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This well known Lemma about $I$-stable filtrations asserts:

Lemma (Artin-Rees) Let $A$ be a Noetherian ring and $E$ a finitely generated $A$-module. Let $F$ be a submodule of $E$ and $\{E_i\}$ an $I$-stable filtration. Then the induced filtration on $F$ is $I$-stable.

One of its immediate corollaries is the following:

Corollary Let $A$ be a Noetherian ring and $I$ an ideal. Then the following equality holds: $$I \left( \bigcap_{i=0}^{\infty} I^i\right)=\left( \bigcap_{i=0}^{\infty} I^i\right)$$

I'm looking for a counterexample to the Artin-Rees Lemma in one of the two forms above, when the ring A is not Noetherian.

Does anyone have some ideas or references?

Thank you!

rschwieb
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Sabino Di Trani
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1 Answers1

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Just to bring everything together and answer this question, here's how you can adapt one of the counterexamples shown in the Math Overflow post. We start with the Noetherian ring, not finitely generated module case first, and then show how to transform that into the non-Noetherian ring, finitely generated module case.

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 $pG_\omega=0\ne G_\omega$.

(Change of notation with respect to the question: $R$ is the ring, $M$ is the module, $N \subseteq M$ is the submodule.)

Take $R = \mathbb{Z}$, $M = G$ and $N = G_\omega =\langle{x}\rangle$ with the example above. ($G$ can be seen as a $\mathbb{Z}$-module because $G$ is an abelian group.) Note that $R$ is Noetherian but $M$ is not finitely generated as an $R$-module. Let $I = p\mathbb{Z} \triangleleft R$ for the fixed prime $p$.

Take the $I$-adic filtration on $M$, which is $(M_n)_{n \ge 0} := (p^nG)_{n \ge 0}$. This is automatically $I$-stable. However, the induced subpsace filtration on $N$ is $(M_n \cap N)_{n \ge 0} = (N)_{n \ge 0}$ since by construction $N \subseteq M_n$ for all $n \ge 0$. This is clearly not $I$-stable as $IN = p\langle{x}\rangle = 0 \ne N$.

Now to get the example with $R$ not Noetherian, again as suggested by the above answer on MO,

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 $I\bigcap I^n\ne\bigcap I^n$.

we can take $R = \mathbb{Z} \times G$, $M = R$, and $N = 0 \times G_\omega = 0 \times \langle{x}\rangle$. You can see that $R$ is not Noetherian because the ideal $0 \times G$ is not finitely generated. Now apply reasoning very similar to the above to get the same result.

You can also show that $R = \mathbb{Z}, M = G, N = G_\omega$ gives you a counterexample to the exactness of $I$-adic completion when the hypotheses of Artin-Rees are not satisfied. Indeed, we have the following exact sequence: $$ 0 \to N \to M \to M/N \to 0$$ If $\widehat{\phantom{A}}$ denotes completion with respect to the $I$-adic topology, then $\widehat{N} \cong N$ because with the $I$-adic topology, $IN = 0$ is an open neighbourhood of $0$, so this is the discrete topology, and similarly $\widehat{M/N} \cong M/N$. But the sequence $$ 0 \to \widehat{N} \to \widehat{M} \to \widehat{M/N} \to 0$$ is not exact, that is, $$ 0 \to {N} \to \widehat{M} \to {M/N} \to 0$$ is not exact because the map $N \to \widehat{M}$ is not injective. Indeed, under this map, $x \in N$ maps to the coherent sequence $(x, x, x, \ldots)$ in $\widehat{M}$ which is equal to $(0, 0, \ldots)$ because $x = py_1 = p^2y_2 = \cdots$ and so $x \in p^nG = I^nM$ for all $n \ge 0$.

Anakhand
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