Intersection of powers of an ideal in a Noetherian ring

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 8.7) 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!