[proofplan] For each member of the finite basic cover, choose finitely many localized generators and replace them by their numerators in $M$. These finitely many numerators generate an $R$-submodule $N \subset M$ whose localization at every $f_i$ is all of $M_{f_i}$. We then prove that every class in $M/N$ is killed locally by powers of the $f_i$, and the unit-ideal hypothesis upgrades these local annihilations to global vanishing. Hence $M=N$, so $M$ is finitely generated. [/proofplan] [step:Choose global numerators for the localized generators] For each $i \in \{1, \ldots, n\}$, the $R_{f_i}$-module $M_{f_i}$ is finitely generated. Choose an integer $r_i \ge 0$ and elements $m_{i1}, \ldots, m_{ir_i} \in M$ such that the localized elements $m_{i1}/1, \ldots, m_{ir_i}/1 \in M_{f_i}$ generate $M_{f_i}$ as an $R_{f_i}$-module. Indeed, starting from any finite generating family in $M_{f_i}$, each generator has the form $a^{-1}m$ with $a$ a power of $f_i$ and $m \in M$. Such a generator is an $R_{f_i}$-multiple of $m/1$, so replacing every chosen generator by its numerator still gives a finite generating family. Define $N \subset M$ to be the $R$-submodule generated by all the finitely many elements $m_{ij}$: \begin{align*} N := \sum_{i=1}^{n} \sum_{j=1}^{r_i} R m_{ij}. \end{align*} Since the index set $\{(i,j) : 1 \le i \le n,\ 1 \le j \le r_i\}$ is finite, $N$ is a finitely generated $R$-module. [guided] For each fixed $i$, finite generation of $M_{f_i}$ means that there are finitely many elements of the localized module from which every element of $M_{f_i}$ can be obtained by $R_{f_i}$-linear combinations. A typical element of $M_{f_i}$ is represented by data $m \in M$ and an integer $a \ge 0$ as \begin{align*} m/f_i^a. \end{align*} If this element appears among a chosen list of localized generators, then it is already generated by the simpler localized element $m/1$, because \begin{align*} m/f_i^a = (1/f_i^a)(m/1) \end{align*} inside the $R_{f_i}$-module $M_{f_i}$. Thus, after replacing every localized generator by its numerator, we may assume that the generators of $M_{f_i}$ have the form $m_{i1}/1, \ldots, m_{ir_i}/1$ with $m_{ij} \in M$. We then collect all these numerators, for all $i$, into one submodule: \begin{align*} N := \sum_{i=1}^{n} \sum_{j=1}^{r_i} R m_{ij}. \end{align*} The finiteness of the cover is used here: there are only finitely many indices $i$, and for each $i$ there are only finitely many chosen generators. Therefore the displayed sum is generated by finitely many elements of $M$, so $N$ is finitely generated as an $R$-module. [/guided] [/step] [step:Show the quotient vanishes after localizing at every $f_i$] Let $Q := M/N$ be the quotient $R$-module, and let \begin{align*} p: M \to Q \end{align*} be the quotient homomorphism. Fix $i \in \{1, \ldots, n\}$. Since $m_{i1}/1, \ldots, m_{ir_i}/1$ generate $M_{f_i}$ and each $m_{ij}$ lies in $N$, every element of $M_{f_i}$ is represented by an element of the localized submodule $N_{f_i}$. Hence the localized quotient $Q_{f_i}$ is the zero $R_{f_i}$-module. Equivalently, for every $q \in Q$ and every $i \in \{1, \ldots, n\}$, there exists an integer $k_i(q) \ge 0$ such that \begin{align*} f_i^{k_i(q)} q = 0 \end{align*} in $Q$. [/step] [step:Use the unit ideal condition to annihilate each quotient class] Let $q \in Q$ be arbitrary. From the previous step, for each $i \in \{1, \ldots, n\}$ choose an integer $k_i \ge 0$ such that \begin{align*} f_i^{k_i}q = 0. \end{align*} We claim that the ideal generated by $f_1^{k_1}, \ldots, f_n^{k_n}$ is all of $R$. If not, it is a proper ideal, so by the [maximal ideal existence](/theorems/8159) theorem it is contained in some maximal ideal $\mathfrak{m} \subset R$. Then $f_i^{k_i} \in \mathfrak{m}$ for every $i$, and since $\mathfrak{m}$ is prime, $f_i \in \mathfrak{m}$ for every $i$ with $k_i > 0$. If some $k_i=0$, then $f_i^{k_i}=1_R$, so the generated ideal is already $R$, contrary to the assumption. Thus all $f_i$ lie in $\mathfrak{m}$, so $(f_1, \ldots, f_n) \subset \mathfrak{m}$, contradicting $(f_1, \ldots, f_n)=R$. Therefore there exist elements $a_1, \ldots, a_n \in R$ such that \begin{align*} 1_R = \sum_{i=1}^{n} a_i f_i^{k_i}. \end{align*} Multiplying this identity by $q$ in the $R$-module $Q$ gives \begin{align*} q = 1_R q = \sum_{i=1}^{n} a_i f_i^{k_i} q = 0. \end{align*} Since $q \in Q$ was arbitrary, $Q=0$. [/step] [step:Conclude that the chosen finitely generated submodule is all of $M$] Since $Q=M/N=0$, the quotient map $p:M \to Q$ has kernel all of $M$. But $\ker p=N$, by definition of the [quotient module](/page/Quotient%20Module). Hence $M=N$. The submodule $N$ was generated by the finite set \begin{align*} \{m_{ij} : 1 \le i \le n,\ 1 \le j \le r_i\}. \end{align*} Therefore $M$ is finitely generated as an $R$-module. [/step]