Vipul: 2 revisions

2008-05-12T16:34:50Z

2 revisions

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Vipul at 20:46, 15 March 2008

2008-03-15T20:46:02Z

← Older revision		Revision as of 20:46, 15 March 2008
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	==Statement==		==Statement==

			===Topological statement===

	The [[spectrum of a commutative unital ring]], with the usual topology, is a [[sober space]]: in other words, any irreducible closed subset is the closure of a one-point subset.		The [[spectrum of a commutative unital ring]], with the usual topology, is a [[sober space]]: in other words, any irreducible closed subset is the closure of a one-point subset.

			===Ring-theoretic statement===

			Any irreducible closed subset in the spectrum of a commutative unital ring corresponds to a [[prime ideal]] under the [[Galois correspondece between a ring and its spectrum]].

			==Proof==

			''Given'': A commutative unital ring <math>R</math>, its spectrum <math>Spec(R)</math>, an irreducible closed subset <math>A</math> of <math>Spec(R)</math>. Let <math>\mathcal{Z}(I)</math>, for an ideal <math>I</math>, denote the set of prime ideals containing <math>I</math>.

			''To prove'': There is a [[prime ideal]] <math>P</math> such that <math>A</math> is precisely the set of all prime ideals containing <math>P</math>

			''Proof'': By definition of closed set, there exists a unique radical ideal <math>P</math> such that <math>A</math> is precisely the set of prime ideals containing <math>P</math>. We need to show that <math>P</math> is in fact prime. Suppose not. Then there exist ideals <math>J</math> and <math>K</math> of <math>R</math> such that <math>JK \subseteq P</math> but neither <math>J</math> nor <math>K</math> is contained in <math>P</math>.

			We now argue the following:

			* <math>\mathcal{Z}(J) \cup \mathcal{Z}(K) = \mathcal{Z}(P) = A</math>: If <math>Q</math> is a prime ideal containing <math>P</math>, then <math>JK \subset Q</math>. But primeness of <math>Q</math> forces <math>J \subset Q</math> or <math>K \subset Q</math>, thus forcing <math>Q \in \mathcal{Z}(J) \cup \mathcal{Z}(K)</math>.
			* Both are closed subsets: This is by definition
			* Both are proper subsets: If <math>\mathcal{Z}(J) = \mathcal{Z}(P)</math>, then <math>P</math> being a radical ideal, must be the radical of <math>J</math>. But that'd force <math>J \subset P</math>, a contradiction to assumption. Hence <math>\mathcal{Z}(J)</math> must be a proper subset of <math>A</math>. A similar argument holds for <math>\mathcal{Z}(K)</math>.

			Thus, we have expressed <math>A</math> as a union of two proper closed subsets, a ''contradiction''. Hence, our original assumption was false, and <math>P</math> is prime.

Vipul at 14:32, 20 January 2008

2008-01-20T14:32:16Z

New page

==Statement==

The [[spectrum of a commutative unital ring]], with the usual topology, is a [[sober space]]: in other words, any irreducible closed subset is the closure of a one-point subset.

Spectrum is sober - Revision history

Vipul: 2 revisions

Vipul at 20:46, 15 March 2008

Vipul at 14:32, 20 January 2008