finitely generated module over $R[U^{-1}]$ is the localization of a finitely generated module over $R$ This is Exercise 2.10 of Eisenbud's $\textit{Commutative Algebra}$:
Show that every finitely generated module over $R[U^{-1}]$ is the localization of a finitely generated module over $R$. 
I'm not sure how to start because I don't see what a natural choice is for the module that it is a localization of. 
 A: Let $M = \text{span}_{R[U^{-1}]} (x_1, \cdots, x_k)$ be a finitely generated $R[U^{-1}]$-module. From the injection $R \hookrightarrow R[U^{-1}]$, we have an $R$-module structure on $M$. Now, let $N \le M$ denote the $R$-submodule $N = \text{span}_{R} (x_1, \cdots, x_k)$. We claim that $M \cong R[U^{-1}] \otimes_{R} N$ as $R[U^{-1}]$-modules.
Proof: Consider the $R[U^{-1}]$-module map $R[U^{-1}] \otimes_{R} N \to M$ given by $$\sum (r_i/u_i) \otimes x_i \to \sum (r_i / u_i) x_i.$$ Surjectivity is clear. Suppose $\sum (r_i/u_i) x_i = 0$ in $M$ and let $u = u_1 u_2 \cdots u_k \in U$. 
Multiplying the relation through by $u$ implies $\sum r_i (u_1 \cdots \hat{u_i} \cdots u_k) x_i = 0$ in $N$, hence $$r_1 (u_2 \cdots u_k) x_1 = - \sum_{i\ge 2} r_i (u_1 \cdots \hat{u_i} \cdots u_k) _i.$$ It follows $$u \cdot ((r_1/u_1) \otimes x_1) = 1 \otimes (r_1 (u_2 u_3 \cdots u_k) x_1) = - \sum_{i\ge 2} 1 \otimes (r_i (u_1 \cdots \hat{u_i} \cdots u_k) x_i) = u \cdot \left( - \sum_{i\ge 2} (r_i/u_i) \otimes x_i \right).$$ Since $u$ is a unit in $R[U^{-1}]$, we conclude $\sum (r_i/u_i) \otimes x_i = 0$ in $R[U^{-1}] \otimes N$, i.e. the map is injective. $\blacksquare$
