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 $I \circ J \subseteq P$. {{AimForCont}} that both $I \not \subseteq P$ and $J \not \subseteq P$. 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. {{qed|lemma}} === $(2)$ 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: {{begin-eqn}} {{eqn | l = r \circ s       | r = \paren {m \circ a} \circ \paren {n \circ b} }} {{eqn | r = \paren {m \circ n} \circ \paren {a \circ b}       | c = {{Defof|Commutative Ring}} }} {{eqn | o = \in       | r = P       | c = {{Defof|Ideal of Ring}}: $m \circ n \in R$, $a \circ b \in P$ }} {{end-eqn}} 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. {{qed|lemma}} === $(1)$ 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 $b \in P$ {{AimForCont}} $R \setminus P$ is not multiplicatively closed. That is: {{begin-eqn}} {{eqn | lo= \exists a, b \in R \setminus P:       | l = a \circ b       | o = \notin       | r = R \setminus P       | c =  }} {{eqn | ll= \leadsto       | l = a \circ b       | o = \in       | r = P       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \in       | r = P       | c =  }} {{eqn | lo= \lor       | l = b       | o = \in       | r = P       | c =  }} {{end-eqn}} 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. {{qed|lemma}} === $(3)$ 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 $\circ$. {{AimForCont}} 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. {{qed}} Category:Prime Ideals of Rings l2gz7sdmg0jxc6btjim8xli8z2bdzbl