Let $C$ be a curve (so a $1$-dimensional, proper $k$-scheme).
A point $a \in C$ is associated - therefore $a \in Ass(\mathcal{O}_C)$ iff for the unique maximal ideal $m_a$ of the stalk $\mathcal{O}_{C,a}$ we have
$$m_a \in Ass_{\mathcal{O}_{C,a}}(\mathcal{O}_{C,a})$$
therefore $m_a = Ann_{\mathcal{O}_{C,a}}(r)$ for a $r \in \mathcal{O}_{C,a}$.
My question is why do we have following equivalence:
$$a \not \in Ass(\mathcal{O}_C) \Leftrightarrow {a} \text{ is a Cartier divisor }D_{a} \text{ with } Supp(D_a) = \{a\} $$
What could going wrong if $a \in Ass(\mathcal{O}_C)$?
Suppose we have a Cartier divisor $D_a$ with $Supp(D_a) = \{a\}$. Therefore the corresponding ideal sheaf $\mathcal{I}(-D_a)$ vanishes in every $c \in C$ with $c \neq a$.
Wlog the problem local, therefore $C =Spec(R)$ and $\mathcal{I}(-D_a)$ is given by localisations of principal ideal $dR$ for some $d \in R$
If we donote by $d_c$ the canonical image of $d$ in the localisation map $R \to R_{p_c}$ where the prime ideal $p_c$ corresponds to the point $c \in C$ then the condition $Supp(D_a) = \{a\}$ means
$$d_c \in m_c \Leftrightarrow c \neq a$$
What is going wrong here if $m_a = Ann_{\mathcal{O}_{C,a}}(r)$ for a $r \in \mathcal{O}_{C,a}$?