# General Ring Theory/Prime ideals

Definition (prime ideal):

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

Definition (prime ring):

A ring ${\displaystyle R}$ is called a prime ring iff its zero ideal is a prime ideal.

Proposition (characterisation of prime ideals):

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

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

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