A scheme admitting a surjective integral map from an affine scheme is affine.

Proposition 32.11.2. Let $f : X \to S$ be a morphism of schemes. Assume $X$ is affine and that $f$ is surjective and universally closed1. Then $S$ is affine.

Proof. By Morphisms, Lemma 29.41.11 the scheme $S$ is separated. Then by Morphisms, Lemma 29.11.11 we find that $f$ is affine. Whereupon by Morphisms, Lemma 29.44.7 we see that $f$ is integral.

By the preceding paragraph, we may assume $f : X \to S$ is surjective and integral, $X$ is affine, and $S$ is separated. Since $f$ is surjective and $X$ is quasi-compact we also deduce that $S$ is quasi-compact.

By Lemma 32.7.3 we can write $X = \mathop{\mathrm{lim}}\nolimits _ i X_ i$ with $X_ i \to S$ finite. By Lemma 32.4.13 we see that for $i$ sufficiently large the scheme $X_ i$ is affine. Moreover, since $X \to S$ factors through each $X_ i$ we see that $X_ i \to S$ is surjective. Hence we conclude that $S$ is affine by Lemma 32.11.1. $\square$

[1] An integral morphism is universally closed, see Morphisms, Lemma 29.44.7.

Comment #3025 by Brian Lawrence on

Suggested slogan: A scheme admitting a surjective integral map from an affine scheme is affine.

Comment #6678 by Xuande Liu on

There is an interesting fact told by my friend Longke Tang. In this lemma, we only need $f$ to be surjective and universally closed. And affineness can be easily deduced from them.

Notice the diagram $\xymatrix{ X \ar[d]^{} \ar[r]^{} & X\times X \ar[r] & X\times Y\ar[d]\\ Y \ar[rr] & & Y\times Y}$. The map $X\to Y\times Y$ is a closed morphism since it is the composition of such morphisms. Hence $Y\to Y\times Y$ has closed image since $X\to Y$ is surjective, which implies $Y$ is separated.

Now that $X\to Y$ is the composition of $X\to X\times Y$ and $X\times Y\to Y$, we know $f$ is an affine morphism. And the rest follows from the proposition.

Comment #6679 by on

This is nice, thanks! I will add this the next time I go through all the comments. The fact that $Y$ is separated, if it is the target of surjective, universally closed morphism from a separated scheme $X$, is already contained in Lemma 29.41.11. Thanks again.

Comment #6889 by on

OK, I have no finally added this in this commit. Please let me know if I spelled your name(s) wrong.

Comment #6997 by Laurent Moret-Bailly on

Could this be true if $S$ is only assumed to be an algebraic space? I could not find this in later chapters.

Comment #7000 by Laurent Moret-Bailly on

# 6998: In such situations, it would greatly help users to announce future generalizations: in this case the reference to 07VT could be given here but also at 01ZT (case where $f$ is finite), 01YQ ($f$ finite and $Y$ noetherian) and their algebraic space versions 07VP and 07VS.

Comment #7001 by Laurent Moret-Bailly on

Can't guess why my last comment appears like this!

Comment #7002 by on

It's kind of funny how the font is so large all of a sudden! I didn't know this was even possible. Sorry about that!

Anyway, this is a general question. The problem is that there are too many cases of this that it is very hard to do this. Most of the lemmas on schemes have a corresponding lemma on algebraic spaces. In addition, we also want to state and prove the corresponding result for algebraic stacks. The best way to find these is to search for them. In this specific case:

1. On google search for "site:stacks.math.columbia.edu surjective integral affine space" or "site:stacks.math.columbia.edu surjective universally closed affine space" (unfortunately I only just changed the statement of 07VT from integral to universally closed so google may not yet know about this). In general google is very good at finding stuff in the Stacks project!
2. Look in the same place in the corresponding chapter on algebraic spaces.
3. Use the search bar of the stacks project: there are several things you can enter that will find the results you mention. One that gives you 07VT and 07VS is surjective "is affine" algebraic space* (and click the search only statements button) but that is kind of a cheat. I have found that I get better at making sure I find all results pertaining to a topic by using the wild card and just taking some time going through a mildly long list of results.

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