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The more general form of Krull intersection theorem says:

Let $R$ be local and Noetherian and $I \subset R$ a proper ideal. If $M$ is finitely generated over $R$, and $N=\cap_1^{\infty} I^iM$, then $IN=N$.

What is the simplest counter-examples when one (and only one) condition among: $R$ local, $R$ Noetherian or $M$ finitely generated is dropped? So this is three questions I guess.

Sorry if this is too easy for this site. It has been a while, you know!

LATER: Andrea's answer gave a counter-example to the stronger statement: there is an element $r \in I$ such that $N(1-r)=0$. I believe it is not a counter-example to the form stated above, see David Eisenbud's book on commutative algebra, the Example after Corollary 5.5 and Exercise 5.6.

However, Dustin Cartwright pointed out that one can safely drop the "local" hypothesis. So there are only two questions left.

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    $\begingroup$ You have more hypotheses than necessary: $R$ doesn't have to be local and $I$ doesn't have to be a proper ideal. These (or other) hypotheses are necessary for versions which conclude that the intersection is $0$. For example, Theorem 74 in Kaplansky's Commutative Rings is the statement you quote without the local or proper hypotheses. $\endgroup$ Jul 31 '11 at 20:16
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For $R$ noetherian and $M$ not finitely generated you can take the following example from Kaplansky, Infinite Abelian Groups: The abelian group $G$ with generators $x$ and $y_k$ for $k=1,2,\dots$ and relations $px=0$, $x=py_1=p^2y_2=\dots=p^ky_k=\dots$ ($p$ some fixed prime) satisfies $G_\omega=\bigcap p^kG=\langle x\rangle$, but $pG_\omega=0\ne G_\omega$.

Building upon this example, you also get an example for the case $R$ non-noetherian and $M$ finitely generated: with the same $G$, you can set $R=\mathbb Z\times G$ with $G$ as a square-zero ideal, $M=R$ and $I=pR$. Now, $\bigcap I^n=0\times G_\omega$ again satisfies $I\bigcap I^n\ne\bigcap I^n$.

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Here's a variation on user2035's answer.

Let $R=\mathbb Q[x, z, y_1, y_2, \ldots]/ \langle x-zy_1, x-z^2y_2, \ldots \rangle$. Then take $M=R$ and $I=\langle z \rangle$, and then $\cap_{i=1}^n I^n = \langle x \rangle$, but $z\langle x \rangle \neq \langle x \rangle$. Obviously, in this case, $R$ is not Noetherian.

Now keep $M$ as above, but replace $R$ with the subring $\mathbb Q[z]$. The intersection theorem fails in the same way as above. In this case, $R$ is Noetherian, but $M$ is not a finitely generated $R$-module.

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If $R$ is not Noetherian, you could read remark 2 at page 110 of Atiyah, Macdonald - Introduction to Commutative Algebra.

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  • $\begingroup$ This is not a counter example to this form of Krull's theorem. See my edit. $\endgroup$ Aug 1 '11 at 1:07

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