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?
Best Answer
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.$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ._{proof \ from \ sharp}$