The Stacks project

We need to say something about the coefficients $a$, e.g., that they are in $F$.

In the end of proof of lemma 59.82.2, why does "$s\mid_{Z'}$ Zariski locally comes from a section of $\mathcal{F}\mid_{X'}$..." induces it also globally comes from a section? You seem to use those injections to glue them all together, but note that Zariski local means it won't be finite over $X$, do we still have those injections?

You use $x$ in the proof both to denote a point and an element of ideal $I$. There is no confusion, but I guess it would be nice to change one of those.

I think this proof is a little tricky whenever $S^{-1}\mathcal{C}$ is not locally small (for non-small $\mathcal{C}$) because of the size issues I discuss here. Nevertheless, one can avoid these size issues and show existence of finite products via this argument.

Let $s\in\{+,-,b\}$. I think that when proving $K^s(\mathcal{A})$ is a triangulated category from (a) showing that it is a triangulated subcategory of $K(\mathcal{A})$ we obtain less information than from (b) adapting the proof of Proposition 13.10.3. Namely, with (a) we learn that the triangulated structure of $K^s(\mathcal{A})$ is the distinguished triangles in $K(\mathcal{A})$ whose objects belong to $K^s(\mathcal{A})$, whereas with (b) we define the triangulated structure on $K^s(\mathcal{A})$ to be the triangle isomorphism class in $K^s(\mathcal{A})$ spawned by the triangles $(A,B,C(\alpha),\alpha,i,-p)$, with $\alpha\in\operatorname{Mor}(K^s(\mathcal{A}))$. With (a), all we can say is that a triangle $(X,Y,Z,f,g,h)$ in $K^s(\mathcal{A})$ is distinguished iff it is isomorphic to a triangle $(A,B,C(\alpha),\alpha,i,-p)$ with $\alpha$ in $K(\mathcal{A})$, but not necessarily in $K^s(\mathcal{A})$. From here, I don't see how one can conclude that $\alpha$ may be picked in $K^s(\mathcal{A})$. Note that $K^s(\mathcal{A})$ is not strictly full in $K(\mathcal{A})$, since the map from the zero chain complex to $\cdots \xrightarrow{1_A} A\xrightarrow{0}A\xrightarrow{1_A}A\xrightarrow{0}\cdots$ is a chain homotopy equivalence.

"question inverse" in the statement of Lemma 09B1 is probably a typo for "quasi-inverse".

In the definition of the ring $B_i$ it should be $s(b) = t_i(r)$ instead of $s(b) = t_i(t)$.

In the statement, second sentence, one can weaken "$\mathcal{D}'$ is an additive full subcategory of $\mathcal{D}$" to just "$\mathcal{D}'$ is a non-empty preadditive full subcategory of $\mathcal{D}$" (where a preadditive subcategory of a preadditive category is a subcategory with a preadditive structure such that the inclusion functor is additive). Condition (2) implies additivity for $\mathcal{D}$:

Pick any object $X$ in $\mathcal{D}$. By (TR1), the triangle $X\xrightarrow{1_X}X\to 0\to X[1]$ is distinguished. Thus, (2) implies that $\mathcal{D}'$ has a zero object. On the other hand, by Lemma 13.4.11(3) and (TR2), the triangle $Y[-1]\xrightarrow{0}X\to X\oplus Y\to Y$ is distinguished. Now (2) implies that $\mathcal{D}'$ has the direct sum for $X$ and $Y$. This shows that $\mathcal{D}'$ is additive.

We need replace "$w = a_ n x^ n + a_{n + 1} x^{n + 1} + \ldots )$" into "$w = a_ n x^ n + a_{n + 1} x^{n + 1} + \ldots$".

It seems to me that Lemma 0ALJ and Lemma 0CT7 both are special cases of Lemma 0DC7. Also the proofs of all three lemmas are quite similar, they all reduce to Lemma 0ALI.

Proof of Lemma 08LR, last part of the proof. It says ''commutes and if $W\in \operatorname{Ob}(\mathcal{C})$ and $\alpha,\beta\colon X\to Y$ such that..." It should say $\alpha,\beta\colon W\to X$. Also, maybe I'm not seeing it, but saying that $X\cong X \times_Y X$ isn't enough? Should it be $(X,\operatorname{id}_X,\operatorname{id})\cong (X\times_Y X,\text{pr_1},\text{pr_2})$? Thanks in advance.

It seems that $\operatorname{Spec}(A)$ and $\operatorname{Spec}(B)$ have been reversed in the statement of the lemma. It should say $U=\operatorname{Spec}(B) \subset X$ and $V=\operatorname{Spec}(A) \subset Y$.

My previous comment was incorrect. The point is whenever $M_d$ vanishes for all $d$ suffciently large, we get $\widetilde M = 0$.

It might be nice to add the rephrasing of (2):

(3) $U$ contains every associated point of $X$.

Sorry, I'm caring about lemma 031G. Since if all $R_{i}=R$ are invariant, then we have $\text{colim}\ (S\bigotimes_{S_{i}}\Omega_{S_{i}/R})=\Omega_{S/R}$ by universal property and Yoneda lemma, where the canonical map comes from lemma 00RS. Briefly, given a projective system of schemes over a base, if we consider(via pull back) all those differential modules on the limit scheme, then we have a correspondence. But if we consider(via push forward) all those differential modules on the base scheme, do we still have a correspondence?

At the end of the first proof, it would be helpful to say that we are done in the case $\varphi(s)=0$ thanks to the Leibnitz rule.

The following sentence (right before Lemma 0EVP) is not well parsed:

"The following lemma tells us that the pro-h topology is equal to the pro-ph topology is equal to the V topology."

Or is it intentional?

Since now $[1]$ is only assumed to be an autoequivalence, in sequence 13.3.2.1 not all four consecutive terms are honest triangles, right? (For instance $T^{-1}Z\to X\to Y\to Z$ is not an actual triangle, since $TT^{-1}Z$ might not be strictly equal to $Z$.)

In case it's worth to point this out, actually TR1+TR2+TR4 implies TR3. See Lemma 2.2 in <cite authors="May, J.P.">May, J.P., The additivity of traces in triangulated categories, Adv. Math. 163, No.1, 34-73 (2001). ZBL1007.18012.</cite>

Under the risk of sounding nitpicking, I want to ask: when in the statement we say "and such that the localization functor $Q:\mathcal{D}\to S^{-1}\mathcal{D}$ is exact", shouldn't we speficy the natural isomorphism $Q\circ [1]\cong [1]\circ Q$? (Which just some automorphism of $[1]_{S^{-1}\mathcal{D}}$, granted that $Q\circ [1]=[1]\circ Q$.) Under each choice of $\xi\in\operatorname{Aut}([1]_{S^{-1}\mathcal{D}})$, I think we do get actually different triangulated structures on $S^{-1}\mathcal{D}$ by imposing exactness of $(Q,\xi):\mathcal{D}\to S^{-1}\mathcal{D}$ (a unique one for each $\xi$, yeah, but still different.)

I know that most of the time in these sections when we say that some functor is triangulated we do not specify the natural transformation (that is part of the data of a triangulated functor), although maybe in this case it's worth to do so.