Let $A$ be a ring and $M$ be an $A$-module. Suppose that the collection $\{ g_{i} \}_{i=1}^{m}$ generate the unit ideal of $A$. I want to show that if for each $g_{i}$, the localization $M_{g_{i}}$ is finitely generated as an $A_{g_{i}}$-module, then $M$ is finitely generated as an $A$-module.
I wanted to see if someone could check if what I have seems ok. The proof here is from EGA II (6.1.4.1) (which can be found here) and I just wanted to make sure I haven't made any mistakes in translation. In fact, one part I am particularly uncomfortable with is the summation I have labeled $(*)$. Grothendieck has this sum over $i$, and I am not sure why. It seems like it should be over $j$, since we are in a particular localization and generated by the generators indexed by $j$, right? Or have I missed something?
Suppose we have a system of generators for each $M_{g_{i}}$ given by a collection $$ \left\lbrace \frac{m_{ij}}{g_{i}^{n_{ij}}} \right\rbrace_{i=1}^{m} $$ But since the index sets of both $i$ and $j$ are finite, we can choose a sufficiently large integer $N$ so that we we have generating sets $$ \left\lbrace \frac{m_{ij}}{g_{i}^{N} } \right\rbrace_{ j \in J_{i} } $$ for each $M_{g_{i}}$. We claim that the collection of all the $\{ m_{ij} \}$, which is a finite collection, generates $M$ as an $A$-module. Let $M'$ be the $A$-submodule of $M$ generated by this collection. Now let $m \in M$ be any element. We will show that $m \in M'$. Denote the localization maps $$ \phi_{i}: M \longrightarrow M_{g_{i}}. $$ Then since we have generating sets for any $M_{g_{i}}$, we have that $$ \phi_{i}(m) = \frac{m}{1} = \frac{\left( \sum_{j} a_{ij}m_{ij} \right)}{g_{i}^{p}} \qquad (*) $$ Then for a sufficiently large power $r$, we have $g_{i}^{r}m \in M'$. But note that the collection $\{ g_{1}^{r}, g_{2}^{r}, \ldots , g_{m}^{r} \}$ also generates $A$. This has the geometric interpretation that if the $D(g_{i})$ cover $\text{Spec }A$, then so do the $D(g_{i}^{r})$ as distinguished open affines. So then we can find elements $b_{i} \in A$ so that $$ 1 = \sum_{i=1}^{m} b_{i}g_{i}^{r}. $$ But then we have $$ m = 1 \cdot m= \left( \sum_{i=1}^{m} b_{i}g_{i}^{r} m \right) \in M'. $$ So we have shown that $M \subseteq M'$ which proves the result.
Looks good to me! As far as I can tell you are correct and EGA has a typo in equation $(*)$.
Here's a slightly different way to do the end of the proof which might be enlightening. Consider the quotient module $K=M/M'$. Since localization is exact and $M'$ contains generators of $M$ after localizing at any $g_i$, $K_{g_i}=0$ for each $i$. We wish to show $K=0$, and to do so it suffices to show that $K_\mathfrak{m}=0$ for any maximal ideal $\mathfrak{m}$ (since if $x\in K$ is nonzero, its image in $K_{\mathfrak{m}}$ is nonzero for any $\mathfrak{m}$ which contains the annihilator of $x$). But $\mathfrak{m}$ is a proper ideal, and therefore there must be some $g_i$ such that $g_i\not\in\mathfrak{m}$. Since $g_i$ is among the elements of $A$ that are inverted in the localization $K_{\mathfrak{m}}$ and $K_{g_i}=0$, we have $K_{\mathfrak{m}}=0$, as desired.