Let \[ f:X\longrightarrow Y \] be a quasi-finite morphism of affine varieties over the algebraically closed ground field \(k\). Show that \[ \dim X\leq\dim Y. \]
Solution
Let \[ Z=\overline{f(X)}\subseteq Y \] with its reduced induced structure.
1. The closed subset \(Z\) is an affine variety, and the induced morphism\[ g:X\longrightarrow Z \]is dominant and quasi-finite.
The variety \(X\) is irreducible, so its continuous image \(f(X)\) is irreducible. The closure of an irreducible subset is irreducible, hence \(Z\) is irreducible. Since \(Z\) is closed in the affine variety \(Y\), it is affine. Thus \(Z\) is an affine variety.
By definition of \(Z\), the factor map \[ g:X\longrightarrow Z \] has dense image and is therefore dominant.
For every point \(z\in Z\), its fibre under \(g\) is the same set as its fibre under \(f\). Since \(f\) is quasi-finite, that fibre is finite. Hence \(g\) is quasi-finite.
2. One has\[ \dim X\leq\dim Z. \]
Choose any point \[ z\in g(X). \] The fibre \[ g^{-1}(z) \] is nonempty and finite. Hence every irreducible component of this fibre is a point and has dimension \(0\).
By the fibre-dimension theorem for the dominant morphism \[ g:X\longrightarrow Z, \] every irreducible component of every nonempty fibre has dimension at least \[ \dim X-\dim Z. \] For the chosen fibre this gives \[ 0 \geq \dim X-\dim Z. \] Therefore \[ \dim X\leq\dim Z. \]
3. Since \(Z\subseteq Y\) is closed,\[ \dim Z\leq\dim Y. \]
Every chain of irreducible closed subsets of \(Z\) is also a chain of irreducible closed subsets of \(Y\). Therefore the supremum of chain lengths defining \(\dim Z\) cannot exceed the corresponding supremum defining \(\dim Y\).
4. Therefore\[ \boxed{\dim X\leq\dim Y.} \]
Steps 2 and 3 give \[ \dim X \leq \dim Z \leq \dim Y. \]
5. Q.E.D.
Steps <1>1–<1>4 prove that a quasi-finite morphism of affine varieties cannot raise dimension.