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Can someone please go through a proof of the fact that the ray class group of a number field is finite?

I just can't find a nice readable elementary one on the internet...

Thanks in advance.

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Just to put it out there... have you considered, I don't know, a book and/or a library? There is a world outside the internet, after all. –  Arturo Magidin Oct 20 '11 at 4:56
You've looked at Theorem 1.7 of Milne's notes? I thought this proof was pretty clear. A more specific question on a specific proof of this fact would be warmly received here, I think. Good luck! –  Dylan Moreland Oct 20 '11 at 5:10
@Arturo: I didn't figure my question was so cripted; yes, I was asking for a source recommendation (of course, books fit into that description). –  Anna Oct 20 '11 at 5:41
@Dylan: Thanks! Didn't know of Milne's CFT book although at the moment I'm reading his ANT :-) –  Anna Oct 20 '11 at 5:42
@Anna: Ah, you were asking for references... That would be a different kettle of fish, indeed. But what you actually wrote is "can someone please go through a proof..." Looks like you were asking for someone to present and walk you through a proof, not to direct you to a source where you can read one. –  Arturo Magidin Oct 20 '11 at 5:45

1 Answer 1

The ray class group of conductor $\mathfrak m$ is defined to be the group of fractional ideals prime to $\mathfrak m$, modulo those fractional ideals which are principal, with a generator congruent to $1 \bmod \mathfrak m$. (Here, if a real place is contained in $\mathfrak m$, "congruenct to $1$ at such a place" is interpreted as "positive at this place".) In symbols, write this as $I_{\mathfrak m}/P_{\mathfrak m}^1$.

It is easy to see that any ideal class admits a representative that is coprime to $\mathfrak m$, and so the ray class group surjects onto the ideal class group. In symbols, write $P_{\mathfrak m}$ for the principal ideals in $I_{\mathfrak m}$; then we have the short exact sequence $$1 \to P_{\mathfrak m}/P^1_{\mathfrak m} \to I_{\mathfrak m} /P^1_{\mathfrak m} \to I_{\mathfrak m}/P_{\mathfrak m} \to 1.$$ The middle term is the ray class group, and, as noted, it surjects onto the class group, which is the third term.

Since the class group is finite, to show that the ray class group is finite, we just have to show that the first term $P_{\mathfrak m}/P^1_{\mathfrak m}$ is finite. Luckily, this kernel is easy to compute:

If $\mathfrak a$ is a principal ideal, we can map it to its principal generator $a$. If $a$ is $1$ mod $\mathfrak m$, then $\mathfrak a$ lies in $P^1_{\mathfrak m}$, and so we apparently get an identifiation between $P_{\mathfrak m}/P^1_{\mathfrak m}$ and $(\mathcal O/\mathfrak m)^{\times}.$ Actually, this is not quite true, because $a$ is not well-defined; it is only determined up to multiplication by a unit. Thus actually we get an isomorphism $$P_{\mathfrak m}/P^1_{\mathfrak m} \cong (\mathcal O/\mathfrak m)^{\times}/(\text{reduction mod }\mathfrak m \text{ of } \mathcal O^{\times}).$$ In any event, this is finite, since it is a quotient of the finite group $(\mathcal O/\mathfrak m)^{\times}$.

[Note: if $\mathfrak m$ contains some real places $v$, then we interpret $\mathcal O/\mathfrak m$ as denoting the product of copies of $\{\pm 1\}$ (one for each real place in $\mathfrak m$) and the usual quotient $\mathcal O/\mathfrak m'$, where $\mathfrak m'$ is the finite part of $\mathfrak m$. And of course, we interpret reduction modulo $\mathfrak m$ as being the product of the maps which attach the sign with respect to $v$ as an element in $\{\pm 1\}$, for each real place $v$ dividing $\mathfrak m$, and as the usual reduction mod $\mathfrak m'$ for the finite part $\mathfrak m'$ of $\mathfrak m$.]

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