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It is clear that if $p_1, p_2, \dots, p_k$ are all the distinct prime divisors of $n$ then maximal ideals of $\mathbb{Z}/n\mathbb{Z}$ are $p_i \mathbb{Z}/n\mathbb{Z}$ as the quotient field will be $\mathbb{Z}/p_i\mathbb{Z}$ which is field. Now, I am interested to find the intersection of all of these maximal ideals.

$$J(R) = \bigcap p_i \mathbb{Z}/n\mathbb{Z}$$

However I couldn't manage to find. I saw one more claim which is $\bigcap p_i \mathbb{Z}/n\mathbb{Z}=p_1p_2\dots p_n \mathbb Z /n\mathbb Z$. How can I show this argument.

Fuat Ray
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Let $R$ be a commutative ring, and let $I,J\subseteq R$ be ideals.

If $I+J=R$ then $$I\cap J=IJ$$

Proof: It is clear that $IJ\subseteq I\cap J$, to show the reverse inclusion notice that $$\begin{align}I\cap J&\subseteq (I\cap J)I+(I\cap J)J\\&\subseteq IJ\end{align}$$

cansomeonehelpmeout
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  • Don't post answers to duplicate questions, as this is against the site guidelines. Also your answer here has been written elsewhere 10 times on this site. Instead invest the time to use site search (when the OP didn't do it) to find the corresponding duplicate and mark it as such. – Martin Brandenburg Nov 18 '24 at 17:11
  • See here for more details. Meanwhile I found the duplicate, took one minute. – Martin Brandenburg Nov 18 '24 at 17:18