Is EGA II. 5.6.1's Statement of Chow's Lemma Correct? It states that if $X\rightarrow S$ is a separated morphism of finite type, with $S$ quasi-compact and $X$ having finitely many irreducible components, then there is a surjective and projective $S$-morphism $f: X'\rightarrow X$, such that

*

*$X'$ is quasi-projective over $S$;

*for some open subscheme $U\hookrightarrow X$, $U'=f^{-1}(U)$ is dense in $X'$, and $f$ induces an isomorphism $U'\xrightarrow{~}U$.

Morever if $X$ is reduced (resp. irreducible, integral), then it can be arranged so that $X'$ is such.
The first step in EGA's proof is giving me a hard time as discussed in
Proof of Chow's lemma in EGAII
Takumi's writeup (accepted answer) as well as Görtz/Wedhorn 13.100 assumes $S$ quasi-compact and quasi-separated. What if $S$ is quasi-compact but NOT quasi-separated?
 A: Found my restatement of EGA II.5.6.1 a bit off. The original assumption reads “ is a quasi-compact scheme”. In EGA, schemes are assumed separated (as opposed to pre-schemes), hence quasi-separated.
In short, EGA II.5.6.1 DOES assume  to be quasi-separated and quasi-compact.
One can still ask what if  is quasi-compact but not quasi-separated. Probably (imo) less interesting though as such situations rarely occur in practice.
A: I think I’ve repaired the proof. But this looks a bit too simple so I may very well be missing something…
In the accepted answer to the linked question, the question is to find whether there exists an open subset $U \subset X$ contained in a given irreducible component $C$ of $X$ (and let $C’$ be the reunion of all the other irreducible components of $X$), such that we have a quasi-compact morphism $i: U \rightarrow X$.
Let $V_i, i \in I$ be a finite affine open cover of $S$. Let $U_0=C \backslash C’$, which is an irreducible open subset of $X$, contained in $C$ and dense in $C$. Let $J$ be the set of the $i \in I$ such that $X_{V_i}$ meets $U_0$ (ie such that $V_i$ contains the image in $S$ of the generic point of $C$). Then the intersection $U_1$ of $U_0$ and all the $X_{V_i}$ for $i \in J$ is an open subset of $X$, contained in $C$ and dense in $C$.
Then let $U_2 \subset U_1$ be an affine open subset. We’ll show that $U_2 \rightarrow X$ is affine. By Stacks, Part 2, Lemma 29.11.11  (2) (https://stacks.math.columbia.edu/tag/01SG), it is enough to show that $U_2$ is affine over $S$. But the inverse image in $U_2$ of each $V_i$ is either $U_2$ (affine) or $\emptyset$, so we’re done.
