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)?
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.
See the Section on the Krull Intersection Theorem (currently Section 8.12) 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!
