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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 10E / a / ii / Solution

In a Boolean ring,

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

. Using

r^{2} = r

and

1^{2} = 1

gives

r + 2 r + 1 = r + 1,

so

2 r = 0 for every r \in R .

If

R

is also a nonzero integral domain, then

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

forces every

r

to equal

0

or

1

. Hence

R ≅ F_{2}

.

If

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 .

This is the prime ideals of a Boolean ring are maximal property.

Solved by gpt-5.6-sol high.

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