A complete variety $X$ over $k$ is defined to be one where the projection map $X\times_{\operatorname{spec} k} Y\to Y$ is closed for every variety $Y$. It is stated on Wikipedia and various places that complete varieties are proper over $k$. But that means $X\to \operatorname{spec} k$ is a proper morphism, which means $X\times_{\operatorname{spec} k} Y\to Y$ is closed for any scheme $Y$. How to see the two are equivalent?
A complete variety is proper over $k$: why does it suffice to check universally closed on varieties instead of all schemes
algebraic-geometryschemes
Best Answer
This is a neat trick: if $f:X\to S$ is a quasi-compact morphism of schemes, it suffices to check whether $f$ is universally closed on morphisms locally of finite presentation $S'\to S$ (in particular, over a noetherian base, i.e. a field, locally of finite presentation = finite type). I'll outline the proof from Stacks 05JX below.
Suppose $X_T \to T$ is not closed for some $T\to S$. This means that there is some specialization $t_1\rightsquigarrow t$, a point $\xi$ in $X_T$ mapping to $t_1$, and no specialization of $\xi$ which lands in $(X_T)_t$. But after a suitable reduction to the affine case, we can write $T$ as a colimit of finitely presented schemes over $S$, and the emptiness of the fiber over $t$ must actually occur at some scheme in our colimit.