3
$\begingroup$

I am reviewing Galois theory for my algebra prelim, and I got stuck on the following problem.

Prove that $\mathbb{Q}(\sqrt{5+2\sqrt{6}}) = \mathbb{Q}(\sqrt{2},\sqrt{3})$. Deduce that $K$ is a normal extesnsion over $\mathbb{Q}$ and hence a Galois extention.

Now, I can show both are degree 4 extensions, so if can I show one is contained in the other, I'll be done (two vector spaces of equal dimension such that one is contained in the other are equal). My approach so far has been to show that $\mathbb{Q}$ and $\sqrt{5+2\sqrt{6}}$ generates $\sqrt{2}$ and $\sqrt{3}$. I can show that $\sqrt{6}$ is in $\mathbb{Q}(\sqrt{5+2\sqrt{6}})$ by squaring $\sqrt{5+2\sqrt{6}}$. I've also tried

$$\frac{1}{\sqrt{5+2\sqrt{6}}}=\sqrt{5-2\sqrt{6}}$$ But I cannot show $\sqrt{2}$ or $\sqrt{3}$ is in $\mathbb{Q}(\sqrt{5+2\sqrt{6}})$. Is there an elementary algebraic trick I'm missing or should I approach the problem differently?

$\endgroup$
5
$\begingroup$

Use $$\sqrt{5+2\sqrt6}=\sqrt{3+2\sqrt{3\cdot2}+2}=\sqrt{(\sqrt3+\sqrt2)^2}=\sqrt3+\sqrt2.$$

$\endgroup$
  • $\begingroup$ Does this work as well? I see it now: $(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$ and $(\sqrt{5+2\sqrt{6}})^2= 5+2\sqrt{6}$. So this implies that $\sqrt{2}+\sqrt{3} = \sqrt{5+2\sqrt{6}}$. Thanks. $\endgroup$ – MEG Aug 14 at 2:22
  • $\begingroup$ @MEG Yes, of course. $\endgroup$ – Michael Rozenberg Aug 14 at 2:28

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.