Compatibility of two definitions of the projective class group of a group ring

Question

Set-up:

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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 $\mathcal{A}.$ $\Big]$

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Question:

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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!

Answer 1

One important ingredient is the following theorem of Swan (Theorem 8.1. in Induced representations and projective modules): If $P$ is any projective f.g. $RG$-module, then $P \otimes_R K$ is free.

With that understood, we can do that following: We actually have a splitting $K_0 (\mathcal P_R ) \cong \mathbb{Z} \oplus \widetilde{K_0 RG}$, where $\mathbb Z$ corresponds to the free modules. (You can define $\widetilde{K_0 RG}$ as the cokernel of the map $K_0 \mathbb{Z} \rightarrow K_0 (\mathcal P_R )$ that sends $\mathbb Z^n$ to $R[G]^n$ and then show that this map splits). The theorem from above guarantees that $\widetilde{K_0 RG}$ is actually the kernel of the homomorphism $K_0 (\mathcal P_R ) \rightarrow K_0 (\mathcal P_K )$ you considered. The group $\widetilde{K_0 RG}$ is called the reduced $K$-theory group of $RG$.

Now the other ingredient is to know that for the group ring $RG$, the locally free class group and the reduced $K$-theory group agree. This is actually non-trivial and is covered for example in Curtis, Reiner Methods of Representation Theory Volume 2, 49.11. Note that what you call the kernel of $\rho_R$ is what they write as $Cl(RG)$.

As a side remark: The reason people considered locally free class groups and reduced algebraic $K$-theory separately is that they are in fact different when applied to an arbitrary order in $KG$. Oftentimes, to compute $K_0(RG)$ it is helpful to go from $RG$ to a maximal order $\Lambda$, compute its $K$-theory (which is often easier) and then check which information got lost along the way.