Confusion about local properties + trying to show a finite morphism is quasi-finite Let $A$ be a commutative ring and let $B$ be a finite $A$-algebra. Let $f:A \to B$ be a ring homomorphism. I want to show that whenever $\mathfrak{p} \subseteq A$ is a prime ideal, then there are finitely many prime ideals $\mathfrak{q} \subseteq B$ such that $f^{-1}(\mathfrak q) = \mathfrak p$. Now I am a bit confused because I have been told that this is a local property so I can assume that $A$ is a local ring. However I am not sure why it is a local property or even what the corresponding statement is for each localisation of A at a prime ideal.
would really appreciate any advice on this, thank you.
 A: You can suppose that $A$ is not only a local ring, but actually a field. Let $f \colon A \to B$ be a finite ring homomorphism, let $\mathfrak{p}$ be a prime ideal of $A$, let $f^* \colon \mathrm{Spec} B \to \mathrm{Spec} A$ the associated map, i.e. $f^*(\mathfrak{q}) = f^{-1}(\mathfrak{q})$ for every prime ideal $\mathfrak{q}$ of $B$.
(1) You can suppose that $A$ is local and $\mathfrak{p}$ is the unique maximal ideal of $A$. Consider $S = A \setminus \mathfrak{p}$. $S^{-1}f \colon A_\mathfrak{p} \to S^{-1}B$ is a finite homomorphism of rings. You should check that $(f^*)^{-1}(\mathfrak{p})$ is in bijection with $((S^{-1}f)^*)^{-1}(\mathfrak{p}A_{\mathfrak{p}})$. Hence you can replace $A$ and $\mathfrak p$ with $A_\mathfrak{p}$ and $\mathfrak{p} A_\mathfrak{p}$.
(2) You can suppose that $A$ is a field and $\mathfrak p = 0$. $f$ induces a finite homomorphism $\bar{f} \colon A/\mathfrak{p} \to B/\mathfrak{p}B$. Check that the fiber of $f^*$ over $\mathfrak{p}$ is in bijection with the fiber of $\bar{f}^*$ over $\overline{\mathfrak p} = 0$. You can replace $A$ with the field $A/\mathfrak{p}$ and $B$ with $B/\mathfrak{p}B$.
Steps (1) and (2) can be unified in a unique step: replace $A$ with $k(\mathfrak{p})$, and $B$ with $B \otimes_A k(\mathfrak{p})$, where $k(\mathfrak{p})$ is the residue field of $\mathfrak{p}$. This is described in Proposition 3.1.16 of Liu's Algebraic geometry and arithmetic curves.
The proof continues in my answer here.
A: This $f$ had better be the homomorphism giving $B$ its $A$-algebra structure. If it is allowed to be arbitrary then I don't think the statement is true: let $k$ be an infinite field, and define $f\colon k[x] \to k[x]$ by $p(x) \mapsto p(0)$. Then $f^{-1}((x - a)) = (x)$ for all $a \in k$. [Geometrically, this is the map from the affine line to itself sending all closed points to the origin.]
To attack the problem, localize at $\mathfrak p$ and get a finite homomorphism $f_\mathfrak p\colon A_\mathfrak p \to B_\mathfrak p$. This fits into a commutative diagram of rings
\begin{array}{ccc}
A_\mathfrak p & \stackrel{f_\mathfrak p}\rightarrow & B_\mathfrak p \\
\uparrow & & \uparrow \\
A & \stackrel f\to & B
\end{array}
and there is a corresponding commutative diagram of spectra. I think the key is to draw that diagram and figure out what you know about the four maps involved.
The ring $B_\mathfrak p$ is canonically isomorphic to $T^{-1}B$, where $T$ is the multiplicative set $f(A - \mathfrak p)$. So the primes of $B_\mathfrak p$ correspond to the primes $\mathfrak q$ of $B$ such that $f^{-1}(\mathfrak q) \subset \mathfrak p$. [See Proposition 6.1 of Milne] And after staring at your diagram for a while you'll discover that $f_\mathfrak p^{-1}(\mathfrak qB_\mathfrak p) = \mathfrak pA_\mathfrak p$ if and only if $f^{-1}(\mathfrak q) = \mathfrak p$.
Now you can reduce to the case where $(A, \mathfrak p)$ is local, in which you never need worry that $f^{-1}(\mathfrak q) \not\subset \mathfrak p$. Can you think of a ring whose spectrum describes the primes of $B$ containing $\mathfrak pB$? Then you can apply the answers given by Andrea and I here.
