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

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

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

Proof.

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

$\begin{array}[]{ll}\operatorname{Spec}(R)&\displaystyle=\bigcup_{\lambda\in% \Lambda}U_{\lambda}\\ &\displaystyle=\bigcup_{\lambda\in\Lambda}\bigg{(}\operatorname{Spec}(R)% \setminus V\left(I_{\lambda}\right)\bigg{)}\\ &\displaystyle=\operatorname{Spec}(R)\setminus\bigcap_{\lambda\in\Lambda}V% \left(I_{\lambda}\right)\\ &\displaystyle=\operatorname{Spec}(R)\setminus V\left(\sum_{\lambda\in\Lambda}% I_{\lambda}\right),\end{array}$

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

Let $r\in R$. Then $\displaystyle r=r\cdot 1_{R}=r\sum_{\ell\in L}i_{\ell}=\sum_{\ell\in L}r\cdot i% _{\ell}\in\sum_{\ell\in L}I_{\ell}$. Thus, $\displaystyle\sum_{\ell\in L}I_{\ell}=R$. Therefore, $\displaystyle V\left(\sum_{\ell\in L}I_{\ell}\right)=\emptyset$. Since

$\begin{array}[]{ll}\operatorname{Spec}(R)&\displaystyle=\operatorname{Spec}(R)% \setminus V\left(\sum_{\ell\in L}I_{\ell}\right)\\ &\displaystyle=\operatorname{Spec}(R)\setminus\bigcap_{\ell\in L}V\left(I_{% \ell}\right)\\ &\displaystyle=\bigcup_{\ell\in L}\bigg{(}\operatorname{Spec}(R)\setminus V% \left(I_{\ell}\right)\bigg{)}\\ &\displaystyle=\bigcup_{\ell\in L}U_{\ell},\end{array}$

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

Title proof that $\operatorname{Spec}(R)$ is quasi-compact ProofThatoperatornameSpecRIsQuasicompact 2013-03-22 16:07:40 2013-03-22 16:07:40 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Proof msc 14A15 VIemptysetImpliesIR