Let $I$ and $J$ be prime ideals. Then is $I\cdot J=I\cap J$?

Question

Let $I$ and $J$ be two prime ideals in a commutative ring $R$ with unity. Is it true that $I\cdot J=I\cap J$? I know that it is not true if the ideals under consideration are not prime: consider the ideals $(6)$ and $(8)$ in the ring $\Bbb{Z}$. Their intersection is $(24)$, but their product is $(48)$.

However, I cannot think of a counter-example in the case of prime ideals. I tried proving the assertion in the following way: We know that $I\cdot J\subset I\cap J$. Hence, we shall try and prove the inclusion in the other direction.

Let $a$ be an element in $I\cap J$ that is not present in $I\cdot J$. Then $a^2\in I\cdot J$. How will this then go on to prove that $a\in I\cdot J$?

Answer 1

One has $I\cdot J=I\cap J$ if the $I$ and $J$ are coprime ideals, i.e. if $I+J=R$. In general, one can only say $\;I\cdot J\subseteq I\cap J$.

This is the basis of the Chinese remainder theorem: if $I$ and $J$ are coprime ideals then $I\cdot J=I\cap J$, and $$R/(I\cdot J)=R/I\times R/J.$$

Answer 2

This will be true if $I + J = R$. To see why, note that if $I$ and $J$ are coprime ideals, then $1 = x + y$ for some $x \in I$, $y \in J$. Consider $z \in I \cap J$. Then

$$ z = z \cdot 1 = z \cdot (x+y) = \underbrace{z}_{\text{in $J$}} \cdot \underbrace{x}_{\text{in $I$}} + \underbrace{z}_{\text{in $I$}} \cdot \underbrace{y}_{\text{in $J$}} \in I \cdot J $$

Hence $I \cap J \subseteq I \cdot J$. In general, $I \cdot J \subseteq I \cap J$, from which equality follows.

However, this condition $I + J = R$ is not absolutely necessary. For example, consider $R = \mathbb{Z}/6$ with ideals $I = J = (2)$. Then $I \cdot J = (2) = I \cap J$, even though $I + J \ne R$.