Iwasawa mu-invariant for abelian extensions of quadratic number fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T13:03:18Z http://mathoverflow.net/feeds/question/59294 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59294/iwasawa-mu-invariant-for-abelian-extensions-of-quadratic-number-fields Iwasawa mu-invariant for abelian extensions of quadratic number fields Chris Wuthrich 2011-03-23T12:17:27Z 2012-04-11T12:10:37Z <p>Let K be a number field and $p$ an odd prime. Let $\mu$ be the Iwasawa $\mu$-invariant of the class group of the <strong>cyclotomic</strong> $\mathbb{Z}_p$-extension of $K$. If $K$ is abelian over $\mathbb{Q}$ then it is known that $\mu=0$ (Ferrero-Washinton, see Washington 7.5). Iwasawa conjectured that $\mu=0$ for all $K$. </p> <p>Is something known for the case when $K$ is abelian over an imaginary quadratic field $k$ ?</p> http://mathoverflow.net/questions/59294/iwasawa-mu-invariant-for-abelian-extensions-of-quadratic-number-fields/93758#93758 Answer by Filippo Alberto Edoardo for Iwasawa mu-invariant for abelian extensions of quadratic number fields Filippo Alberto Edoardo 2012-04-11T12:10:37Z 2012-04-11T12:10:37Z <p>Have you tried looking at Sinnott's paper where he re-proves Ferrero-Washington for $\mathbb{Q}$? It is in Invent. Math, 1984 vol. 75 (2) pp. 273-282.</p> <p>He proves that to compute the $\mu$-invariant of a function that can be expressed as $\Gamma$-transform of a power series, it is enough to know the $\mu$-invariant of the series. He then applies this to the construction of the $p$-adic $L$-function of Iwasawa where an <em>explicit</em> expression (page 282 and equation (4.3) on page 280) of the paper can be found. Since this ''explicit expression'' comes from the Euler system of cyclotomic units and we now dispose of the Euler system of elliptic units (<em>i.e.</em> we now call it in such a way) for the cyclotomic extension of your imaginary quadratic field, it is plausible that Sinnot's argument applies. But if Coates and Rubin have doubts there must be something tricky behind. </p>