prime idealPrimeIdeal2001-10-20 01:40:082005-07-24 04:03:14Definition\usepackage{amssymb}
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\usepackage{xypic}Let $R$ be a ring. A two-sided proper ideal $\mathfrak{p}$ of a ring $R$ is called a prime ideal if the following equivalent conditions are met:
\begin{enumerate}
\item If $I$ and $J$ are left ideals and the product of ideals $IJ$ satisfies $IJ \subset \mathfrak{p}$, then $I \subset \mathfrak{p}$ or $J \subset \mathfrak{p}$.
\item If $I$ and $J$ are right ideals with $IJ \subset \mathfrak{p}$, then $I \subset \mathfrak{p}$ or $J \subset \mathfrak{p}$.
\item If $I$ and $J$ are two-sided ideals with $IJ \subset \mathfrak{p}$, then $I \subset \mathfrak{p}$ or $J\subset \mathfrak{p}$.
\item If $x$ and $y$ are elements of $R$ with $xRy \subset \mathfrak{p}$, then $x \in \mathfrak{p}$ or $y \in \mathfrak{p}$.
\end{enumerate}
$R/\mathfrak{p}$ is a prime ring if and only if $\mathfrak{p}$ is a prime ideal. When $R$ is commutative with identity, a proper ideal $\mathfrak{p}$ of $R$ is prime if and only if for any $a,b \in R$, if $a\cdot b \in \mathfrak{p}$ then either $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$. One also has in this case that $\mathfrak{p} \subset R$ is prime if and only if the quotient ring $R/\mathfrak{p}$ is an integral domain.