4
$\begingroup$

This question is motivated by a Quora post and the top answer to it.

The post wishes to make explicit the difference between units in cyclotomic fields and cyclotomic units.

One problem with answering such a question is the difficulty in computing fundamental units in cyclotomic fields or subfields thereof.

However, there is a case when we know the fundamental unit---real quadratic fields.

This led me to the answer to a question on MO which looks at whether the norm of a cyclotomic unit is a fundamental unit in a real quadratic field. The answer is given in the negative.

However, while checking whether this answers the original question, one notices the gap: it is not clear that the cyclotomic units contained in the real quadratic field are norms from the cyclotomic field that contain this field.

In summary, here are the questions:

Let $K$ be a real quadratic field and $E$ be a cyclotomic field that contains it. Let $C_E$ be the group of (i.e. generated by) cyclotomic units in $E$ and $C_K$ be the intersection of $C_E$ with $K$.

  1. Is there a formula for the generator of $C_K$?

  2. Is there a situation where this generator is not the norm of a cyclotomic unit from $C_E$? (If not, then (1) is already answered!)

  3. What is the smallest $K$ for which the fundamental unit is not in $C_K$?

In (3), I am guessing that there is always such a $K$ and so it would answer the original question. If not:

  1. What is the smallest Abelian extension of $\mathbb{Q}$ in which one can give an example of a unit which is not a cyclotomic unit?
$\endgroup$
2
  • $\begingroup$ Have you looked at Section 8.1 of Washington's book on cyclotomic fields? It doesn't directly answer your question, but maybe some of the information there could help, such as Theorem 8.2 and Section 4 of the tables at the end of the book. $\endgroup$
    – KConrad
    Commented Feb 23, 2023 at 17:00
  • $\begingroup$ On page 143 of this book Washington refers to a paper of Sinott to say "it is perhaps better to take norms" to define $C_K$. I assume that this means that this image is easier to compute than the intersection. $\endgroup$
    – Kapil
    Commented Feb 24, 2023 at 8:25

0

You must log in to answer this question.

Browse other questions tagged .