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There's a small theorem in algebra that I've been toying with. Let $R$ be integrally closed in its quotient field, and $E$ a splitting extension of the quotient field, and $S$ be the integral closure of $R$ in $E$. Then for a maximal ideal $p$ of $R$, there are only finitely many maximal ideals of $S$ whose intersection with $R$ is precisely $p$.

I was trying to formulate a small example. Let $f(X)=X^3+X+1$, and $E\supseteq\mathbb{Q}$ a splitting extension for $f$. So taking $R=\mathbb{Z}$ in this case, let $S$ be the integral closure of $\mathbb{Z}$ in $E$. I choose say $(3)$ as a maximal ideal of $\mathbb{Z}$. So there must be finitely many maximal ideals $q$ of $S$ such that $q\cap\mathbb{Z}=(3)$.

This sparked my curiosity, if we know that there are only finitely many such maximal ideals $q$, how many are there actually, at least in this case? I couldn't think of a way to actually enumerate them all to be able to say "There are 9 such maximal ideals!," or something along those lines. Thank you kindly.

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For the case of the integers and the rationals, these are well-known. If $r$ is the number of distinct primes lying over $(p)$, and $f$ is the degree of the extension(s) $S/\mathfrak{q}$ over $\mathbb{Z}/(p)$ (where $\mathfrak{q}$ is any prime lying over $(p)$), and $e$ is the largest integer such that $\mathfrak{q}^e$ contains $(p)$ (the "ramification"), then $erf = [E:\mathbb{Q}]$. –  Arturo Magidin Jan 21 '12 at 1:24
    
Thank you @Arturo, so I would want to solve $e=\frac{[E:\mathbb{Q}]}{rf}=\frac{6}{rf}$? I don't see a way to solve for $r$ and $f$ in general for arbitrary $\mathfrak{q}$. –  Sofie JG Jan 21 '12 at 2:08
    
You would find $f$ by computing the size of $S/\mathbb{q}$ (which will give you the dimension of the extension); the value of $e$ can be found in any number of ways for specific $(p)$ and $E$, but if you have any particular $\mathfrak{q}$ you can simply compute $\mathfrak{q}$, $\mathfrak{q}^2$, etc until you find which one fails to contain $(p)$ (it will be at most $[E:\mathbb{Q}]$). Once you have $e$ and $f$, you can find $r$. I second the advice to take a look at a book on algebraic number theory; Marcus's is very good. –  Arturo Magidin Jan 21 '12 at 4:17

1 Answer 1

Let $R$ be a noetherian ring with fraction field $K=Frac(R)$.
Let $K\subset E$ be a finite separable extension field of dimension $[E:K]=n$ and let $S$ be the integral closure of $R$ in $E$: it is also a Dedekind ring.

Now given a non-zero prime $\mathfrak p\subset R$, decompose the extended ideal $\mathfrak p\cdot S\subset S$ as :

$$\mathfrak p\cdot S=\Pi \mathfrak P^{e_\mathfrak P}$$

(you can do that, since $S$ is Dedekind)
You then have the fundamental formula $$n=\Sigma e_{\mathfrak P } \cdot f_ {\mathfrak P} $$
where $f_ {\mathfrak P}=[S/\mathfrak P:R/\mathfrak p]$.
The $\mathfrak P$'s in the above formulae are exactly the primes of $S$ over $\mathfrak p$ and there are thus at most $n$ of them. This is the bound you were looking for.

In your example, the splitting field $E$ of $X^3+X+1$ over $\mathbb Q$ has dimension 6 over $\mathbb Q$ and so you have at most 6 prime ideals over the prime ideal $\mathfrak p = (3)$ .

Bibliography J.-P. Serre, Local Fields, Chapter 1, §4.

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Thank you Georges. What is $e_\mathfrak{P}$? I regret that I don't understand most of the language in this post. Is there by any chance a simpler approach to this particular case of integers and rationals? –  Sofie JG Jan 21 '12 at 2:04
    
Dear @Sophie:1) A non-zero ideal in a Dedekind ring is a product of prime ideals, but some of them may be repeated, in which case the number of times the prime ideal appears is indicated by the exponent (here $e_\mathfrak P$). 2) Even if you don't yet understand everything, you can use the very simple bound $n$ that I give. It often happens in mathematics that we provisionally use a formula without having yet understood how it is obtained. –  Georges Elencwajg Jan 21 '12 at 2:34
    
Thanks again Georges. So there are at most 6 prime ideals over $(3)$, and thus at most 6 maximal ideals over $(3)$. Is it possible to pinpoint how many there are exactly, or is a bound the best one can do? –  Sofie JG Jan 21 '12 at 2:51
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I haven't done it, but it is possible to calculate everything explicitly in your example, in particular the number of prime ideals over $(3)$. Check any book on algebraic numner theory, like Number Fields by Marcus and you will see how it is done in a fairly elementary way. –  Georges Elencwajg Jan 21 '12 at 3:17

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