# Does closed immersion imply projective morphism?

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 $\mathbb{P}_{Y}^{m}$.

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Are you asking whether a closed immersion is a projective morphism? In that case, you could just take $m = 1$, where projective space over $Y$ is just $Y$. – Akhil Mathew Aug 26 '11 at 1:13
@Akhil: you mean zero, no? – Mariano Suárez-Alvarez Aug 26 '11 at 2:14
@Mariano: Thanks, you're right. – Akhil Mathew Aug 26 '11 at 2:15
@Akhil: Really? I know $Proj\mathbb{Z}[x_0]$ is an one element scheme, but it seems $Y\times_{\mathbb{Z}}Proj\mathbb{Z}[x_0]$ is not $Y$(though I don't know what it should be)? – Li Zhan Aug 26 '11 at 9:16
I 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. – Li Zhan Aug 27 '11 at 9:38

Alternatively, just take $m = 0$ and then $$\mathbf{P}^0 = \textrm{Proj } \mathbf{Z} = \textrm{Spec } \mathbf{Z}$$