How to show 2 cannot be totally ramified in $\mathbb{Q}(\sqrt{6},\sqrt{10})$? I am trying to show that 2 cannot be totally ramified in $\mathbb{Q}(\sqrt{6},\sqrt{10})$. I know that it is totally ramified in $\mathbb{Q}(\sqrt{6})$ and $\mathbb{Q}(\sqrt{10})$ since $2O_{\mathbb{Q}(\sqrt{6})}=(2,\sqrt{6})^2$ and similarly $2O_{\mathbb{Q}(\sqrt{10})}=(2,\sqrt{10})^2$. 
Any help would be appreciated.

For a prime $\mathfrak{p}$ in $\mathbb{Q}(\sqrt{6},\sqrt{10})$ above $2$, let $I$ be the inertia group and $F$ the inertia field. 
From basic ramification theory, $|I|=e_{\mathfrak{p}/2}$ and $2$ does not ramify in $F|\mathbb{Q}$. 
We will prove $|I|=e_{\mathfrak{p}/2}=4$.
Suppose $|I|=2$. Then $F$ is a proper subextension, so it must be one of $\mathbb{Q}(\sqrt{6}), \mathbb{Q}(\sqrt{10})$ or $\mathbb{Q}(\sqrt{15})$. These have discriminants $24$, $40$ and $60$ respectively, so $2$ ramifies in $F|\mathbb{Q}$ (absurd).
Finally, suppose $|I|=1$. This means $F=\mathbb{Q}(\sqrt{6},\sqrt{10})$ and that $2$ does not ramify in $\mathbb{Q}(\sqrt{6},\sqrt{10})$. This is clearly absurd, since we saw that $2$ ramifies in the intermediate extensions, so it necessarily ramifies in $\mathbb{Q}(\sqrt{6},\sqrt{10})$.
