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单词 ProofThatoperatornameSpecRIsQuasicompact
释义

proof that Spec(R) is quasi-compact


Note that most of the notation used here is defined in the entry prime spectrum.

The following is a proof that Spec(R) is quasi-compact.

Proof.

Let Λ be an indexing set and {Uλ}λΛ be an open cover for Spec(R). For every λΛ, let Iλ be an ideal of R with Uλ=Spec(R)V(Iλ). Since

Spec(R)=λΛUλ=λΛ(Spec(R)V(Iλ))=Spec(R)λΛV(Iλ)=Spec(R)V(λΛIλ),

V(λΛIλ)=. Thus, by this theorem (http://planetmath.org/VIemptysetImpliesIR), λΛIλ=R. Since 1RR=λΛIλ, there exists a finite subset L of Λ such that, for every L, there exists an iI with 1R=Li.

Let rR. Then r=r1R=rLi=LriLI. Thus, LI=R. Therefore, V(LI)=. Since

Spec(R)=Spec(R)V(LI)=Spec(R)LV(I)=L(Spec(R)V(I))=LU,

{Uλ}λΛ restricts to a finite subcover. It follows that Spec(R) is quasi-compact.∎

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