The Stacks project

In the proof, a period needed just before the setence defining $\delta_j$. Also in the proof, a redundant period in "which shows that we get a morphism of complexes."

Here's an example to answer your question from 2021 on the blog: let $X=X_1\cup X_2$, $X_1=X_2=\mathrm{Spec} k[t_1,t_2,\dots]$ be the non-qs scheme from Example 01KL, and $Y = X_1 \sqcup \mathrm{Spec} \mathcal{O}_{X_2,0}$. Then the obvious morphism $f:Y\to X$ is flat and surjective, and (since $Y$ is qc) trivially satisfies (4). But it isn't an fpqc covering: $f^{-1}(X_2)= (X_1\smallsetminus \{0\}) \sqcup \mathrm{Spec} \mathcal{O}_{X_2,0}$, so $X_2$ isn't the image of any qc open of $Y$.

I am not familiar with the notation $\mathcal{U}:U$, which is not defined here. What does it mean precisely? Is it standard?

The reference to EGA should be to Exposé VIII, instead of VII.

The proof can actually be simplified: after we've shown that $K' \otimes _{R'}^\textbf{L} R/I=K' \otimes _{R'} ^\textbf{L} R'/J=0$, we can show directly that $K'=0$. We do this by showing inductively that $K' \otimes _{R'} ^\textbf{L} R'/J^i=0$ for every positive integer $i$. A method similar to that of Lemma 09AR may work, i.e. we represent $K'$ by a K-flat complex with flat terms, and then tensor it with the short exact sequence $0\rightarrow J^i/J^{i+1}\rightarrow R'/J^{i+1}\rightarrow R'/J^i\rightarrow 0$.

Maybe it could be interesting to add the following statement to this section.

Lemma. Let $X$ be a scheme with locally finitely many irreducible components. The normalization morphism $\nu:X^\nu\to X$ is an isomorphism if and only if $X$ is reduced and normal.

Proof. If $\nu$ is an isomorphism, then $X$ is reduced and normal by Lemma 29.54.5. Conversely, suppose $X$ is reduced and normal. We give two proofs that this implies $\nu$ is an isomorphism.

First proof. By Lemma 29.53.6 and Properties, Lemma 28.7.5, we can assume $X$ is integral. By Lemma 29.54.5, $\nu$ is integral and $X^\nu$ has a unique irreducible component (so it is an integral scheme). We conclude from Lemmas 29.54.7 and 29.54.8. $\square$

Second proof. By Divisors, Lemma 31.25.2, it suffices to show that $\mathcal{O}_X$ is integrally closed in $\mathcal{K}_X$ (for $\underline{\operatorname{Spec}}_X\mathcal{O}_X=X$). This amounts to seeing that $\mathcal{O}_X(U)$ is integrally closed in $\mathcal{K}_X(U)$ for $U\subset X$ open affine. But for such $U$, the ring $\mathcal{K}_X(U)$ is the total ring of fractions of $\mathcal{O}_X(U)$, as $X$ is reduced (Divisors, Lemma 31.25.1). We conclude by Algebra, Lemma 10.37.12. $\square$

Just to mention it (it took me some time to find it): normalization and relative normalization are treated in EGA II, §6.3.

In the proof of $(4) \implies (1)$, should the indexing of the naive cotangent complex be from $j = 1$ to $n$, instead of $j = 1$ to $m$?

This theorem should be formulated for sufficiently negative $n_0$ and $n<n_0$ as well for the sake of the next theorem.

In the proof, $-r-1$ should be $-r+1$ (twice) and $-r$ should be $-r+2$

Also in the second last and the last sentences, $-1$ should be changed to $n-1$.

The use of Lemma 15.64.3, i.e. Lemma 064S is to $H^{n-1}$, not $H^{-1}$. In the next sentence, $\beta$ should map to $C^{n-1}$.

In the statement, "of perfect complexes" should be "of complexes" as $E'$ is not necessarily perfect.

If X is not Hausdorff, then being extremally disconnected doesn't imply being totally disconnected. For instance: if X is an indiscrete space, then X is clearly extremally disconnected and it's not Hausdorff. But X is connected and, if X has at least two points, then X is not totally disconnected. Is this what you asked?

"If $X$ is Hausdorff and extremally disconnected, then $X$ is totally disconnected (this isn't true in general)." I do not understand what is meant by "this isn't true in general".

In the definition of Hom complexes after Lemma 36.9.2, in the equation $\oplus_{p+q=n}Hom_A(I^p, M^q)$, one of $p$ should be negated.

$\mathcal{K}$ in the proof is undefined.

$L$ and $M$ are not defined in the statement.

I am not sure if the last equality in the proof reflects the fact that the diffential of $I$ is the negative of that of $K$. I am sorry if I am wrong.

5 lines above 36.9.0.1, you need an additional ) to $H^{i-1}$.