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If $R$ is a commutative ring with identity and $M_1, \dots, M_r$ are distinct maximal ideals in $R$, then show that $M_1\cap M_2 \cap \cdots \cap M_r = M_1M_2\cdots M_r$. Is this true if "maximal" is replaced by "prime"?

$M_1M_2\cdots M_r \subset M_1\cap M_2 \cap \cdots \cap M_r$ is trivial. Can you help me?

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    $\begingroup$ In a commutative ring $R$ with identity it is always true that, for any two ideal $I$ and $J$, $IJ \subseteq I \cap J.$ Also if $I + J = R,$ then $IJ = I \cap J.$ $\endgroup$ – Krish Dec 13 '14 at 8:40
  • $\begingroup$ I explain * Krish*'s comment as an answer $\endgroup$ – user 1 Dec 13 '14 at 8:50
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Let $I$ and $J$ be comaximal ($I+J=R$). then $IJ = I \cap J$, because:
$IJ \subseteq I \cap J.$ and $$I \cap J=(I \cap J)R=(I \cap J)(I+J)=(I \cap J)I+(I \cap J)J$$
but $ (I \cap J)I\subseteq JI $ and $ (I \cap J)J\subseteq IJ $. It follows that $ I \cap J\subseteq IJ.$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ._{proof \ from \ sharp}$

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