Quillens higher Kgroups of rings can be realized as π_{n}K(C)  the Waldhausen KTheory of a suitable Waldhausen category C. Is this also true for Milnor KTheory of Rings? Is there a functor F from rings to waldhausen categories s.t. $K^M_n(R)\cong \pi_n(K(F(R))$?

1$\begingroup$ Interesting question. I would also like to know this. $\endgroup$ – user717 Dec 18 '09 at 22:57
I don't know if there any evidence for this to be true. Note that Quillen Kgroups are defined as homotopy groups of some space (+construction, Qconstruction, Waldhausen construction etc), whereas Milnor Kgroups were defined in terms of generators and relations, which generalize generators and relations for classical K_2.
More invariantly Milnor Kgroups can be constructed using homology of GL_n (paper of Suslin and Nesterenko) or as certain motivic cohomology groups of a field (SuslinVoevodsky). However, these constructions are unrelated to any homotopy groups.
Also, I'm not sure how you define Milnor Ktheory for a general ring R? (I was interpreting your question with "ring R" replaced by "field F".)

1$\begingroup$ There is a paper on Milnor KTheory of Rings from ElbazVincent and MüllerStach, but the definition is certainly much older (they cite a 1977 paper of Guin). At least for a field,as you say, Milnor KTheory is on the diagonal of the motivic cohomology of that field and sice motivic cohomology is representable as homotopy classes of maps into the motivic EilenbergMacLane spaces one can write someting like $K^M_n(k)=[\mathrm{spec}(k),\tilde U_tr\tilde\mathbb Z_tr(S^q_t)]_{sPre_*}$. I guess one can play around a little bit and write that as homotopy groups, but at all I don't know the answer. $\endgroup$ – user2146 Dec 22 '09 at 17:38

$\begingroup$ Oops, for n=q of course. So I mean $K^{M}_{q}(k)=[\mathrm{spec}(k)_{+},\tilde{U}_{tr}\tilde{\mathbb{Z}}_{tr}(S^q_t)]_{sPre_*}$ $\endgroup$ – user2146 Dec 22 '09 at 17:48

$\begingroup$ There is a definition for Milnor Kgroups for local rings: math.uiuc.edu/Ktheory/0865 math.uiuc.edu/Ktheory/0791 $\endgroup$ – user19475 Jan 18 '10 at 20:00
Bob Thomason proved that there is no Milnor Ktheory functor for schemes, with a reasonable map to Quillen Ktheory, in:
Le principe de scindage et l'inexistence d'une $K$theorie de Milnor globale. [The splitting principle and the nonexistence of a global Milnor $K$theory] Topology 31 (1992), no. 3, 571588.
 John
The original question seems not to have been answered yet. One answer might be that it would be unnatural to expect all the Milnor Kgroups of a field R to arise as the homotopy groups of a single space $K(F(R))$, because the natural way they currently arise is as homotopy groups of separate spaces, or better, of separate spectra. The spectra are the EilenbergMacLane spectra $\mathbb Z(n)$ associated to the chain complexes that compute motivic cohomology of $R$, namely, $K_n^M R = \pi_{n} \mathbb Z(n)$.