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2. Base For the Zariski Topology of Spectrum of a ring Let X be a topological space. A family of open sets B is a base for the topology on X if every open set of X is a union of elements of B. Proposition 2.1. The family of basic open sets B = {D(f ) : f ∈ A} forms a base for the Zariski topology on Spec A. Proof. Let U be an open subset of Spec A. Then there exists an ideal I of A so that U = Spec A \ V (I). For each x ∈ U, we know x 6∈ V (I). In other words, I is not contained in x. Then there exists fx ∈ I \ x. Hence fx 6∈ x and thus x 6∈ V (fx ). It is equivalent to say that x ∈ D(fx ). On the S other hand, (fx ) ⊂ I and hence V (I) ⊂SV (fx ). We see that D(fx ) ⊂ U. Notice that U ⊂ x∈U D(fx ) ⊂ U. We conclude that U = x∈U D(fx ). This shows that every open set of X is a union of elements of B. By definition, B forms a basis for the Zariski topology on Spec A. Definition 2.1. A topological space is quasi compact if and only if every open covering has a finite sub cover. Proposition 2.2. The spectrum Spec A of a ring is quasi-compact. Proof. Let {Uλ } be an open covering of X = Spec A. Since {D(f ) : f ∈ A} forms a base for the Zariski topology of X, we can in fact assume PD(fλ ) for some T that Uλ is of the form S V (f ) = ∅, i.e. V ( D(f ) = X. This implies that fλ ∈ A. Then λ λ λ (fλ )) = V (1). λ λ pP pP P n ∈ (f ) = (1). Hence 1 ∈ (f ). By definition, 1 (f ) for some n. But Hence λ λ λ λ λ λ P n aλn ∈ A and fλ1 , · · · , fλn so 1 =P 1. We see that λ (fλ ) = (1). We can choose aλ1 , · · · , P that S ni=1 aλi fλi = 1. This implies that the unit ideal (1) = ni=1 (fλi ). This implies that X = ni=1 D(fλi ). Hence Spec A is quasi-compact. Corollary 2.1. Suppose A is a ring and f ∈ A. Then D(f ) is also quasi-compact. Proof. We will prove that D(f ) is homeomorphic to Spec Af , where Af is the localization Sf−1 A, where Sf = {f n : n ≥ 0}. Hence D(f ) is the spectrum of a ring. By the previous proposition, D(f ) is quasi-compact. 1