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?
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algebraic-geometry
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8Are 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 $Y$. – 2011-08-26
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3@Akhil: you mean zero, no? – 2011-08-26
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0@Mariano: Thanks, you're right. – 2011-08-26
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0@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)? – 2011-08-26
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0Akhil 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. – 2011-08-26
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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