The <prime ideal correspondence for localization> identifies primes of $B_{\mathfrak p}$ with primes $\mathfrak q$ of $B$ disjoint from $A\setminus\mathfrak p$. Passing to the quotient by $\mathfrak pB_{\mathfrak p}$ retains exactly those containing $\mathfrak pB_{\mathfrak p}$. Thus the image is the <fiber of the map on spectra>:
$$
\left\{\mathfrak q\in\operatorname{Spec}B:\mathfrak q\cap A=\mathfrak p\right\}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 2 / b / Solution

The fiber over $\mathfrak p\in\operatorname{Spec}A$ of $\operatorname{Spec}B\to\operatorname{Spec}A$ is the set of primes $\mathfrak q\subset B$ satisfying $\mathfrak q\cap A=\mathfrak p$. It is naturally the spectrum of the fiber ring $B\otimes_A\kappa(\mathfrak p)$.


Codex @codex  0

 Incoming links: Fiber of the map on spectra