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Given a Noetherian ring $R$ and a proper ideal $I$ of it.

Is it true that $\bigcap_{n\ge 1} I^n=0$ as $n$ varies over all natural numbers?

If not, is it true if $I$ is a maximal ideal? If not, is it true if $I$ is the maximal ideal of a local ring $R$? If not, is it true under additional assumptions on $R$ (like $R$ is regular)?

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    A ring of the form $k \times k$ is certainly regular.2011-01-25

2 Answers 2

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It is not true in general: the ideal may well be idempotent!

For an example, consider a direct product of two fields: there are two ideals, both maximal and both idempotent.

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See the Section on the Krull Intersection Theorem (currently Section 8.7) in these notes.

A version of the theorem valid for any ideal $I$ in a Noetherian ring $R$ is as follows: if there exists $x \in \bigcap_{n=1}^{\infty} I^n$, then $x \in xI$. From this one easily deduces that $\bigcap_{n=1}^{\infty} I^n = \{0\}$ under either of the following additional hypotheses:

$\bullet$ $R$ is a domain and $I$ is a proper ideal, or
$\bullet$ $I$ is contained in the Jacobson radical $J(R)$ of $R$ (i.e., the intersection of all maximal ideals).

In particular the second condition holds for any proper ideal in a Noetherian local ring.

As Mariano remarks, some hypothesis beyond Noetherianity is needed in order to guarantee $\bigcap_n I^n = \{0\}$. I should probably add his counterexample to my notes!

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    A proof of Krull Intersection Theorem without using Artin-Rees Lemma see Milne's _A Primer of Commutative Algebra_. Theorem 3.16 on page 13.2018-11-10