Let $M$ be a finitely generated $R$-module. Prove that the following conditions on $M$ are equivalent:
(a) $M$ is locally free over $R$ (i.e. $M_m$ is free over $R_{m}$ for all maximal ideals $m\subset R$).
(b) For every maximal ideal $m\subset R$, there is an $f \not \in m$ such that $M_f$ is free over $R_f$.
(c) There exist $f_1,\dots, f_n \in R$ such that $(f_1, \dots, f_n) =1 $ and $M_{f_i}$ is free over $R_{f_i}$ for all $i$.
I was able to prove $(a) \Rightarrow (b)$. For $(b) \Rightarrow (c)$, this is what I have: Let $S = \big\{f \in R:M_f\text{ is free over }R_f\big\}$. Let $(S)$ be the ideal generated by the elements of $S$. Suppose that $(S) \neq (1)$. It follows that $(S)$ is contained in some maximal ideal $m$ of $R$. By our assumption, there exists $f \not \in m$ such that $M_f$ is free over $R_f$. By the definition of $S$, this implies that $f \in S$. Thus, $f \in (S)\subset m$, a contradiction. We therefore conclude that $(S) = (1)$. How does $M$ being finitely generated as an $R$-module imply that there is some finite generating set $\big\{f_1,\dots, f_k\big\} \subset S$ such that $(f_1,\dots, f_k) =(S) = (1)$? I imagine I have to use some sort of an argument involving z-families.
For $(c) \Rightarrow (a)$, this is what I have: Let $m \subset R$ be a maximal ideal. Note that since $m$ is a proper ideal, and $(f_1,\dots, f_n) = R$, there exist some $f_k $ such that $f_k \not \in m$ and $M_{f_k}$ is free over $R_{f_k}$. Since free implies locally free and $m_{f_k}$ is maximal in $R_{f_k}$, this implies that $(M_{f_k})_{m_{f_k}}$ is free over $(R_{f_{k}})_{m_{f_k}}$. How does $f_k \in R\setminus m$ imply that $(M_{f_k})_{m_{f_k}} = M_m$ and $(R_{f_{k}})_{m_{f_k}} =R_m$? This would prove that $M_m$ is free over $R_m$.
Best Answer
For $(b) \Rightarrow (c)$, since $(S) = (1)$, we have $1 \in (S)$, so there is some finite expression $r_1f_1 + \ldots + r_kf_k = 1$, where the $r_i \in R$ and $f_i \in S$. Clearly then $1 \in (f_1,\ldots, f_k)$, so $(f_1,\ldots, f_k) = R$.
For $(c) \Rightarrow (a)$, let $f_k \not \in m$ such that $M_{f_k}$ is free over $R_{f_k}$. You want to show that $(M_{f_k})_{m_{f_k}} = M_m$ and $(R_{f_k})_{m_{f_k}} = R_m$. Strictly speaking, these equalities are not true, since an element of $(M_{f_k})_{m_{f_k}}$ looks like $(\frac{x}{s})/(\frac{y}{t})$, while an element of $M_m$ just looks like $\frac{x}{t}$. What you really want are natural isomorphisms. This is kind of a fuzzy notion, but in practice, it usually means that two things are "basically the same thing".
So, define $S = \{1,f_k,f_k^2,\ldots\}$ and $T = \{\frac{x}{s} : x \not \in m, s \in S\}$. Then $M_{f_k} = S^{-1}M$ and $(M_{f_k})_{m_{f_k}} = T^{-1}(S^{-1}M)$.We will explicitly define isomorphisms between $T^{-1}(S^{-1}M)$ and $M_m$. I'll sketch the proof and leave the details to you, but let me know in the comments if you want me to fill anything in explicitly.
Define $\varphi: M_m \to T^{-1}(S^{-1}M)$ by $$ \varphi(\frac{x}{t}) = \frac{x}{1}/\frac{t}{1} $$ (you need to check that $\frac{t}{1} \in T$ and also that this map is well-defined). Define $\psi:T^{-1}(S^{-1}M) \to M_m$ by $$ \psi(\frac{x}{s}/\frac{y}{t}) = \frac{xt}{sy}. $$ Since $s \in S$, clearly $s \in R\setminus m$. Since $\frac{y}{t} \in T$, we have $y \in R\setminus m$ by definition of $T$, so indeed $sy \in R\setminus m$, and therefore $\frac{xt}{sy}$ is an element of $M_m$. Again, you need to check that the map is well-defined. Once you have verified that these maps are well-defined, you need to show that they are mutual inverses. So, there are now four exercises for you:
(4) is easy. The other three are more involved, but they all follow from playing with the definitions of localisation. I found the following trick very useful for solving them: if you want to show that two things are equal, subtract one from the other and show that the result is $0$.
Finally, $(R_{f_k})_{m_{f_k}} = R_m$ is just the special case $M=R$.
Edit
Proving that $(M_{f_k})_{m_{f_k}} = M_m$ is fiddly, but the intuition is quite simple. You obtain $M_{f_k}$ by forcing $f_k$ to be a unit (more precisely, and automorphism). You then obtain $(M_{f_k})_{m_{f_k}}$ by further forcing everything outside $m_{f_k}$ to be unit. But since the denominators in $R_{f_k}$ are already units in $M_{f_k}$, this is just the same as forcing the rest of $R\setminus m$ to be units. T