Partition of Unity in Algebraic Geometry (Eisenbud 2.19) Problem:
Suppose $R$ is a commutative ring and $f_1, f_2 \in R$ generate the unit ideal. Let $M$ be an $R$-module, and assume we are given elements $$m_i \in M[f_i^{-1}]$$ such that $$m_1 = m_2 \in M[f_1^{-1}f_2^{-1}].$$
Show that there exists $m \in M$ so that $m = m_i$ when you localize at $f_i$.
My attempt:
We write $m_i \in M[f_i^{-1}]$ as a fraction $$m_i = \frac{\tilde{m_i}}{f_i^{e_i}},$$ and the compatibility assumption becomes $$(f_1f_2)^N (f_2^{e_2} \tilde{m_1} - f_1^{e_1} \tilde{m_2}) = 0, N \gg 0$$ in $M$.
The assumption that $f_1$ and $f_2$ generate the unit ideal is expressed by $af_1+bf_2=1$. By replacing $f_i$ with as high a power of $f_i$ as necessary, we may assume without loss of generality that $e_1=e_2=N=1$; it is a standard trick that the Bezout equation still holds.
Let $m = a \tilde {m_1} + b \tilde {m_2}$. This definition is motivated by what it should be if $\tilde{m_i} = f_i m$.
So we wish to show that $$m = m_1 = \frac{\tilde{m_1}}{f_1} \in M[f_1^{-1}],$$ i.e.,  $$f_1^N(f_1m-\tilde{m_1}) = 0$$ for sufficiently large $N$. Substituting the definition of $m$ back in and simplifying using the Bezout equation gives $$f_1^N(-b f_2 \tilde{m_1} + bf_1 \tilde{m_2})=0.$$ I know from the above that $f_2 \tilde{m_1}-f_1 \tilde{m_2}$ is killed by $f_1f_2$, but I don't know that it's killed by a high power of $f_1$.
This appears to be the approach outlined in the solution. What am I missing?
 A: Since $af_1+bf_2=1$, we can find $c$ and $d$ such that $$cf_1^2+df_2^2=1$$
Instead of letting $m = a \tilde {m_1} + b \tilde {m_2}$, we set
$$m = c f_1 \tilde {m_1} + d f_2 \tilde {m_2}$$ 
We need to show 
$$f_1^M(f_1m-\tilde{m_1}) = 0$$ for sufficiently large M.
But this is 
$$f_1^M(-d f_2^2 \tilde{m_1} + d f_1 f_2 \tilde{m_2})=0$$
$$f_1^M f_2(-d f_2 \tilde{m_1} + d f_1 \tilde{m_2})=0$$
A: To motivate refining the definition of $m$ as WWK has done, we note that by part (a) of Eisenbud's exercise, $m$ must necessarily be unique. The original version $m = a \tilde{m_1} + b \tilde{m_2}$ may not be unique. This can be seen either on the geometry side of the picture or on the commutative algebra side:
From a sheaf-theoretic point of view, the $m_i$ define sections of a sheaf of modules on $\operatorname{Spec}(R[f_i^{-1}])$, and $\tilde{m_i}$ extends $f_i m_i$ to a global section on $\operatorname{Spec}(R) = \operatorname{Spec}(R[f_1^{-1}]) \cup \operatorname{Spec}(R[f_2^{-1}])$. This extension may include some junk on $\operatorname{Spec}(R) - \operatorname{Spec}(R[f_i^{-1}])$, so for a well-defined $m$, it makes sense to cancel the junk by multiplying by $f_i$ again.
For a more elementary way to understand this, we recall that $\tilde{m_i}$ is only well-defined up to the annihilator of $f_i^N$, for $N \gg 0$, acting on $M$. My choice to restrict the compatibility assumption on $m_1$ and $m_2$ to $$f_1f_2(f_2\tilde{m_1} - f_1 \tilde{m_2})=0$$ actually restricts what can be added to $\tilde{m_i}$ though; this equation isn't necessarily preserved by adding an annihilator for $f_i^N$ for $N \ge 2$. It is, however, preserved by adding an annihilator for $f_i$, so it is necessary (though, a priori, not sufficient) that our formula for $m$ be well-defined up to that. If we write $\operatorname{Ann}_i := \operatorname{Ann}_M(f_i)$ for this annihilator, then my original definition of $m$ yields the set $$a(\tilde{m_1} + \operatorname{Ann}_1)+b(\tilde{m_2}+\operatorname{Ann}_2),$$ which is only well-defined up to the additive subgroup $$a \operatorname{Ann}_1 + b \operatorname{Ann}_2,$$ and there's no reason to believe that this additive subgroup should be $\{0\}$. WWK's refinement yields the set $$cf_1(\tilde{m_1}+\operatorname{Ann}_1)+df_2(\tilde{m_2} + \operatorname{Ann}_2),$$ which is well-defined up to the additive subgroup $$cf_1 \operatorname{Ann}_1 + df_2 \operatorname{Ann}_2,$$ which is $\{0\}$. As I said, this doesn't a priori imply that $m$ is well-defined as a function of $m_1$ and $m_2$; it just gets us a step closer to believing that $m$ is probably well-defined. WWK's argument, of course, implies that $m$ is well-defined.
