Why are there finitely many ramified prime ideals?

Question

Reference: Neukirch - ANT p.49

Let $R$ be a Dedekind domain with the quotient field $K$ and $L/K$ be a finite separable field extension and $A$ be the integral closure of $R$ in $L$.

Let $\theta\in A$ such that $L=K(\theta)$ and define $J:=\{\alpha\in A: \alpha A\subset R[\theta]\}$ and $m$ be the minimal polynomial for $\theta$ over $K$.

Let $d$ be the discriminant of the polynomial $m$, so that $d=\prod_{0\leq i<j\leq n-1} (\theta^i -\theta^j)$ where $n=[L:K]$.

The text shows that any prime ideal satisfying $d\notin P$ and $PA+J=A$ is unramified in $L$.

How do I conclude that there are only finitely many ramified prime ideals in $L$? That is, how do I prove that there are finitely many prime ideals satisfying $d\in P$ or $PA+J\neq A$?

Since $R$ is a Dedekind domain, of course, there are finitely many prime ideals satisfying $d\in P$. However, why are there finitely many prime ideals $P$ such that $PA+J\neq A$?

Answer 1

I'm a bit unsure of where you're stuck, since I think one can show that the set of primes $\mathfrak{p}$ such that $\mathfrak{p}A + \mathfrak{F} \neq A$ is finite in much the same way that one shows the set of primes such that $d \in \mathfrak{p}$ is finite.

Since you didn't state this explicitly, let me first note that $A$ is also a Dedekind domain: its Krull dimension is $1$ by the Cohen-Seidenberg theorems, it is integrally closed, and one can use Proposition 5.17 of Atiyah-Macdonald to show that it is Noetherian.

Since $R$ is Dedekind, then $dR$ has a unique prime factorization $$ dR = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_s^{e_s} \, . $$ Since to contain is to divide, then the only prime ideals $\mathfrak{p}$ such that $d \in \mathfrak{p}$ are the primes $\mathfrak{p}_1, \ldots, \mathfrak{p}_s$ appearing in the factorization of $dR$.

Similarly since $A$ is Dedekind, then $\mathfrak{F}$ has a unique prime factorization $$ \mathfrak{F} = \mathfrak{P}_1^{a_1} \cdots \mathfrak{P}_t^{a_t} \, . $$ If $\mathfrak{p}$ is an ideal of $R$ such that $\mathfrak{p}A + \mathfrak{F} \neq A$, then there exists a prime $\mathfrak{P}$ of $A$ such that $\mathfrak{P} \mid \mathfrak{p}A$ and $\mathfrak{P} \mid \mathfrak{F}$. Then $\mathfrak{P} = \mathfrak{P}_i$ for some $i$ by the uniqueness of factorization of ideals. Thus we see that an ideal $\mathfrak{p}$ of $R$ satisfies $\mathfrak{p}A + \mathfrak{F} \neq A$ iff $\mathfrak{p} = \mathfrak{P_i} \cap A$ for some $i$, i.e., for one of the primes appearing in the factorization of $\mathfrak{F}$.

Answer 2

More of an extended comment than an answer:

Ramification of primes in number theory is basically the same picture as ramification which occurs in the theory of branched covering spaces (for example, see my answer to this question).

The basic philosophy is: prime ideals are points. For an integral domain, its "dimension" (ie, Krull dimension) is the longest chain of prime ideals you can find, minus 1. For rings of integers, $(0)$ is prime, and every nonzero prime is maximal, so the longest chain you can find is 2, so the dimension is 1. Under this philosophy, the prime ideal $(0)$ should be thought of as a "generic point", which is a point whose closure (in the Zariski topology) is the whole space - it is also called a codimension 0 point. The nonzero primes are then codimension 1 points (in this case, they are closed points).

It turns out that in nice situations (which includes the case of rings of integers), ramification is a closed property which happens purely in codimension 1. However, in the Zariski topology on an 1-dimensional integral domain, the open sets are the cofinite sets - ie, the closed sets are the finite sets, and hence ramification can only happen at closed points (ie, nonzero primes), and only had finitely many of them.