Here $X\to Y$ is a projective morphism means: $X\to Y$ factors through a closed immersion $X\to \mathbb{P}_{Y}^{m}$, and then followed by the projection $\mathbb{P}^{m}_{Y}\to Y$. I have no idea how to find this $\mathbb{P}_{Y}^{m}$.
Does closed immersion imply projective morphism?
2
$\begingroup$
algebraic-geometry
-
1I guess this follows from the fact: for any ring $R$,Proj$R[x]=D_{+}(x)=$Sepc$(R[x]_{(x)})=$Spec$(R)$. This also says every affine scheme is a projective scheme, which contradicts the intuition in varieties. – 2011-08-27
1 Answers
3
A finite morphism is proper. So we know that a closed immersion is finite and therefore is proper.
Alternatively, just take $m = 0$ and then $\mathbf{P}^0 = \textrm{Proj } \mathbf{Z} = \textrm{Spec } \mathbf{Z}$