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.
You should definitely try checking the definition of prime ideal. Let $\displaystyle P,Q$ be prime ideals of the ring. Suppose $\displaystyle ab\in P\cap Q$. Then $\displaystyle ab\in P$ and $\displaystyle ab\in Q$. Is it possible to conclude that either $\displaystyle a\in P\cap Q$ or $\displaystyle b\in P\cap Q$?
if $\displaystyle P,Q$ are prime ideals, then $\displaystyle P \cap Q$ is a prime ideal if and only if either $\displaystyle P \subseteq Q$ or $\displaystyle Q \subseteq P.$ here is why: one side is obvious. now suppose that $\displaystyle P \cap Q$ is a prime ideal and $\displaystyle P \nsubseteq Q$ and $\displaystyle Q \nsubseteq P.$ choose $\displaystyle a \in P \setminus Q$ and $\displaystyle b \in Q \setminus P.$ then, since $\displaystyle P,Q$ are two-sided ideals, we have $\displaystyle aRb \subseteq P \cap Q.$ therefore, since we assumed that $\displaystyle P \cap Q$ is a prime ideal, we must have either $\displaystyle a \in P \cap Q$ or $\displaystyle b \in P \cap Q,$ 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.