# $\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?

Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by finite group.

Denote the (Banach) space (in fact an algebra under convolution), $l^1(G)=\{\sum_{g\in G}\lambda_g g| \sum_{g\in G}|\lambda_g|<\infty, \lambda_g\in \mathbb{C}\}$.

Note that $\mathbb{Z}G\subset l^1(G)$ and we can consider $l^1(G)$ as a right $\mathbb{Z}G$-module, my question is:

$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $l^1(G)$ is a flat $\mathbb{Z}G$-(right) module?

If it is true, any reference (maybe for related topics)? Otherwise, any counterexample? Especially, is it true for the Heisenberg group?

RK: I have asked a general one in MSE but no answer appeared, so I asked here a more specific one.

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Just a comment - any group $G$ is canonically isomorphic to the opposite group $G^{op}$, where multiplication is reversed (i.e. $g\cdot h := hg$). So $\mathbb{Z} G$ is left-Noetherian iff it is right-Noetherian. But I don't know how the $l^1$ condition interacts with standard homological algebra. Can you prove what you want for $G = \mathbb Z$? –  Peter Samuelson Jul 21 '13 at 0:51
I am not sure it has been proved when $G=\mathbb{Z}$, but perhaps you may find it helpful to read theorem 3.1(4) and corollary 3.5 in the following paper:arxiv.org/abs/1103.1567 –  Jiang Jul 21 '13 at 2:36
And "how the $l^1$ condition interacts with standard homological algebra" is more or less what I want to know. –  Jiang Jul 21 '13 at 2:37

Here are a few observations that are too long for a comment but don't provide a full answer by any means.

As Peter Samuelson observed left-right questions aren't significant in this setting since $\mathbb{Z}G$ is isomorphic to $\mathbb{Z}G^{op}$.

Also, note that $\mathbb{C}G$ is a flat $\mathbb{Z}G$-module since $\mathbb{C}$ is a flat $\mathbb{Z}$ module. Thus, by standard base change arguments, to prove that $l^1(G)$ is flat over $\mathbb{Z}G$ it would suffice to prove that it is flat over $\mathbb{C}G$. At least in the case $G$ is polycyclic-by-finite, $\mathbb{C}G$ will also be (left and right) Noetherian.

Now I'm going to restrict to the case $G=\mathbb{Z}$, so $\mathbb{C}G\cong \mathbb{C}[x,x^{-1}]$, the ring of Laurent polynomials in $x$. This ring is a principal ideal domain.

Now I refer to Weibel's Introduction to homological algebra (CUP) Proposition 3.2.4. A left $R:=\mathbb{C}[x,x^{-1}]$-module $B$ is flat as such if and only if $\mathrm{Tor}_1^R(R/I,B)=0$ for every right ideal $I$ of $R$.

But we've already observed that every (right) ideal $I$ is of the form $fR$ in this case, and (for $f\neq 0$), we can compute $\mathrm{Tor}_1^R(R/fR,B)$ explicitly as $[f]B :=\{ b\in B\mid fb=0\}$.

In other words, to answer your question positively for $G=\mathbb{Z}$ it suffices to prove that $l^1(G)$ has no elements killed by $f$ for any non-zero $f\in \mathbb{C}G$. I think it should be possible to prove this by elementary arguments although I haven't even begun to check.

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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. –  Simon Wadsley Sep 17 '13 at 10:54
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. –  Jiang Sep 19 '13 at 2:01
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 $\mathbb{C}G$? –  Jiang Sep 19 '13 at 2:06
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. –  Simon Wadsley Sep 19 '13 at 13:27