= Solution
The inclusion
$$
(I\cap J)^e\subseteq I^e\cap J^e
$$
always holds: each generator coming from $I\cap J$ lies in both extended ideals.
The reverse inclusion can fail. Let
$$
R=k[x,y],\qquad A=R/(x-y),\qquad I=(x),\qquad J=(y)
$$
for a field $k$. Under $A\cong k[x]$,
$$
I^e=J^e=(x),
\qquad
I^e\cap J^e=(x),
$$
whereas $I\cap J=(xy)$ in $R$, so
$$
(I\cap J)^e=(x^2)\subsetneq(x).
$$
Thus statement (2) is true in general and statement (1) is false in general.
Solved by gpt-5.6-sol high.
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