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Let $A$ be a local noetherian domain. Let $M$ be a torsion free $A$-module equipped with an $A$-linear action of a group $G$. Let $\mathfrak{m}$ be the maximal ideal in $A$.

Is the natural map $M^G \otimes_A A/\mathfrak{m} \rightarrow (M/\mathfrak{m})^G$ injective? If not, what is a counterexample?

I already know how to construct examples where the map is not surjective.

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    Hmm. Doesn't the long exact cohomology sequence shed some light to this? Start with the short exact sequence $$0\rightarrow mM\rightarrow M\rightarrow M/mM\rightarrow 0.$$ We get (at the $H^0$-level) that the sequence $$0\rightarrow (mM)^G\rightarrow M^G\rightarrow (M/mM)^G\rightarrow$$ is exact (the next group would be first cohomology). So a counterexample would need to have $(mM)^G$ larger than $m(M^G)$. Can that happen, when $A$ is a domain, and $M$ is torsion free?2012-10-29
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    Yes, of course that it another way to look at it. It is unclear whether this has gained you any insight or not, however.2012-10-29
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    Sorry about mangling the use of frak. Replacing the previous coomment with an edited version.2012-10-30
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    You're, of course, right in that using that exact sequence doesn't really give us a new angle. Anyway, if ${\frak m}$ is principal, say ${\frak m}=A\pi$, then it should be true. If $y\in {\frak m}M$ is a fixed point, then $y=\pi z$ for some $z\in M$. For all $g\in G$ we have $y=g\cdot y=\pi(g\cdot z)$, so $\pi(g\cdot z-z)=0$. As there was no torsion, we must have $g\cdot z=z$, so $y\in {\frak m} (M^G)$, and the claim follows. I don't know about the general case.2012-10-30

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