# 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
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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