$M$ noetherian module over noetherian ring $A$, $I\subset A$ ideal. Then $\exists 0\neq m\in M, Im=0$ iff $\forall x\in I,\exists 0\neq m\in M,xm=0$
Forward implication is obvious. I am kind of having trouble to prove this directly without using associated primes here. It amounts to prove the converse statement by inducting on the number of generators of $I$. It suffices to prove the basic case for 2 generated $I$.
$\textbf{Q:}$ Here is my thought process. Suppose $I=(f_1,f_2)$. By assumption, there is $m_1,m_2\in M-0$ s.t. $f_im_i=0$. $f_1m_1=0$ is good. It suffices to consider $J=(m_2:_Am_1)$ quotient ideal. WLOG, $m_2\not\in (m_1)$. Thus $J\neq A$. However, I could not see $J\neq 0$. Furthermore, there is no guarantee that $Jm_2\neq 0$. Hint or suggestion on how to proceed?
Ref. Serre Local Algebra pg 8, Prop 7.