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Feb 4, 2014 at 11:42 comment added Andrei Smolensky Actualy, what I wrote above is incorrect. The theorem states that $K_1(\Phi,A[x])=0$ if and only if $K_1(\Phi,A)=0$ and $K_1(\Phi,A[x],xA[x])=0$. Not sure what happens if one drops the assumption on relative $K_1$.
Feb 3, 2014 at 15:51 comment added Andrei Smolensky For Chevalley groups, however, the situation is better than for isotropic reductive groups in general. Namely, in a paper "Whitehead groups of Chevalley groups over polynomial rings" by E. Abe Corollary 1.9 states that for any ring $A$ and $\mathop{\mathrm{rank}}\Phi>1$ $K_1(\Phi,A[x])=0$ if and only if $K_1(\Phi,A)=0$. This result is actually referred to on the first page of the paper by Stavrova you mentioned.
Feb 3, 2014 at 15:32 history answered Simon Pepin Lehalleur CC BY-SA 3.0