most recent 30 from http://mathoverflow.net 2013-05-22T07:36:11Z http://mathoverflow.net/feeds/question/109945 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109945/around-a-theorem-of-kronecker GreginGre 2012-10-17T20:13:32Z 2012-10-17T21:25:51Z

<p>Hi,</p> <p>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 $1$ ?</p> <p>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. </p> <p>Any suggestion ?</p> <p>Thanks!</p>

http://mathoverflow.net/questions/109945/around-a-theorem-of-kronecker/109952#109952 Peter Mueller 2012-10-17T21:25:51Z 2012-10-17T21:25:51Z

<p>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. <a href="http://mathoverflow.net/questions/38680/can-an-algebraic-number-on-the-unit-circle-have-a-conjugate-with-absolute-value-d" rel="nofollow">this MO-link</a>), then set $k=\mathbb Q(u)$, so $u$ is the only $k$-conjugate of $u$.</p>