Let $G$ be a finite group.

Let $R$ be a Dedekind domain of zero characteristic with $K=$ Frac$(R).$

Let $\mathcal{P}_R$ denote the category of finitely generated, projective $R[G]$-modules.

Let $\mathcal{L}_R$ denote the category of locally free $R[G]$-modules.

$\Big[$ An $R[G]$-module $X$ is called "locally free" if

  1. $X$ is finitely generated over $R[G],$

  2. $R_P\otimes_R X$ is free over $R_P[G]$ for all the maximal ideals of $R$ (where $R_P$ is completion at $P$).

Moreover, for $X \in \mathcal{L}_R$ we have that $K\otimes_R X$ is free over $K[G]$ and

$$\mathrm{rank}_{K[G]}(K\otimes_R X)=\mathrm{rank}_{R_P[G]}(R_P\otimes_R X)$$

for all maximal ideals $P$ of $R.$ We therefore have a group homomorphism $$\rho_R:K_0(\mathcal{L}_R)\to \mathbb{Z}; \; \rho_R: [X]\mapsto \mathrm{rank}_{K[G]}(K\otimes_R X),$$

where $K_0(\mathcal{A})$ denotes the Grothendieck group of an abelian category $\mathcal{A}.$ $\Big]$


In the literature, I have seen the "projective class group" of $R[G]$ defined both as

  1. the kernel of $\rho_R,$

  2. the kernel of the group homomorphism $K_0(\mathcal{P}_R)\to K_0(\mathcal{P}_K)$ given by $[P] \mapsto [K\otimes_R P].$

However, I am stuck to see why these are naturally isomorphic.

Any help gratefully received!


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