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

Agustí Roig
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Li Zhan
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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 01:13
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    @Akhil: you mean zero, no? – Mariano Suárez-Álvarez Aug 26 '11 at 02:14
  • @Mariano: Thanks, you're right. – Akhil Mathew Aug 26 '11 at 02:15
  • @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 09:16
  • Akhil is correct. In order to see this, you might begin by convincing yourself that Proj $\mathbb Z[x_0]$ is not a one-element scheme. – Matt E Aug 26 '11 at 11:24
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    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 09:38

1 Answers1

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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 $$\mathbf{P}^0 = \textrm{Proj } \mathbf{Z} = \textrm{Spec } \mathbf{Z}$$

Future
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