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2008-05-12T16:28:57Z

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{{curing-ideal property implication}}

==Statement==

===Property-theoretic statement===

The [[property of ideals]] of being a [[primary ideal]] is stronger than the property of being an [[ideal with prime radical]].

===Verbal statement===

The [[radical of an ideal|radical]] of a [[primary ideal]] is [[prime ideal|prime]].

==Converse==

The converse of the statement is not true. {{proofat|[[prime radical not implies primary]]}}

However, any [[ideal with maximal radical]] is primary; in other words, if the [[radical of an ideal|radical]] is a maximal ideal, the ideal we started with is primary. {{proofat|[[maximal radical implies primary]]}}

==Proof==

Suppose <math>R</math> is a [[commutative unital ring]] and <math>Q</math> is a [[primary ideal]] in <math>R</math>. Let <math>P</math> be the radical of <math>Q</math>. We need to show that <math>P</math> is prime in <math>R</math>.

Pick <math>a,b \in R</math> such that <math>ab \in P</math>. Then, since <math>P</math> is the radical of <math>Q</math>, there exists <math>n \in \mathbb{N}</math> such that <math>(ab)^n \in Q</math>. Hence <math>a^nb^n \in Q</math>, so either <math>a^n \in Q</math> or there exists <math>m</math> such that <math>(b^n)^m = b^{nm} \in Q</math>. In the first case, <math>a \in P</math>, and in the second case, <math>b \in P</math>.

Primary implies prime radical - Revision history

Vipul: 1 revision

Vipul at 23:25, 5 January 2008