Characterization of quasi-separated schemes in EGA. Is the statement correct?

Question

$\def\bbA{\mathbb{A}}$I am trying to understand the following result:

This is from EGA IV, première partie (here's the link to Numdam; see p. 228).

Specifically, I am interested on the equivalence with c). I think c) doesn't imply a). A counterexample is given by the following: Let $\bbA_k^\infty=\operatorname{Spec}(k[x_1,x_2,\dots])$ be the infinite-dimensional affine space over $k$, and let $U=\bbA_k^\infty\setminus V(x_1,x_2,\dots)=\bigcup_{i=1}^\infty D(x_i)$. Denote $X$ to the gluing of two copies of $\bbA_k^\infty$ along the identity on $U$ (i.e., $X$ is the infinite-dimensional affine space over $k$ with double origin). Then $X$ is a non-quasi-separated scheme (see here). However, $X$ is quasi-compact (the two copies of $\bbA_k^\infty$, which are quasi-compact, cover $X$), so the trivial cover $(U_\alpha)=(X)$ satisfies c).

My question is: What is the correct characterization that Grothendieck could have had on mind when he stated c)?

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Edit: Here's the proof attempt of c)$\Rightarrow$a) in EGA

See Alex Kruckman's answer for an explanation of why this proof doesn't work.

Answer 1

Here's one idea: if instead of assuming $U_\alpha$ are only quasi-compact we assume $U_\alpha$ are quasi-compact and quasi-separated, then c) is equivalent to a).

a)$\Rightarrow$c). Use b') (the implication a)$\Rightarrow$b') is proven in EGA).

c)$\Rightarrow$a). We want to see that the morphism $\Delta_X:X\to X\times_\mathbb{Z}X$ is quasi-compact. Since “being quasi-compact” is a property of morphisms of schemes which is local on the target, it suffices to see that the restriction to the open subset $U_\alpha\times_\mathbb{Z}U_\beta\subset X\times_\mathbb{Z}X$, which is $\Delta_X^{-1}(U_\alpha\times_\mathbb{Z}U_\beta)=U_\alpha\cap U_\beta\to U_\alpha\times_\mathbb{Z}U_\beta$, is quasi-compact. Since a scheme $Y$ is quasi-compact if and only if $Y\to\operatorname{Spec}\mathbb{Z}$ is quasi-compact, we have that the composite $U_\alpha\cap U_\beta\to U_\alpha\times_\mathbb{Z}U_\beta\to\operatorname{Spec}\mathbb{Z}$ is quasi-compact. By 03GI, it suffices to argue that $U_\alpha\times_\mathbb{Z}U_\beta\to\operatorname{Spec}\mathbb{Z}$ is quasi-separated. This is because (i) the morphism $U_\alpha\times_\mathbb{Z} U_\beta\to U_\beta$ is quasi-separated for it is the base change of the quasi-separated morphism $U_\alpha\to\operatorname{Spec}\mathbb{Z}$ (01KU), and (ii) quasi-separatedness is stable under compositions (01KU), so $U_\alpha\times_\mathbb{Z}U_\beta\to U_\beta\to\operatorname{Spec}\mathbb{Z}$ is quasi-separated.

Answer 2

(c) and (a) are both false for the infinity space with double origin.

(c) is false : The intersection the two copies of $\mathbb{A}^\infty_k$ is $\mathbb{A}^\infty_k - (x_1, x_2, ...)$. This is not compact as the open cover $D(x_1), D(x_2), ...$ does not have a finite subcover.

(a) is false as your link shows.

Answer 3

I agree that this appears to be a mistake in EGA. The direction in question, (c)$\implies$(a), says that a sufficient condition for a scheme $X$ to be quasi-separated is that there exists a quasi-compact open cover of $X$ such that the intersection of any two of these sets is quasi-compact.

The proof seems to suggest that to check a morphism $f$ is quasi-compact, it suffices that there exists a quasi-compact open cover $(U_i)$ of the codomain such that each $f^{-1}(U_i)$ is quasi-compact.

This is not true, as your example shows. Letting $X$ be the infinite-dimensional affine space with doubled origin, $X\times_{\mathbb{Z}} X$ is a quasi-compact open cover of $X\times_{\mathbb{Z}} X$, and its preimage $\Delta^{-1}(X\times_{\mathbb{Z}} X)=X$ is quasi-compact, but $\Delta$ is not quasi-compact. Letting $U$ be the first copy of infinite-dimensional affine space and $V$ the second copy, $U\times_{\mathbb{Z}} V$ is an affine (hence quasi-compact) subset of $X\times_{\mathbb{Z}} X$, but its preimage under $\Delta$ is $U\cap V$, which is not quasi-compact.

However, the assertion is true if we assume the open sets $U_i$ are affine. This is Stacks Project 01K4.

So I think the simplest fix is just to require in condition (c) that the open cover is an affine open cover. Then the proof goes through as written. It also follows from Stacks Project 01KO, taking $S$ to be $\mathrm{Spec}(\mathbb{Z})$ with the trivial open cover.