Question
Suppose $A$ is a commutative ring with identity and $U\subseteq Spec(A)$ is open. Show that $U$ is quasi-
compact in the Zariski topology if and only if $U = Spec(A)\backslash V(I)$ for some finitely generated
generated ideal $I\subseteq A.$
My attempt
Forward Direction:
Suppose $U$ is quasi-compact. Since $U$ is open in the Zariski topology, $U = Spec(A)\backslash V(I)$ for some ideal $I\subseteq A.$ We claim that $I$ is finitely generated. Since the family of basic open sets $B=\{D(f)|f\in A\}$ forms a basis for the Zariski topology on $Spec(A),$ $U=\bigcup_{f\in S} D(f)$ for some subset $S\subseteq A.$ As $U$ is quasi-compact, this cover has a finite subcover. i.e., $U=\bigcup_{i=1}^{n} D(f_i)$ for some $f_1,…,f_n\in S.$ I think we need to claim that $I$ is generated by $f_1,…,f_n.$ (please correct me here if I'm wrong.) So,
$$U=Spec(A)\backslash V(I)=\bigcup_{i=1}^{n} D(f_i)=\bigcup_{i=1}^{n} Spec(A)\backslash V(f_i)=Spec(A)\backslash \bigcap_{i=1}^{n} V(f_i)$$
$$\Longrightarrow V(I)=\bigcap_{i=1}^{n} V(f_i)$$
$$\Longrightarrow \sqrt{I}=\sqrt{\sum_{i=1}^{n}} (f_i).$$
From here, I am not sure if this implies $I$ is generated by $f_1,…,f_n.$
Backward Direction:
I tried to imitate the proof of "$Spec(A)$ is quasi-compact", which I can prove. I proved this by using a metric space X is compact if and only if for each family of closed sets with the finite intersection property have nonempty intersection. However, I could not prove the backward direction successfully.
Can someone please help? Thank you!
Best Answer
For the forward direction, you've got the right idea, and we've cleaned up the argument a bit in the comments.
For the backward direction, assume $U = \operatorname{Spec}A \setminus V(I)$ for $I = (f_1,\ldots,f_n)$ a finitely generated ideal. We make the following observation: $$U = \operatorname{Spec} A \setminus \bigcap_{i=1}^n V(f_i) = \bigcup_{i=1}^n\left( \operatorname{Spec} A\setminus V(f_i)\right) = \bigcup_{i=1}^n \operatorname{Spec} A_{f_i}$$ where $A_{f_i}$ is the ring $A$ with $f_i$ inverted. (These are mostly the same observations you made above.) Now, we know each $\operatorname{Spec}A_{f_i}$ is quasi-compact (since it is affine). We have then that $U$ is a finite union of quasi-compact open subsets, and so it is itself quasi-compact. (This is a result from topology. If it doesn't sound familiar, you should try to prove it :) )