Krull intersection theorem for Noetherian domains: Difference between revisions
(New page: {{applicationof|Krull intersection theorem}} ==Statement== Let <math>R</math> be an integral domain and <math>I</math> a proper ideal in <math>R</math>. Then, we have: <math>\bi...) |
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==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> | ||
Revision as of 19:17, 3 March 2008
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: