Regular local ring implies integral domain - Revision history

Vipul: 4 revisions

2008-05-12T16:34:13Z

4 revisions

← Older revision	Revision as of 16:34, 12 May 2008
(No difference)

Vipul: /* Induction step */

2008-01-05T23:12:12Z

Induction step

← Older revision		Revision as of 23:12, 5 January 2008
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	* By [[Nakayama's lemma]], <math>M^2 \ne M</math>		* By [[Nakayama's lemma]], <math>M^2 \ne M</math>
	* The set of [[minimal prime ideal]]s of <math>R</math> is finite ({{~~justify~~}})		* The set of [[minimal prime ideal]]s of <math>R</math> is finite ({{clarify}})

	Now, suppose <math>M</math> were contained in the union of <math>M^2</math> and the minimal prime ideals. Then, by the [[prime avoidance lemma]], <math>M</math> must be contained either in <math>M^2</math> or in one of the minimal prime ideals. <math>M^2 \ne M</math> thus forces <math>M</math> to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.		Now, suppose <math>M</math> were contained in the union of <math>M^2</math> and the minimal prime ideals. Then, by the [[prime avoidance lemma]], <math>M</math> must be contained either in <math>M^2</math> or in one of the minimal prime ideals. <math>M^2 \ne M</math> thus forces <math>M</math> to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.

	Thus, there exists an element <math>x</math> in <math>M</math> which is outside the union of <math>M^2</math> and all the minimal prime ideals. Let <math>S = R/(x)</math> and <math>N = MS</math>. Clearly <math>N</math> is the unique maximal ideal in <math>S</math>. By the choice of <math>S</math>, <math>dim(S) < dim(R)</math>, and in fact we can conclude that <math>dim(S) = d - 1</math> {{~~justify~~}}.		Thus, there exists an element <math>x</math> in <math>M</math> which is outside the union of <math>M^2</math> and all the minimal prime ideals. Let <math>S = R/(x)</math> and <math>N = MS</math>. Clearly <math>N</math> is the unique maximal ideal in <math>S</math>. By the choice of <math>S</math>, <math>dim(S) < dim(R)</math>, and in fact we can conclude that <math>dim(S) = d - 1</math> {{clarify}}.

	Now <math>N/N^2</math> is a proper homomorphic image of <math>M/M^2</math> so it can be generated by <math>(d-1)</math> elements. By [[Nakayama's lemma]], <math>N</math> can also be generated by <math>(d-1)</math> elements.		Now <math>N/N^2</math> is a proper homomorphic image of <math>M/M^2</math> so it can be generated by <math>(d-1)</math> elements. By [[Nakayama's lemma]], <math>N</math> can also be generated by <math>(d-1)</math> elements.

Vipul: /* Induction step */

2007-08-10T08:30:34Z

Induction step

← Older revision		Revision as of 08:30, 10 August 2007
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	Now, suppose <math>M</math> were contained in the union of <math>M^2</math> and the minimal prime ideals. Then, by the [[prime avoidance lemma]], <math>M</math> must be contained either in <math>M^2</math> or in one of the minimal prime ideals. <math>M^2 \ne M</math> thus forces <math>M</math> to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.		Now, suppose <math>M</math> were contained in the union of <math>M^2</math> and the minimal prime ideals. Then, by the [[prime avoidance lemma]], <math>M</math> must be contained either in <math>M^2</math> or in one of the minimal prime ideals. <math>M^2 \ne M</math> thus forces <math>M</math> to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.

	Thus, there exists an element <math>x</math> in <math>M</math> which is outside the union of <math>M^2</math> and all the minimal prime ideals. {{~~fillin~~}}		Thus, there exists an element <math>x</math> in <math>M</math> which is outside the union of <math>M^2</math> and all the minimal prime ideals. Let <math>S = R/(x)</math> and <math>N = MS</math>. Clearly <math>N</math> is the unique maximal ideal in <math>S</math>. By the choice of <math>S</math>, <math>dim(S) < dim(R)</math>, and in fact we can conclude that <math>dim(S) = d - 1</math> {{justify}}.

			Now <math>N/N^2</math> is a proper homomorphic image of <math>M/M^2</math> so it can be generated by <math>(d-1)</math> elements. By [[Nakayama's lemma]], <math>N</math> can also be generated by <math>(d-1)</math> elements.

			Thus <math>S</math> is a [[regular local ring]], and hence, by the induction step assumption, <math>S</math> is an [[integral domain]]. Hence <math>x</math> is a [[prime ideal]] of <math>R</math>. But since <math>x</math> lies outside every minimal prime ideal, there is a minimal prime ideal properly contained inside <math>x</math>. Call this minimal prime ideal <math>Q</math>.

			If <math>y \in Q</math> is any element, then we may write <math>y = ax</math> for some <math>a \in R</math>. But then since <math>x \notin Q</math>, <math>a \in Q</math>. Thus, <math>Q = xQ</math>. [[Nakayama's lemma]] now yields <math>Q = 0</math>, and hence <math>R</math> is an integral domain.

Vipul at 10:01, 9 August 2007

2007-08-09T10:01:06Z

← Older revision		Revision as of 10:01, 9 August 2007
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			{{curing property implication}}

	==Statement==		==Statement==

Vipul at 10:00, 9 August 2007

2007-08-09T10:00:32Z

New page

==Statement==

Any [[regular local ring]] is an [[integral domain]]. In other words, if a ring has a unique maximal ideal and that ideal is generated by a set whose size is the [[Krull dimension]] of the ring, then the ring is an integral domain.

==Results used==

* [[Nakayama's lemma]]
* [[Prime avoidance lemma]]

==Proof==

Let <math>R</math> be a [[regular local ring]] and <math>M</math> be its unique [[maximal ideal]]. We now prove the result by induction on the [[Krull dimension]] of <math>R</math>.

===Base case for induction===

If the dimension of <math>R</math> is zero, then, by the definition of regular local ring, the maximal ideal must be trivial and hence, the ring must actually be a [[field]], and hence an [[integral domain]].

===Induction step===

Suppose the result is true for dimensions up to <math>d - 1</math>. We need to prove that the result is true for <math>R</math> of Krull dimension <math>d</math>.

We know the following:

* By [[Nakayama's lemma]], <math>M^2 \ne M</math>
* The set of [[minimal prime ideal]]s of <math>R</math> is finite ({{justify}})

Now, suppose <math>M</math> were contained in the union of <math>M^2</math> and the minimal prime ideals. Then, by the [[prime avoidance lemma]], <math>M</math> must be contained either in <math>M^2</math> or in one of the minimal prime ideals. <math>M^2 \ne M</math> thus forces <math>M</math> to be a minimal prime ideal, which would make the Krull dimension zero, contradicting our assumption that the Krull dimension is at least 1.

Thus, there exists an element <math>x</math> in <math>M</math> which is outside the union of <math>M^2</math> and all the minimal prime ideals. {{fillin}}