index of the group of cyclotomic units in the full unit group
The elements of are defined analytically.
The subgroup is of finite index in . Furthermore, the index is : Let be the group of units in and let . Then . Moreover, it can be shown that because (this is exercise 8.5 in ).
The subgroups behave “well” in towers. More precisely, the norm of down to is . This follows from the fact that the norm of down to is .
Let be prime and let . Let be a primitive th root of unity.
The cyclotomic unit group is the group of units generated by and the units
with and .
The cyclotomic unit group is the group generated by and the cyclotomic units of .
Let be an element of . Then:
Notice that in order to show that the index of in is finite it suffices to show that the index of in is finite. Indeed, let . Since is a totally imaginary field and by Dirichlet’s unit theorem the free rank of is . On the other hand, and is totally real, thus the free rank of is also . Therefore the free rank of and are equal. As we claimed before, the index is rather interesting to us.
Theorem 1 (,Thm. 8.2).
Let be a prime and . Let be the class number of . The cyclotomic units of are a subgroup of finite index in the full unit group . Furthermore:
where in the last equality one uses the properties of Gauss sums and the class number formula in terms of Dirichlet L-functions evaluated at . This yields that in non-zero, therefore the index in is finite and moreover
An immediate consequence of this is that if divides then there exists a cyclotomic unit such that is a th power in but not in .
- 1 L. C. Washington, Introduction to Cyclotomic Fields, Second Edition, Springer-Verlag, New York.
|Title||index of the group of cyclotomic units in the full unit group|
|Date of creation||2013-03-22 15:42:49|
|Last modified on||2013-03-22 15:42:49|
|Last modified by||alozano (2414)|