Inverses of Units: Are All Conjugates Always in the Field?

[gonzo]

Okay, if we are looking at a typical extended number field Q(w), and it's corresponding ring of integers, we know that for any given element in this field, it is not necessary that all of it's conjugates are in the same field. A typical example being:

$Q(\theta), \theta=\root 3\of{3}, \theta \in R$

Now, the inverse of a unit is the product of all of it's conjugates (possibly times -1), so if all of it's conjugates are in the field then it's inverse is also an algebraic integer in the ring of integers in question.

I would like to know if it works out so that the conjugates of units are _always_ in the field in question, so that units always have inverses in their ring of integers, even when all the conjugates of an arbitrary element don't.

If so, why?

Thanks.

 

[robert Ihnot]

I think what is going on here has something to do with definition, or exactly what we are trying to do.

Take the cube root of 3, well if we also have cube root of 1=a, then we can obtain in the ring a^2, and trivially a^3=1. But if we look at the 15th root of 3, well, we could also add the cube root of 1, which would give us some units and conjugates, but not all.