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Made isomorphism clearer.
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user26223
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If I remember correctly, it is isomorphic to the group of ordinary (virtual) characters over $K=\mbox{Frac}(\mathcal{O})$, the field of fractions of $\mathcal{O}$. That is, the Grothendieck group of $KG$. An isomorphism is given by simply tensoring with $K$, so that $\mathcal{O}G$-modules $M,N$ are identified in $K_0(\mathcal{O}G)$ if and only if $K\otimes_{\mathcal{O}}M$ and $K\otimes_{\mathcal{O}}N$ afford the same character. In particular, all torsion modules are zero in the Grothendieck group.

This was proved by Swan in 'The Grothendieck ring of a finite group', Topology 2, 85-110, 1963. He proves a more general result over an arbitrary integral domain there, and Theorem 3 is the one you're after.

If I remember correctly, it is isomorphic to the group of ordinary (virtual) characters over $K=\mbox{Frac}(\mathcal{O})$, the field of fractions of $\mathcal{O}$. That is, the Grothendieck group of $KG$. An isomorphism is given by simply tensoring with $K$.

This was proved by Swan in 'The Grothendieck ring of a finite group', Topology 2, 85-110, 1963. He proves a more general result over an arbitrary integral domain there, and Theorem 3 is the one you're after.

If I remember correctly, it is isomorphic to the group of ordinary (virtual) characters over $K=\mbox{Frac}(\mathcal{O})$, the field of fractions of $\mathcal{O}$. That is, the Grothendieck group of $KG$. An isomorphism is given by simply tensoring with $K$, so that $\mathcal{O}G$-modules $M,N$ are identified in $K_0(\mathcal{O}G)$ if and only if $K\otimes_{\mathcal{O}}M$ and $K\otimes_{\mathcal{O}}N$ afford the same character. In particular, all torsion modules are zero in the Grothendieck group.

This was proved by Swan in 'The Grothendieck ring of a finite group', Topology 2, 85-110, 1963. He proves a more general result over an arbitrary integral domain there, and Theorem 3 is the one you're after.

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user26223
  • 298
  • 1
  • 5

If I remember correctly, it is isomorphic to the group of ordinary (virtual) characters over $K=\mbox{Frac}(\mathcal{O})$, the field of fractions of $\mathcal{O}$. That is, the Grothendieck group of $KG$. An isomorphism is given by simply tensoring with $K$.

This was proved by Swan in 'The Grothendieck ring of a finite group', Topology 2, 85-110, 1963. He proves a more general result over an arbitrary integral domain there, and Theorem 3 is the one you're after.