# Tensor product with irreducible representation has no $G$-invariant submodules

Let $\rho: G \to GL(V)$ be a finite dimensional irreducible representation of a group $G$ over an algebraically closed field $\mathbb{F}$ of characteristic $0$, and let $R$ be a commutative ring with unity on which $G$ acts. Suppose also that $\mathbb{F} \subseteq R$. We can form the extension of scalars $$M=R \otimes_{\mathbb{F}} V,$$ making the tensor product into an $R$-module. Note that $G$ acts on $M$ by $g \cdot(r \otimes v)=g \cdot r \otimes g \cdot v$, $\forall g \in G$.

My question is:

Is it always true that if $N \subseteq M$ is a $G$-invariant submodule then $N=\{0\}$ or $N=M$?

My reasoning for this is that a basis for $M$ as an $R$-module is the set $\{1 \otimes v_i\}$, where the $v_i$ form a basis for $V$. Since $V$ is an irreducible representation, it contains no proper subspaces. I don't see how one could form any proper $G$-invariant submodules of $M$ without violating this fact.

If one were to consider $M$ as a (possibly infinite) vector space, then I could see how there might be proper invariant subspaces $N \subseteq M$. However, I am specifically interested in $M$ as an $R$-module.

If $I\subseteq R$ is any $G$-invariant ideal, then $I\otimes V$ is a $G$-invariant submodule of $R\otimes V$. So if $R$ has any nontrivial $G$-invariant ideals (e.g., if $G$ acts trivially on $R$ and $R$ is not a field), then $R\otimes V$ will have nontrivial $G$-invariant submodules.
I think it might be false. Take for instance $R=\mathbb{F}[t]$ with trivial $G$-action, and $N=tM$.