# Equivalence of Definitions of Prime Ideal of Commutative and Unitary Ring

## Theorem

The following definitions of the concept of Prime Ideal of Commutative and Unitary Ring are equivalent:

### Definition 1

A prime ideal of $R$ is a proper ideal $P$ such that:

$\forall a, b \in R : a \circ b \in P \implies a \in P$ or $b \in P$

### Definition 2

A prime ideal of $R$ is a proper ideal $P$ of $R$ such that:

$I \circ J \subseteq P \implies I \subseteq P \text { or } J \subseteq P$

for all ideals $I$ and $J$ of $R$.

### Definition 3

A prime ideal of $R$ is a proper ideal $P$ of $R$ such that:

the complement $R \setminus P$ of $P$ in $R$ is closed under the ring product $\circ$.

## Proof

Let $\struct {R, +, \circ}$ be a commutative and unitary ring throughout.

### $(1)$ implies $(2)$

Let $P$ be a prime ideal of $R$ by definition 1.

Then by definition:

$\forall a, b \in R : a \circ b \in P \implies a \in P$ or $b \in P$

Let $I \circ J \subseteq P$.

Aiming for a contradiction, suppose that both $I \nsubseteq P$ and $J \nsubseteq P$.

Then by definition of subset:

$\exists a \in I \setminus P, b \in J \setminus P$

But by definition of subset product

$a \circ b \in P$ as $I \circ J \subseteq P$

Thus we have $a, b \in P$ such that:

$a \circ b \in P$ where $a \notin P$ and $b \notin P$

But this contradicts the criterion for $P$ to be a prime ideal of $R$ by definition 1

Thus by Proof by Contradiction, either $I \subseteq P$ or $J \subseteq P$,

Thus $P$ is a prime ideal of $R$ by definition 2.

$\Box$

### $(2)$ implies $(1)$

Let $P$ be a prime ideal of $R$ by definition 2.

Then by definition:

$I \circ J \subseteq P \implies I \subseteq P \text { or } J \subseteq P$

for all ideals $I$ and $J$ of $R$.

Let $a, b \in R$ such that $a \circ b \in P$.

Consider $\ideal a$ and $\ideal b$, the ideals generated by $a$ and $b$.

Let $r \in \ideal a, s \in \ideal b$.

Then:

$\exists m, n \in R: r = m \circ a, s = n \circ b$

Therefore:

 $\ds r \circ s$ $=$ $\ds \paren {m \circ a} \circ \paren {n \circ b}$ $\ds$ $=$ $\ds \paren {m \circ n} \circ \paren {a \circ b}$ Definition of Commutative Ring $\ds$ $\in$ $\ds P$ Definition of Ideal of Ring: $m \circ n \in R$, $a \circ b \in P$

This shows that $\ideal a \circ \ideal b \subseteq P$.

By definition 2, $\ideal a \subseteq P$ or $\ideal b \subseteq P$.

This implies $a \in P$ or $b \in P$.

Thus $P$ is a prime ideal of $R$ by definition 1.

$\Box$

### $(1)$ implies $(3)$

Let $P$ be a prime ideal of $R$ by definition 1.

Then by definition:

$\forall a, b \in R : a \circ b \in P \implies a \in P$ or $b \in P$

Aiming for a contradiction, suppose $R \setminus P$ is not multiplicatively closed.

That is:

 $\ds \exists a, b \in R \setminus P: \,$ $\ds a \circ b$ $\notin$ $\ds R \setminus P$ $\ds \leadsto \ \$ $\ds a \circ b$ $\in$ $\ds P$ $\ds \leadsto \ \$ $\ds a$ $\in$ $\ds P$ $\, \ds \lor \,$ $\ds b$ $\in$ $\ds P$

But this contradicts the assertion that $a, b \in R \setminus P$.

Thus by Proof by Contradiction $R \setminus P$ is multiplicatively closed.

Thus $P$ is a prime ideal of $R$ by definition 3.

$\Box$

### $(3)$ implies $(1)$

Let $P$ be a prime ideal of $R$ by definition 3.

Then by definition:

the complement $R \setminus P$ of $P$ in $R$ is closed under the ring product $\circ$.

Aiming for a contradiction, suppose let $a \circ b \in P$ such that $a \notin P$ and $b \notin P$.

Then:

$a, b \in R \setminus P$

by definition of relative complement.

But $R \setminus P$ is closed under the ring product $\circ$.

That means:

$\forall a, b \in R \setminus P \implies a \circ b \in R \setminus P$

But this contradicts the assertion that $a \circ b \in P$.

Thus by Proof by Contradiction either $a \in P$ or $b \in P$ (or both).

Thus $P$ is a prime ideal of $R$ by definition 1.

$\blacksquare$