$\DeclareMathOperator{\rad}{rad}$I know that if $I, J$ are ideals of a commutative ring with unity $R$, then $\rad(I \cap J) = \rad(I) \cap \rad(J)$. It is easy to see that $\rad(I \cap J) \subseteq \rad(I) \cap \rad(J)$. For the other way around, if $x \in \rad(I) \cap \rad(J)$, then $x^m \in I$ and $x^n \in J$, so $x^{m+n} \in I \cap J$.
In the arbitrary intersection case, if $\{J_i\}_{i \in \alpha}$ is a infinite family of ideals, then again it's easy to see that $\rad(\bigcap_i J_i) \subseteq \bigcap_i \rad(J_i)$, however for the other way around I can no longer do the same trick, as I only know that for every $i \in \alpha$ there is a $k_i$ such that $x^{k_i} \in J_i$. However in general the $\operatorname{lcm}(\{k_i \mid i \in \alpha \})$ may not exist. So I am not sure how to prove it or even if it's true.
I came onto this problem trying to prove Atiyah's and MacDonald's Intro to Commutative Algebra exercise $1.9.$, where it is claimed that if $I$ is a proper ideal then $I = \rad(I) \iff I$ is an intersection of prime ideals.
My plan was doing $\rad(I) = \rad(\bigcap_i J_i) = \bigcap_i \rad(J_i) = \bigcap_i J_i=I$, but I'm stucked in the middle equality.
Best Answer
$\DeclareMathOperator{\rad}{rad}$The reverse containment does not always hold for infinite intersections.
For consider $\mathbb{Z}[x]$. Note that $\rad((x^n)) = (x)$ for all $n > 0$.
So we have $\bigcap\limits_{n = 1}^\infty \rad((x^n)) = (x)$.
But we also have $\bigcap\limits_{n = 1}^\infty (x^n) = (0)$. And $\rad((0)) = (0)$.
To complete your proof, you don't need this result.
We already know that $I \subseteq \rad(I)$ by the definition of radical. Thus, the only direction we need is $\rad(I) \subseteq I$. To prove this, we have $\rad(I) = \rad(\bigcap_i J_i) \subseteq \bigcap_i \rad(J_i) = \bigcap_i J_i = I$. We only need the direction you've already proved.