# Finding the degree of $\mathbb{Q}(\sqrt{3+\sqrt{7}},\sqrt{3-\sqrt{7}})/\mathbb{Q}(\sqrt{3+\sqrt{7}})$

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.

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

## 1 Answer

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