$R$ is a noetherian ring. $M$ is a multiplicative system. If $x$ is in the intersection of all primary components of $\{0\}$ disjoint from $M$, then $\exists m\in M$, $mx=0$
My proof: If any primary component of $\{0\}$ is disjoint from $M$, then the statement is true.
If $\exists $ primary components of $\{0\}$ that are disjoint from $M$, then assume $\{0\}=(\cap q_i)\cap(\cap q_i')$ is a minimal decomposition, where $q_i\cap M=\emptyset$ and $q_j'\cap M\ne \emptyset$. If $x$ is in the intersection of all primary components of $\{0\}$ disjoint from $M$, then $x\in \cap q_i$. If $m_j\in q_j'\cap M$ for each $j$, then $m=\Pi m_j\in M\cap (\cap q_j')\Longrightarrow xm\in (\cap q_i)\cap(\cap q_i')=\{0\}$. $\square$
Is my proof correct? If it is, then seems $x$ in finitely many primary components of $\{0\}$ disjoint from $M$ guarantees $x$ annihilates some $m\in M$, much weaker than the condition that $x$ is in every primary components of $\{0\}$ disjoint from $M$. Is there some explanation "in essence" for why the condition can be weakened to that extent?