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.

 

[tacman]

The answer to this question depends on the specific field and ring of integers in question. In general, it is not always the case that the conjugates of units are always in the field. However, there are certain cases where this is true.

One example is when the field is a Galois extension, meaning that it is a field generated by a root of a polynomial with coefficients in the base field and the polynomial has all of its roots in the field. In this case, all of the conjugates of units will be in the field, and therefore the inverse of a unit will also be in the field.

Another example is when the field is a cyclotomic extension, meaning that it is a field generated by a root of unity. In this case, all of the conjugates of units will also be roots of unity, which are by definition in the field. Therefore, the inverse of a unit will also be in the field.

However, there are also cases where the conjugates of units are not always in the field. One example is the field Q(w), where w is a primitive cube root of unity. In this case, the conjugates of units will not always be in the field, so the inverse of a unit may not be in the ring of integers.

In summary, whether or not the conjugates of units are always in the field depends on the specific field and ring of integers in question. In some cases, such as Galois and cyclotomic extensions, this is true, but in general it is not always the case.