properties of radical ring

**student2011** · April 12th 2011, 01:40 AM

Show that $\sqrt{I}=\cap{P}$ , where the intersection is over all prime ideals of $R$ that contain $I$ .

I know how to verify the one inclution $\sqrt{I}\subset\cap{p}$ , but how can I show the second inclution.

Thank you in advance

**FernandoRevilla** · April 12th 2011, 05:02 AM

Hint:

The nilradical of $R$ is the intersection of all prime ideals of $R$ . Apply this result to $R/I$ .

**student2011** · April 12th 2011, 12:46 PM

Thaaaaank you very much, only now I get the point.

$L(R/I)=\cap{(P/I)}$ where $P$ is prime ideal in $R$ contains $I$ . But we know that $\cap{(P/I)}=\cap{P}/I$ . Thus, if $x \in \cap{P}$ , then $x+I \in \cap{P} +I=L(R/I)$ . It is easy to show that $L(R/I)=\sqrt{I}$ .

Thank you again

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