Show that $\mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}})=\mathbb{Q}(\sqrt{3})$. 
Show that $\mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}})=\mathbb{Q}(\sqrt{3})$.

I know that by closure, $\sqrt{3}\in \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}})$, but what about the other containment? I want to use the order of the extensions. I know the degrees of each one are $4$ over the field $\mathbb{Q}$, and I know that $\mathbb{Q}(\sqrt{1+\sqrt{3}})$ and $ \mathbb{Q}(\sqrt{1-\sqrt{3}})$ are not equal to each other. But where can I go from here? How do I show that $[\mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}(\sqrt{3}]=1$?
 A: $ [\mathbb{Q}(\sqrt{1+\sqrt{3}}):(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}})][ \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}]=4$.
You deduce that $[ \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}]$ divides $4$, it is not $4$ since $\mathbb{Q}(\sqrt{1+\sqrt{3}})$ is distinct of $ \mathbb{Q}(\sqrt{1-\sqrt{3}})$, so it is $1$ or $2$, it is not $1$ since $\mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}})$ contains $\mathbb{Q}(\sqrt3)$, so it is $2$, you deduce that:
$[ \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}]=[ \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}(\sqrt3)][\mathbb{Q}(\sqrt3):\mathbb{Q}]$. Since $[ \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}]=[\mathbb{Q}(\sqrt3):\mathbb{Q}]=2$, you deduce that $[ \mathbb{Q}(\sqrt{1+\sqrt{3}})\cap \mathbb{Q}(\sqrt{1-\sqrt{3}}):\mathbb{Q}(\sqrt3)]=1$.
A: Let $\alpha:=\sqrt{1+\sqrt{3}}$ and $\beta:=\sqrt{1-\sqrt{3}}$. You know that
$$[\Bbb{Q}(\alpha):\Bbb{Q}]=[\Bbb{Q}(\beta):\Bbb{Q}]=4,$$
and that $\Bbb{Q}(\sqrt{3})\subset\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)$. Because the degree is multiplicative over towers of field extensions
$$[\Bbb{Q}(\alpha):\Bbb{Q}]=[\Bbb{Q}(\alpha):\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)]\cdot[\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta):\Bbb{Q}(\sqrt{3})]\cdot[\Bbb{Q}(\sqrt{3}):\Bbb{Q}],$$
and because you know the two fields $\Bbb{Q}(\alpha)$ and $\Bbb{Q}(\beta)$ are not the same, you have
$$[\Bbb{Q}(\alpha):\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)]>1,$$
and clearly $[\Bbb{Q}(\sqrt{3}):\Bbb{Q}]=2$. Because $[\Bbb{Q}(\alpha):\Bbb{Q}]=4$ it follows that
$$[\Bbb{Q}(\alpha):\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)]=2
\qquad\text{ and }\qquad
[\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta):\Bbb{Q}(\sqrt{3})]=1,$$
which means that $\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)=\Bbb{Q}(\sqrt{3})$.
