| Version 8 |
Version 7 |
|
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:
|
Let $R$ be a ring with identity. 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} |
\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 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 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 $I$ 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 $y \in \mathfrak{p}$. |
\item If $x$ and $y$ are elements of $R$ with $xRy \subset \mathfrak{p}$, then $x \in \mathfrak{p}$ or $y \in \mathfrak{p}$. |
| \item $R/\mathfrak{p}$ is a prime ring. |
\item $R/\mathfrak{p}$ is a prime ring. |
| \end{enumerate} |
\end{enumerate} |
|
When $R$ is commutative with identity, an 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 $b \in \mathfrak{p}$.
|
When $R$ is commutative, an 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 an ideal $\mathfrak{p} \subset R$ is prime if and only if the quotient ring $R/\mathfrak{p}$ is an integral domain. |
One also has in this case that an ideal $\mathfrak{p} \subset R$ is prime if and only if the quotient ring $R/\mathfrak{p}$ is an integral domain. |
