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?

  • 3
    $\begingroup$ 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 $u\in \mu(K)$). $\endgroup$ – Alex Youcis Sep 24 '13 at 17:33

Let me just point you to the Wikipedia article on Fundamental units. I think it explains most of your questions so there's no point in repeating what's already in there.

Also, this mathworld article has more information and some examples.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.