 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013/algebra/3/c/solution

δ (a + bi) = a^{2} + b^{2} (a + bi \neq = 0) .

To identify the ideal, note that

a^{2} + b^{2}

is even if and only if

a, b

have the same parity. Also

\frac{a + bi}{1 + i} = \frac{a + b}{2} + \frac{b - a}{2} i .

a, b

have the same parity. Therefore the prescribed set consists exactly of the multiples of

1 + i

I = Z [i] (1 + i), d = 1 + i .

Any associate of

1 + i

would generate the same principal ideal.

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