I'm trying to solve the following problem, but I don't know how to do it and it would be great if someone could give me some hints:
For elements $f_1, \ldots, f_n$ of a commutative ring $A$, show that the following statements are equivalent:
1.) $(f_1, \ldots , f_n) = A$
2.) For each nonzero module $M$ over $A$, at least one of the localized modules $M_{f_i}$ is nonzero.
I know that $M=0 \Leftrightarrow M_{p}=0$ for $p \subset A$ prime $\Leftrightarrow M_{m} = 0$ for $m \subset A$ maximal.
But for that, I would have to show that the $f_i$ is prime/maximal $\forall i$. And I guess that 1.) gives somehow a hint for this, but I'm not able to put this together. Could someone please help me?
The $f_i$s have no reason to be prime. Here's a hint:
$M$ is non-zero if and only if there's a $\mathfrak p\in \operatorname{Spec}A$ such that $M_{\mathfrak p}\ne \{0\}$.
Now, consider the Zariski topology on $\operatorname{Spec}A$: $$V\bigl((f_1,\dots,f_n)\bigr)=\smash{\bigcap_{i=1}^n} V(f_i),$$ and note $\;(f_1,\dots,f_n)=A \iff \displaystyle\bigcap_{i=1}^n V(f_i)=\varnothing$, so there's an $f_i$ such that $f_i\notin\mathfrak p$.
Can you take it from here?