Let $R$ be a ring, let $p\subseteq R$ be an ideal and let $M$ be an $R$-module. Let $m\in M$. Let us define $$\begin{align} &\operatorname{ann}(m)=\{r\in R:rm=0\}\\ &\operatorname{Ass}(M)=\{p\in\operatorname{spec}(R):\exists 0\ne m\in M, \operatorname{ann}(m)=p\} \end{align}$$ Show that $p\in\operatorname{Ass}(M)\iff \exists \phi:R/p\to M$ and $\phi$ is injective.
I have shown the direction $\Rightarrow$.
Suppose $\exists \phi: R/p\to M$ where $\phi$ is injective. We want to show that $p\in\operatorname{spec}(R)$ and $\exists 0\ne m\in M$ s.t. $\operatorname{ann}(m)=p$.
Let's start with showing that $p\in \operatorname{spec}(R)$; We want to show that $R/p$ is integral domain. Let $\bar{x},\bar y\in R/p $ s.t. $\bar x\bar y=p$. $$xy+p=(x+p)(y+p)=p \\\Rightarrow 0_M=\phi(p)=\phi(xy+p)=x\phi(y+p) $$ If it's mean that necessarily $x=0_M\vee \phi(y+p)=0_M$ then we're done. But I don't know if this statement is true: $[rm=0_M\Rightarrow r=0_R\vee m=0_M]$.
I assume that $R$ has a unit.
Let $p$ be an element of $Ass(M)$, there exists $m\in M$ such that $Ann(M)=p$. Consider $\phi:R\rightarrow M$ defined by $\phi(r)=rm$, $Ker(\phi)=p$.
On the other hand suppose that $p$ is an element of $Spec(R)$ and there exists an injective morphism of $R$-modules $\phi:R/p\rightarrow M$. Let $m=\phi(1)$. For every $x\in p$, $\phi(x)=\phi(x.1)=x\phi(1)=rx=0$ implies that $p\subset Ann(M)$. Let $x\in Ann(M), \phi(x)=\phi(x.1)=x\phi(m)=0$, since $\phi$ is injective, we deduce that $m\in p$.