A morphism \(f: X \to Y\) is quasi-finite if for every point \(y \in Y\) the fibre \(\inverseof{f}(y)\) is a finite set.
Show that a finite morphism is quasi-finite.
Show that a finite morphism is closed, that is, the image of any closed subset is closed.
Show by example that a surjective, finite-type, quasi-finite morphism need not be finite.
Solution
1. Let \(f:X\to Y\) be finite and let \(y\in Y\).Then the scheme-theoretic fibre\[ X_y=X\times_Y\Spec\kappa(y) \]is the spectrum of a finite-dimensional \(\kappa(y)\)-algebra.
Choose an affine neighborhood \[ V=\Spec B\subseteq Y \] of \(y\). Since \(f\) is finite, Hartshorne II.3.4 gives \[ f^{-1}(V)=\Spec A \] with \(A\) finite as a \(B\)-module.
The fibre is \[ X_y \cong \Spec\bigl(A\otimes_B\kappa(y)\bigr). \] Because \(A\) is generated by finitely many elements as a \(B\)-module, the tensor product is finite-dimensional as a vector space over \(\kappa(y)\).
2. A finite-dimensional algebra over a field has only finitely many prime ideals.
Such an algebra is Artinian as a ring. In an Artinian ring every prime ideal is maximal, and there are only finitely many maximal ideals: otherwise the descending chain obtained from successive finite products of distinct maximal ideals would not stabilize. Hence its spectrum is a finite set.
3. Every finite morphism is quasi-finite.
For each \(y\in Y\), <1>1–<1>2 show that the fibre \(X_y\) has finitely many points. Its underlying set is naturally the set-theoretic fibre \(f^{-1}(y)\). Thus every fibre of \(f\) is finite.
4. Let \(f:X\to Y\) be finite.Then \(f\) is a closed map.
It suffices to check closedness over an affine open cover of \(Y\). Let \[ V=\Spec B\subseteq Y, \qquad f^{-1}(V)=\Spec A, \] with \(A\) finite over \(B\). Let \[ Z\cap f^{-1}(V)=V(J)\subseteq\Spec A \] for an ideal \(J\subseteq A\).
The quotient \(A/J\) is integral over the image of \(B/(J\cap B)\) because finite ring extensions are integral and integrality passes to quotients. The induced map \[ B/(J\cap B)\hookrightarrow A/J \] is injective. By lying over, every prime of \(B/(J\cap B)\) is the contraction of a prime of \(A/J\). Hence the image of \(V(J)\) in \(\Spec B\) is exactly \[ V(J\cap B), \] which is closed.
Thus \(f(Z)\cap V\) is closed in every affine member \(V\) of an open cover of \(Y\). Therefore \(f(Z)\) is closed in \(Y\).
5. Let \(k\) be a field and define\[ X=\mathbb G_{m,k}\amalg\Spec k, \qquad Y=\mathbb A^1_k. \]Map the first component to \(Y\) by the usual open immersion\[ \mathbb G_m=D(t)\hookrightarrow\mathbb A^1 \]and map the second component to the origin.Then\[ f:X\longrightarrow Y \]is surjective, of finite type, and quasi-finite.
The image of the first component is \(Y\setminus\{0\}\), and the second component maps to \(0\), so \(f\) is surjective.
Both component maps are of finite type: the open immersion corresponds to \[ k[t]\longrightarrow k[t,t^{-1}], \] which is generated as a \(k[t]\)-algebra by \(t^{-1}\), while the point inclusion corresponds to the quotient \[ k[t]\longrightarrow k[t]/(t)\cong k. \] Since \(X\) is the finite disjoint union of these two components, \(f\) is of finite type.
For \(a\ne0\), the fibre over \(a\) consists of the single point \(a\in\mathbb G_m\). The fibre over \(0\) consists of the single extra point \(\Spec k\). Hence every fibre is finite, so \(f\) is quasi-finite.
6. The morphism in 5 is not finite.
The component \[ \mathbb G_m\subseteq X \] is closed, because it is a component of a disjoint union. Its image is \[ \mathbb A^1\setminus\{0\}, \] which is not closed in \(\mathbb A^1\).
By 4, every finite morphism is closed. Therefore this morphism cannot be finite.
7. Q.E.D.
Step 3 proves part (a), step 4 proves part (b), and steps <1>5–<1>6 give the counterexample required in part (c).