# Exact sequence of abelian groups - Corolary 11.7 - Neukirch

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?

$$\mathcal{C}\ell(\mathcal{O})$$ 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.
• 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 $a_1=\cdots=a_n=0$? Sep 3, 2021 at 13:59
• @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 $\Bbb Q \otimes_{\Bbb Z} A$ Sep 3, 2021 at 14:37