I'm trying to prove that the intersection of two prime ideals, $P_1$ and $P_2$ (in any ring $R$, can be commutative or not) is prime if and only if $P_1 \subseteq P_2 $ or $P_2 \subseteq P_1 $.
Here's what I have so far, though I feel like I'm going down the wrong direction into a rabbit hole:
Let $P_1$ and $P_2$ be two prime ideals. By definition, this means that if two ideals $AB \subseteq P_1$, then either $A \subseteq P_1$ or $B \subseteq P_1$.
So we let $AB \subseteq P_1 $and $ CD \subseteq P_2$ where A, B, C, D are ideals.
Take the intersection $AB \cap CD$, and assume it is also prime.
Again, this means by definition that if $XY \subseteq AB \cap CD$, then $X \subseteq AB \cap CD$ or $Y \subseteq AB \cap CD$ . Assume the prior ($X \subseteq AB \cap CD$ ). By definition of intersection, this means $X \subseteq AB$ and $X \subseteq CD$. And thus $X \subseteq P_1$ and $X \subseteq P_2$, so $X \subseteq P_1 \cap P_2$
Then I'm stuck, and I feel like that got me nowhere. I don't know how to turn it to prove $P_1 \subseteq P_2 $ or $P_2 \subseteq P_1 $. Any help? Am I even going in the right direction or should I try something else?
Best Answer
Note that $P_1P_2 \subset P_1 \cap P_2$. Thus if $P_1 \cap P_2$ is prime then $P_1 \subset P_1 \cap P_2$ or $P_2 \subset P_1 \cap P_2$.