The Stacks project

Also: instead of the expression at the beginning $$(F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, F(0))$$ shouldn't we write the following expression? $$(F(0), F(0), F(0), 1_{F(0)}, 1_{F(0)}, \xi_0\circ F(0)).$$ I mean, both expressions coincide after we show that $F(0)=0$, but the expression is written before $F(0)=0$ is proven.

I think the notation may be a little bit confusing: The morphism denoted as $(0,1)$ in $(X, X \oplus Y, Y, (1, 0), (0, 1), 0)$ is not the same morphism as the corresponding one in $F(1, 0) \oplus F(0, 1) : F(X) \oplus F(Y) \to F(X \oplus Y)$. To make the notation unambiguous, I would write $(X, X \oplus Y, Y,\binom{1}{0}, (0, 1), 0)$ for the former expression and $F\binom{1}{0}\oplus F\binom{0}{1}:F(X)\oplus F(Y)\to F(X\oplus Y)$ for the latter one. (This is the notation from https://ncatlab.org/nlab/show/matrix+calculus .)

And about the "we omit the rest of the argument": wouldn't it suffice to invoke 12.7.1?

Second paragraph of proof, line 3: "Then for some nonzero $a\in A$...".

I think there is a word missing at the beginning of the second paragraph. It should be "To get a contradiction $\textbf{let}$ $C \subset X$ be a proper curve [...]".

This is not a big deal, but strictly speaking, one should say for any non-zero $x\in K$, blabla as you talk about $x^{-1}$. The same issue persists in the next lemma.

One could more precisely show that $P \mapsto P/IP$ induces a bijection between isomorphism classes of finite projective $R$-modules and isomorphism classes of finite projective $R/I$-modules: The current formulation shows the surjectivity of the map. To show the injectivity, let $P$, $P'$ be finite projective $R$-modules such that $P/IP \cong P'/IP'$. Hence we find an $R$-linear map $P \to P'/IP'$. As $P$ is projective, it lifts to an $R$-linear map $u\colon P \to P'$. As $u$ is an isomorphism modulo $I$, it is surjective by Nakayama's lemma. The ranks of the finite projective $R$-modules $P$ and $P'$ are equal in every open neighborhood of $V(I)$. As $I$ is contained in the radical of $R$, the only open neighborhood of $V(I)$ is $\Spec A$. Hence $u$ is a surjective map of finite projective modules of the same rank. Hence it is an isomorphism.

In the definition of order, there is $\mathcal{O}_{X,z}$ which should be $\mathcal{O}_{X,Z}.$

In the first sentence "There is a notion of regular sequence which is slightly weaker than that of a regular sequence" the word "quasi" is presumably missing.

No, the assumption $n\in \mathcal{O}_S^*$ is the correct one: if $n$ is not invertible in the local rings of $S$ then the sequence is not exact on the right (étale-local surjectivity fails). Think about what kind of extension it needs to extract $p$-roots in characteristic $p$.

Typo in assumption, should be $n \in \mathbf{N}$.

I've found this in Serre's Faisceaux algébriques cohérents. Properties (3) and (4) are, respectively, theorems 13.2 and 13.1. Here is an English translation: http://achinger.impan.pl/fac/fac.pdf On this pdf the theorems are on pp. 18-19

$R^{\oplus t}$ should be $A^{\oplus t}$

While homeomorphisms have "2 out of 3" property, it's unclear if universal homeomorphism also has the property. (In fact, the one being used in your second sentence is the only unclear one, because one needs to look at base changes that doesn't come from the "most downstairs base"? I tried to see this, and easily reduced to the case where the composition is identity, but then got stuck. Maybe this is already discussed somewhere in the SP?)

In any case, it is still true that F_{X/S} is a universal homeomorphism: WLOG one may assume both X and S are affine, then one can check condition (2) in [0CNF] holds true directly.

In the proof (2nd paragraph), there is an "abd", which I think should be an "and".

Is the condition that $f$ is flat redundant? We can use (2) in \href{https://stacks.math.columbia.edu/tag/0B91}{Lemma 0B91}, which does not need flatness of $f$.

In the first part of the proof Why any open set W would work? I think we have to choose W to be the open set for which the minimum is attained.

A little typo: in the first line of the proof, for "triple" read "quadruple".

It can be proved considering constructible topology.

First, ${\rm Spec}(S)$ is quasi-compact in constructible topology by Lemma 5.23.2. Hence $T$ is also quasi-compact in constructible topology. Now ${\rm Spec}(R)$ is Hausdorff in constructible topology, again by Lemma 5.23.2. So $T$ is closed in constructible topology. By Lemma 5.23.6, $T$ is closed in usual topology.

In Step 1, it says "Hence we may form the pushout along the injective map $A^{\bullet} \rightarrow I^{\bullet}$''. But is this map really injective? Maybe what was meant is "Hence we may form the pushout along $A^{\bullet} \to I^{\bullet}$ to get...'' And then "Since $A^{\bullet} \to B^{\bullet}$ is an injection, the pushout square implies that $B^{\bullet} \to E^{\bullet}$ is a quasi-isomorphism." (Assuming that was the intention...also was the latter fact proved in the Stacks project?)

In the definition of $s^n_j$ before 08NC, you used the notations i and F. Do you mean V and U?