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I have to prove the following proposition. Let $A$ be a commutative ring with $1$, $I$ an ideal of $A$ and $P_0$ a prime minimal ideal of $I$. If $A$ is a local Noetherian ring with maximal ideal prime equal to $P_0$, then $\sqrt{I}=P_0$.

The inclusione $\sqrt{I}\subseteq P_0$ is clear. I have no idea about how to prove the other inclusion. Can anyone help me, please?

Grace53
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    This question should help.. After all, if $P_0$ is both minimal and maximal among primes containing $I$, which primes contain it? – Daniel Oct 16 '22 at 16:53
  • @Daniel There is no prime $P$ such that $I\subsetneq Q\subsetneq P_0$ and there is no ideal $J$ such that $P_0\subsetneq J\subseteq A$. – Grace53 Oct 16 '22 at 18:17
  • I agree: there is no prime $Q$ so that $I \subseteq Q \subsetneq P_0$, and no prime ideals strictly containing $P_0$, since $(A, P_0)$ is a local ring. Moreover, $P_0$ is the only maximal ideal. Now what does the the linked question imply about $\sqrt{I}$? – Daniel Oct 17 '22 at 02:13

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The radical of an ideal is the intersection of all the prime ideals which contains it.

(For this you can see the first chapter of the Atiyah, but you may just observe that the nilradical is the intersection of all prime ideals, and you can apply this to the quotient $A/I$)

If $P_{0}$ is both a minimal and maximal prime ideal then it is the only prime ideal of the ring. Since $I$ is an ideal, it must be contained in a maximal ideal, but there is only $P_{0}$, hence the one and only prime ideal which contains $I$ is $P_{0}$.

To conclude, the radical is just $P_{0}$ because the intersection is taken over just one ideal.

Mascara
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