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Search options not deleted user 4008
23 votes

Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left mo...

Here is an alternative to Andreas proof (which if you unfold it is not so different): We have a functor $M\mapsto \mathrm{Hom}_R(M,R)=:M^\ast$ which gives both a contravariant functor from left $R$-mo …
Torsten Ekedahl's user avatar
11 votes

Is a field uniquely determined by its multiplicative group/how much knows K_1 about fields?

$\mathbb Q^\ast$ is isomorphic to $\{\pm1\}$ times a free abelian group of countable rank. The same is true for an imaginary quadratic field of class number $1$ and different from $\mathbb Q(\sqrt{-3} …
Torsten Ekedahl's user avatar