This is the Zariski topology on Spec(R) obviously, let R be a commutative ring with unit.
I want to show that if $P$ is a prime ideal, and $J, K$ are radical ideals, such that $$ P \supseteq J \cap K$$ we have that $$P \supseteq J\text{ or } P\supseteq K. $$ Obviously the reverse implication is true, but I am not sure how to show this one.
I tried a proof via contradiction, but if $x \in J\setminus P$ and $y\in K\setminus P$, there is no reason why we should have that $xy \in J \cap K$, so that one could use the definition of prime ideal to derive a contradiction.
Context: Given Zariski-closed $X, Y \subset Spec(R)$, I want to show that $$X \cup Y = Z(I(X) \cap I(Y)), \quad X \cap Y = Z(I(X)+I(Y)). $$ Right now I am only working on the first part, I don't really have a clue for the second part yet (Exercise 4.12.11 in Algebraic Geometry: A Problem-Solving Approach by Garrity et al). We have that for arbitrary subsets $X,Y$ of Spec(R) that $I(X \cup Y)=I(X)\cap I(Y)$. I have shown that a set is Zariski closed if and only if it equals its Zariski closure, i.e. if and only if $W=Z(I(W))$. So basically I need to show that $X \cup Y$ is Zariski closed, because then $X \cup Y = Z(I(X \cup Y))= Z(I(X)\cap I(Y))$. Now we have that for any Zariski closed set, $X=Z(J)$ where $J$ is a prime ideal (I proved this from a weaker definition). So $X=Z(J), Y=Z(K)$, so $$X \cup Y = \{ P\in Spec(R): P\supseteq J\ \lor\ P\supseteq K\}.$$ Now if $X=Z(J)$, we have that $I(X)=I(Z(J))=J$, because $I(Z(J))=J$ for any radical ideal, and for an arbitrary ideal $L$, $I(Z(L))=Rad(L)$. Therefore when $X,Y \subseteq Spec(R)$ are Zariski-closed: $$Z(I(X \cup Y))=Z(I(X) \cap I(Y))=Z(J \cap K)=\{P \in Spec(R): P \supseteq J\cap K \}.$$ So showing that $X \cup Y$ is Zariski closed reduces to the problem I've given above.
Definitions: Let $S \subseteq R$. Then $Z(S)=\{P \in Spec(R): P \supseteq S \}\subseteq Spec(R)$. A set $X \subseteq Spec(R)$ is said to be Zariski-closed if it is of the form $X=Z(S)$ for some $S \subseteq R$. We have that $Z(S)=Z(\langle S \rangle) = Z(Rad(\langle S \rangle))$. Thus equivalently, a set $X \subseteq Spec(R)$ is Zariski-closed if and only if it is of the form $X=Z(J)$ for $J$ a radical ideal.
Let $X \subseteq Spec(R)$ be arbitrary, i.e. not necessarily Zariski-closed. Then $I(X)=\bigcap_{P \in X} P \subseteq R$, the intersection of all prime ideals $P$ which are elements of the set $X$.
The Zariski closure of a set $X \subseteq Spec(R)$ is defined to be $Z(I(X))$. $X$ is Zariski if and only if it equals its Zariski closure. Let $X$ be Zariski closed. Then $X=Z(J)$ for a radical ideal $J$. Thus $Z(I(X))=Z(I(Z(J)))=Z(J)=X$ because $I(Z(J))=J$ for any radical ideal $J$. Let $X$ equal its Zariski closure. Then $X=Z(I(X))$, so $X=Z(J)$ for the radical ideal $I(X)$.
I just realized how silly this question was -- I should have thought more carefully about it earlier.
Anyway, we don't even need that $J,K$ be radical ideals -- it suffices if they are arbitrary ideals.
Assume that $P$ is a prime ideal, and $J,K$ are ideals.
We want to show that $$P \supseteq J \cap K \implies P \supseteq J \quad or \quad P \supseteq K. $$ This is equal via contraposition to: $$P \not\supseteq J \quad and \quad P \not\supseteq K \implies P \not\supseteq J \cap K. $$ I will now show the contrapositive statement as follows.
Let $x \in J, x \not\in P$, and $y \in K, y \not\in P$. This is possible because of our assumptions.
Now because $J$ is an ideal, and $x\in J$ and $y \in K \subseteq R$, we have that $xy \in J$.
Likewise, because $K$ is an ideal, $y \in K$, $x \in J \subseteq R$, we have that $xy \in K$.
So then clearly $xy \in J \cap K$ by definition of intersection.
Assume by means of contradiction that $xy \in P$, then $ x \in P$ or $ y \in P$ because $Y$ is a prime ideal. This is a contradiction of course because we are assuming that $x \not\in P, y \not\in P$. Therefore $xy \not\in P$.
Since there exists an element $xy \in J \cap K, xy \not\in P$, we have that $P \not\supseteq J \cap K$. $\square$
In case anyone is curious, I also have the proof for $$X \cap Y = Z(I(X)+I(Y)),$$ for $X,Y$ Zariski-closed. Because they are Zariski-closed, $X=Z(J), Y=Z(K)$ for radical ideals $J,K$, so the statement is equivalent to $$Z(J) \cap Z(K) = Z(J +K), $$ since $I(Z(J))=J$ for any radical ideal $J$. Applying definitions, this is equivalent to: $$\{P \in Spec(R): P\supseteq J \quad and \quad P \supseteq K \} = \{ P \in Spec(R): P\supseteq J + K \}. $$ Now because $0 \in J, 0 \in K$, we clearly have $J \subseteq J+K, K \subseteq J+K$. So if $P \supseteq J+K$, we have that both $P \supseteq J$ and $P \supseteq K$, proving one direction. For the other direction, let $P \supseteq J$ and $P \supseteq K$. Let $z \in J+K$, then $z=x+y$, for $x \in J \subseteq P$, $y \in K \subseteq P$. Because $P$ is an ideal, and $x,y \in P$, $z=x+y \in P$. So $J+K \subseteq P$. This completes the other direction and thus the proof. $\square$