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I'm trying to show the following:

Let $I,J$ be ideals of a commutative ring (with 1) $A$ such that $I_{P}=J_{P}$ for every prime ideal P of R. Here $I_{P}$ means the localization of $I$ in $P$. Then $I=J$.

Well I was thinking in using the following result: let $M$ be an $A$-module. If $M_{P}=0$ for every prime ideal then $M=0$.

But don't we need some kind of assumption like $J \subset I$?

Because if for instance, say $J \subset I$ then $(I/J)_{P} \cong I_{P}/J_{P}$ so $(I/J)_{P}$ is the trivial module for every prime ideal $P$ so $I/J=0$ hence $I=J$.

Is this wrong? How do we proceed?

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    @user6495:$I$had deleted it because I had some nonsense in it which I realized it was nonsense after posting it, and then forgot to fix it...2011-04-01

1 Answers 1

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There's been no activity, so let me post this as an answer, then.

You know what to do if $J\subseteq I$ or if $I\subseteq J$.

How about considering the ideals $I$ and $I\cap J$, then? If you can show that $(I\cap J)_P = I_P\cap J_P$, then your argument will go through for $I$ and $I\cap J$, showing $I=I\cap J$. Then you can repeat the argument with $J$ and $I\cap J$.

Localization is pretty nice. They commute with finite sums, finite intersections, and quotients; and localization is an exact functor. This is in Atiyah-MacDonald, Prop. 3.3 and Corollary 3.4.