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For an integral extension $A\subseteq B$ and $\mathfrak{p}\in \mathrm{Spec}A$, localization gives a bijectionIt combines the prime ideal correspondence for localization with the contraction of a maximal ideal under an integral extension.

Put $S=A\setminus \mathfrak{p}$ and definewhere $A$ acts on $B$ through the given inclusion. By the prime ideal correspondence for localization, primes of $B_{\mathfrak{p}}$ correspond to primes $\mathfrak{q}$ of $B$ satisfying $\mathfrak{q}\cap S=\varnothing$, equivalently $\mathfrak{q}\cap A\subseteq \mathfrak{p}$.

The localized extension $A_{\mathfrak{p}}\subseteq B_{\mathfrak{p}}$ remains integral. If $\mathfrak{q}\cap A=\mathfrak{p}$, thenis an integral domain integral over the field $A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$. An integral domain integral over a field is a field, so $S^{-1}\mathfrak{q}$ is maximal.

Conversely, if $\mathfrak{n}$ is maximal in $B_{\mathfrak{p}}$, the contraction of a maximal ideal under an integral extension is maximal in the local ring $A_{\mathfrak{p}}$, hence equals $\mathfrak{p}{A}_{\mathfrak{p}}$. Contracting once more to $A$ gives $\mathfrak{p}$. Extension and contraction are inverse under localization, proving the required fiber primes of an integral extension bijection.

Let $\mathfrak{m}$ be maximal in $R/I$ and let $\mathfrak{n}\subseteq R$ be its contraction. The ideal $\mathfrak{n}$ is maximal and contains $I$. It is disjoint from $S=1+I$: if $1+i\in \mathfrak{n}$ with $i\in I\subseteq \mathfrak{n}$, then $1\in \mathfrak{n}$. The prime ideal correspondence for localization therefore defines the proper ideal $S^{-1}\mathfrak{n}$, andis a field. Thus $f(\mathfrak{m})=S^{-1}\mathfrak{n}$ is maximal.

Conversely, let $\mathfrak{q}$ be the contraction to $R$ of a maximal ideal of $S^{-1}R$. Then $\mathfrak{q}$ is maximal among ideals disjoint from $S$. Since $\mathfrak{q}+I$ is disjoint from $S$—otherwise an equation $q+i=1+j$ would put $1+(j-i)\in \mathfrak{q}\cap S$—maximality gives $I\subseteq \mathfrak{q}$. If a proper ideal strictly contained $\mathfrak{q}$ in a maximal ideal $\mathfrak{n}$ of $R$, then $\mathfrak{n}$ would also contain $I$ and hence remain disjoint from $S$, a contradiction. Thus $\mathfrak{q}$ is maximal and contains $I$.

The two standard extension-contraction bijections, first for $R\to R/I$ and then for $R\to S^{-1}R$, now give inverse maps

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.