Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

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 $\mathfrak P$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.