3
$\begingroup$

Let $B$ be a ring, and $G$ a finite group acting faithfully on $B$, where the order of $G$ is invertible on $B$. Let $B^G$ be the ring of $G$-invariants, and suppose $B^G\rightarrow B$ is flat. Let $M$ be a module over $B^G$, not necessarily free. There is a natural map $$M\rightarrow (M\otimes_{B^G}B)^G$$ sending $m\mapsto m\otimes 1$. Is this map an isomorphism?

1 Answers 1

3

If the order of $G$ is invertible in $B$, then the $G$-invariants of $M \otimes_{B^G} B$ can be identified with the image of the averaging operator

$$\frac{1}{|G|} \sum_{g \in G} g(-).$$

Hence every element of $(M \otimes_{B^G} B)^G$ is a sum of elements of the form

$$\frac{1}{|G|} \sum_{g \in G} m \otimes gb = m \otimes \left( \frac{1}{|G|} \sum_{g \in G} gb \right)$$

meaning that it lies in the image of the natural map $M \cong M \otimes_{B^G} B^G \to M \otimes_{B^G} B$. So this map is surjective. That it's injective follows from the hypothesis that $B^G \to B$ is flat. Hence it's an isomorphism.

  • 0
    Ah! Neat! Do you have a proof (or reference) of the fact that when the order of $G$ is invertible on $B$, then the $G$ invariants are described by the image of the averaging operator? – 2017-02-22
  • 1
    @Amy: it's a straightforward exercise; just observe that the averaging operator 1) fixes all $G$-invariants and 2) has image contained in $G$-invariants. – 2017-02-22
  • 0
    ah..so I guess in general if $|G|$ is invertible over a ring $R$, then the ring of $G$-invariants of any $R$-module $M$ is given as the image of that operator. Cool! – 2017-02-23