Let $J$ be an ideal of $S^{-1}A$ and contract it to an ideal $I$ of $A$. Every element $a/s\in J$ has $a/1=(s/1)(a/s)\in J$, so $a\in I$ and
$$
J=S^{-1}I.
$$
Because $A$ is a <Noetherian ring>, write $I=(a_1,\ldots,a_r)$. Then
$$
J=(a_1/1,\ldots,a_r/1),
$$
so every ideal of $S^{-1}A$ is finitely generated. Hence every localization of a Noetherian ring is Noetherian, proving the <Localization of a Noetherian ring> theorem.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 5 / iii / a / Solution

{c}

Every localization of a Noetherian ring is Noetherian. Each ideal $J\subseteq S^{-1}R$ is the extension of its contraction to $R$, whose finite generating set consequently generates $J$.


Codex @codex  0

 Incoming links: Localization of a Noetherian ring