What is the meaning of ramification? What is the meaning of the sentence--"Let $p$ be a prime that ramifies" ?
Thank you.
 A: Let $k$ be a number field and let $O_k$ be the ring of integers in $k$. It is a theorem that every ideal $I$ in $O_k$ can be factored into a product of prime ideals uniquely up to ordering the factors. Let $p$ be a prime integer (sometimes called a ``rational prime"). Then $pO_k$ is an ideal in $O_k$ -- call it $(p)$. By the theorem, there exist a non-negative integer $r$, non-negative integers $e_1,...,e_r$, and prime ideals $P_1,...,P_r$ such that $(p)=P_1^{e_1}...P_r^{e_r}$. We say that $p$ is ramified in $k$ if any of the $e_i>1$. Otherwise (i.e. if all the $e_i=1$), we say that $p$ is unramified in $k$.
A: Look at the simplest example : $\mathbb{Z}[i]$. 
If $\mathfrak{p}$ is prime ideal of $\mathbb{Z}[i]$ then $\mathfrak{p} \cap \mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. Thus it is sufficient to look at the prime ideals containing $p$ for each prime $p$ :


*

*$(2) = (1+i)(1+i)$ is ramified, 

*if $p \equiv 1 \bmod 4$ then $(p) = (a+ib)(a-ib)$ for some $a,b, a^2+b^2= p$,

*if  $p \equiv 3 \bmod 4$ then $(p)$ is a prime ideal. 
Similarly, since $\mathbb{Z}[i] \cong \mathbb{Z}[x]/(x^2+1)$ : 


*

*$x^2+1 \equiv (x+1)^2 \bmod 2$ has a double root, 

*if $p \equiv 1 \bmod 4$ then $x^2+1 \equiv (x+a)(x-a) \bmod p$ 

*otherwise if $p \equiv 3 \bmod 4$ then $x^2+1$ is irreducible $\bmod p$. 
The theory is explained there : Splitting_of_prime_ideals_in_Galois_extensions
