Projective morphism

A morphism $X\to S$ is projective when it factors as a closed immersion $X\hookrightarrow {\mathbb{P}}_S^n$ followed by the projection to $S$. Every projective morphism is proper.

For every scheme $X$, projection $p:X\times {\mathbb{P}}^n\to X$ is a closed map, and remains so after every base change. Hence projective space is universally closed over its base. The assertion concerns the underlying scheme points, not just rational points over a non-algebraically closed field.^[ref]

The existence part of the valuative criterion explains the role of projective space. Given a valuation ring $V$ with fraction field $K$, a $K$-point $[a_0:\cdots :a_n]$ can be rescaled so that every coordinate belongs to $V$ and one coordinate is a unit. Indeed, among the finitely many nonzero coordinates choose $a_j$ whose principal fractional ideal contains all the others; then $a_i/a_j\in V$ and $a_j/a_j=1$. These coordinates define the required morphism of schemes $\mathrm{Spec}V\to {\mathbb{P}}^n$. Since projective space is also separated and of finite type, this proves that projection is a proper morphism.

On a Noetherian scheme, one can instead use finitely many homogeneous equations and the Projective Nullstellensatz to express the image as closed rank-defect conditions on graded multiplication map for a projective fiber. An arbitrary affine base need not give a finitely generated defining ideal, so that finite-equation hypothesis must not be assumed in the general statement.

Over the real numbers, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ projects on real points to $\{t\ge 0\}$, which is not Zariski closed. Projective properness concerns scheme or geometric images, not the image of rational points over an arbitrary field. Similarly, $y_0^2+y_1^2$ has no real projective zero, but its homogeneous ideal contains no power of the irrelevant ideal of projective space, since it has a complex projective zero.