Graded modules over $k[t,t^{-1}]$ 
If $R=k[t,t^{-1}]$ is a graded ring where $R_0=k$ is a field and $t\in R$ is a homogeneous element of positive degree which is transcendental over $k$, how can I prove that every graded $R$-module is free?

 A: A low-tech way:
Let $M$ be a graded $k[t^{\pm1}]$-module, so that in particular $M=\bigoplus_{n\in\mathbb Z}M_i$ as a vector space. For each $n\in\mathbb Z$ the map $M_n\to M_{n+1}$ given by multiplication by $t$ is a linear bijection, with inverse given by multiplication by $t^{-1}$, of course. It follows that for all $n\in\mathbb Z$ we have $M_n=t^nM_0$. Moreover,  it is easy to see that the map $k[t^{\pm1}]\otimes_k M_0\to M$ obtained by restricting the multiplication map $k[t^{\pm1}]\otimes_k M\to M$ to the subspace $k[t^{\pm1}]\otimes_k M_0$ is in fact an isomorphism of $k[t^{\pm1}]$-modules, if we see its domain as a $k[t^{\pm1}]$-module in the obvious way. But this obvious module is free: any basis of $M_0$ gives a basis.
N.B.: This is, I guess, what the high-tech proof amounts to in this concrete situation...
A: One way:
A graded $H=k[t^{\pm1}]$-module $M$ is the same thing as Hopf $H$-module (so it is an $H$-module, and $H$-comodule, and both structures are compatible). The Fundamental Theorem of Hopf modules tells you, then, that such a beast is free as an $H$-module.
The best reference for this is, in my opinion, Sweedler's book [Sweedler, Moss E. Hopf algebras. Mathematics Lecture Note Series W. A. Benjamin, Inc., New York 1969 vii+336 pp. MR0252485]
