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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!

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  • $\begingroup$ 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
    Commented Oct 17, 2012 at 20:53
  • $\begingroup$ You should register an account to prevent duplicate identities. $\endgroup$
    – S. Carnahan ♦
    Commented Nov 2, 2012 at 7:36

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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 $u$.

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