Because $0\notin S$, the localization $S^{-1}A$ remains an integral domain. Part (a) makes it Noetherian. Integral closedness is preserved by localization: if $x$ in the common fraction field is integral over $S^{-1}A$, clearing the finitely many denominators in a monic equation shows that $sx$ is integral over $A$ for some $s\in S$, whence $sx\in A$ and $x\in S^{-1}A$.

The <prime ideal correspondence for localization> shows that every chain of primes in $S^{-1}A$ comes from a chain in $A$, so
$$
\dim S^{-1}A\leq1.
$$
If its dimension is one, it is a Noetherian integrally closed domain of dimension one and hence a <Dedekind domain>. If its dimension is zero, its zero ideal is maximal, so the domain is a field. This proves the <Localization of a Dedekind domain> alternative.

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 / b / Solution

{c}

If $R$ is a Dedekind domain and $0\notin S$, then $S^{-1}R$ is an integrally closed Noetherian domain of dimension at most one. It is therefore either a Dedekind domain or, when its dimension is zero, a field.


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