# When is $\mathbb{Z}$ a flat $\mathbb{Z}G$-module?

Suppose that $\mathbb{Z}$ is a flat $\mathbb{Z}G$-module for a group $G$.

Question: Is $G$ the trivial group ?

Nb. I know that the question can be answered affirmatively if $G$ is finitely generated.

Edit: I think the following lemma solves the problem. I would be grateful if someone could have a look on its proof and give some feedback whether it looks ok. Thanks.

Lemma: Let $R \le S$ be rings with unit such that $S$ is flat as left $R$-module. Then every flat left $S$-module is also flat as left $R$-module.

Now let $G\neq 1$ be any group. If $G$ is abelian then $\mathbb{Z}$ isn't flat by Georges' argument. If $G$ is not abelian, we can find an abelian subgroup $1 \neq H \le G$. Now, if $\mathbb{Z}$ were flat as $\mathbb{Z}G$-module, it would also be flat as $\mathbb{Z}H$-module by the lemma. But we just saw that this isn' true. Hence $\mathbb{Z}$ isn't flat as $\mathbb{Z}G$-module. We have therefore shown:

For a group $G$ the following is equivalent:

1. $G=1$
2. $H_i(G,-)=0\,$ for all $i > 0$
3. $H^i(G,-)=0\,$ for all $i > 0$

Proof of the Lemma: Let $E$ be a flat left $S$-module and let $i: M \to N$ be an embedding of right $R$-modules. We have to show that $i \otimes id_E: M \otimes_R E \to N \otimes_R E$ is also an embedding.

Since $S$ is a flat left $R$-module, tensoring with $S$ from the right yields an embedding $i \otimes id_S: M \otimes_R S\to N \otimes_R S$ of right $S$-modules. Similarly, as $E$ is a flat left $S$-module, we obtain the embedding $$(i \otimes id_S)\otimes id_E: (M \otimes_R S)\otimes_S E \to (N \otimes_R S) \otimes_S E$$ which, by associativity of the tensor product, is equivalent to $$i \otimes (id_S\otimes id_E): M \otimes_R (S\otimes_S E) \to N \otimes_R (S \otimes_S E)$$ which, by the natural isomorphism $S \otimes_S E \cong E$ is equivalent to $$i \otimes id_E: M \otimes_R E \to N \otimes_R E.$$ Hence $i \otimes id_E$ is an embedding and thus $E$ is flat as left $R$-module. QED

• That would make all its homology trivial then?
– user641
Aug 14, 2012 at 1:42
• Yes (in positive degrees). Aug 14, 2012 at 2:14
• Sorry I don't see why being flat makes the homologies trivial; this is only if the module is projective (in which case Ralph's last sentence implies that he knows the result). Aug 14, 2012 at 2:53
• Chris, this is just the definition of flatness!!! Aug 14, 2012 at 3:27
• Of course! Completely overlooked the basic construction: $H_n(G;\mathbb{Z})=H_n(F_G)$ where $F$ is a particular exact sequence (projective resolution of $\mathbb{Z}$) and is tensored by $\mathbb{Z}G$, preserving exactness for $F_G$. Aug 14, 2012 at 8:11

Result If $G$ is not perfect (i.e. if $G\neq[G,G]$) , then the $\mathbb Z[G]$-module $\mathbb Z$ is not flat.
Example If $G\neq 0$ is commutative, then the $\mathbb Z[G]$-module $\mathbb Z$ is not flat

Proof
1) If $I\subset A$ is an ideal of a ring and $A/I$ is $A$-flat, then $I/I^2=0$
Indeed, tensor the injection $0\to I\to A$ with $A/I$ and obtain the injection $0\to I/I^2\to A/I: [i]\mapsto \bar i$ .
This last map is zero and can only be injective if $I/I^2=0$

2) In our case $A=\mathbb Z[G]$ and $I$ is the augmentation ideal consisting of the $\sum a_gg$ with $\sum a_g=0$.
Weibel's Introduction to Homological Algebra assures us (in an exercise page 164) that $I/I^2=G/[G,G]$.
Hence flatness of $\mathbb Z$ implies $I/I^2=0$ by 1) which in turn forces $G=[G,G]$ : the Result follows.

• Thanks for your reply (I think your proof can be shortened by simply noting that $H_1(G,\mathbb{Z})=G_{ab}$). So a counter-example to the question must be not f.g. and not perfect. Aug 14, 2012 at 8:42
• Dear Ralph: well, maybe, but as a matter of personal taste I would prefer not to invoke homology theory and give proofs from first principles. Amusingly you confirm my point of view that you only really have integrated a result if you find all the presentations by others unsatisfying! Aug 14, 2012 at 8:54