Show that if \(X\) is projective and \(Y\) is affine over \(k = \CC\), then the projection \(\pi_2: X\cross Y\to Y\) is proper and closed in the Zariski topology.
Solution
Choose a projective embedding \[ i:X\hookrightarrow\PP^n_\CC. \]
1. The product embedding\[ i\cross\id_Y: X\cross Y \hookrightarrow \PP^n_\CC\cross Y = \PP^n_Y \]is a closed immersion.
The map \(i\) is a closed immersion because \(X\) is projective. Closed immersions are preserved by arbitrary base change. Base-changing \(i\) along \[ Y\longrightarrow\Spec\CC \] gives exactly \[ i\cross\id_Y: X\cross Y\hookrightarrow\PP^n_Y. \] Hence this map is a closed immersion.
2. The projection\[ \pi_2:X\cross Y\longrightarrow Y \]is projective.
By step 1, \(\pi_2\) factors as \[ X\cross Y \hookrightarrow \PP^n_Y \longrightarrow Y, \] where the first map is a closed immersion and the second is the standard projective-space projection. This is precisely the definition of a projective morphism in D-MORPROJ.
3. The projection \(\pi_2\) is proper.
Projective morphisms over a Noetherian base are proper by D-MORPROJ. The affine variety \(Y\) is Noetherian, so step 2 implies that \(\pi_2\) is proper.
4. The projection \(\pi_2\) is a closed map in the Zariski topology.
A proper morphism is universally closed by definition (D-8XX95). In particular, without any further base change, it sends closed subsets to closed subsets. Step 3 therefore implies that \(\pi_2\) is closed.
5. Q.E.D.
Steps <1>1–<1>3 prove properness, and step 4 proves Zariski closedness.