 Solution

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

Solved by gpt-5.6-sol high.