Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?)

share|improve this question

2 Answers 2

up vote 5 down vote accepted

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.

share|improve this answer
    
Dear @Zev, +1 for this answer. The only thing I would change is when you say "in particular finite." This sounds like the extension of fraction fields being algebraic implies that it is finite. Maybe you meant to say that you usually see ramification theory when the extension of fraction fields is finite, and in particular, algebraic? –  Keenan Kidwell Jun 10 '13 at 20:18
    
Thanks for catching that Keenan, my choice of phrasing was poor; I meant "particularly" when I said "in particular", as I didn't want to rule out the possibility that ramification theory could apply to a non-finite algebraic extension, but wanted to emphasize that I'd only seen it for finite extensions personally. I've edited to hopefully make my meaning clearer. –  Zev Chonoles Jun 10 '13 at 20:21
    
No problem. It is clear now. –  Keenan Kidwell Jun 10 '13 at 20:24

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.

share|improve this answer

Your Answer

 
discard

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.