Let $G$ be a finite group acting on a commutative ring $R$ via ring maps. In doubt, one can assume $R$ to be noetherian or regular if one wants. Let $P$ be a $1$-dimensional free $R$-module with a $G$-action satisfying $g(r\cdot x) = g(r)\cdot g(x)$ for $r\in R$ and $x\in P$, i.e. what some people call a $R$-semilinear $G$-action. Examples include the following, where always $G= \mathbb{Z}/2 = \langle t\rangle$:

1) $R= \mathbb{Z}[i]$ with $t(i) = -i$. We act on $P \cong \mathbb{Z}[i]$ via interchanging $1$ and $i$.

2) $R = \mathbb{C}[[X,Y]]$, $t(X) = Y$, $t(Y) = X$. We act on some power series $f\in P \cong \mathbb{C}[[X,Y]]$ via $t(f) = f^{op} \cdot e^{X-Y}$, where $f^{op}$ denotes interchanging $X$ and $Y$.

3) $R = k[X^{\pm 1}]$ for any field $k$ and $t(X) = X^{-1}$. We act on $P \cong k[X^{\pm 1}]$ by $t(1) = X$ (or, more generally, $t(X^n) = X^{-n+1}$).

These representations are quite different, but they have all in common that a tensor power of them is "trivial", i.e. isomorphic to $R$ as an $R$-module with semilinear $G$-action. In the first example, $P^{\otimes_R 2}$ (with the diagonal action) is trivial in this sense as $t(i\otimes 1) = 1 \otimes i = i\otimes 1$ and thus $1\mapsto i$ is a $G$-equivariant isomorphism $R\to P$. The second example is even trivial itself as $1+e^{X-Y}$ is an invariant generator of $P$. In the third example, $X(\otimes 1)$ is an invariant element of $P\otimes_R P$.

This suggests the following question:

Question: Is there always some tensor power $P^{\otimes_R n}$ (with diagonal action) which is "trivial", i.e. isomorphic to $R$ as an $R$-module with semilinear $G$-action? Can $n$ be chosen to be $|G|$?

This is certainly true in the special case that $G$ acts trivially on $R$: Then $G$ acts $R$-linearly on $P$, i.e. each element $g\in G$ acts via multiplication by an element $r_g$. As $g^{|G|} = e_G$, all elements $r_g$ are $|G|$-th roots of unity in $R$. Thus, $P^{\otimes_R |G|}$ has trivial $G$-action.

Note also the following relationship to algebraic geometry: An $1$-dimensional free $R$-module $P$ with semilinear $G$-action is equivalent to a line bundle $L$ on the stack quotient $Spec R//G$. The line bundle $L$ is trivial exact if $P$ is "trivial" in the sense above.

Edit: Although the answer to the question seems to be rather straightforward, I decided not to delete the question since someone might still find it helpful.


1 Answer 1


My current office mate and I found out that this question has a rather simple answer, namely: Yes.

I use the same notation as in the question. Let $(x)$ be a basis of $P$. Consider the $|G|$-fold tensor product $P^{\otimes_R G}$. Clearly $x' = x \otimes x \otimes \cdots \otimes x$ is a basis of $P$. Consider furthermore an element $y \in P^{\otimes_R G}$ defined by $\otimes_{g\in G} g(x)$.


  1. $y$ is $G$-invariant.
  2. $y = r \cdot x'$ for a unit $r\in R$.

These are indeed enough: The element $y$ defines an $R$-linear map $$ \alpha: R \to P^{\otimes_R G},\, 1\mapsto y.$$ Since $y$ is $G$-invariant, the map $\alpha$ is $G$-equivariant. By the second claim, $\alpha$ is an isomorphism.

Proofs of Claims:

  1. Let $h\in G$. Let $r_h \in R$ be the element with $h(x) = r_h \cdot x$. Then $y = (\prod_{g\in G} r_g)\cdot x'$ and $h(y) = \otimes_{g\in G} (hg)(x) = (\prod_{g\in G} r_{hg})\cdot x'$. These are equal.

  2. It is enough to show that $r_h$ is a unit in $R$ for every $h\in H$. We have $$h^m(x) = h^{m-1}(r_h) \cdots h(r_h) \cdot r_h \cdot x$$

    If the order of $h$ is $m$, this equals $1$. Thus $r_h$ is a unit in $R$.


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