# Product of two maximal ideals in a commutative ring with unity is equal to their intersection. [closed]

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, AWertheimDec 25 '13 at 18:53

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Grigory M, Jonathan, Davide Giraudo, AWertheim
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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
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. $k[x,y,z]$.
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