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Let $R$ be a commutative ring, $G$ a finite group with an action over $R$. Let $G_1, G_2 \subset G$ be two subgroups. Then the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$ is surjective, where $R^{H}$ means the $H$-invariants of $R$.

(Add: I omit a condition that $G$ is a subgroup of $Aut(R)$.)

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  • $\begingroup$ I think this is correct at least in many cases although I don't know how to prove it in general case or give a counter-example. $\endgroup$
    – Zheming Xu
    Commented Aug 13 at 6:50

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I think that statement is false.

Let $R=\mathbb{Z}[\varepsilon]/\varepsilon^2$ and let $G_1=\mathbb{Z}/2\times 1$ and $G_2=1\times \mathbb{Z}/2$ and $G=\mathbb{Z}/2\times\mathbb{Z}/2$. Let each $G_i$ act on $R$ via $\varepsilon\mapsto -\varepsilon$. Then the map above is given by $\mathbb{Z}\otimes \mathbb{Z}\to R$, which is not surjective.

With the additional assumption that $G$ is a subgroup of $\operatorname{Aut}(R)$, we can modify the counterexample as follows:

Let $R=\mathbb{Z}[\varepsilon_1,\varepsilon_2]/(\varepsilon_1^2,\varepsilon_2^2)$ and let $G_1$ send both $\varepsilon_i$'s to their negatives and $G_2$ send only $\varepsilon_1$ to its negative. Then the map is given by $\mathbb{Z}[\varepsilon_1\varepsilon_2]\otimes \mathbb{Z}[\mathbb{\varepsilon_2}]\to R$, which is not surjective.

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  • $\begingroup$ Thank you. Maybe I drop a condition that $G$ is a subgroup of $Aut(R)$. $\endgroup$
    – Zheming Xu
    Commented Aug 13 at 7:04
  • $\begingroup$ I wonder if it is necessary to consider "G is a subgroup of $Aut(R)$? I mean that according to the definition of action, it is implicity contained in the definition that $x\mapsto g.x$ is a Ring automorphism. On the other hand I think the first version of your answer satisfies the Ring automorphism property: the conjugation is a ring automorphism on the trancated polynomial ring $\frac{Z[x]}{x^2}$. Am I mistaken? $\endgroup$
    – Ali Taghavi
    Commented Aug 13 at 13:48
  • $\begingroup$ @AliTaghavi the question is whether the map $G \to \operatorname{Aut}(R)$ is injective, which is not true in the first example. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 14 at 5:03

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