Kuratowski closure operator on the set of prime ideals of a ring

Question

Let $R$ be a ring; a two-sided ideal $I$ is called a prime ideal if whenever $J$ and $K$ are two-sided ideals such that $JK = \{ab : a \in J, b\in K \} \subseteq I,$ at least one of them is contained in $I.$

I am attempting to show that the following closure operation on the set of all prime ideals of $R$ satisfies Kuratowski's axioms for a closure to determine a topology.

Let Prime($R$) be the set of all prime ideals of $R.$ Define the closure of a subset $A \subseteq \textrm{Prime(}R \textrm{)},$ $cl(A),$ to be the collection of all prime ideals $I \in \textrm{Prime(}R \textrm{)}$ that contain the ideal that is the intersection of all the elements of $A$.

I have completed the first three of the following four axioms for a closure to determine a topology on a set $X$:

1. $cl(\emptyset) = \emptyset.$

2. For every subset $A \subseteq X,$ we have $A \subseteq cl(A).$

3. For every subset $A \subseteq X,$ we have $cl(cl(A)) = cl(A).$

4. For any to subsets $A, B \subseteq X,$ we have $cl(A) \cup cl(B) = cl(A \cup B).$

Can anyone help me out with some sort of hint as to how I can demonstrate that the closure defined satisfies this fourth property?

Answer 1

Some ideas:

From 2. and the fact that $A \subseteq A \cup B$ we get $\operatorname{cl}(A) \subseteq \operatorname{cl}(A \cup B)$ and similarly $\operatorname{cl}(B) \subseteq \operatorname{cl}(A \cup B)$ and combining those

$$\operatorname{cl}(A) \cup \operatorname{cl}(B) \subseteq \operatorname{cl}(A \cup B)$$

is always true, so the thing to prove is the reverse inclusion.

So let $I$ be a prime ideal that contains $\bigcap\{J: J \in A \cup B\}$ so lies in $\operatorname{cl}(A \cup B)$.

If $I \notin \operatorname{cl}(A)$ this means $\bigcap \{J: J \in A\} \nsubseteq I$ by definition, and also if $I \notin \operatorname{cl}(B)$, we know that $\bigcap\{J : J \in B\} \nsubseteq I$ as well. So assume both of these hold and get a contradiction. So there is some $x \in J_0 \in A$ with $x \notin I$ and some $y \in J_1 \in B$ with $y \notin I$ (just unpacking definitions). Can you find a contradiction from these facts, maybe using that we have prime ideals?