On the ramification index in Dedekind extensions Citing from my textbook...
Definition If $R\subseteq R'$ are Dedekind domains and $\mathfrak{P}$ is a nonzero prime ideal of $R'$ and $\mathfrak{p}=\mathfrak{P}\cap R$ then the ramification index of $\mathfrak{P}$ over $R$ is the power of $\mathfrak{P}$ appearing in the prime factorization of $\mathfrak{p}R'$.
My question is: it couldn't happen that $\mathfrak{P}\cap R=0$, the zero ideal? If yes, could you provide some (easy) examples? If yes is the answer, how should the definition be modified? (i.e. what is the ramification index in this case?)
 A: Yes, it could happen - for example with $R=\mathbb{Z}$, $R'=\mathbb{C}[x]$, and $\mathfrak{P}=(x)$. However, the setting for ramification theory is (as far as I have seen) when $\mathrm{Frac}(R)\subseteq\mathrm{Frac}(R')$ is a finite, or at the very least algebraic, extension. In this setting, there is an induced injective map
$$R/\mathfrak{p}\hookrightarrow R'/\mathfrak{P},$$
$R'/\mathfrak{P}$ is a field, and the extension $R'/\mathfrak{P}$ over $R/\mathfrak{p}$ is integral, so that $R/\mathfrak{p}$ must be a field as well by Proposition 5.7 of Atiyah-Macdonald.
A: In the book Algebraic Number Fields by Janusz, one finds the lemma VI.II stating that  

If $A\subset B$ are integral domains with $A$ integrally closed and $B$ integral over $A$, then the intersection of a non-zero prime ideal in $B$ and $A$ is a non-zero prime ideal of $A$.  

This answers your doubt negatively, when $A$ is integrally closed.
Now look at the corollary III.VII saying that  

If $B$ is a commutative nötherian ring in which every prime ideal is maximal, then there are distinct prime ideals $\mathfrak P_1,\ldots,\mathfrak P_n$ and positive integers $a_1,\ldots,a_n$ such that $\mathfrak P_1^{a_1},\ldots,\mathfrak P_n^{a_n}=0$.  

This might give us a way of constructing a simple example:  

Let $A=\mathbb Z/(n)$ for some integer $n=p_1^{a_1}\ldots p_k^{a_k}$, and let $B$ be the integral closure of $\mathbb Z$ in a finite extension of $\mathbb Q$. And let $C=B/nB$. If $\mathfrak P_1,\ldots,\mathfrak P_m$ are all the prime ideals above the primes dividing $n$, with respective ramification indices $e_i,i=1,\ldots,m$, then $\Pi_{i=1}^m\overline {\mathfrak P}_i^{e_ia_j}=(0)$ for $\mathfrak P_i$ dividing $p_j$, where the overline means the image in $C$.So we still can define ramificatino indices as in the non-zero case.  

Maybe the above example is inconsistent at some point, or per chance the above arguments are ridiculous somewhere. If a similar situation occurs, please inform me. Thanks in advance.  
