Past exam of the algebraic number theory course of the University of Oxford / 2023 / 2 / b / ii

Solution

For every

i \geq 0

, the inclusion

P^{i + 1} \subseteq P^{i}

is strict: equality, followed by multiplication by the inverse of

P^{i}

, would give

P = R

. Hence one can choose

π \in P^{i} ∖ P^{i + 1} .

Put

J = (π) + P^{i + 1}

. Then

P^{i + 1} ⊊ J \subseteq P^{i} .

The classification just proved, applied modulo

P^{i + 1}

, leaves no ideal strictly between consecutive powers. It follows that

P^{i} = (π) + P^{i + 1} .

For

i = 0

, the same argument says

(π) + P = R

, using

π \in / P

and maximality.

The extension to every

r \geq 1

P^{i} = (π) + P^{i + r}

. Multiplication of the initial equality by

P^{r}

gives

P^{i + r} = π P^{r} + P^{i + r + 1} .

Substitution yields

P^{i} = (π) + π P^{r} + P^{i + r + 1} = (π) + P^{i + r + 1} .

Thus

P^{i} = (π) + P^{i + r} for every r \geq 1 .

Taking

r = n - i

for

0 \leq i < n

gives

P^{i} / P^{n} = (π + P^{n}) .

R / P^{n}

. Its zero ideal is generated by zero, so every ideal of this quotient ring is a principal ideal.

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