Equivalence of Definitions of Primary Ideal

Theorem

Let $R$ be a commutative ring with unity.

The following definitions of the concept of Primary Ideal are equivalent:

Definition 1

A proper ideal $\mathfrak q$ of $R$ is called a primary ideal if and only if:

$\forall x,y \in R :$

$x y \in \mathfrak q \implies x \in \mathfrak q \; \lor \; \exists n \in \N_{>0} : y^n \in \mathfrak q$

Definition 2

A proper ideal $\mathfrak q$ of $R$ is called a primary ideal if and only if:

each zero-divisor of the quotient ring $R / \mathfrak q$ is nilpotent.

Proof

Definition 1 implies Definition 2

Let $x + \mathfrak q$ be a zero-divisor of $R / \mathfrak q$.

That is, there is a $y \not \in \mathfrak q$ such that:

$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$

Thus:

$xy + \mathfrak q = 0 + \mathfrak q$

which means:

$xy \in \mathfrak q$

Then, by hypothesis:

$\exists n \in \N_{>0} : x^n \in \mathfrak q$

so that:

$\paren {x + \mathfrak q}^n = x^n + \mathfrak q = 0 + \mathfrak q$

Therefore $x + \mathfrak q$ is nilpotent.

$\Box$

Definition 2 implies Definition 1

Let $xy \in \mathfrak q$ but $x \not \in \mathfrak q$.

That is:

$\paren {x + \mathfrak q} \paren {y + \mathfrak q} = 0 + \mathfrak q$

Then, by hypothesis, $\paren {y + \mathfrak q}$ is nilpotent in $R \ \mathfrak q$.

Thus:

\(\ds \exists n \in \N_{>0}: \, \)

\(\ds \paren {y + \mathfrak q}^n\)

\(=\)

\(\ds y^n + \mathfrak q\)

\(\ds \)

\(=\)

\(\ds 0 + \mathfrak q\)

so that:

$y^n \in \mathfrak q$

$\blacksquare$