I am trying to understand the proof from this paper, Lemma 2.4. Here's what I've got so far.
Lemma: Let $A$ be Noetherian, let $M$ be finitely generated, and let $Q$ be an irreducible $A$-submodule of $M$. Then $Q$ is primary.
Proof: Since $Q\neq M$, ${\rm Ass}(M/Q)\neq\emptyset$. Suppose that ${\rm Ass}(M/Q)$ contains two distinct prime ideals, $\mathfrak p_1={\rm Ann}(\bar x_1)$ and $\mathfrak p_2={\rm Ann}(\bar x_2)$. Clearly, $\bar x_1$ and $\bar x_2$ are nonzero. We claim that $A\bar x_1\cap A\bar x_2=\{0\}$. Indeed, if $a\bar x_1=b\bar x_2$ for nonzero $a,b\in R$, then $a\not\in{\rm Ann}(\bar x_1)$ and $b\not\in{\rm Ann}(\bar x_2)$...
I am not sure why the last line above holds. If $a\bar x_1=b\bar x_2$, why can we not have both $a\in{\rm Ann}(\bar x_1)$ and $b\in{\rm Ann}(\bar x_2)$? I fail to see why this cannot hold.
The paper doesn’t say that $a, b$ are nonzero, but $a\bar x_1=b\bar x_2$ is nonzero.