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2008-05-12T16:24:08Z

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Vipul: /* Relation with other results */

2007-08-09T02:22:57Z

Relation with other results

← Older revision		Revision as of 02:22, 9 August 2007
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	==Relation with other results==		==Relation with other results==

	* [[Irreducible implies prime (~~PIR~~)]] uses a very similar approach		* [[Irreducible implies prime (PID)]] uses a very similar approach
	* [[Irreducible implies prime (Dedekind)]]		* [[Irreducible implies prime (Dedekind)]]

Vipul at 02:22, 9 August 2007

2007-08-09T02:22:11Z

New page

==Statement==

In a [[Noetherian ring]], any [[irreducible ideal]] is [[primary ideal|primary]].

==Relation with other results==

* [[Irreducible implies prime (PIR)]] uses a very similar approach
* [[Irreducible implies prime (Dedekind)]]

==Proof==

===Hands-on proof===

Let <math>R</math> be a Noetherian ring and <math>P</math> an irreducible ideal in <math>R</math>. Suppose <math>ab \in P</math> but <math>a \notin P</math>. We need to show that there exists a <math>n</math> such that <math>b^n \in P</math>.

Define <math>P_i</math> as the ideal of all elements <math>x</math> such that <math>xb^i \in P</math>. Clearly we get an ascending chain of ideals in <math>R</math>:

<math>P = P_0 \le P_1 \le P_2 \le \ldots</math>

Since <math>R</math> is Noetherian, there exists a natural number <math>n</math> such that <math>P_n = P_m</math> for all <math>m \ge n</math>. In other words, there exists a <math>n</math> such that if <math>b^{n+1}x \in P</math>, then <math>b^nx \in P</math> for any <math>x \in R</math>.

Now let <math>Q = P + (a)</math> and <math>R = P + (b^n)</math>. Clearly, <math>Q \cap R</math> contains <math>P</math>. We want to show that <math>Q \cap R = P</math>.

For this, suppose <math>y \in Q \cap R</math>. Then we can write:

<math>y = p_1 + ua = p_2 + vb^n</math>

where <math>p_i \in P</math>. This tells us that:

<math>ua - vb^n \in P</math>.

multiplying both sides by <math>b</math> we get:

<math>uab - vb^{n+1} \in P</math>

By our assumption, <math>ab \in P</math> and hence <math>uab \in P</math>.

This gives us:

<math>vb^{n+1} \in P</math>

but from the property of <math>n</math>, <math>vb^n \in P</math> and hence <math>y = p_2 + vb^n \in P</math>.

Thus any element of <math>Q \cap R</math> is in <math>P</math>, so <math>Q \cap R = P</math>.

===Comparison with the proof for principal ideal rings===

In the case of principal ideal rings, we employ a very similar proof, except that the <math>n</math> is replaced by 1.

Irreducible implies primary (Noetherian) - Revision history

Vipul: 2 revisions

Vipul: /* Relation with other results */

Vipul at 02:22, 9 August 2007