General Ring Theory/Prime ideals

Definition (prime ideal):

Let $R$ be a ring. A prime ideal of $R$ is an ideal $p\leq R$ such that whenever $I,J\leq R$ are ideals of $R$ so that $I\cdot J\subseteq p$, then either $I\subseteq p$ or $J\subseteq p$.

Definition (prime ring):

A ring $R$ is called a prime ring iff its zero ideal is a prime ideal.

Proposition (characterisation of prime ideals):

Let $R$ be a ring, and let $P\leq R$ be an ideal. The following are equivalent:

1. $P$ is a prime ideal of $R$
2. $R/P$ is a prime ring
3. Whenever $I,J\leq R$ are left ideals in $R$, then $I\cdot J\subseteq P\Rightarrow (I\subseteq P\vee J\subseteq P)$
4. Whenever $I,J\leq R$ are right ideals in $R$, then $I\cdot J\subseteq P\Rightarrow (I\subseteq P\vee J\subseteq P)$
5. Whenever $x,y\in R$ are such that $xRy\in P$, then either $x\in P$ or $y\in P$.

Proof: We'll prove $1.\Rightarrow 2.\Rightarrow 5.\Rightarrow 3.\Rightarrow 1.$, since $3.\Leftrightarrow 4.$ follows by symmetry. Suppose first that $P$ is a prime ideal. Let ${\overline {I}},{\overline {J}}\leq R/P$ so that ${\overline {I}}\cdot {\overline {J}}\subseteq \{0+P\}$, the zero ideal of $R/P$. Then if $\pi :R\to R/P$ is the projection, consider $I:=\pi ^{-1}({\overline {I}})$, $J:=\pi ^{-1}({\overline {J}})$. Then $I\cdot J\subseteq \pi ^{-1}({\overline {I}}\times {\overline {J}})\subseteq \{0+P\}$ (since $\pi$ is a ring homomorphism), so that without loss of generality $I\subseteq P$, and hence ${\overline {I}}\leq \{0+P\}$. Suppose now that $R/P$ is a prime ring. Let then $x,y\in R$ such that $xRy\in P$. We use the bar notation ${\overline {z}}:=\pi (z)$ ($\pi :R\to R/P$ being the projection) for $z\in R$. Then we get that ${\overline {xzy}}\in R/P$ is zero for all $z\in R$. Then define the ideals $K:=\langle {\overline {x}}\rangle \leq R/P$ and $L:=\langle {\overline {y}}\rangle$, so that then $K\cdot L\leq \{0+P\}$, hence without loss of generality $K=\{0+P\}$ and hence $\pi ^{-1}(K)\leq P$ and thus $x\in P$. Suppose now that 5. holds, and let $I,J\leq R$ be left ideals such that $I\cdot J\leq P$. Suppose that there existed $x,y\in R$ so that $x\in I\setminus P$ and $y\in J\setminus P$. Then still $xRy\in P$, a contradiction. Note finally that $3.\Rightarrow 1.$ is trivial. $\Box$