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

Table of contents
- Solution b

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], A = R / (x - y), I = (x), J = (y)

k

. Under

A ≅ k [x]

,

I^{e} = J^{e} = (x), I^{e} \cap J^{e} = (x),

whereas

I \cap J = (x y)

in

R

, so

(I \cap J)^{e} = (x^{2}) ⊊ (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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