The Stacks project

In the second paragraph it should be $f^nym$ istead of $f^tym$ in two places

Is the definition of "finite étale" here just finite and étale? Since finite flat is not the same as finite locally free, could you explain why, in the last paragraph, "since Y → Z is finite étale and hence finite locally free ..." without any noetherian assumption? Thanks.

We have an extra ")" in the first equation.

I am rather puzzled by the identity $\omega_{B/A}=Hom_A(B,A)$ for a finite flat A module B. I also notice that for $f:X\to Y$ finite and flat you claim $f^!(\mathcal O_X)=\omega_{Y/X}$, but this is not in accordance with the case that $X=Y$ is a smooth variety over an algebraically closed field of characteristic $p$ and $f$ the absolute Frobenius morphism where we have $f^!(\mathcal O_X)=\omega_X^{1-p}$, see Sect 1.3, p21 in the book on Frobenius splittings by Brion and Kumar. Note that in that case $\omega_{X/X}=\varpi_X$.

The case $n=1$ also works as $X'$ would be empty hence connected.

Shouldn't this work just as well for a morphisim locally of finite type?

If $A \rightarrow B$ is ring map of finite type, $M$ is a finitely generated $B$-module, and $\mathfrak{p}$ is a prime of $B$ such that $M_\mathfrak{p}$ is flat over $A$, then take $B'$ a finite polynomial ring surjecting onto $B$. We still have that $M$ is a finite $B'$-module, and if $\mathfrak{q}$ is the preimage of $\mathfrak{p}$ in $B'$, then $M_\mathfrak{q} = M_\mathfrak{p}$ is still flat over $A$. So proposition 05I5 guarantees that $M_\mathfrak{p}$ is finitely presented over $B'_\mathfrak{q}$. But then $M_\mathfrak{p}$ is also finitely presented over $B_\mathfrak{p}$.

Typo: the $7$th line from the below "Note that if $(T,r)$ is a special ..." should be "Note that if $(f,r)$ is a special ..."

For 5.21.2 a proof could be :

Naturally it suffices to show it for the union of two nowhere dense sets as the result will follow by induction. So, let $A,B\subset X$ be nowhere dense subsets. As for every pair of subsets of $X$, $\overline{A\cup B}=\overline{A}\cup\overline{B}$ so if $\mathcal{U}\subset\overline{A\cup B}$ is an open subset of $X$, then, as $\mathcal{U}=\left(\mathcal{U}\cap\overline{A}\right)\bigcup\left(\mathcal{U}\cap(X-\overline{A})\right)$ but $\mathcal{U}\cap(X-\overline{A})=\emptyset$ as $B$ is nowhere dense and $\mathcal{U}\cap(X-\overline{A})\subset\overline{B}$, we have that $\mathcal{U}\subset\overline{A}$ and since $A$ is nowhere dense, $\mathcal{U}=\emptyset$.

For 5.21.5 a proof could be :

As $f(X)$ is closed in $Y$, $\overline{f(T)}\subset f(X)$ and as $f:X\rightarrow f(X)$ is an homeomorphism, $f(\overline{T})$ is closed in $f(X)$ and since it contains $f(T)$, $\overline{f(T)}\subset f(\overline{T})$ so $(\text{interior of }\overline{f(T)})\subset f(\overline{T})$ which then gives $f^{-1}((\text{interior of }\overline{f(T)}))\subset f^{-1}(f(\overline{T}))=\overline{T}$. As $T$ is nowhere dense, $f^{-1}((\text{interior of }\overline{f(T)}))=\emptyset$ so $(\text{interior of }\overline{f(T)})=\emptyset$ as $(\text{interior of }\overline{f(T)})\subset f(X)$.

Lemma 007Z. I'm confused by last two lines of proof. And I wonder if $s\mid_V$ means $s_v$? Also I noticed the textbook by Gortz, saying colimit of $\mathscr F^{\sharp}$ is precisely $\mathscr F_x$.

In parts (7) and (8), "section" should be "retraction".

I suggest that we put the equivalent condition (8) to be "there exists a unique section $\tau:S\rightarrow R$ of $R\rightarrow S$ such that $\tau^{-1}(\mathfrak{m})=\mathfrak{q}$". The proof of $(1) \Rightarrow (8)$ in fact shows that this $\tau$ is unique by using Lemma 10.153.2 as pointed out by comment 8561. Also the proof of Lemma 10.153.11 used the uniqueness part.

One final comment (which I learnt from Yuchen Wu, all mistakes are solely mine).

It seems that at least in the absolute case i.e. when $A'=R'$ and therefore $A=R$ then one can show that tor-dimension can be checked after base change to $R$ without the seemingly implicit pseudocoherence assumption (apologies if this is already obvious in the proof above).

The reason is that if $M'$ is an $R'$-module so that $IM'=0$ then $M'$ is already an $R$ module. Therefore $M'=R\otimes^L_R M'$.

Now the equation $K'\otimes^L_{R'}M'=K\otimes^L_{R}M'$ seems to hold without passing through the explicit resolutions $P$ and $E$ because $$K'\otimes^L_{R'}M'=K'\otimes^L_{R'}(R\otimes^L_R M')=(K'\otimes^L_{R'}R)\otimes^L_RM'=K\otimes_R^L M'.$$

Then one reasons as you do.

If this is fine, I would suggest adding this in tag 0651.

I guess that the slogan I suggested seems to be a bit stronger than what the lemma is doing (eg. as a $k[x]/x^e$-module $k$ has unbounded tor-dimension). Maybe a better slogan is that 'perfectness and tor-amplitude can be checked after base change along a nilpotent surjection'

I apologise if this is my misunderstanding, but this seems to be a useful technical lemma. Maybe one can add a slogan 'Perfect complexes lift along nilpotent immersions with same tor-amplitude'

In the third line of the first paragraph, should $M$ be a finite $A$-module, not finite type?

Should the $S$ in the diagram be a $Y$ instead?

Remark 10.63.12. Associate primes of module $S$ in ring $S$ could be taken as $(x_i)+(x_i^2)$, which is annihilator of $x_i+(x_i^2)\in S$? So it's actually non-empty?

I am sorry if this is my misunderstanding, but perhaps it would be nice to have a slogan here 'tor amplitude is stable under base change'

I'm a bit confused about the diagram before Lemma 59.33.3. Is it a typo? Given the natural ring morphisms $\mathcal{O}_ {T,t}\rightarrow \mathcal{O}^h_ {T,t} \rightarrow \mathcal{O}^{sh}_ {T,t}$, we would get $\mathrm{Spec}(\mathcal{O}^{sh}_ {T,t})\rightarrow \mathrm{Spec}(\mathcal{O}^{h}_ {T,t}) \rightarrow \mathrm{Spec}(\mathcal{O}_ {T,t})\rightarrow T$ natural in $(T,t)$.

Is it proven somewhere in the Stacks Project that this definition for $\operatorname{Ext}^i(A,B)$, $A,B\in\mathcal{A}$ indeed yields the $i$-th derived functor of $\operatorname{Hom}$? I cannot find it neither in 13.27 nor with the searcher.

Right after 13.27.1 it is commented that we have the expected long exact sequence, but why should this $\delta$-functor be universal?