Timeline for $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat (right) $\mathbb{Z}G$-module?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 19, 2013 at 13:27 | comment | added | Simon Wadsley | I'm not even sure that it is easy to extend my strategy to $G=\mathbb{Z}^n$ for $n>1$, although it may be easier than I think. Classifying right ideals in $\mathbb{C}G$ for $G$ a discrete Heisenberg group is probably hard. To get an idea why, see arxiv.org/abs/math/0102190. The first sentence of section 7 of your reference suggests that $Tor^1_{\mathbb{C}G}(\mathbb{C}G/f\mathbb{C}G,l^1(G))=0$ for all non-zero $f\in \mathbb{C}G$ whenever $G$ is torsionfree polycyclic. However, it isn't clear to me how hard this result was to prove already in that generality. | |
| Sep 19, 2013 at 2:06 | comment | added | Jiang | How to modify your reduction to the case such that $G$ is a non-commutative polycyclic-by-finite group, say the discrete Heisenberg group? In particular, what does a general right ideal $I$ look like in $\mathbb{C}G$? | |
| Sep 19, 2013 at 2:01 | comment | added | Jiang | in this paper ams.org/journals/proc/1998-126-03/S0002-9939-98-04025-8/… the author mentioned at the last line of page 721 that $0\neq f\in \mathbb{C}(\mathbb{Z})$ is a uniform nonzero divisor. | |
| Sep 17, 2013 at 10:54 | comment | added | Simon Wadsley | It occurs to me that in the final paragraph one might as well assume that $f$ is irreducible in $R$. Since (by the fundamental theorem of algebra) such an $f$ is a unit times $(x-\lambda)$ for $\lambda\in \mathbb{C}$ non-zero it should be very easy to complete the case $G=\mathbb{Z}$ by hand. | |
| Sep 17, 2013 at 8:45 | history | edited | Simon Wadsley | CC BY-SA 3.0 |
moved some stray commas.
|
| Sep 17, 2013 at 8:38 | history | answered | Simon Wadsley | CC BY-SA 3.0 |