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Jun 3, 2019 at 6:47 comment added naf $\pi$ and $\pi'$ will almost never be flat (since the maps are birational). Also, the conjecture is now known to be false! Explicit counterexamples were given by Rob de Jeu.
Jun 3, 2019 at 5:17 comment added user127776 My assumption was to replace the proper regular with proper flat and all $K$ with $K'$ and repeat the same argument. In order the pullback to make sense in $K'$-theory the morphism needs to be flat. So $\pi$ has to be flat. I thought it is clear but right now I can't figure it out, probably that's the problem. I'm not even sure whether proper flatness of $\mathcal{X}$ and $\mathcal{X}'$ implies the flatness of $\mathcal{X}''$
Jun 3, 2019 at 2:26 history asked user127776 CC BY-SA 4.0