We know that for any prime P, the radical R(P)=P. However is the converse of this Statement true. That is, if we know that radical of an ideal I is itself, i.e. R(I)=I, is I prime? I presume it is not but couldn't come with a counterexample.
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You are right, this is not true. For a counterexample, consider the ideal $I = (xy) \subset k[x,y]$ for any field $k$.
red_trumpet
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1Nice answer. I think you can generalize this by taking $R$ to be any UFD, and $I$ to be the principal ideal generated by any squarefree (nonunit) element (e.g., $R = \mathbb{Z}, I = 6\mathbb{Z}$). In general, $I$ is a radical ideal of $R$ if and only if $R/I$ is reduced, which is a fruitful way to produce examples. – Alex Wertheim Mar 15 '20 at 06:22
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For $I$, $J$ ideals we have
$$R(I\cap J)\subset R(I)\cap R(J) \subset R(I J)$$ so $$R(I\cap J)=R(I)\cap R(J)$$
Therefore, if $R(I)=I$, $R(J)=J$, then $R(I\cap J) = I\cap J$.
That should provide other examples of ideals with $R(I)=I$ (radical ideals), for instance finite intersections of prime ideals.
Other examples: since $R(R(I)) = R(I)$, then $R(I)$ is radical.
Note that if in the given ring every ideal is an intersection of primary ideals ( true for every notherian ring) then every primary ideal is an intersection of prime ideals.
orangeskid
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