Let $R$ be a commutative unital ring. Let $\mathrm{Spec}(R) = \{ \mathfrak p \subset R \mid \mathfrak p \text{ a prime ideal of } R \}$. We define a set $C$ to be closed in this space if and only if there is an ideal $I$ such that $C(I) = \{\mathfrak p \mid I \subset \mathfrak p, \mathfrak p \text{ a prime ideal of } R \}$.
Now I'd like to show that these sets form a topology on $\mathrm{Spec}(R)$:
(i) For the zero ideal we get $C(0) = \mathrm{Spec}(R)$ and for $R$ we get $C(R) = \varnothing$.
(ii) Arbitrary intersections are closed again: $\bigcap_\alpha C(I_\alpha) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, I_\alpha \subset \mathfrak p \text{ for all } \alpha \} = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \sum_\alpha I_\alpha \subset \mathfrak p \} = C(\sum_\alpha I_\alpha)$.
(iii) We want to show that finite unions are closed again. It's enough to show it for two ideals $I,J$: $C(I) \cup C(J) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \text{ such that either } I \subset \mathfrak p \text{ or } J \subset \mathfrak p \}$
Now I'm stuck. How do I express the "either or" in terms of operations on ideals? Thanks for your help.
Best Answer
Extension of my hint above.
If $I\not\subset \mathfrak p$ and $J\not\subset\mathfrak p$, is it possible that $IJ\subset\mathfrak p$?
There's actually an interesting parallel here. While you can take arbitrary sums of ideals, you can only take the product of a finite number of ideals.