Spectrum of Noetherian ring is Noetherian - Revision history

Vipul: 3 revisions

2008-05-12T16:34:50Z

3 revisions

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

Vipul at 21:24, 15 March 2008

2008-03-15T21:24:50Z

← Older revision		Revision as of 21:24, 15 March 2008
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	==Converse==		==Converse==

	The converse is not in general true. In other words, there can be non-Noetherian rings whose spectrum is Noetherian. For instance, if the quotient of a ring by its [[nilradical]] is Noetherian, then the spectrum is Noetherian, even if the ring itself is not Noetherian~~. Thus, for instance, any [[local ring]] has, for its spectrum, a Noetherian space, although it is ''not necessarily Noetherian''~~.		The converse is not in general true. In other words, there can be non-Noetherian rings whose spectrum is Noetherian. For instance, if the quotient of a ring by its [[nilradical]] is Noetherian, then the spectrum is Noetherian, even if the ring itself is not Noetherian.
	==Proof==		==Proof==

	The key idea is this: a strictly descending chain of closed subsets in the spectrum, gives rise to a strictly ascending chain of radical ideals in the ring. Thus, if the spectrum had an infinite strictly descending chain of closed subsets, then the Noetherian ring would have an infinite strictly ascending chain of radical ideals: a contradiction.		The key idea is this: a strictly descending chain of closed subsets in the spectrum, gives rise to a strictly ascending chain of radical ideals in the ring. Thus, if the spectrum had an infinite strictly descending chain of closed subsets, then the Noetherian ring would have an infinite strictly ascending chain of radical ideals: a contradiction.

Vipul at 21:22, 15 March 2008

2008-03-15T21:22:45Z

← Older revision		Revision as of 21:22, 15 March 2008
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	{{further\|[[Noetherian space]]}}		{{further\|[[Noetherian space]]}}

			==Converse==

			The converse is not in general true. In other words, there can be non-Noetherian rings whose spectrum is Noetherian. For instance, if the quotient of a ring by its [[nilradical]] is Noetherian, then the spectrum is Noetherian, even if the ring itself is not Noetherian. Thus, for instance, any [[local ring]] has, for its spectrum, a Noetherian space, although it is ''not necessarily Noetherian''.
	==Proof==		==Proof==

	The key idea is this: a strictly descending chain of closed subsets in the spectrum, gives rise to a strictly ascending chain of radical ideals in the ring. Thus, if the spectrum had an infinite strictly descending chain of closed subsets, then the Noetherian ring would have an infinite strictly ascending chain of radical ideals: a contradiction.		The key idea is this: a strictly descending chain of closed subsets in the spectrum, gives rise to a strictly ascending chain of radical ideals in the ring. Thus, if the spectrum had an infinite strictly descending chain of closed subsets, then the Noetherian ring would have an infinite strictly ascending chain of radical ideals: a contradiction.

Vipul: 1

2008-03-15T21:19:24Z

New page

{{spectrum topology fact}}

==Statement==

The [[spectrum]] of a [[Noetherian ring]] is a [[Noetherian space]].

==Definitions used==

===Spectrum===
{{further|[[Spectrum]]}}

===Noetherian ring===
{{further|[[Noetherian ring]]}}

===Noetherian space===
{{further|[[Noetherian space]]}}

==Proof==

The key idea is this: a strictly descending chain of closed subsets in the spectrum, gives rise to a strictly ascending chain of radical ideals in the ring. Thus, if the spectrum had an infinite strictly descending chain of closed subsets, then the Noetherian ring would have an infinite strictly ascending chain of radical ideals: a contradiction.