The Proposition $(11.6)$ of Neukirch's Algebraic Number Theory states that if $\mathcal O$ is a dedekind domain with field of fractions $K$ and $X$ is a set of nonzero prime ideals of $\mathcal O$ with finite complement, there is a canonical exact sequence:

$\begin{eqnarray*} 1 \rightarrow U(\mathcal O) \rightarrow U(\mathcal O(X)) \rightarrow \bigoplus_{\mathfrak p\not\in X}K^*/U(\mathcal O_{\mathfrak p}) \rightarrow \mathscr C\ell(\mathcal O) \rightarrow \mathscr C\ell(\mathcal O(X)) \rightarrow 1, \end{eqnarray*}$

and that $K^*/U(\mathcal O_{\mathfrak p})\cong \mathbb Z$, for all prime $\mathfrak p\not\in X$.

Now let $\mathcal O_K$ be the ring of integers of $K$, let $S$ denote a finite set of prime ideals of $\mathcal O_K$, and let $X$ be the set of all prime ideals that do not belong to $S$. We put $\mathcal O_K^S = \mathcal O_K (X)$. The units of this ring are called the $S$-units.

(11.7) Corollary. For the group $K^S = (\mathcal O_K^S)*$ of $S$-units of $K$ there is an isomorphism $K^S\cong \mu(K) \times \mathbb Z^{\# S+r+s-1}$ where $r$ and $s$ are the number of real imersions and pairs of complex imersions of $K$.

The proof of Neukirch is as follows:

Proof: The torsion subgroup of $K^S$ is the group $\mu(K)$ of roots of unity in $K$. Since $\mathscr C\ell(\mathcal O)$ is finite, we obtain the following identities from the exact sequence above and from Dirichlet Units Theorem:

$rank(K^S)=rank(\mathcal O_K^*)+rank(\bigoplus_{\mathfrak p\in S}\mathbb Z)=\#S+r+s-1$.

I suppose the proof uses that result that relates exact sequences of $\mathbb Z$-modules and the ranks of these modules. But I only find:

$rank~U(\mathcal O)-rank~U(\mathcal O(X))+rank~\mathbb Z^{\# S}-rank~\mathscr C\ell(\mathcal O)+rank~\mathscr C\ell(\mathcal O(X))=0$.

Dirichlet Unit Theorem gives $rank~U(\mathcal O)=r+s-1$, but do I have some information about the ranks of the class groups?


1 Answer 1


$\mathcal{C}\ell(\mathcal{O})$ is finite and hence so is $\mathcal{C}\ell(\mathcal{O}(X))$, being a quotient of the former. A finite abelian group has rank zero.

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    $\begingroup$ This is because every element is of torsion? I thought that linear independence was defined like that: a set $x_1,\dotsc,x_n$ is linearly independent over $\mathbb Z$ iff $a_1x_1+\cdots+a_n x_n=0\Rightarrow a_1x_1=\cdots=a_nx_n=0$. The right definition is $a_1=\cdots=a_n=0$? $\endgroup$ Sep 3, 2021 at 13:59
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    $\begingroup$ @Lorenzo yes, the second definition is correct one. I personally think that the best way to define the rank of an abelian group $A$ is the dimension of $\Bbb Q \otimes_{\Bbb Z} A$ $\endgroup$ Sep 3, 2021 at 14:37

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