Interpretation of hint for Exercise 2.19b of Eisenbud I am doing exercise 2.19b of Eisenbud's Commutative Algebra with a View Towards Algebraic Geometry. Here we have an $R$-module $M$ and elements $\{f_i\}$ which generate the unit ideal. The exercise asks to prove

If $m_i \in M[f_i^{-1}]$ are elements such that $m_i$ and $m_j$ go to the same element of $M[f_i^{-1}f_j^{-1}]$, then there is an element $m \in M$ such that $m$ goes to $m_i$ in $M[f_i^{-1}]$ for each $i$.

Not seeing how to do this, I looked at the hint in the back of the book. The hint claims "$(f_i f_j)^N(m_i - m_j) = 0$ for large $N$". I don't know how this is even meaningful since $m_i$ and $m_j$ are in different $R$-modules. You could view it as an equality in $M[f_i^{-1}f_j^{-1}]$, but there we would simply have $m_i = m_j$ and the multiplier $(f_if_j)^N$ would be superfluous.
I'm not asking for a complete solution to the exercise. I just want to know, how should this equality be understood?
 A: You are right that the hint does not make much sense. The proof of the exercise can be found in textbooks on algebraic geometry because it is actually the sheaf condition for $\widetilde{M}$. The following proof is a little bit more simple than the usual proof because it makes reductions whenever possible and in the end the problem is quite trivial to solve.
First of all, we may restrict to finitely many $f_i$. This will be important in the proof, namely to chose exponents uniformly. Write $m_i = n_i/f_i^k$, where $k$ may be chosen uniformly. Since $n_i/f_i^k = n_j / f_j^k$ in $M[f_i^{-1} f_j^{-1}]$, there is some $m \geq 0$ (which may be chosen uniformly) such that $f_i^m f_j^m f_j^k n_i = f_i^m f_j^m f_i^k n_j$ holds in $M$. But we also have $m_i = (f_i^m n_i)/f_i^{m+k}$. After this replacement, we may assume that $f_j^k n_i = f_i^k n_j$ holds in $M$ (this is probably what Eisenbud's hint means). Since the $f_i^k$ also generate the unit ideal and localizing at $f_i^k$ is the same as localizing at $f_i$, we may in fact assume that $k=1$. Thus, all we have is simply a sequence of elements $n_i \in M$ satisfying $f_j n_i = f_i n_j$. Now it is easy to finish ...
