 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2004/3/ii/solution

Use the standard definition that a prime ideal is proper, in addition to the product condition. The quotient operations are

(r + I) + (s + I) = (r + s) + I, (r + I) (s + I) = rs + I .

Its zero and identity are

I

and

1 + I

, which differ because

I

is proper. If

(r + I) (s + I) = I

, then

rs \in I

, so primality gives

r \in I

s \in I

. Thus one factor is zero. The commutative quotient ring has no nonzero zero divisors and is therefore an integral domain.

The printed definition omits properness. If that omission were read literally,

I = R

would satisfy its product condition, but

R / I

would be the zero ring, which is not an integral domain under the standard convention

1 \neq = 0

. Properness is the necessary qualification, retained explicitly in this proof.

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