# Prime decomposition in ring extensions

Let $R\subseteq R'$ be Dedekind domains, let $\mathfrak{p}$ be a nonzero prime ideal of $R$. Then $\mathfrak{p}R'$ is an ideal of $R'$ and it has a factorization $$\mathfrak{p}R'=\mathfrak{P}_1^{e_1}\cdots\mathfrak{P}_g^{e_g}$$ in which $\mathfrak{P}_1,\ldots\mathfrak{P}_g$ are distinct prime ideals of $R'$ and $e_1,\ldots,e_g$ are positive integers.

At this point, my textbook says something obscure (in my opinion). It says: note thet $\mathfrak{P_i}\cap R=\mathfrak{p}$ for each $i$. $\textbf{Thus}$ the integer $e_i$ is completely determined by $\mathfrak{P}_i$.

I can't understand the sentence in itself: what does it mean that the exponent is completely determined by the ideal?? Secondly, how the previous argument should show this?

Any suggestion/comment/help would be appreciated

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For an ideal $\mathfrak{P}$ appearing the in the decomposition of $\mathfrak{p}$, one often writes the suggestive notation $\mathfrak{P} | \mathfrak{p}$. Note that this means $\mathfrak{p}$ is contained in the ideal $\mathfrak{P}$. Then the power $e$ appearing in the decomposition is the highest power of $\mathfrak{P}$ in which $\mathfrak{p}$ is contained. In other words,
$\mathfrak{p} \in \mathfrak{P}^e$ but $\mathfrak{p} \notin \mathfrak{P}^{e+1}$.
This is the sense in which the $e$ is determined by this decomposition.
The point of the sentence is that given $\mathfrak P$, you can determine $\mathfrak p$ (in other words, there is a unique prime ideal of $R$ that contains $\mathfrak P$ in its factorization in $R'$, something that may not be obvious a priori, but which follows from the formula $\mathfrak p = R \cap \mathfrak P$). So given $\mathfrak P$, you can determine $\mathfrak p$, and hence (by factoring $\mathfrak p$ in $R'$) determine $e$. Thus $e$ is an invariant of $\mathfrak P$ and the extension $R'/R$ alone, whereas a priori it is an invariant of the triple $(R'/R, \mathfrak p, \mathfrak P)$. It is called the ramification index of $\mathfrak P$.