The fundamental unit in the ring of algebraic integers.

I can't understand my own notes in number theory....I've written: let $K$ be a number field, $\mathscr{O}_K$ its ring of integers and $U_K=\mathscr{O}_K^{\star}$ the multiplicative group of units in $\mathscr{O}_K$. Denote by $u$ the fundamental unit in $\mathscr{O}_K$, i.e. the unit greater than $1$ which generates the $\textbf{free}$ part of the units group $U_K$.....

What is $u$?! Why greater than $1$? I know Dirichlet theorem, which says that $U_K$ is the direct product of a finite cyclic group and a free abelian group of rank $r+s-1$ with $r$ the number of real embeddings and $s$ the number of pairs of non real embeddings in $\mathbb{C}$.

So, every unit $u$ in $\mathscr{O}_K$ can be expressed uniquely as $u=wu_1^{b_1}\cdots u_{r+s-1}^{b_{r+s-1}}$, with $u_j$ units in $\mathscr{O}_K$, $b_j$ integers and $w$ some root of unity.

So what is the so-called fundamental unit?

• Well, "the fundamental unit" is, in general, non-sensical for a general number field, unless this is non-standard terminology. If, as you point out, $r+s-1>1$, then there will be more than one fundamental unit. As for the unit greater than one, I have no idea. If $K$ is Galois you can always find a fundamental unit with absolute value greater than one (just consider the conjugates, and use the fact that $|\sigma(u)|$ can't be $1$ for all $u$, else $u\in \mu(K)$). – Alex Youcis Sep 24 '13 at 17:33