 Solution

Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq Y$ containing $y$. Since $f$ is an affine morphism, $f^{-1}V=\mathrm{Spec}B$; 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 isIf $b_1,\dots ,b_r$ generate the $A$-module $B$, their images span the algebra $B/\mathfrak{m}B$ 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 $X_y$, soThe 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.