Show that if \(f:X\to Y\) is finite and \(\dim X = \dim Y\), then \(f\) is closed and surjective.
Solution
Write \[ A=\mco(Y), \qquad B=\mco(X), \] and let \[ \varphi=f^*:A\longrightarrow B \] be the comorphism. Since \(f\) is finite, \(B\) is a finite \(A\)-module via \(\varphi\).
1. Every finite morphism of affine varieties is closed.
Let \[ Z=V(J)\subseteq X \] be closed. Since \(B\) is finite over \(A\), it is integral over \(\varphi(A)\). Passing to quotients, \(B/J\) is integral over the image of \[ A/\varphi^{-1}(J)\longrightarrow B/J. \] After replacing the source ring by its image, this is an injective integral extension. By lying over, every maximal ideal of \(A\) containing \(\varphi^{-1}(J)\) is the contraction of a maximal ideal of \(B\) containing \(J\). Hence \[ f(Z)=V\bigl(\varphi^{-1}(J)\bigr), \] which is closed in \(Y\).
Thus \(f\) is a closed map.
2. If\[ I=\ker\varphi, \]then\[ \dim B=\dim(A/I). \]
The ring \(B\) is finite over \(A\), hence integral over the image \[ A/I\hookrightarrow B. \] Integral extensions preserve Krull dimension: lying over and going up lift prime chains from \(A/I\) to \(B\), while incomparability prevents a strict prime chain in \(B\) from contracting with repetitions. Therefore \[ \dim B=\dim(A/I). \]
3. Under the hypothesis\[ \dim X=\dim Y, \]the comorphism \(\varphi:A\to B\) is injective.
The varieties \(X\) and \(Y\) are irreducible, so \(A\) and \(B\) are finitely generated integral algebras over the ground field. In particular, \[ I=\ker\varphi \] is a prime ideal of the domain \(A\).
Suppose \(I\ne0\). Then \[ \height I\geq1. \] The affine dimension formula P-AGH2320DIMENSION gives \[ \dim(A/I)+\height I=\dim A, \] so \[ \dim(A/I)<\dim A. \] By step 2, \[ \dim B=\dim(A/I)<\dim A. \] Equivalently, \[ \dim X<\dim Y, \] contrary to the hypothesis. Hence \[ I=0, \] and \(\varphi\) is injective.
4. The morphism \(f\) is surjective.
By step 3, the comorphism is injective, so \(f\) is dominant. Thus \[ \overline{f(X)}=Y. \] By step 1, the image \(f(X)\) is closed. Hence \[ f(X)=\overline{f(X)}=Y, \] so \(f\) is surjective.
5. Q.E.D.
Step 1 proves closedness for every finite morphism, and steps <1>2–<1>4 show that the equal-dimension hypothesis forces surjectivity.