# Math Help - Prime ideals.

1. ## Prime ideals.

Hi:
Let P, Q be prime ideals in a ring R. Of course P intersection Q is an ideal in R. But I can't prove it is a prime ideal. Any hint would be greatly appreciated. Thanks.

2. You should definitely try checking the definition of prime ideal. Let $P,Q$ be prime ideals of the ring. Suppose $ab\in P\cap Q$. Then $ab\in P$ and $ab\in Q$. Is it possible to conclude that either $a\in P\cap Q$ or $b\in P\cap Q$?

3. Originally Posted by ENRIQUESTEFANINI
Hi:
Let P, Q be prime ideals in a ring R. Of course P intersection Q is an ideal in R. But I can't prove it is a prime ideal. Any hint would be greatly appreciated. Thanks.
Use the second isomorphism theorem (the one with the intersection), and apply the fact that an ideal is prime if and only if the factor ring is an integral domain.

4. Originally Posted by roninpro
You should definitely try checking the definition of prime ideal. Let $P,Q$ be prime ideals of the ring. Suppose $ab\in P\cap Q$. Then $ab\in P$ and $ab\in Q$. Is it possible to conclude that either $a\in P\cap Q$ or $b\in P\cap Q$?
If it were possible then R would be a commutative ring, I think. But my ring does not need to be commutative.

5. if $P,Q$ are prime ideals, then $P \cap Q$ is a prime ideal if and only if either $P \subseteq Q$ or $Q \subseteq P.$ here is why: one side is obvious. now suppose that $P \cap Q$ is a prime ideal and $P \nsubseteq Q$ and $Q \nsubseteq P.$ choose $a \in P \setminus Q$ and $b \in Q \setminus P.$ then, since $P,Q$ are two-sided ideals, we have $aRb \subseteq P \cap Q.$ therefore, since we assumed that $P \cap Q$ is a prime ideal, we must have either $a \in P \cap Q$ or $b \in P \cap Q,$ which is obviously absurd.

6. Originally Posted by NonCommAlg
if $P,Q$ are prime ideals, then $P \cap Q$ is a prime ideal if and only if either $P \subseteq Q$ or $Q \subseteq P.$ here is why: one side is obvious. now suppose that $P \cap Q$ is a prime ideal and $P \nsubseteq Q$ and $Q \nsubseteq P.$ choose $a \in P \setminus Q$ and $b \in Q \setminus P.$ then, since $P,Q$ are two-sided ideals, we have $aRb \subseteq P \cap Q.$ therefore, since we assumed that $P \cap Q$ is a prime ideal, we must have either $a \in P \cap Q$ or $b \in P \cap Q,$ which is obviously absurd.
Thanks NonCommAlg. Hope the site won't be unavailable again.
It's perfectly clear. But then, what did swlabr mean?
I had already found a counterexample: if M(Z) is the ring of all matrices of order 2 over Z, then M(2Z) and M(3Z) are prime ideals in M(Z) but M(2Z) intersection M(3Z) = M(6Z) is not.
Also, even when R is commutative, the statement is false in general. For example, 2Z and 3Z are prime in Z but 6Z is not because 6 is not prime. Thanks again.

P.S.: MHF won't accept any Latex expression. I do not why.