Does exponentiation of ideals factor through intersection? I am trying to prove that $(I \cap J)^2 = I^2 \cap J^2$. So far I have established the forward direction:
$(\subseteq)$ Let $a \in (I \cap J)^2$. Then $ a = a_1a_2 + a_3a_4 + a_5a_6 + \dots + a_{n-1}a_n$, where $a_i \in I \cap J.$ Then  $a_i \in I, a_i \in J \implies a \in I^2, a \in J^2 \implies a \in I^2 \cap J^2$, as desired.
I am struggling with the reverse direction. I have evidence in the case of a few checked ideals in $\mathbb{Z}$, but I am not sure this holds in general (maybe not outside of a PID?). Any help would be appreciated.

I am not sure that this holds in general, but I know that it is true for Dedekind Domains (so therefore also true for a PID) due to the unique factorisation of prime ideals. Here is a proof of that:
Write $I=\prod\limits_{i}\mathfrak{p_{i}}^{e_{i}}$ and $J=\prod\limits_{i}\mathfrak{p_{i}}^{f_{i}}$ where the $\mathfrak{p}_{i}$ are prime ideals and $e_{i},f_{i}\in\mathbb{Z}_{\geq 0}$. Then
$$I^{2}=\prod\limits_{i}\mathfrak{p}_{i}^{2e_{i}}\hspace{5mm}\text{and}\hspace{5mm}J^{2}=\prod\limits_{i}\mathfrak{p}_{i}^{2f_{i}}.$$
Recall that $I\cap J=\prod\limits_{i}\mathfrak{p_{i}}^{\max(e_{i},f_{i})}.$ Therefore we have
$$(I\cap J)^{2}=\prod\limits_{i}\mathfrak{p}_{i}^{2\max(e_{i},f_{i})}=\prod\limits_{i}\mathfrak{p}_{i}^{\max(2e_{i},2f_{i})}=I^{2}\cap J^{2},$$
as desired.
