 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-3/10e/a/ii/solution

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

In a Boolean ring,

(r + 1)^{2} = r + 1

. Using

r^{2} = r

and

1^{2} = 1

gives

r + 2 r + 1 = r + 1,

2 r = 0 for every r \in R .

R

is also a nonzero integral domain, then

r (r - 1) = r^{2} - r = 0

forces every

r

to equal

0

1

. Hence

R ≅ F_{2}

I

is prime in a Boolean ring, then

R / I

is a nonzero Boolean integral domain by part (i), and hence is

F_{2}

. Since the quotient is a field,

I is maximal .

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