| Version current |
Version 9 |
| 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. A two-sided proper ideal $\mathfrak{p}$ of a ring $R$ is called a prime ideal if the following equivalent conditions are met: |
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| \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 $y \in \mathfrak{p}$. |
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\item $R/\mathfrak{p}$ is a prime ring. |
| \end{enumerate} |
\end{enumerate} |
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$R/\mathfrak{p}$ 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.
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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. |
