Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a finite group. We know that the $K$-group $K_0(QG)$ of the rational group ring $QG$ is a free abelian group generated by the irreducible representations of $G$ over $Q$. Now let $R$ be a subring of the rationals where the order $|G|$ is invertible. What is the relation between $K_0(RG)$ and $K_0(QG)$?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

I would guess that the map on $K_0$ is an isomorphism, butI could only show the surjectivity right now:

The inclusion of rings $RG\rightarrow QG$ induces a map on $K_0$. Given a projective $RG$-module - say it is a submodule of $RG^n$ - to its $Q$-span. It is a projective submodule of $QG^n$.

So let us first show that this map is surjective, ie. every projective $QG$ module arises this way. Given any such $P'$ the obvious candidate for a preimage would be $RG^n\cap P'$.

First note that it is as a $R$-module a direct summand of $RG^n$. $R$ is a PID and hence one just has to verify that the quotient is $R$-torsionfree. But $RG^n/(RG^n\cap P')$ embeds into the $Q$-vectorspace $QG^n/P'$ and hence it is $R$-torsionfree.

So we have a section of $R$-modules $s:RG^n/(RG^n\cap P')\rightarrow RG^n$. It need not be a $RG$-map. So let us make it equivariant by setting $s'(x):=\frac{1}{|G|}\sum_{g\in G}gs(g^{-1}x)$. Note that it is still a section (project down again; it is a $RG$ map).

So we have found a $RG$-complement of $RG^n\cap P'$; hence $RG^n\cap P'$ is a projective $RG$-module. So the map $K_0(RG)\rightarrow K_0(QG)$ is surjective.

share|improve this answer
2  
It seems to me that it should not be injective in general but might always have finite kernel. Think of the case when $G$ has prime order $p$ and $R$ has only $p$ inverted: In this case the kernel ought be more or less the ideal class group of the $p$th cyclotomic field. –  Tom Goodwillie Jun 13 '12 at 14:09
    
Did I get the argument correctly: So in this case we have $Z[1/p][X]/(X^p-1)\cong Z[1/p] \times Z[1/p][X]/(1+..+x^{p-1})$ and the same holds for $Q$ instead of $Z[1/p]$. But $Q[X]/(1+..+x^{p-1})$ is a field so its $K_0$-group is the integers and this does not always hold for $K_0(Z[1/p][X]/(1+..+x^{p-1}))$. So there has to be a Kernel. –  HenrikRüping Jun 13 '12 at 15:02
    
Another thing is that if the (stable) isomorphism type of $P'\cap RG^n$ did only depend on the isomorphism type of $P'$ (and not of the chosen embedding) the map would be injective. If the map is not injective, it should be possible to write down an explicit example of two different embeddings. This might be interesting. Maybe it is also possible to express the Kernel as a set of certain equivalence classes of embeddings .... –  HenrikRüping Jun 13 '12 at 15:03
    
Hi, Henrik. Thanks for your comments. That's really nice. Now the only problem is whether the kernel is a finite group, whose order depending on the order of $|G|$? –  yeshengkui Jun 20 '12 at 2:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.