1
$\begingroup$

Hi,

let $k/\mathbb{Q}$ be a number field. Assume that $u$ is an algebraic integer such that all $k$-conjugates have modulus $1$. Is $u$ a root of $1$ ?

If $k=\mathbb{Q}$, the answer is YES (this is Kronecker's theorem). I am pretty sure that this result is false if $k$ is an arbitrary number field, but I don't see any obvious counter-example.

Any suggestion ?

Thanks!

$\endgroup$
  • $\begingroup$ If $u\in k$ and $k$ is Galois over $\bf Q$, then this is true. See Lemma 1.6 in Washington, "Introduction to cyclotomic fields". $\endgroup$ – Damian Rössler Oct 17 '12 at 20:53
  • $\begingroup$ You should register an account to prevent duplicate identities. $\endgroup$ – S. Carnahan Nov 2 '12 at 7:36
4
$\begingroup$

The answer depends on the number field $k$. Of course, it cannot hold for all fields $k$, for if $u$ is an algebraic integer of modulus $1$ which is not a root of unity (there are plenty of them, see e.g. this MO-link), then set $k=\mathbb Q(u)$, so $u$ is the only $k$-conjugate of $u$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.