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.
 A: In fact, $2$ does totally ramify in $\mathbb{Q}(\sqrt{6},\sqrt{10})$. 
There are various ways to see this. Here is a nice one:

Theorem: Suppose that $L/K$ is a finite Galois extension of number fields. Suppose that $\mathfrak{p}$ is a prime of $\mathcal{O}_K$ which totally ramifies in every proper subextension of $L/K$, but does not totally ramify in $L$. Then, $\text{Gal}(L/K)\cong \mathbb{Z}/q\mathbb{Z}$ for some prime $q$.

Roughly, the reason is as follows. Take the inertia field $E$ of some prime $\mathfrak{P}$ over $\mathfrak{p}$. Note then that $E=L$. Indeed, let $\mathfrak{q}=\mathfrak{P}\cap \mathcal{O}_E$. Assume that $E\subsetneq L$ then, note that by assumption $e(\mathfrak{q}\mid\mathfrak{p})=[E:K]$ and by definition of the inertia field we have that $e(\mathfrak{P}\mid\mathfrak{q})=[L:E]$. Then, we have that $e(\mathfrak{P}\mid\mathfrak{p})=[L:K]$ contradictory to assumption. So, $E=L$ as desired. So, now suppose that $K\subseteq M\subsetneq L$ is any proper subextension of $L/K$. Then, by assumption
$$[M:K]=e(\mathfrak{P}\cap\mathcal{O}_M\mid \mathfrak{p})$$
but since $M\subseteq E=L$ we know that $e(\mathfrak{P}\cap\mathcal{O}_M\mid\mathfrak{p})=1$ and so $[M:K]=1$ or $M=K$. Thus, we see that $\text{Gal}(L/K)$ has no non-trivial proper subgroups, which trivially implies that $\text{Gal}(L/K)\cong\mathbb{Z}/q\mathbb{Z}$ for some prime $q$ as desired.
So, how does this help in our case? Well $\mathbb{Q}(\sqrt{6},\sqrt{10})/\mathbb{Q}$ is Galois, whose only proper subextensions are $\mathbb{Q}(\sqrt{6}),\mathbb{Q}(\sqrt{10})$, and $\mathbb{Q}(\sqrt{15})$. Note that $2$ ramifies in all of these (since all of their discriminants are even), but for quadratic extensions ramifying is the same thing as totally ramifying and so $2$ totally ramifies in every proper subextension of $\mathbb{Q}(\sqrt{6},\sqrt{10})$. 
So, by our theorem, if $2$ did not ramify in $\mathbb{Q}(\sqrt{6},\sqrt{10})$ then $\text{Gal}(\mathbb{Q}(\sqrt{6},\sqrt{10})/\mathbb{Q})$ would be cyclic of prime order, which it clearly is not.
A: 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 $\mathbb{Q}(\sqrt{6},\sqrt{10})$.
