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Math Help - Prime ideals.

  1. #1
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    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.
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  2. #2
    Senior Member roninpro's Avatar
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    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?
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    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.
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  4. #4
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    Quote Originally Posted by roninpro View Post
    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.
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  5. #5
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    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.
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    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.
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