Zariski topology on prime $\mathrm{Spec}$ of a ring $R$ 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.
 A: 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.
A: Suppose $\mathfrak p \in C(IJ)$ then $IJ \subset \mathfrak p$. Since $\mathfrak p$ is prime, $I \subset \mathfrak p$ or $J \subset \mathfrak p$. Thus $\mathfrak p \in C(I) \cup C(J)$.
Suppose $\mathfrak{p} \in C(I) \cup C(J)$ then $I \subset \mathfrak p$ or $J \subset \mathfrak p$. Since ideals are closed under products $IJ \subset \mathfrak p$. Hence $\mathfrak p \in C(IJ)$.
A: Since I'm stuck on which one of the two helpful answers to accept, I'm posting an answer myself:
Claim: $C(I) \cup C(J)$ is closed again
Proof: $C(I) \cup C(J) = \{\mathfrak p \mid \mathfrak p \text{ a prime ideal of } R, \text{ such that either }  I \subset  \mathfrak p  \text{ or } J  \subset \mathfrak p \}$. 
We claim that: $\text{ either }  I \subset  \mathfrak p  \text{ or } J  \subset \mathfrak p $ if and only if $IJ \subset \mathfrak p$.
$\implies$ Is clear since $IJ \subset I$ and $IJ \subset J$.
$\Longleftarrow$ Let $IJ \subset \mathfrak p$, that is, $\{ \sum^n i_k j_k \mid i_k \in I, j_k \in J \} \subset \mathfrak p $, in particular, $ij \in \mathfrak p$ for all $i \in I$ and all $j \in J$. Now assume that $\text{ neither }  I \subset  \mathfrak p  \text{ nor } J  \subset \mathfrak p $. Then there are $i$ and $j$ such that $i \notin \mathfrak p$ and $j \notin \mathfrak p$. But we have $ij \in \mathfrak p$ and since $\mathfrak p$ is prime, either $i \in \mathfrak p$ or $j \in \mathfrak p$, which is a contradiction. Hence $\text{ either }  I \subset  \mathfrak p  \text{ or } J  \subset \mathfrak p $.
Hence  $C(I) \cup C(J)$ is closed since $C(I) \cup C(J) = C(IJ)$.
