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

A discrete valuation on a field $K$ is a surjective group homomorphismsatisfyingwhenever $x+y\ne 0$. Its discrete valuation ring is

One standard characterization defines a Dedekind domain as a Noetherian integrally closed domain of Krull dimension one. Equivalently, all its localizations at nonzero prime ideals are discrete valuation rings.

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Choose $0\ne a\in \mathfrak{m}$. Since the one-dimensional local domain has no nonzero prime ideal other than $\mathfrak{m}$, one has $\sqrt{(a)}=\mathfrak{m}$. Finite generation of $\mathfrak{m}$ therefore gives some $n\ge 1$ with ${\mathfrak{m}}^n\subseteq (a)$. Choose $n$ minimal andIn the fraction field of $A$, put $y=x/a$. Then $y\notin A$ but

If $y\mathfrak{m}\subseteq \mathfrak{m}$, multiplication by $y$ would preserve the nonzero finitely generated faithful $A$-module $\mathfrak{m}$. The determinant trick would make $y$ integral over $A$, contradicting that $A$ is integrally closed and $y\notin A$. Hence some $t\in \mathfrak{m}$ satisfies $yt\notin \mathfrak{m}$. Since $yt\in A$ and $A$ is local, $yt$ is a unit.

For any $z\in \mathfrak{m}$, one has $yz\in A$, and thereforeThus $\mathfrak{m}\subseteq (t)$, while the reverse inclusion follows from $t\in \mathfrak{m}$. ConsequentlyThis proves the principal maximal ideal in a one-dimensional normal local domain result.

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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 $a\in I$ andBecause $A$ is a Noetherian ring, write $I=(a_1,\dots ,a_r)$. Thenso 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.

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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 $A$, soIf 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.