Krull intersection theorem for Noetherian domains: Difference between revisions
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{{applicationof|Krull intersection theorem}} | {{applicationof|Krull intersection theorem}} | ||
==Statement== | ==Statement== | ||
Let <math>R</math> be an [[integral domain]] and <math>I</math> a [[proper ideal]] in <math>R</math>. Then, we have: | Let <math>R</math> be a [[Noetherian domain]] (i.e. a [[Noetherian ring]] that is also an [[integral domain]]) and <math>I</math> a [[proper ideal]] in <math>R</math>. Then, we have: | ||
<math>\bigcap_{j=1}^\infty I^j = 0</math> | <math>\bigcap_{j=1}^\infty I^j = 0</math> |
Latest revision as of 16:26, 12 May 2008
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
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: