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This question already has an answer here:

The title says it. I want to find an element $\alpha$ such that

$\mathbb Q(\alpha)=\mathbb Q(\sqrt{2},\sqrt{3})$.

I tried something like $\sqrt{2}+\sqrt{3}$ but that didn't help...

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marked as duplicate by Dietrich Burde, user26857 abstract-algebra Feb 22 '16 at 14:36

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  • $\begingroup$ $\sqrt2 +\sqrt3$ will work $\endgroup$ – sqtrat Feb 22 '16 at 13:04
  • $\begingroup$ But I cannot see how I can show that $\sqrt{2}$ is in $\mathbb Q(\alpha)$ $\endgroup$ – Marc Feb 22 '16 at 13:05
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    $\begingroup$ Look at $\dfrac{1}{\sqrt{2} + \sqrt{3}}$. $\endgroup$ – Daniel Fischer Feb 22 '16 at 13:11
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Consider the field $\mathbb{Q}(\sqrt2 +\sqrt3)$. Then, all powers of $\alpha = \sqrt2+\sqrt3$ must be in $\mathbb{Q}(\alpha)$. Thus, $\frac{\alpha^{2}-5}{2} = \sqrt 6$ as well and thus $\sqrt6 \cdot \alpha = 2\sqrt 3 + 3\sqrt2$ as well. Hence $\sqrt 6 \alpha -2 \alpha = \sqrt 2 \in \mathbb{Q}(\alpha)$. Similarly $\sqrt 3 \in \mathbb{Q}(\alpha)$ and it follows that $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt2,\sqrt3)$. It is easy to show that $\mathbb{Q}(\sqrt2,\sqrt 3) \subset \mathbb{Q}(\alpha).$

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A hint: The proof (I know) that such an element exists gives also way determining this element (and for "small" extension this can be carried out by hand). However, let us write $$\alpha = \sqrt{2} + \sqrt{3}$$ and compute all powers $\alpha^j$ for $j=0,1,2,3$. They can be written in terms of $\sqrt{2}, \sqrt{3}$ and $\sqrt{2} \sqrt{3} = \sqrt{6}$. I am confident that you will now see how to write $\sqrt{2}$ and $\sqrt{3}$ in terms of $\alpha$. In particular you should e.g. get that $\sqrt{2} = \frac{\alpha^3-9\alpha}2.$

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