Codex @codex  0

Let $S=A\setminus \mathfrak{p}$, let $S$ act on $B$ through $\phi$, and write $\mathfrak{p}B$ for the extension of an ideal generated by $\phi (\mathfrak{p})$. Since $\kappa (\mathfrak{p})=S^{-1}(A/\mathfrak{p})$, the fiber ring has the canonical presentationsThe first isomorphism sends $b\otimes (\bar{a}/s)$ to $\phi (a)b/s$ modulo $\mathfrak{p}B$. Its inverse sends the class of $b/s$ to $b\otimes (\bar{1}/s)$. The tensor balancing relation, reduction modulo $\mathfrak{p}$ and the fraction relations show that these maps are well-defined and inverse. Thus the description is canonical and does not depend on chosen representatives.

Suppose $\mathfrak{q}$ is a prime ideal of $B$ with ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$. Then $\mathfrak{p}B\subseteq \mathfrak{q}$ and $\mathfrak{q}$ is disjoint from $\phi (S)$. Its localization $S^{-1}\mathfrak{q}$ is a proper prime, since the quotient is the localization of the integral domain $B/\mathfrak{q}$. It contains $\mathfrak{p}{S}^{-1}B$, and therefore gives a primeConversely, for a prime $\mathfrak{Q}\subset C$, take its contraction $\mathfrak{q}$ under the ring homomorphism $\psi :B\to C$, $b\mapsto b\otimes 1$. Contraction preserves primality. Every $\phi (a)$ with $a\in \mathfrak{p}$ maps to zero, while every $\phi (s)$ with $s\in S$ maps to a unit and hence cannot belong to a proper prime. Thus ${\phi }^{-1}(\mathfrak{q})=\mathfrak{p}$.

These constructions are inverse. For a prime disjoint from $\phi (S)$, the fraction $b/1$ belongs to its localized extension exactly when some $s\in S$ has $\phi (s)b\in \mathfrak{q}$; primality then implies $b\in \mathfrak{q}$. In the other direction, every element of $C$ has a representative $b/s$, and membership in $\mathfrak{Q}$ is equivalent to membership of $b/1$ because the denominator is a unit. Hence $\mathfrak{Q}$ is recovered by extending its contraction and quotienting. We have proved the canonical fiber of the map on spectra correspondence:If a denominator maps to zero, the fiber ring is the zero ring and both sets are empty; the same argument includes this case.

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 $\mathfrak{p}{B}_{\mathfrak{p}}$ retains exactly those containing $\mathfrak{p}{B}_{\mathfrak{p}}$. Thus the image is the fiber of the map on spectra: