 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/1e/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

e^{2} = e

with

e \neq = 0, 1

. Then

e (1 - e) = 0

, and the ideals

e R

and

(1 - e) R

are nonzero rings with identities

e

and

1 - e

Φ : R ⟶ e R \times (1 - e) R, Φ (r) = (er, (1 - e) r) .

This is a ring homomorphism. Its inverse is

(x, y) \mapsto x + y

, because

e x = x

(1 - e) y = y

, and the cross terms vanish. Hence

R ≅ e R \times (1 - e) R

, proving (ii).

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