proof that $\mathrm{Spec}(R)$ is quasi-compact

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

The following is a proof that $\mathrm{Spec}(R)$ is quasi-compact.

Proof.

Let $\mathrm{\Lambda }$ be an indexing set and ${\left\{U_{\lambda }\right\}}_{\lambda \in \mathrm{\Lambda }}$ be an open cover for $\mathrm{Spec}(R)$. For every $\lambda \in \mathrm{\Lambda }$, let $I_{\lambda }$ be an ideal of $R$ with $U_{\lambda }=\mathrm{Spec}(R)\setminus V\left(I_{\lambda }\right)$. Since

$V\left({\displaystyle \sum _{\lambda \in \mathrm{\Lambda }}}{I}_{\lambda }\right)=\mathrm{\varnothing }$. Thus, by this theorem (http://planetmath.org/VIemptysetImpliesIR), ${\displaystyle \sum _{\lambda \in \mathrm{\Lambda }}}{I}_{\lambda }=R$. Since $1_R\in R={\displaystyle \sum _{\lambda \in \mathrm{\Lambda }}}{I}_{\lambda }$, there exists a finite subset $L$ of $\mathrm{\Lambda }$ such that, for every $\mathrm{\ell }\in L$, there exists an $i_{\mathrm{\ell }}\in I_{\mathrm{\ell }}$ with $1_R={\displaystyle \sum _{\mathrm{\ell }\in L}}{i}_{\mathrm{\ell }}$.

Let $r\in R$. Then $r=r\cdot 1_R=r{\displaystyle \sum _{\mathrm{\ell }\in L}}{i}_{\mathrm{\ell }}={\displaystyle \sum _{\mathrm{\ell }\in L}}r\cdot i_{\mathrm{\ell }}\in {\displaystyle \sum _{\mathrm{\ell }\in L}}{I}_{\mathrm{\ell }}$. Thus, ${\displaystyle \sum _{\mathrm{\ell }\in L}}{I}_{\mathrm{\ell }}=R$. Therefore, $V\left({\displaystyle \sum _{\mathrm{\ell }\in L}}{I}_{\mathrm{\ell }}\right)=\mathrm{\varnothing }$. Since

${\left\{U_{\lambda }\right\}}_{\lambda \in \mathrm{\Lambda }}$ restricts to a finite subcover. It follows that $\mathrm{Spec}(R)$ is quasi-compact. ∎

Title	proof that $\mathrm{Spec}(R)$ is quasi-compact
Canonical name	ProofThatoperatornameSpecRIsQuasicompact
Date of creation	2013-03-22 16:07:40
Last modified on	2013-03-22 16:07:40
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 14A15
Related topic	VIemptysetImpliesIR