elements of absolute value one in cyclotomic fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:50:04Z http://mathoverflow.net/feeds/question/107786 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107786/elements-of-absolute-value-one-in-cyclotomic-fields elements of absolute value one in cyclotomic fields Binzhou Xia 2012-09-21T17:09:27Z 2012-09-24T16:25:07Z <p>Let $p$ be a rational prime and $\zeta_p$ a primitive $p$th root of unity. </p> <p>What do we know about the set ${z\in\mathbb{Q}(\zeta_p):|z|=1}$? </p> http://mathoverflow.net/questions/107786/elements-of-absolute-value-one-in-cyclotomic-fields/107787#107787 Answer by François Brunault for elements of absolute value one in cyclotomic fields François Brunault 2012-09-21T17:28:19Z 2012-09-21T17:28:19Z <p>For any $u \in L^\times = \mathbf{Q}(\zeta_p)^\times$, letting $z=u/\bar{u}$, we have $|z|=1$. The converse is true : let $G=\mathbf{Z}/2\mathbf{Z}$ act on $L^\times$ by complex conjugation, then $G$ is the Galois group of $L$ over $K = L \cap \mathbf{R}$. By Hilbert 90, $H^1(G,L^\times) = {1}$, which says precisely that any $z \in L^\times$ such that $|z|^2 = z \bar{z}=1$ is of the form $z=u/\bar{u}$ for some $u \in L^\times$.</p> <p>Let $U \subset L^\times$ be the subgroup of elements $z$ satisfying $|z|=1$. Then the map $u \mapsto u/\bar{u}$ induces an isomorphism $U \cong L^\times / K^\times$. In particular $U$ is not of finite type.</p>