Around a theorem of Kronecker - MathOverflow 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 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 Answer by Peter Mueller for Around a theorem of Kronecker 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>