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Answer by Chris H for Grothendieck group of coherent sheaves is not a ring?
I think the reason that we don't have a natural multiplication is that the tensor product is not exact, which can be seen in the affine case. So given $0\rightarrow M_1\rightarrow M_2\rightarrow...
View ArticleGrothendieck group of coherent sheaves is not a ring?
My question is motivated by the fact that the Grothendieck group $K^0(X)$ of vector bundles on $X$ can be given a ring structure via the tensor product. But it seems to me that the Grothendieck group...
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