Krull intersection theorem for Noetherian domains
From Commalg
This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
View more results involving Noetherianness or Read a survey article on applying Noetherianness
This fact is an application of the following pivotal fact/result/idea: Krull intersection theorem
View other applications of Krull intersection theorem OR Read a survey article on applying Krull intersection theorem
Contents
Statement
Let be a Noetherian domain (i.e. a Noetherian ring that is also an integral domain) and
a proper ideal in
. Then, we have:
Proof
Applying Krull intersection theorem for modules
The Krull intersection theorem states that if is a finitely generated module over a Noetherian ring
and
is an ideal inside
, then there exists
such that:
We apply this to the case where , to get that there exists
, such that:
Applying the integral domain condition and properness of the ideal
Since is a proper ideal,
. Hence
, so the element
cannot be zero.
Thus, by the fact that we are in an integral domain, and by the above equation, we get: