You should definitely try checking the definition of prime ideal. Let be prime ideals of the ring. Suppose . Then and . Is it possible to conclude that either or ?
if are prime ideals, then is a prime ideal if and only if either or here is why: one side is obvious. now suppose that is a prime ideal and and choose and then, since are two-sided ideals, we have therefore, since we assumed that is a prime ideal, we must have either or which is obviously absurd.
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.