Ramification of primes Let $K \subset L$ be two fields with ring of integers $\mathcal O_K$ and $\mathcal O_L$.
If a prime $p$ is totally ramified in $\mathcal O_K$, is it true that $p$ is also ramified in $\mathcal O_L$?
 A: If $p$ is (not necessarily totally) ramified in $K$ then there is a prime ideal $\mathfrak p$ of $\mathcal O_K$ s.t. $p \in \mathfrak p^2$. Let $\mathfrak P$ be any prime ideal of $\mathcal O_L$ lying over $\mathfrak p$. Then $p \in \mathfrak p^2 \subseteq \mathfrak P^2$, so $p$ is also ramified in $\mathcal O_L$.
In general, $(p) = p\mathcal O_K$ can be uniquely written as a product of prime ideals of $\mathcal O_K$:
$$(p) = \mathfrak p_1^{e_1} \cdots \mathfrak p_r^{e_r}.$$
The number $e(\mathfrak p_i | p) := e_i$ is called the ramification index of $\mathfrak p_i$ over $p$. Similarly, each $\mathfrak p_i$ can be written as a unique product of prime ideals of $\mathcal O_L$. Substitute this into the product of $\mathfrak p_i$'s to find 
$$e(\mathfrak P | p) = e(\mathfrak P | \mathfrak p) \cdot e(\mathfrak p | p)$$
whenever $\mathfrak P \subset\mathcal O_L$ lies over $\mathfrak p$ and $\mathfrak p \subset \mathcal O_K$ lies over $p$. Now $p$ is ramified in $K$ iff there is a prime ideal $\mathfrak p $ of $K$ s.t. $e(\mathfrak p | p) > 1$. By the multiplicativity of the ramification index, this implies $e(\mathfrak P|p) > 1$ for a prime ideal $\mathfrak P$ of $L$ lying over $\mathfrak p$, so $p$ is also ramified in $L$.
