Equivalence for rings with localization property

Question

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.

Answer 1

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.