= Solution
{c}
Choose an <affine open subscheme> $V=\operatorname{Spec}A\subseteq Y$ containing $y$. Since $f$ is an <affine morphism>, $f^{-1}V=\operatorname{Spec}B$; by hypothesis $B$ is a <finitely generated module> over $A$. If $\mathfrak m\subset A$ corresponds to the <closed point> $y$, then $\kappa(y)=A/\mathfrak m$. The <scheme-theoretic fiber> is
$$
X_y=X\times_Y\operatorname{Spec}\kappa(y)
\cong\operatorname{Spec}\bigl(B\otimes_A\kappa(y)\bigr)
=\operatorname{Spec}(B/\mathfrak mB).
$$
If $b_1,\ldots,b_r$ generate the $A$-<module> $B$, their images span the <algebra> $B/\mathfrak mB$ as a <vector space> over $\kappa(y)$. It is therefore finite-dimensional. By the allowed fact about the <spectrum of a finite-dimensional commutative algebra>, it has only finitely many <prime ideals>. These are exactly the points of $X_y$, so
$$
\boxed{|f^{-1}(y)|=|\operatorname{Spec}(B/\mathfrak mB)|<\infty.}
$$
The points of the <scheme-theoretic fiber> correspond to the <prime ideals> of $B$ contracting to $\mathfrak m$, which explains the equality with the underlying set-theoretic <fiber>. An empty <fiber> is allowed. In fact, replacing $A/\mathfrak m$ by $\kappa(y)$ gives the same finite-dimensional argument for every point, without requiring it to be a <closed point>.
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