Ykhjuy

Let $A subseteq B$ be a ring extension where $A,B$ are both finitely generated $mathbb C$-domain of the same Krull dimension. Also assume $A$ is regular (i.e. $A_{ mathfrak p}$ is regular local ring for every prime ideal $mathfrak p$ of $A$ ).

If $P$ is a prime ideal of $A$ with finitely many prime ideals of $B$ lying over it , then does there exist $f in A setminus P$ such that every prime ideal of $A_f$ has finitely many prime ideals of $B$ lying over it ?

The comment of @MatthieuRomagny gives a link to some positive answers under additional hypotheses. Nonetheless, the answer is negative without further hypotheses. Here is one counterexample.

Let $k$ be a field, and let $A$ be the polynomial ring $k[s,t,u]$. Let $B$ be the ring $k[x,y,z,w]/langle zw-1 rangle$. Both of these are finitely generated $k$-algebras that are regular, in fact $k$-smooth. Now consider the $k$-algebra homomorphism, $$p:Ato B, p(s) = x(1-x), p(t) = xy, p(u) = z(1-xz).$$ If we invert the element $s$ in $A$ and the image element $x(1-x)$ in $B$, then the induced ring homomorphism is finite and flat of rank $4$. In particular, the localized ring homomorphism is injective. Since the natural map from $A$ to $A[1/s]$ is injective, also $p$ is injective.

Finally, consider the prime ideal $P=langle s,t,urangle$ in $A$. There is precisely one prime lying above this prime, namely $langle x-1,y,z-1,w-1rangle$. However, the prime ideal $Q=langle s,t rangle$ has infinitely many primes lying over it, e.g., $langle x,q(y) rangle$ where $q(y)in k[y]$ is an arbitrary nonzero, noninvertible element that is irreducible. Since $Q$ is contained in $P$, for every $fin Asetminus P$, also $f$ is in $Asetminus Q$. Thus, the ideal $QA_f$ is a prime ideal of $A_f$. So the induced ring homomorphism $A_f to B_{p(f)}$ still admits a prime ideal $QA_f$ that has infinitely many primes lying over it in $B_{p(f)}$.

There is an even easier counterexample if we allow that the set of primes lying over $P=langle s,t,u rangle$ is empty, namely the self-map, $$r:Ato A, r(s) = s, r(t) = st, r(u) = 1-su.$$ This ring homomorphism is even birational.

Please note, there is a partial positive answer in the birational case. Even assuming that $A$ is only normal, the connectedness part of Zariki's Main Theorem implies that if the fiber over $P$ intersects the quasi-finite locus of the morphism inside $text{Spec} B$, then the morphism becomes an isomorphism after inverting some $fin Asetminus P$. Perhaps this is the result that motivated this question. However, the example above shows that this need not be true if the generically finite ring homomorphism is not birational.