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Given a number field $K$ (i.e. $\mathbb Q\le\ K\le\mathbb C$, $[K:\mathbb Q]=n$), the relative number ring is $R=\mathbb A\cap K$, where $\mathbb A$ is the ring of the algebraic integers in $\mathbb C$.

Let's consider the group of unit of $R$, call it $U$.

Consider a map $\log:U\longrightarrow\mathbb R^{r+s}$, a multiplicative-to-additive group homomorphism defined as in When does $V/\operatorname{ker}(\phi)\simeq\phi(V)$ imply $V\simeq\operatorname{ker}(\phi)\oplus\phi(V)$? (at the end).

$\log$ maps $U$ in the hyperplane of $\mathbb R^{r+s}$ $H:y_1+\dots+y_{r+s}=0$, i.e. $\log(U)\subseteq H$.

Suppose to know that $\log(U)$ is a $d$-dimensional lattice, call it $\Lambda_U$. Hence fix $u_1,\dots,u_d\in U$ mapping to a $\mathbb Z$-basis of $\Lambda_U$ and consider then the subgroup of $U$ generated by these $u_i$, call it $W$.

What does it mean that "the $u_i$ generate $W$ freely"?

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Joe
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This means that $u_1$, $u_2$, $\ldots$, $u_d$ forms a basis for $W$ as a $\mathbf{Z}$-module. In other words, every element in $W$ can be written as $\mathbf{Z}$-linear combination of $u_1$, $u_2$, $\ldots$, $u_d$ uniquely. Of course, we interpret $\sum_{i=1}^d a_iu_i$ as $\prod_{i=1}^{d}u_i^{a_i}$. This also means that there are no non-trivial relations, i.e. we can't find $a_1$, $a_2$, $\ldots$, $a_d$, not all them zero, such that $\prod_{i=1}^du_i^{a_i}=1$.

user157986
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  • Let me know if I understood: the $u_i$ generate $W$, which is a multiplicative subgroup of $U$, hence (all is abelian here) an element of $W$ can be written as $u_1^{k_1}\dots u_d^{k_d}$, with $k_i\in\mathbb Z$ in a unique way (in is in this sense we say that $u_i$ generate $W$ freely). Hence seeing this in additive notation, we have that $W\simeq\mathbb Z^d$. Am I right? – Joe Aug 25 '14 at 11:54
  • Yes, You are right. – user157986 Aug 26 '14 at 09:42