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Past exam of the mathematics course of the University of Cambridge
/
2025
/
iii
/
Paper 101
/
1
/
v
/
b
/
Solution
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2025
iii
Paper 101
1
v
b
Created
2026-09-24
Updated
2026-09-24
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The inclusion
(
I
∩
J
)
e
⊆
I
e
∩
J
e
(1)
always holds: each generator coming from
I
∩
J
lies in both extended
ideals
.
The reverse inclusion can fail. Let
R
=
k
[
x
,
y
]
,
A
=
R
/
(
x
−
y
)
,
I
=
(
x
)
,
J
=
(
y
)
(2)
for
a
field
k
. Under
A
≅
k
[
x
]
,
I
e
=
J
e
=
(
x
)
,
I
e
∩
J
e
=
(
x
)
,
(3)
whereas
I
∩
J
=
(
x
y
)
in
R
, so
(
I
∩
J
)
e
=
(
x
2
)
⊊
(
x
)
.
(4)
Thus statement (
2
) is true in general and statement (
1
) is false in general.
Solved by
gpt-5
.
6
-sol high.
Ancestors
(12)
b
v
1
Paper 101
iii
2025
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
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