# Equivalence for rings with localization property

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.

• "By assumption a), the maximal ideals are precisely of the form $(f_1, \dots f_{j-1}, f_{j+1}, \dots, f_m), j= 1, \dots, m$." This is false, and certainly does not follow from a). Have you considered any examples? How about $R = \mathbb{C}[x]$ and $(x, x+1) = \mathbb{C}[x]$? So you claim the only maximal ideals of $\mathbb{C}[x]$ are $(x)$ and $(x+1)$? May 16, 2019 at 15:05
• You are right, that is a really stupid mistake. May 16, 2019 at 15:06
• @bavor42 now I give the example...
– Jian
May 16, 2019 at 15:09

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.

• One more question: Why is there $N$ large enough s.t. $f_i^Nm = 0$? May 16, 2019 at 15:10
• @bavor42 definition of null elements in localization
– Jian
May 16, 2019 at 15:11
• And why is $R = (f_1^N, \dots, f_m^N)$? May 16, 2019 at 15:33
• But thanks so far, I appreciate your help :) May 16, 2019 at 15:33
• @bavor42 if not, $(f_1^N,...,f_m^N)$ contains in a maximal ideal.So this maximal ideal must contains $(f_1,...,f_m)$.This is impossible.
– Jian
May 17, 2019 at 0:09