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For two unital rings $R,S$, one can form the ring $R\otimes_\mathbb{Z} S$ (this is the coproduct in the category of rings). I was wondering whether the following formula holds for the Grothendieck group of projective modules over these rings: $K_0(R\otimes_\mathbb{Z} S ) \cong K_0(R) \otimes_\mathbb{Z} K_0(S)$ as abelian groups (or as rings if $R,S$ are commutative rings).

Since there are ring maps $R \rightarrow R\otimes_\mathbb{Z} S$ and $S \rightarrow R\otimes_\mathbb{Z} S$, there are induced maps $K_0(R) \rightarrow K_0(R\otimes_\mathbb{Z} S ), K_0(R) \rightarrow K_0(R\otimes_\mathbb{Z} S )$; if $R,S$ are commutative rings, these are maps of commutative rings and hence there is an induced map $K_0(R) \otimes_\mathbb{Z} K_0(S) \rightarrow K_0(R\otimes_\mathbb{Z} S )$. Is this an isomorphism? Note that $K_0(R \times S) \cong K_0(R) \times K_0(S)$ but I am interested in the tensor product of rings.

More generally, is there a description of $Proj(R \otimes_\mathbb{Z} S)$ in terms of $Proj(R \otimes_\mathbb{Z} S)$. In another direction, what about if there is an extra ring $T$: is $K_0(R \otimes_T S) \cong K_0(R) \otimes_{K_0(T)} K_0(S)$?

user39598
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1 Answers1

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No, this is not true in general. For instance, suppose $R=S=\mathbb{Q}(i)$. Then $K_0(R)=K_0(S)\cong\mathbb{Z}$, but $R\otimes S\cong \mathbb{Q}(i)\times \mathbb{Q}(i)$ so $K_0(R\otimes S)\cong\mathbb{Z}^2$.

In that example, the natural map $K_0(R)\otimes K_0(S)\to K_0(R\otimes S)$ fails to be surjective. It can also fail to be injective. For instance, if $R$ and $S$ are fields of different characteristic, then $R\otimes S=0$ so $K(R\otimes S)=0$.

Eric Wofsey
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