{Q}(\alpha\beta)$ or what? $\endgroup$ – Jyrki Lahtonen Feb 17 '15 at 17:51
  • $\begingroup$ @JyrkiLahtonen Yes, this is the point where I don't know how to do this $\endgroup$ – namsap Feb 17 '15 at 17:52
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    $\begingroup$ Ok. Your intuition is well calibrated in the sense that $am+bn=1$ comes in handy. Hint: $\beta^m\in\Bbb{Q}(\alpha\beta)\implies \beta^{am}\in\Bbb{Q}(\alpha\beta)$. $\endgroup$ – Jyrki Lahtonen Feb 17 '15 at 17:55
  • $\begingroup$ Great! I think, I can post an answer now. $\endgroup$ – namsap Feb 17 '15 at 18:00
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First of all, since $n,m$ are coprime, there exist $a,b \in \mathbb{Z}$ sucht that $am+bn=1$. As stated above, we can conclude $(\alpha\beta)^m = \alpha^m\beta^m = 2 \beta^m \in \mathbb{Q}(\alpha\beta) \implies \beta^m \in \mathbb{Q}(\alpha\beta) \implies (\beta^m)^a = \beta^{ma} \in \mathbb{Q}(\alpha\beta)$ (because $a \in \mathbb{Z}$). And this implies $3^b\beta^{ma} \in \mathbb{Q}(\alpha\beta)$, because $3^b \in \mathbb{Q}$. Then we can use the fact that $am+bn=1$ to see that $3^b\beta^{ma} = (\beta^n)^b\beta^{ma} = \beta^{nb}\beta^{ma}=\beta^{ma+nb} = \beta$, i.e. $\beta \in \mathbb{Q}(\alpha\beta)$. This implies $\alpha = \alpha\beta / \beta \in \mathbb{Q}(\alpha\beta)$. So we know $\mathbb{Q}(\alpha,\beta) \subseteq \mathbb{Q}(\alpha\beta)$. The other inclusion is obvious, so we get $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha\beta)$.

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    $\begingroup$ Looks good to me! $\endgroup$ – Jyrki Lahtonen Feb 17 '15 at 18:16

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