Showing that the fiber of a morphism of schemes over a point is homeomorphic to the preimage

Solution 1:

The primes in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ are the primes in $B_{\mathfrak{p}}$ containing $\phi(\mathfrak{p})B_{\mathfrak{p}}$. And the primes in $B_{\mathfrak{p}}$ are the primes in $B$ not meeting $S:=\phi(A\setminus \mathfrak{p})$. Thus, a prime in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ is of the form $\mathfrak{q}B_{\mathfrak{p}}$ with $\mathfrak{q}\subset B$ not meeting $S$ such that $\phi(\mathfrak{p})B_{\mathfrak{p}}\subset \mathfrak{q}B_{\mathfrak{p}}$. Contracting yields $\phi(\mathfrak{p})\subset \mathfrak{q}$. On the other hand, $$ \mathfrak{q}\subset B\setminus \phi(A\setminus \mathfrak{p}). $$ Applying $\phi^{-1}$ and using $\phi(\mathfrak{p})\subset \mathfrak{q}$ yields $$ \mathfrak{p}\subset \phi^{-1}(\mathfrak{q}) \subset \phi^{-1}(B\setminus \phi(A\setminus \mathfrak{p})) \subset \mathfrak{p} $$ and consequently $\phi^{-1}(\mathfrak{q})=\mathfrak{p}$.