Definition. A topological space $X$ is said to be quasi-separated if the intersection of two quasi-compact open subsets $U,V\subset X$ is quasi-compact.

In Definition 11.14 from these notes from a course in Algebraic Geometry taught at Bonn some years ago, they define a "quasi-separated morphism of schemes" as one such that the inverse image of any quasi-separated open subset is quasi-separated. On the other hand, on p. 207 of Vakil's Rising Sea, a "quasi-separated morphism of schemes" is defined as one such that the pre-image of an affine open subset is quasi-separated.

My question is: are these definitions equivalent?

Clearly, Bonn's implies Vakil's (as every affine scheme is quasi-separated). But I'm not sure about the converse. Specifically, one can consider the canonical map from the infinite dimensional space with doubled origin to the affine line with doubled origin (induced by $k[x_1]\to k[x_1,x_2,\dots]$). The target of this scheme morphism is a quasi-separated scheme for it is Noetherian. But the source is not quasi-separated (see Example 4 here. So this morphism wouldn't be Bonn-quasi-separated). However, I think this morphism is Vakil-quasi-separated?

Edit: Grothendieck already gave a positive answer in EGA IV, première partie, Propositions (1.2.6) and (1.2.7) (it's on p. 227, here's the link to Numdam). See my answer below for another argument.

  • 1
    $\begingroup$ I wrote it on this follow-up MO question, but let me write it here again: your example is not a well-defined morphism. To which of the origins are points $(0,a_2,a_3,\ldots) \neq (0,0,0,\ldots)$ mapped? $\endgroup$
    – Remy
    Commented Jun 8, 2023 at 16:16
  • $\begingroup$ @Remy Thank you. I made an edit on my answer which now includes your obstruction. $\endgroup$ Commented Jun 10, 2023 at 9:46

1 Answer 1


$\def\bbA{\mathbb{A}}$I think the idea for a counterexample I sketched cannot be so and that, in fact, Vakil-quasi-separatedness implies Bonn-quasi-separatedness. I will explain the argument for this implication. First note the following fact:

$$ \tag{*} \text{If $S$ is a scheme, then $S$ is a quasi-separated topological space}\\\text{ if and only if $S\to\operatorname{Spec}\mathbb{Z}$ is a Vakil-quasi-separated morphism.} $$ The implication to the left is obvious. The implication to the right follows from the fact that open subsets of a quasi-separated topological space are also quasi-separated.

Now let $f:X\to Y$ be a Vakil-quasi-separated morphism of schemes and let $U\subset Y$ be an open subset that is a quasi-separated topological space. Using the facts:

  • $f^{-1}(U)\to U$ is Vakil-quasi-separated,
  • $U\to\operatorname{Spec}\mathbb{Z}$ is Vakil-quasi-separated, by $(*)$,
  • Vakil-quasi-separatedness is stable under composition (see 01KU; note that in the Stacks Project "quasi-separated morphism" means "Vakil-quasi-separated morphism", by 01KO),

we deduce that the composite $f^{-1}(U)\to U\to\operatorname{Spec}\mathbb{Z}$ is Vakil-quasi-separated, and by $(*)$ we conclude that $f^{-1}(U)$ is a quasi-separated topological space and that therefore $f$ is Bonn-quasi-separated.

EDIT: The map of the counterexample idea cannot be actually constructed:

If $\bbA_k^\infty=\operatorname{Spec}k[x_1,x_2,\dots]$ is the infinite dimensional affine space over $k$, then the infinite dimensional affine space over $k$ with double origin is the glueing of two copies of $\bbA_k^\infty$ along the identity on $U=\bbA_k^\infty\setminus V(x_1,x_2,\dots)=\bigcup_{i=1}^\infty D(x_i)$. That is, it is the pushout in schemes of $$ \label{1}\tag{1} \bbA_k^\infty\leftarrow U\to\bbA_k^\infty. $$ Denote it $Y$. If $X$ is the affine line over $k$ with double origin, then defining a map $Y\to X$ is the same as defining maps $\bbA_k^\infty\rightarrow X\leftarrow\bbA_k^\infty$ that turn the diagram \eqref{1} into a commutative square. Looking at the hint “[the map] is induced by $k[x_1]\to k[x_1,x_2,\dots]$” from my original post (it's been more than one year since I wrote it, I don't remember what I was thinking about when I wrote it) I interpret it to define enter image description here

where $\bbA_k^1\to X\leftarrow\bbA_k^1$ are the two projections from the pushout of $\bbA_k^1\leftarrow D(t)\to\bbA_k^1$ (here $\bbA_k^1=\operatorname{Spec}k[t]$), and the arrows $\bbA_k^\infty\to\bbA_k^1$ are induced by $t\in k[t]\mapsto x_1\in k[x_1,x_2,\dots]$. However, this diagram doesn't commute, for it doesn't commute on underlying sets. This is because of what Remy pointed out on a comment: the points $(0,a_1,a_2,\dots)\neq(0,0,0,\dots)$ in $U$ are sent to a different origin in $X$ through each path of the diagram.


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