Lemma 1.84
Let $R$ be a PID. Suppose that $M\simeq R/(a)$ where $a$ is a nonzero element of $R$. Moreover, suppose $p$ is an irreducible element of $R$. Then
(a) If $p\nmid a \implies M_p=\{0\}$ and $pM=M\simeq R/(a)$.
(b) If $p\mid a \implies M_p\simeq R/(p)$ and $pM\simeq R/(a/p)$
The proof for part (a) is:
Proof. Suppose $p\nmid a$. Then since $p$ is irreducible, $p$ and $a$ are relatively prime. Also $aM=\{0\}$. Hence, $paM=\{0\}$. Thus $$ aM=M_p \text{ and } pM=M_a $$ and $$ M=M_a\oplus M_p$$ as proven before if $a$ and $p$ are relatively prime.
So $$M_p=aM=\{0\}$$ Also $M=M_a\oplus M_p=M_a\oplus\{0\}=M_a$ and so $$pM=M_a=M\simeq R/(a).$$
The definition for $aM$ and $M_a$ is: $$aM=\{ax:x\in M\} \text{ and } M_a=\{x\in M:ax=0\}$$
My only question for the above proof is how come $aM=\{0\}$? I was playing around with the assumptions, and I've failed to conclude why we can say that.
Any help is much appreciated.