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a characterization of the radical of an ideal

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\begin{document}
\begin{prop} Let $I$ be an ideal in a ring $R$, and $\sqrt{I}$ be its radical.  Then $\sqrt{I}$ is the intersection of all prime ideals containing $I$. \end{prop}

\begin{proof}  Suppose $x\in \sqrt{I}$, and $P$ is a prime ideal containing $I$.  Then $R-P$ is an \PMlinkname{$m$-system}{MSystem}.  If $x\in R-P$, then $(R-P)\cap I\ne \varnothing$, contradicting the assumption that $I\subseteq P$.  Therefore $x\notin R-P$.  In other words, $x\in P$, and we have one of the inclusions.

Conversely, suppose $x\notin \sqrt{I}$.  Then there is an $m$-system $S$ containing $x$ such that $S\cap I=\varnothing$.  Enlarge $I$ to a prime ideal $P$ disjoint from $S$, so that $x\notin P$ (we can do this; for a proof, see the second remark in \PMlinkname{this entry}{MSystem}).  By contrapositivity, we have the other inclusion.
\end{proof}

\textbf{Remark}.  This shows that every prime ideal is a radical ideal: for $\sqrt{P}$ is the intersection of all prime ideals containing $P$, and if $P$ is itself prime, then $P=\sqrt{P}$.
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