# Is morphism from projective space projective morphism？

Suppose $X$ is a projective variety , $Y$ any variety, $f:X\to Y$, then is $f$ a projective morphism? Namely, factor through some closed immersion into $\mathbb{P}^n_Y$.

Since $\operatorname {Spec}(k)\to \operatorname {Spec}(O_{Y,y})\to Y$ is closed immersion, we can take the fiber product and compose the immersions: $X\to \mathbb{P}^n_k\to \mathbb{P}^n_Y$ (suppose $k=\bar{k}$)

However, if we consider schemes over $S$, since not every $Y$ over $S$ has $S$ as closed subscheme, the argument will not apply. Or in the case of varieties, when $Y$ do not have a $k$ point.

1)Does the statement "morphism from projective variety is projective" still hold in the cases?

2)Also I wonder if there is a property like the one for proper variety:

$gf:X \to Y \to Z$ proper and $g$ separated implies $f$ proper.

Does the statement hold if we replace "proper" by "projective" here?

• Your (2) holds if $Y$ is quasicompact [maybe this is part of your definition for varieties, but I think what I'm saying is more general] using the definition of projective from EGA. I'd have to sit down and write some stuff to see if that works here. – Hoot Apr 27 '14 at 20:15
• @mqx: both are true and are classical facts that you can find in any reasonable textbook or online texts (as Stacks project or Vakil) on algebraic geometry. – user143488 Apr 27 '14 at 23:16

## 1 Answer

Since projectiveness is stable under base extension, $X$ projective over $S$ implies $X\times_S Y$ projective over $Y$. Meanwhile, there is a morphism $g:X\rightarrow X\times_S Y$ given by $id: X\rightarrow X$ and $f: X\rightarrow Y$ over S, which is the graph $\Gamma_f$ of the morphsim $f$. When $Y$ is separated, $\Gamma_f$ is a closed imersion inside $X\times_S Y$. Therefore we arrive at the conclusion 1).