Show that a projective morphism with finite fibres, that is, a quasi-finite projective morphism (II, Ex. 3.5), is a finite morphism.
Solution
We use the standing Noetherian hypotheses of Chapter III.
1. It is enough to prove the assertion after replacing the target by an arbitrary affine open subset.
Finiteness of a morphism is local on the target. If \(V\subseteq Y\) is open, then the base change \[ f_V:f^{-1}(V)\longrightarrow V \] is again projective and has finite fibres. Thus it suffices to prove that \(f_V\) is finite for every affine open \(V\subseteq Y\).
2. Assume henceforth that \(Y\) is affine. For every coherent ideal sheaf\(\mci\subseteq\mco_X\),\[ R^1f_*\mci=0. \]
Every fibre of \(f\) is finite, hence has dimension \(0\) when nonempty. Therefore the maximum fibre dimension in [@Har10a, Corollary III.11.2] is \[ r=0. \] That corollary says that for a projective morphism and every coherent sheaf \(\mcf\), \[ R^if_*\mcf=0 \qquad (i>r). \] Applying it to \(\mcf=\mci\) and \(i=1\) gives the claim.
3. For every coherent ideal sheaf \(\mci\subseteq\mco_X\),\[ H^1(X,\mci)=0. \]
The low-degree part of the Leray spectral sequence gives \[ 0 \longrightarrow H^1(Y,f_*\mci) \longrightarrow H^1(X,\mci) \longrightarrow H^0(Y,R^1f_*\mci). \] Because \(f\) is projective and \(\mci\) is coherent, \(f_*\mci\) is coherent by the coherence theorem for projective direct images [@Har10a, Theorem III.8.8]. Since \(Y\) is affine, \[ H^1(Y,f_*\mci)=0 \] by affine vanishing. Step 2 gives \[ H^0(Y,R^1f_*\mci)=0. \] Hence the middle group is zero.
4. The scheme \(X\) is affine.
Step 3 proves \[ H^1(X,\mci)=0 \] for every coherent ideal sheaf \(\mci\) on the Noetherian scheme \(X\). Therefore Serre's cohomological criterion for affineness gives that \(X\) is affine.
5. Writing\[ Y=\Spec A, \qquad X=\Spec B, \]the \(A\)-module \(B\) is finite.
The structure sheaf \(\mco_X\) is coherent. Since \(f\) is projective, the coherence theorem for projective direct images gives that \[ f_*\mco_X \] is a coherent \(\mco_Y\)-module. On the affine scheme \(Y=\Spec A\), its module of global sections is therefore a finite \(A\)-module.
But \[ \Gamma(Y,f_*\mco_X) = \Gamma(X,\mco_X) =B. \] Thus \(B\) is finite as an \(A\)-module.
6. Q.E.D.
By steps 4 and 5, over every affine open \(V=\Spec A\subseteq Y\) the inverse image is affine, \[ f^{-1}(V)=\Spec B, \] with \(B\) a finite \(A\)-module. This is exactly the definition of a finite morphism. Hence \(f\) is finite.