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.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 1 / v / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-101/1/v/b/solution