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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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    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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(I lost my login info) A counter example to the surjectivity is the following. Look at $A= \mathbf{Z}_p$ and $G = \mathbf{Z}$ the cyclic subgroup of $\mathrm{GL}_2(\mathbf{Z}_p)$ generated by the matrix $g = \begin{pmatrix}1 & p \\ 0 & 1\end{pmatrix}$. This acts on the rank two free module $M = \mathbf{Z}_p^2= Ae_1 \oplus Ae_2$ and it is clear that $M^G = e_1$. Yet, $M/pM = \mathbf{F}_p^2$ and the group $G$ is the cyclic subgroup of $\mathrm{GL}_2(\mathbf{F}_p)$ generated by $\overline{g} = 1$. Thus, $(M/pM)^G = M/pM = \mathbf{F}_p^2$.