Timeline for Projective modules over non-rational group rings
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 20, 2012 at 2:46 | comment | added | yeshengkui | 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 $|G|$? | |
| Jun 20, 2012 at 2:41 | vote | accept | yeshengkui | ||
| Jun 13, 2012 at 15:03 | comment | added | HenrikRüping | 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 .... | |
| Jun 13, 2012 at 15:02 | comment | added | HenrikRüping | 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. | |
| Jun 13, 2012 at 14:09 | comment | added | Tom Goodwillie | 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. | |
| Jun 13, 2012 at 13:39 | history | answered | HenrikRüping | CC BY-SA 3.0 |