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I just tried to prove Kronecker-Weber and I know the first step is to show that since any modulus in $\mathbb{Q}$ must divide some modulus of the form $\mathfrak{m}=(n)\infty$, so we just need to show that the ray class field corresponding to that is $\mathbb{Q}(\zeta_n)$.

My issue is the following. We have that the only unit of $\mathbb{Z}$ in $\mathbb{Q}_\mathfrak{m}$ is $1$, because the infinite prime restricts us to positive elements. We also know that the class group is trivial. Hence, the exact sequence for the ray class group gives us the sequence

$$1\to \mathbb{Q}_\mathfrak{m}/\mathbb{Q}_{\mathfrak{m},1}\to C_\mathfrak{m}\to 1$$

It follows that the ray class group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^\times\times\{\pm 1\}$. But this would not be what I'm looking for since the $\textrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\simeq (\mathbb{Z}/n\mathbb{Z})^\times$. What am I missing here?

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up vote 3 down vote accepted

The sequence is $$1\to U / U_{\mathfrak{m},1} \to \mathbb{Q}_\mathfrak{m}/\mathbb{Q}_{\mathfrak{m},1}\to C_\mathfrak{m}\to C=1 \to 1 $$

$U_{\mathfrak{m},1} = U \cap \mathbb{Q}_{\mathfrak{m},1} = \{+1\}$, so the first quotient is $\{\pm 1\}$ and not the trivial group. The two units are sent to different element of $\mathbb{Q}_\mathfrak{m}/\mathbb{Q}_{\mathfrak{m},1}$

Then, $\mathbb{Q}_\mathfrak{m}/\mathbb{Q}_{\mathfrak{m},1}$ is indeed $(\mathbb{Z}/n\mathbb{Z})^\times\times\{\pm 1\}$. The image of $U / U_{\mathfrak{m},1}$ inside this group is $\{(1 \mod n, +1), (-1 \mod n, -1)\}$

So the quotient of the two, $C_\mathfrak{m}$ is $\{\{(a \mod n, +1) ; (-a \mod n, -1)\}, a \in (\mathbb{Z}/n\mathbb{Z})^\times\}$, and is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^\times$

Its neutral element is $\{(1 \mod n, +1) ; (-1 \mod n, -1)\}$, which means that a (principal) prime ideal $(p)$ splits completely in the class field when $p=1 \mod n$ and $p>0$, or when $p= -1 \mod n$ and $p<0$

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Ahh... I seem to have misread the sequence. The first quotient is $U/U_{\mathfrak{m},1}$ not $U_{\mathfrak{m}}/U_{\mathfrak{m},1}$. Thanks. – pki Nov 7 '11 at 16:09

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