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Let $R$ be a commutative ring with unity and let $m_1$ and $m_2$ be two different maximal ideals in $R$. Prove: $m_1 m_2 = m_1 \cap m_2$. Find an example of two different prime ideals $P_1$ and $P_2$ such that: $P_1 P_2 \neq P_1 \cap P_2$.

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closed as off-topic by YACP, Grigory M, Jonathan, Davide Giraudo, AWertheim Dec 25 '13 at 18:53

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Welcome! Please share your thoughts on the problem, and explain what you've tried. For example: Do you know the definitions of maximal / prime ideals, and have you tried to apply them here? Please edit your question to include your efforts. –  T. Bongers Dec 25 '13 at 17:11
    
@T.Bongers fortunately An answer was given to this question, so I don't edit this question any more.But I'll do as you said in next questions that I'll ask.Thanks. –  pardis Dec 25 '13 at 17:49
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1 Answer 1

up vote 4 down vote accepted

Since $m_1,m_2$ are maximal, we have $m_1+m_2=R$. It is a standard result that for ideals $I,J$ such that $I+J=R, IJ=I\cap J$ since $IJ\subseteq I\cap J$ trivially and $$I\cap J=(I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+J(I\cap J)\subseteq IJ+IJ=IJ$$ thus $I\cap J=IJ$.

To find the prime ideal counterexamples, you want to find primes such that $P_1+P_2\ne R$. So you are going to want to look at primes of low height in a high-dimensional ring, e.g. $k[x,y,z]$.

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So for counter example if i take $R=6Z$ and $I=2Z$ and $J=3Z$ where clearly $I+J\neqR$ will this counterexample work?? thanks for responding! –  d13 Mar 3 at 15:37
    
I have no damn clue how to get the counter example. it seems like a mission impossible for me:(( –  d13 Mar 3 at 15:49
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