$X$ is a quasi-separated scheme iff the intersection $U_a \cap U_b$ is quasicompact for $U_a, U_b$

Question

I don't understand the argument in (c) implies (a) of this proof. Why is it that "it suffices that ... is quasicompact."

This is Proposition I.2.7 of EGA IV, Chapter IV, Etude Locale des Schemas et des Morphismes de Schemas.

I don't exactly know what 1.1.1 is a reference to here. The most relevant thing I see in that section is the point that if a map $f : X\to Y$ is locally quasicompact (i.e. $Y$ admits an open cover by $\left\{U_i\right\}$ such that $f^{-1}(U_i)\to U_i$ is quasicompact) it is quasicompact. But I'm not sure how this helps us here.

Answer 1

I agree that the proof has a gap, and this has been complained about on this website at least once before here (though there are some extra assumptions in that post which are not made here, so the proposed solution is not quite applicable for you).

If you assume that the $U_\alpha$ are affine, then this is proven at the Stacks Project in tag 01K4, though as you note in your answer, any quasi-compact but not quasi-separated scheme like $\Bbb A^\infty_k$ with a doubled origin is a counterexample to this claim by taking the trivial open cover.

Here's the lemma and proof transcribed for posterity:

Lemma (Stacks 01K4): Let $f:X\to S$ be a morphism of schemes. The following are equivalent:

1. $f:X\to S$ is quasi-compact,
2. the inverse image of every affine open is quasi-compact,
3. there exists some affine open covering $S=\bigcup_{i\in I} U_i$ such that $f^{-1}(U_i)$ is quasi-compact for all $i$.

Proof. Suppose we are given a covering $S=\bigcup_{i\in I} U_i$ as in (3). First, let $U\subset S$ be any affine open. For any $u\in U$ we can find an index $i(u)\in I$ such that $u\in U_{i(u)}$. As standard opens form a basis for the topology on $U_{i(u)}$ we can find $W_u\subset U\cap U_{i(u)}$ which is standard open in $U_{i(u)}$. By compactness we can find finitely many points $u_1,\ldots,u_n\in U$ such that $U=\bigcup^n_{j=1} W_{u_j}$. For each $j$ write $f^{−1}(U_{i(uj)})=\bigcup_{k\in K_j} V_{jk }$ as a finite union of affine opens. Since $W_{uj}\subset U_{i(uj)}$ is a standard open we see that $f^{−1}(W_{u_j})\cap V_{jk}$ is a standard open of $V_{jk}$, see Algebra, Lemma 10.16.4. Hence $f^{−1}(W_{u_j})\cap V_{jk}$ is affine, and so $f^{−1}(W_{u_j})$ is a finite union of affines. This proves that the inverse image of any affine open is a finite union of affine opens.

Next, assume that the inverse image of every affine open is a finite union of affine opens. Let $K\subset S$ be any quasi-compact open. Since $S$ has a basis of the topology consisting of affine opens we see that $K$ is a finite union of affine opens. Hence the inverse image of $K$ is a finite union of affine opens. Hence $f$ is quasi-compact.

Finally, assume that $f$ is quasi-compact. In this case the argument of the previous paragraph shows that the inverse image of any affine is a finite union of affine opens.

Answer 2

The theorem is incorrect as stated. Condition (c) does not imply condition (a).

Suppose $X$ is a quasi-compact scheme, and let $\{X\}$ be the open cover of $X$ which has only one element, the whole space $X$. Then certainly $X\cap X$ is quasi-compact, so this is a cover satisfying the hypotheses of (c). So according to the theorem, $X$ should be quasi-separated.

However, this is incorrect. It is known that there exist quasi-compact schemes which are not quasi-separated.