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

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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 $Y$.2011-08-26
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    @Akhil: you mean zero, no?2011-08-26
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    @Mariano: Thanks, you're right.2011-08-26
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    @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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    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.2011-08-26
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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.2011-08-27

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