Modules for twisted polynomial ring and Galois descent Let $\mathbb{F}_q$ be a finite field with algebraic closure $\overline{\mathbb{F}}_q$ and consider the twisted polynomial ring $\overline{\mathbb{F}}_q\{ \tau \}$, where multiplication satisfies the relation $\tau x = x^q \tau$ for $x \in \overline{\mathbb{F}}_q$. This ring looks an awful lot like the twisted group ring $\overline{\mathbb{F}}_q[\text{Gal}(\overline{\mathbb{F}}_q/\mathbb{F}_q)]$. By Galois descent, modules over the latter ring are almost the same thing as $\mathbb{F}_q$-vector spaces, except there is a continuity condition: any element of the module must be fixed by a sufficiently large power of the Frobenius.
So here is my question: given an $\overline{\mathbb{F}}_q\{ \tau \}$-module, is there any reason a priori that the action of $\tau$ should extend to a (continuous) action of $\text{Gal}(\overline{\mathbb{F}}_q/\mathbb{F}_q)$? If not, then the category of $\overline{\mathbb{F}}_q\{ \tau \}$-modules contains $\mathbb{F}_q$-vector spaces as a proper subcategory. So what do $\overline{\mathbb{F}}_q\{ \tau \}$-modules correspond to?
Edit: Upon reflection, it is clear that $\tau$ need not act invertibly on an arbitrary $\overline{\mathbb{F}}_q\{ \tau \}$-module: for instance, consider $\overline{\mathbb{F}}_q[t]$ with $\tau \cdot f = f^q$. So replace all occurrences of $\overline{\mathbb{F}}_q\{ \tau \}$ in the previous paragraph with the twisted Laurent polynomials $\overline{\mathbb{F}}_q\{ \tau,\tau^{-1} \}$.
 A: The answer is affirmative for $\overline{\mathbb{F}}_q\{ \tau, \tau^{-1} \}$-modules which are finite-dimensional $\overline{\mathbb{F}}_q$-vector spaces, meaning that in this case the action of $\tau$ automatically extends to an action of $\widehat{\mathbb{Z}}$, and hence descent theory applies.
To see why, let $V$ be an $\overline{\mathbb{F}}_q\{ \tau, \tau^{-1} \}$-module which is finite-dimensional over $\overline{\mathbb{F}}_q$ and choose a basis $V \cong \overline{\mathbb{F}}_q^n$. Then we can write $\tau$ as a matrix $A \in \text{GL}_n(\overline{\mathbb{F}}_q)$ in the sense that $\tau \cdot v = A \ \text{Fr}(v)$ for all $v \in V$ (here $\text{Fr}$ raises coordinates to the $q^{\text{th}}$ power). By a theorem of Lang, the map $\text{GL}_n(\overline{\mathbb{F}}_q) \to \text{GL}_n(\overline{\mathbb{F}}_q)$ given by $B \mapsto B^{-1}\text{Fr}(B)$ is surjective, and in particular we can write $A = B^{-1}\text{Fr}(B)$ for some $B \in \text{GL}_n(\overline{\mathbb{F}}_q)$. Now fix $v \in V$ and compute $$\tau^n \cdot v = A \ \text{Fr}(A) \ \text{Fr}^2(A) \cdots \text{Fr}^{n-1}(A) \ \text{Fr}^n(v) \\ = B^{-1} \ \text{Fr}(B) \ \text{Fr}(B^{-1}) \ \text{Fr}^2(B) \cdots \text{Fr}^{n-1}(B^{-1}) \ \text{Fr}^n(B) \ \text{Fr}^n(v) \\ = B^{-1} \ \text{Fr}^n(B) \ \text{Fr}^n(v),$$ noting that for sufficiently large $n$ the right side is equal to $v$.
I would still like to see a proof or counterexample when $V$ is infinite-dimensional.
