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That is if $f :P^n_ {X} \rightarrow P^m_{Y}$ is a proper morphism, then it is a projective morphism. Projective morphism means that it factors through projective scheme of last term and the first component is a closed immersion.

Jian
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1 Answers1

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Let $f:X\to Y$ be a (not necessarily proper) morphism of projective schemes over a field $k$ (the same works over a scheme $S$). Then $f$ is projective.

Proof: $X\subseteq \mathbb{P}^n_k$ (closed immersion) for some $n$. Then consider the closed immersion $X\subseteq X\times_k Y$ in the graph, which is a closed subscheme of $\mathbb{P}^n_k\times_kY=\mathbb{P}^n_Y$ (here I am using that projective morphisms are proper, which is a standard application of the valuative criterion for properness). This gives the required factorization $X\to \mathbb{P}^n_Y\to Y$.

Sonner
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  • you are right.Every morphism $P^n_{k} \rightarrow Y$ is projective by using the graph.So using the projective morphism is proper,we can show every morphism $P^n_{X} \rightarrow Y $is projective morphism where $Y$ is arbitrary scheme.Thank you,Sonner! – Jian Jun 09 '17 at 10:56
  • Sorry ,Sonner you are right. Every morphism $P^n_{S}\rightarrow Y$where $Y $is a $S$-scheme is a projective morphism.Y can't be arbitrary! – Jian Jun 09 '17 at 11:21
  • You are welcome! :) – Sonner Jun 09 '17 at 16:17