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.

  • $\begingroup$ $(\sqrt{3+\sqrt{7}}*(\sqrt{3-\sqrt{7}})=\sqrt(2)$ $\endgroup$ – 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

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 $8$.


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