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We have $\alpha = \sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ and $\beta = \sqrt{(2-\sqrt{2})(3+\sqrt{3})}$. We want to show that $\beta \in \mathbb{Q}(\alpha)$. We have $\alpha \beta = \sqrt{(2+\sqrt{2})(3+\sqrt{3})}\sqrt{(2-\sqrt{2})(3+\sqrt{3})} = \sqrt{(2+\sqrt{2})(3+\sqrt{3})(2-\sqrt{2})(3+\sqrt{3})} = \sqrt{2(3+\sqrt{3})^2} = \sqrt{2}(3+\sqrt{3})$.
We write $$\sqrt{2}(3+\sqrt{3}) = 3\sqrt{2} +\sqrt{6} = 3\sqrt{2} + \sqrt{2}\sqrt{3}.$$ We know that $3\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ and $\sqrt{2} \in \mathbb{Q}(\sqrt{2})$. This implies that $\alpha \beta \in \mathbb{Q}(\sqrt{2},\sqrt{3})$. But I'm confused about how to go from here to show that $\beta \in \mathbb{Q}(\alpha)$. Please help! Thanks!

Bowei Tang
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Miranda
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  • Can you prove that $\mathbb{Q}(\alpha^2) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$? Hint: Show that $\alpha^2$ is a primitive element in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ using the action of Galois on the basis ${1, \sqrt{2}, \sqrt{3}, \sqrt{6}}$. – Haran Apr 03 '25 at 01:35
  • @Haran Yes I'm able to show that $\mathbb{Q}(\alpha^2) = \mathbb{Q}(\sqrt{2},\sqrt{3})$. And I also know that $\alpha \notin \mathbb{Q}(\alpha^2)$. But what should I do next? – Miranda Apr 03 '25 at 01:44
  • You've shown $\mathbb{Q}(\alpha^2)=\mathbb{Q}(\sqrt2,\sqrt3)$. Therefore $\alpha\beta\in\mathbb{Q}(\alpha^2)$, which implies $\alpha\beta\in\mathbb{Q}(\alpha)$, hence $\beta\in\mathbb{Q}(\alpha)$. – Michael Hartley Apr 03 '25 at 02:10
  • This is a somewhat famous field extension, and reappears in exercises regularly. The reason being that this is probably the simplest to describe Galois extension with Galois group $Q_8$. Anyway, we have a lot of material related to this 1, 2, 3,4. – Jyrki Lahtonen Apr 03 '25 at 07:23
  • Feel free to check out ApproachZero for more. – Jyrki Lahtonen Apr 03 '25 at 07:24

1 Answers1

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You can check that

$$24\frac{\beta}{\alpha}+(\alpha^6 - 20 \alpha^4 + 60 \alpha^2 + 24 )=0$$

Hence $\beta\in\mathbb Q(\alpha)$. (The trick to do this is that the powers of $\alpha^2$ is a $\mathbb Q$-linear combination of $1,\sqrt2,\sqrt3,\sqrt6$. )

cybcat
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