1
$\begingroup$

I would like to find the degree of the field extension $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})/\mathbb{Q}(\sqrt{3+\sqrt{7}})$. Here's my thoughts on this problem.

I suspect the result is 2. To be able to show that, it would be enough to show that $\sqrt{3-\sqrt{7}} \notin \mathbb{Q}(\sqrt{3+\sqrt{7}})$, looking at the irreducible polynomial of $\sqrt{3+\sqrt{7}}$, given by $x^4-6x^2+2$, for which $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})$ is the splitting field over $\mathbb{Q}$, and since it is biquartic, the extension must be of degree $1$ or $2$, since $[\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}}):\mathbb{Q}]$ is $4$ or $8$. I think I should find a $\mathbb{Q}$-basis for $\mathbb{Q}(\sqrt{3+\sqrt{7}})$ but I don't know how to do that. After that, I should show that $\sqrt{3-\sqrt{7}}$ is not a rational combination of such basis and I would be done, but I wouldn't know how to tackle that either. Any help would be appreciated.

$\endgroup$
  • $\begingroup$ $(\sqrt{3+\sqrt{7}}*(\sqrt{3-\sqrt{7}})=\sqrt(2)$ $\endgroup$ – i. m. soloveichik Jan 9 at 14:38
  • $\begingroup$ Your degree is indeed 2. See my answer to your other question. $\endgroup$ – nguyen quang do Jan 10 at 5:30
0
$\begingroup$

From my answer there : Prove that $\sqrt{3\pm\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3\mp\sqrt{7}})$.

$\mathbb{Q}(\sqrt{3 \pm \sqrt{7}})=\mathbb{Q}(\sqrt{7},\sqrt{3-\sqrt{7}},\sqrt{2})$ and all extensions have degree $2$ exactly. Thus the total degree is $8$.

$\endgroup$

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.