Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $R$ is a Noetherian commutative ring, and $M$ a finite free $R$-module, with an action of a finitely generated discrete group $G$ by $R$-linear maps.

Is there any homological condition on this data which would ensure that taking $G$-invariants commutes with base change?

That is, for any finite type map $R \rightarrow S$, we have $$(M \otimes S)^G = (M^G)\otimes S.$$

For instance, suppose $M^G = 0$?

share|improve this question
Does $G$ operate on $M \otimes S$ via $g \cdot (m \otimes s) = (gm) \otimes s$ ? –  Ralph Sep 5 '11 at 21:48
I imagine we're assuming trivial action on the coefficients. It took me a moment to realize that we do need a condition: take $G={\mathbb Z}/2$, and let it act on $M={\mathbb Z}$ by multiplication by $-1$. Then $M^G$ over $R={\mathbb Z}$ is zero. However, the invariants over $S={\mathbb Z}/2{\mathbb Z}$ are all of $M\otimes S$. –  Graham Denham Sep 5 '11 at 22:17
add comment

2 Answers

Under the assumption, that $G$ operates as guessed in my comment, one can proceed as follows:

First note, that the conditions make $M$ a $RG$-module. For every $RG$-module $N$, it holds:

$$N^G \cong Hom_R(R,N)^G\cong Hom_{RG}(R,N)= Ext^0_{RG}(R,N) =: H^0(G;N)$$

with natural $R$-module isomorphisms. Since $M$ is $R$-torsion-free, the universal coefficient theorem applies, yielding a short exact sequence of $R$-modules (that's where to assumption about the $G$-operation on the tensor product comes into play): $$0 \to H^0(G;M) \otimes_R S \to H^0(G;M \otimes_R S) \to Tor_1^R(H^1(G;M),S) \to 0.$$

Thus $M^G \otimes S = (M \otimes S)^G$ is equivalent to the vanishing of the Tor-term. Hence we have as a homological criterion:

$$Tor_1^R(H^1(G;M),S) = 0 \text{ for every } R \to S.$$

A sufficient condition is therefore: $H^1(G;M)$ is a flat $R$-module.

Note: $G$ can be an arbitrary group, the finitely generated assumption isn't used.

Edit: In order to apply the universal coefficient theorem (UCT), one has to require $R$ to be hereditary and $M$ to be a projective $R$-module (if $M$ is supposed to be a finitely generated projective $R$-module, it sufficies if $R$ is semi-hereditary).

Remark 1: Hereditary means that submodules of projective modules are again projective; semi-hereditary means that submodules of finitely generated projective modules are again projective. For instance, Dedekind domains are hereditary and Prüfer domains are semi-hereditary.

Remark 2: For the convinience of the reader let me include the way, I use UCT: According to Weibel's homological algebra book (Theorem 3.6.1): If $P$ is a chain complex of flat $R$-modules such that for each $n$, $d(P_n)$ is a flat submodule of $P_{n-1}$, then the following sequence is exact for every $R$-module $S$: $$0 \to H_n(P) \otimes_R S \to H_n(P \otimes_R S) \to Tor_1^R(H_{n-1}(P),S) \to 0.$$ Now, assume $M$ is a projective $R$-module, $X$ is a free resolution of $R$ over $RG$ such that each $X_n$ is a free $RG$-module of finite rank (for example one can take the bar resolution) and set $P := Hom_{RG}(X,M)$. Then, as $R$-modules: $$P_n \cong Hom_{RG}((RG)^k,M) \cong \oplus_1^k Hom_{RG}(RG,M) \cong M^k$$ is a projective, thus flat $R$-module and by hereditary the conditions of UCT are fullfilled. Futhermore, one has to note that $$Hom_{RG}(X,M) \otimes_R S \cong Hom_{RG}(X,M \otimes_R S)$$ holds, since $X_n$ is a free $RG$-module of finite rank and my assumption about the $G$-operation on the tensor product (!).

share|improve this answer
The ideas sounds fine, but since $R$ isn't necessarily a PID, you need to be careful with universal coefficients. I suspect you want to $Tor_i(H^i(G,M),S)=0$ to get the spectral sequence to collapse... –  Donu Arapura Sep 6 '11 at 2:44
Thanks for your hint. In fact I was a little bit careless about the assumptions needed to apply UCT. –  Ralph Sep 6 '11 at 7:51
I'm sorry, to dig this up, but you show here that $H^0(G;M)\otimes_R S \rightarrow H^0(G;M\tensor_R S)$ is always injective. Is that true without the higher vanishing of the $Tor$-groups mentioned in Donu Arapura's comment? –  tkr Oct 29 '12 at 22:58
As explained in the "Edit"-part, I assume in addition to the assumptions of the OP that $R$ is hereditary. –  Ralph Oct 30 '12 at 9:17
add comment

Choose generators $g_1,\dots,g_n$ of $G$ and let $f\colon M\to M^n$, $f(m)=(g_im-m)_{i=1,\dots,n}$. This gives an exact sequence $$0\to M^G\to M\to M^n\to\mathrm{coker}(f)\to0$$ which can be interpreted (EDITED) as part of a flat resolution of $\mathrm{coker}(f)$. Since $\ker(f\otimes1_S)=(M\otimes S)^G$, the quotient $(M\otimes S)^G/\mathrm{im}(M^G\otimes S)$ may be identified with $\mathrm{Tor}_1(\mathrm{coker}(f),S)$, so they both vanish for all $S$ iff $\mathrm{coker}(f)$ is flat. If $\mathrm{coker}(f)$ is flat, we also get that $M^G\otimes S\to(M\otimes S)^G$ is injective, so we get the equivalence $M^G\otimes S\to(M\otimes S)^G$ bijective iff $\mathrm{coker}(f)$ is flat.

share|improve this answer
Needs attention to hypotheses, again. $M$ is free over $R$, but $M^G$ need not be flat, I think, unless (say) $R$ is a PID. –  Graham Denham Sep 6 '11 at 20:21
$\mathrm{coker}(f)$ is just $H^1(G;M)$ in disguise. This follows easily from [Hilton, Stammbach: A Course in Homological Algebra] VI.4 equation (4.5). I also agree with Graham's concern about the flatness of $M^G$. Nevertheless, nice explication (+1). –  Ralph Sep 6 '11 at 22:26
I guess I got confused by the fact that $M^G$ is flat in the case we are trying to characterize. My problem with using $f_1\colon M\to\mathrm{Hom}_G(IG,M)$ (so that $\mathrm{coker}(f_1)=H^1(G,M)$) is that I do not see why the kernel of $M\otimes S\to\mathrm{Hom}_G(IG,M)\otimes S$ should still be $(M\otimes S)^G$. –  user2035 Sep 7 '11 at 6:30
add comment

Your Answer


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.