Spectrum is compact - Revision history

Vipul: 3 revisions

2008-05-12T16:34:48Z

3 revisions

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

Vipul: /* Proof */

2008-03-15T21:08:12Z

Proof

← Older revision		Revision as of 21:08, 15 March 2008
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	<math>\mathcal{I}(\bigcap_i V_i) = \bigoplus \mathcal{I}(V_i)</math>		<math>\mathcal{I}(\bigcap_i V_i) = \bigoplus \mathcal{I}(V_i)</math>

			(the right side denotes the ideal generated in <math>R</math>, and not the abstract direct sum).

	In other words, a prime ideal contains ''all'' the <math>\mathcal{I}(V_i)</math>s if and only if it contians their sum.		In other words, a prime ideal contains ''all'' the <math>\mathcal{I}(V_i)</math>s if and only if it contians their sum.

Vipul: /* Proof */

2008-03-15T21:07:43Z

Proof

← Older revision		Revision as of 21:07, 15 March 2008
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	Clearly, we have, by definition:		Clearly, we have, by definition:

	<math>\mathcal{I}(\bigcap_i V_i) = \~~bigplus~~ \mathcal{I}(V_i)</math>		<math>\mathcal{I}(\bigcap_i V_i) = \bigoplus \mathcal{I}(V_i)</math>

	In other words, a prime ideal contains ''all'' the <math>\mathcal{I}(V_i)</math>s if and only if it contians their sum.		In other words, a prime ideal contains ''all'' the <math>\mathcal{I}(V_i)</math>s if and only if it contians their sum.

Vipul: New page: {{spectrum topology fact}} ==Statement== The spectrum of a commutative unital ring is always a compact space. ==Proof== We shall prove...

2008-03-15T21:07:20Z

New page: {{spectrum topology fact}} ==Statement== The spectrum of a commutative unital ring is always a compact space. ==Proof== We shall prove...

New page

{{spectrum topology fact}}

==Statement==

The [[spectrum of a commutative unital ring|spectrum]] of a [[commutative unital ring]] is always a [[compact space]].

==Proof==

We shall prove the statement using the ''finite intersection property'' formulation of compactness.

''Given'': <math>R</math> is a [[commutative unital ring]], and <math>X = Spec(R)</math> is its spectrum. Suppose <math>V_i</math> is a collection of closed subsets (where <math>i \in I</math>, an indexing set) such that any finite intersection of the <math>V_i</math>s is nonempty.

''To prove'': The intersection of all the <math>V_i</math>s is nonempty.

''Proof'': Let <math>\mathcal{I}(V)</math> denote the [[radical ideal]] corresponding to the closed set <math>V</math> under the [[Galois correspondence between a ring and its spectrum]].

Clearly, we have, by definition:

<math>\mathcal{I}(\bigcap_i V_i) = \bigplus \mathcal{I}(V_i)</math>

In other words, a prime ideal contains ''all'' the <math>\mathcal{I}(V_i)</math>s if and only if it contians their sum.

Suppose the intersection of the <math>V_i</math>s was empty. Then the left side would be the whole ring <math>R</math>, hence so would the right side. In other words, we'd have that <math>R</math> is the sum of the <math>\mathcal{I}(V_i)</math>s.

But this would imply that <math>1 \in R</math> is in the sum of finitely many <math>\mathcal{I}(V_i)</math>s, and that in turn would yield that the intersection of those finitely many <math>V_i</math>s must be empty, contradicting the assumption of finite intersection property.