# A question on Modules

Let $M$ be a module over a ring $A$ and let $f_{1},...,f_{n}$ be elements of $A$ generating the unit ideal. Show that $M=0$ iff $M_{f_{i}}=0$ for $i=1,...,n$.

I feel that this is closely related to saying that $X=Spec A$ is quasi-compact. That is, we can consider a basic open covering ${X_{f_i}}$ of $X$. Then, we show the $f_{i}$ with $i\in I$ generate the unit ideal $(1)$ by writing $1=\sum_{i\in J}a_{i}f_{i}$ with $a_{i}\in A$ where $J$ is some finite subset of $I$. Then we have a finite subcovering.

Any help would be appreciated! Thanks!

It is not related to the quasi-compactness. It is related to the sheaf of modules $\tilde{M}$ on $\mathrm{Spec}(A)$ defined by $\tilde{M}(D(f))=M_f$, namely to the property of being separated. But of course you don't have to know this background, you can just prove it:
If $m \in M$, we have $m/1=0$ in $M_{f_i}=0$, hence $f_i^{r_i} m = 0$ for some $r_i \in \mathbb{N}$, in other words $f_i^{r_i} \in \mathrm{Ann}(m)$. Since the $f_i^{r_i}$ also generate the unit ideal (SE/243679), we see $1 \in \mathrm{Ann}(m)$, i.e. $m=0$.
• Martin, isn't the proof that $\widetilde{M}(D(f))=M_f$ is a sheaf on the basis $\{D(f)\}$, require proving this? Namely, isn't the OPs question precisely the proof that $\widetilde{M}$ is a separated $\{D(f)\}$-presheaf? Seems a bit circular. – Alex Youcis Oct 2 '13 at 23:27
• What you wrote as your proof isn't circular, but the suggestion in your first paragraph that we could somehow deduce this from the fact that $\widetilde{M}$ is a sheaf is circular. We use this fact to show that $\widetilde{M}$ is separated. – Alex Youcis Oct 3 '13 at 21:20