Intersection of two Prime Ideals Must be an Ideal So I know that the "intersection of two prime ideals being a prime ideal" is false. There are some simple examples to disprove that. But does that mean that the intersection of two prime ideals has no possibility of being a plain ideal? Can it result in a different type of ideal? 
 A: For the ring $\mathbb{Z}$, consider the prime ideals $(2)$ and $(3)$.  Their intersection is $(6)$, which is a non-prime ideal.
The intuition is that intersection of ideals corresponds roughly to multiplication of elements.  That is why the intersection of prime ideals is typically not prime.
A: The intersection of two ideals $I$ and $J$ is always an ideal, regardless of whether $I$ and $J$ are prime or not. If $x,y\in I\cap J$, then $x+y\in I$ and $x+y\in J$ since $I$ and $J$ are ideals, hence $x+y\in I\cap J$.
Similarly, if $x\in I\cap J$ then $rx\in I$ and $rx\in J$ since $I$ and $J$ are ideals, hence $rx\in I\cap J$. If $R$ is commutative then this is enough to show that $I\cap J$ is an ideal, and if $R$ isn't commutative then use the same argument for right multiplication. (For non-commutative rings I'm assuming that ideal means two-sided ideal).
A: The intersection of two ideals $I$ and $J$ is the kernel of the homomorphism $R \to R/I \times R/J$ given by $x \mapsto (x \bmod I, x \bmod J)$ and so is an ideal.
If $I$ and $J$ are both prime ideals, then $R/I \times R/J$ is a product of two domains, which is never a domain except in trivial cases. So, it is unlikely that $R/(I \cap J)$ is a domain and $I \cap J$ is a prime ideal.
