Why do the Zariski distinguished open subsets form a base?

Let $R$ be commutative ring with unit. I have to prove that the distinguished open sets form a base for the Zariski topology i.e. any non-empty open set is a union of distinguished ones. We have that for any non-empty open set $U,$ $$U = \operatorname{Spec}(R) - \operatorname{V}(S) = \operatorname{Spec}(R) - \bigcap_{f \in S} \operatorname{V}(f) = \bigcup_{f \in S} \operatorname{Spec}(R)_f.$$ How can I prove that the second and third equalities hold?

• The second equality relies on $\operatorname{V}(S) = \bigcap_{f\in S} \operatorname{V}(f)$ which follows by plaing with the definitions and the third relies on $\operatorname{Spec}(R) - \operatorname{V}(f) = \operatorname{Spec}(R_f).$ Have you seen a proof of the last fact? – Ragib Zaman Sep 14 '13 at 15:06
• @RagibZaman No... I don't – ArthurStuart Sep 14 '13 at 15:07
• Sorry, I was mistaken in my edit of your post and wrote $\operatorname{Spec}(R_f)$ where you had originally written $\operatorname{Spec}(R)_f.$ As Martin says, $\operatorname{Spec}(R)_f = \operatorname{Spec}(R) - \operatorname{V}(f)$ by definition and you don't need the fact that I mentioned. – Ragib Zaman Sep 14 '13 at 15:15

$V(S) = \cap_{f \in S} V(f)$ is trivial from the definition of these sets. Now use $D(f) = V(f)^c$ (complement) which holds by definition, as well as the set-theoretic law that complements interchange intersections with unions. You don't need $D(f)=\mathrm{Spec}(R_f)$.