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Suppose that $G$ is a group acting on a commutative ring $R$, inducing an action on each $R$-module. For any $R$-module $M$, let $M^G$ denote the collection of elements of $M$ invariant under the $G$-action. Then, $M^G$ is an $R^G$-module.

Here is my question. Do we have:

$$ (M\otimes_RN)^G=M^G\otimes_{R^G}N^G $$

for $R$-modules $M$ and $N$?

At least does it hold if $M$ and $N$ are projective? Clearly, this holds when they are free modules.

Are there conditions on the group $G$ that could create this situation? For example, $G$ could be a finite group or an abelian group or a linear reductive group... etc etc

Or perhaps, there could be conditions on the ring $R$ for which we get the above equality?

Improve this question
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    $\begingroup$ When you say $G$ acts on $R$, do you just mean on the underlying additive group, or what? $\endgroup$
    – Todd Trimble
    Commented Sep 22, 2015 at 11:11
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    $\begingroup$ How does an action of $G$ on $R$ induce an action on an $R$-module? $\endgroup$
    – Jeremy Rickard
    Commented Sep 22, 2015 at 11:21
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    $\begingroup$ There is no meaning to "$G$ is a group acting on a commutative ring $R$, inducing an action on each $R$-module". I advise you to read about "$G$-linearizations" in a reference such as "Geometric Invariant Theory", Mumford, Fogarty, Kirwan. For an action of $G$ on $R$, there are some $R$-modules that will admit no compatible $G$-action, e.g., if the support of the module is not a $G$-invariant subset of $\text{Spec}(R)$. On the other hand, modules may admit many different compatible $G$-actions. $\endgroup$
    – Jason Starr
    Commented Sep 22, 2015 at 11:22
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    $\begingroup$ @Suresh. For an $R$-module homomorphism $\phi:N\to N$, there is an induced $R$-module homomorphism $1\otimes \phi:M\otimes_R N \to M\otimes_R N$. However, the ring automorphism induced by $g$ is not an $R$-module homomorphism of $R$. $\endgroup$
    – Jason Starr
    Commented Sep 22, 2015 at 11:25
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    $\begingroup$ Simple example: if $S$ is a ring, then $C_2$ acts on the ring $S\times S$ by swapping the two factors. $M=S\times\{0\}$ is an $S\times S$-module, but I think it's fairly clear that there's no action of $C_2$ on it that is "induced" in any reasonable way by the action on $S\times S$. $\endgroup$
    – Jeremy Rickard
    Commented Sep 22, 2015 at 12:05

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