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Let K be a number field and $p$ an odd prime. Let $\mu$ be the Iwasawa $\mu$-invariant of the class group of the cyclotomic $\mathbb{Z}_p$-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 $K$.

Is something known for the case when $K$ is abelian over an imaginary quadratic field $k$ ?

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Jonathan Sands has a 1991 paper that discusses some Iwasawa features of imaginary quadratic base but I don't know what he really shows. ams.org/journals/proc/1991-112-03/S0002-9939-1991-1057961-4/… –  Junkie Apr 2 '11 at 6:11
    
Interesting. Unfortunately, - as far as I see- he considers non-cyclotomic extensions over the base-field $k$. I am interested in cyclotomic extensions of a $K$, which is itself an abelian extension of $k$. –  Chris Wuthrich Apr 4 '11 at 15:10
    
Update : I asked Karl Rubin and John Coates and both seem to think that the problem is open. Maybe it is within reach, I don't know. –  Chris Wuthrich May 9 '11 at 14:31
    
There is a post of the arxiv today claiming to prove the vanishing of cyclotomic mu for CM fields -- front.math.ucdavis.edu/1105.1970 . –  Robert Pollack May 11 '11 at 13:58
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1 Answer 1

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.

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 explicit 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 (i.e. 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.

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By my above comment I meant that they told me that they did not know of any result that would prove it. I am not sure that they have doubts about how difficult it is. I have not tried myself to prove it, but your idea may be a route to attack the problem... –  Chris Wuthrich Apr 11 '12 at 13:51
    
Let's say that if it were to work, then I guess Sinnott itself would have tried...and succeeded ;-) Filippo –  Filippo Alberto Edoardo Apr 11 '12 at 13:59
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