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 $A$.

The following definitions of the concept of Associated Prime of Module are equivalent:

Definition 1

$\mathfrak p$ is an associated prime of $M$ if and only if:

$\exists x \in M : \map {\operatorname {Ann}_A} x = \mathfrak p$

where $\map {\operatorname {Ann}_A} x$ is the annihilator of $x$.

Definition 2

$\mathfrak p$ is an associated prime of $M$ if and only if:

$M$ contains a submodule which is isomorphic to the quotient ring $A/\mathfrak p$.

Proof

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:

$N \cong A / \mathfrak p$

$\Box$

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:

\(\ds a\)

\(\in\)

\(\ds \map {\operatorname {Ann}_A} x\)

\(\ds \leadstoandfrom \ \ \)

\(\ds 0\)

\(=\)

\(\ds a x\)

Definition of Annihilator

\(\ds \)

\(=\)

\(\ds a \map \psi {a_0 + \mathfrak p}\)

\(\ds \)

\(=\)

\(\ds \map \psi {a a_0 + \mathfrak p}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds a a_0 + \mathfrak p\)

\(=\)

\(\ds 0 + \mathfrak p\)

since $\psi$ is bijective

\(\ds \leadstoandfrom \ \ \)

\(\ds a a_0\)

\(\in\)

\(\ds \mathfrak p\)

Definition of Quotient Ring

\(\ds \leadstoandfrom \ \ \)

\(\ds a\)

\(\in\)

\(\ds \mathfrak p\)

since $\mathfrak p$ is prime and $a_0 \not \in \mathfrak p$

$\blacksquare$