Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
That would make all its homology trivial then? –  user641 Aug 14 '12 at 1:42
    
Yes (in positive degrees). –  Ralph Aug 14 '12 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). –  Chris Gerig Aug 14 '12 at 2:53
2  
Chris, this is just the definition of flatness!!! –  Ralph Aug 14 '12 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 $F_G$. –  Chris Gerig Aug 14 '12 at 8:11

1 Answer 1

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.

share|improve this answer
    
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. –  Ralph Aug 14 '12 at 8:42
1  
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! –  Georges Elencwajg Aug 14 '12 at 8:54

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.