---
date: 2022-03-14 10:22
modification date: Monday 14th March 2022 10:22:54
title: complete
aliases: [complete variety]
---

---

- Tags 
	- #AG/basics 
- Refs:
	- #todo/add-references
- Links:
	- [proper](Unsorted/proper%20morphism.md)

---

# complete

A **complete** variety ([integral](Unsorted/integrally%20closed.md) [separated](Unsorted/separated.md) scheme of [finite fields](finite%20fields)) is a [proper](Unsorted/proper%20morphism.md) variety viewed as a $k\dash$scheme, so [separated](Unsorted/separated.md), [finite type](Unsorted/finite%20type.md), and [universally closed](Unsorted/universally%20closed.md). 

Idea: analog of compactness, e.g. $\AA^n$ is not complete but $\PP^n$ is. Projective implies complete for varieties over a field, and it's hard to produce complete varieties which are not projective. A finite-type scheme $X$ is complete (i.e. proper) over $\CC$ iff $X(\CC)$ is compact Hausdorff.

Idea for universally closed: a variety is complete when every projection with $X$ as a fiber is a closed map. Equivalently, every map $X\to \pt$ is universally closed, i.e. the structure morphism is proper.

Idea: completeness wants to be an algebraic version of compactness, and [properness](Unsorted/proper%20morphism.md) wants to be a *relative* version of completeness.

Idea: you can't miss limit points in $X$ -- but $X$ itself couldn't intrinsically know that. So "embed" $X$ in a bigger space. How do you know if it's big enough? Allow an arbitrary $Z$ to tell $X$ that it's missing limit points by having something with a limit in $Z$ come from something in $X\times Z$ without a limit, i.e. the projection $\pi:X\times Z\to Z$ is not closed.

How to check: the [valuative criterion of properness](Unsorted/valuative%20criterion%20of%20properness.md).

Completeness for a [toric variety](Unsorted/toric.md): $\supp \Delta = \EE^n \da N_\RR$, so the fan covers the ambient Euclidean space.

An affine complete variety over $k = \bar k$ is necessarily finite.

Example: $\AA^1$ is not complete, since $V(xy-1) \to \AA^1$ projecting onto the $x\dash$axis is not closed, since its image is $\GG_m = \AA^1\smts{\pt}$.