Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
5  
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
2  
@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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.