Re: skew fields

From: Mark Sapir (msapir_at_comcast.net)
Date: 03/24/05 

Date: Thu, 24 Mar 2005 03:00:04 +0000 (UTC)

> > Is it known if the multiplicative group of a skew field is residually
> > finite?
> > Any references are welcome.
> > Pete
> >
>
> If I understaned the summary correctly, the following paper
> should supply a counterexample (information taken from MathSciNet):
>
> Kegel, Otto H, Zur Einfachheit der multiplikativen Gruppe eines
> existentiell abgeschlossenen Schiefkoerpers. (German. English summary)
> [On the simplicity of the multiplicative group of an existentially
> closed skew field]
> Results Math. 35 (1999), no. 1-2, 103--106.
>
> Summary: "Let $E$ be an existentially closed skew field in the class of
all
> skew fields with given central subfield $Z$. For the multiplicative
> group $E^*$ it is shown that the group $G\coloneq E^*/Z^*$ is simple."
>

This is too complicated: the multiplicative group of the field R (real
numbers) is not residually finite. Thus you do not need non-commutativity.
Here is a proof that R^* is not residually finite. Suppose that there exists
a homomorphism f from R^* to a finite group separating 2 from 1. Suppose
that f(2) has order k in f(R^*).Then there exist natural numbers m>n such
that a root a of 2 of degree k^m and a root b of 2 of degree k^n have the
same images under f. Thus f(a)=f(b). Raising to the power k^m gives
f(2)=f(b^(k^m))=f(2^(k^(m-n)))=1, a contradiction.

Perhaps a more appropriate question is: are there skew fields K for which
not every finitely generated subgroup of K^* is residually finite?

Mark