Equivalence of Definitions of Associated Prime of Module

Theorem
Let $A$ be a commutative ring with unity.

Let $M$ be a module over $A$.

Let $\mathfrak p$ be a prime ideal in $A$.

Definition 1 implies Definition 2
Suppose that $x \in M$ satisfies:
 * $\map {\operatorname {Ann}_A} x = \mathfrak p$

Define a submodule of $M$ by:
 * $N := \set { a x : a \in A }$

Define a module homomorphism $\phi : A \to N$ by:
 * $a \mapsto a x$

Then the kernel of $\phi$ is:
 * $\map \ker \phi = \map {\operatorname {Ann}_A} x = \mathfrak p$

By First Isomorphism Theorem:
 * $N \cong A / \mathfrak p$

Definition 2 implies Definition 1
Let $N \subseteq M$ be a submodule.

Suppose that there is an isomorphism:
 * $\psi : A / \mathfrak p \to N$

Let $a_0 \in A \setminus \mathfrak p$.

Let:
 * $x := \map \psi {a_0 + \mathfrak p}$

We claim that:
 * $\map {\operatorname {Ann}_A} x = \mathfrak p$

Indeed: