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. $m=0$.

  • $\begingroup$ 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. $\endgroup$ – Alex Youcis Oct 2 '13 at 23:27
  • $\begingroup$ My proof isn't circular. $\endgroup$ – Martin Brandenburg Oct 3 '13 at 9:05
  • $\begingroup$ 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. $\endgroup$ – Alex Youcis Oct 3 '13 at 21:20
  • $\begingroup$ He didn't say anything about deducing this result from sheaf properties; that seems to be something you read into it (perhaps due to the ambiguous inflection of "you don't have to"?). He was pointing out exactly the same thing you did when you said that this fact is "precisely" separatedness. $\endgroup$ – Slade Oct 5 '13 at 22:45

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