PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: High Entry average rating: No information on entry rating
Dirichlet's unit theorem (Theorem)

Let $ K$ be a number field, and let $ \mathcal{O}_K$ be its ring of integers. Then

$\displaystyle \mathcal{O}_K^*\cong \mu(K)\times\mathbb{Z}^{r+s-1}. $
Here $ \mathcal{O}_K^*$ is the group of units of $ \mathcal{O}_K$, $ \mu(K)$ is the finite cyclic group of the roots of unity in $ \mathcal{O}_K^*$, $ r$ is the number of real embeddings $ K\rightarrow \mathbb{R}$, and $ 2s$ is the number of non-real complex embeddings $ K\rightarrow \mathbb{C}$ (which occur in complex conjugate pairs, so $ s$ is an integer).



"Dirichlet's unit theorem" is owned by yark. [ full author list (2) | owner history (1) ]
(view preamble)

View style:

See Also: regulator


Attachments:
units of quadratic fields (Application) by pahio
units of real cubic fields with exactly one real embedding (Application) by Wkbj79
characterizing CM-fields using Dirichlet's unit theorem (Theorem) by rm50
Log in to rate this entry.
(view current ratings)

Cross-references: integer, complex conjugate, complex embeddings, real embeddings, roots of unity, cyclic group, finite, group of units, ring of integers, number field
There are 7 references to this entry.

This is version 7 of Dirichlet's unit theorem, born on 2003-01-20, modified 2007-12-14.
Object id is 3911, canonical name is DirichletsUnitTheorem.
Accessed 2974 times total.

Classification:
AMS MSC11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)
 11R27 (Number theory :: Algebraic number theory: global fields :: Units and factorization)

Pending Errata and Addenda
None.
[ View all 4 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | prove | add result | add corollary | add example | add (any)