I feel like I need a hint for the following exercise:
Let $R$ be some commutative unitary ring. If $M$ is a $R$-module, let $M[f^{-1}]$ denote the localization of $M$ with respect to the set $\{ f^k: k \in \mathbb N_0 \}.$ Then the following are equivalent:
a) $R = (f_1, \dots, f_m)$,
b) $M=0 \Longleftrightarrow M[f_i^{-1}] = 0$ for all $i = 1, \dots , m$.
I think I solved "$\implies$" :
Assume there is some $0 \neq m \in M$. Define $\mathfrak a := \operatorname{Ann }(m) = \{r\in R: rm = 0\} \triangleleft R$. Then $\mathfrak a$ is contained in some maximal ideal $\mathfrak m$. By assumption a), the maximal ideals are precisely of the form $(f_1, \dots f_{j-1}, f_{j+1}, \dots, f_m), j= 1, \dots, m$ [edit: I realized that this is not true]. Then $R\setminus \mathfrak m = (f_j)$ for some $j$ and hence the localization of $M$ at $\mathfrak m$ is $M[f_j^{-1}]$. It follows that
$$\frac{m}{1} = 0 \in M_{\mathfrak m},$$
so $\exists v\in (R\setminus \mathfrak m): vm = 0 \implies v \in \operatorname{Ann}(m) \subseteq \mathfrak m$, contradiction. So $m=0$. Is this correct so far?
For the converse, I thought about considering $R$ as an $R$-module and work with that but got stuck. Any help appreciated.
In your argument,every maximal ideal is of the form you gave? It seems that this is not correct.Such that take two coprime elements in the ring of integers $\mathbb Z=(4,9)$.But $(4),(9)$ are not prime.
1.If $R=(f_1,...,f_m)$,$\forall m\in M$,there exists $N$ large enough such that $f_i^Nm=0$.Remark that $R=(f_1^N,...,f_m^N)$ since $R=(f_1,...,f_m)$.So $1=\sum r_if_i^N$.Hence $m=0$.
2.If $(f_1,...,f_m)$ is not equal to $R$.Then consider $M=R/(f_1,...,f_m)$.$M[f_i^{-1}]=0$ is a contradiction.