The Stacks project

Comments 1 to 20 out of 8433 in reverse chronological order.

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

On ZL left comment #9052 on Definition 50.15.1 in de Rham Cohomology

Typo: " is -torsion free for all " should be " is -torsion free for all ".


On ZL left comment #9051 on Section 20.31 in Cohomology of Sheaves

The global derived hom is only defined until Section 20.44. Maybe one can add a referrence to that section in the second paragraph?


On andy left comment #9050 on Lemma 5.23.3 in Topology

Ah oops I guess compact map only asks for inverse of compact open is compact, sorry!


On andy left comment #9049 on Lemma 5.23.3 in Topology

Doesn't (2) just follow b/c the point is compact and the fibre is inverse image of a point and the map is compact? Also very minor thing--might be helpful to say that the last part is proving (2)


On left comment #9048 on Section 4.3 in Categories

Going to leave as is.


On left comment #9047 on Section 10.29 in Commutative Algebra

Thanks and fixed here.


On left comment #9046 on Lemma 13.14.13 in Derived Categories

OK, I changed the proof to use the better characterization of essentially constant systems. An alternative would be to add the Karoubian completion of a (pre)additive category and then deduce this lemma from Lemma 13.14.7. Changes are here.


On left comment #9045 on Lemma 29.29.5 in Morphisms of Schemes

Thanks very much. I decided to leave in the proof we have now and add yours as a second clearly shorter one. Changes are here.


On left comment #9044 on Lemma 13.23.3 in Derived Categories

This section is a bit obsolete, but I do think the statement of the lemma is what it should be. The idea is that the natural "inclusion" functor is the "same" as the functor one gets from having a resolution functor.


On left comment #9043 on Lemma 13.20.2 in Derived Categories

Going to leave as is.


On left comment #9042 on Section 13.18 in Derived Categories

Since I think it is reasonable to assume that the person who reads this knows how to take pushouts of short exact sequences in abelian categories, I am going to leave this as is.


On left comment #9041 on Lemma 13.18.8 in Derived Categories

@#8417. It seems to me that the statement "there exists...homotopic to zero" is true just by how we localize in a category (as I said in my comment on 05RW).


On left comment #9040 on Lemma 13.18.6 in Derived Categories

OK, thanks. One can also prove (2) using Lemmas 13.18.4 and Remark 13.18.5 and avoid the diagram chasing in the current proof.


On left comment #9039 on Lemma 13.18.4 in Derived Categories

@#8414: I think this mathoverflow post discusses a different thing because here the proof as it is written now shows there exists a homotopy to go from a map zero in degrees to a map zero in degrees where moreover the homotopy is zero except for the map and that clearly does work (in general). Anyway, I have changed it as you suggested here so the point I just made is now moot (or will be when the site is updated). Thanks.


On left comment #9038 on Lemma 13.18.3 in Derived Categories

Yes, but I think it is fine as is too.


On KD left comment #9037 on Section 52.28 in Algebraic and Formal Geometry

Perhaps this is a silly confusion, but why the notation in https://stacks.math.columbia.edu/tag/0EL1? Right before was a section of and in the proof we view as an element . But in the proposition is some number.


On Pat left comment #9036 on Lemma 100.9.4 in Properties of Algebraic Stacks

There seems to be a grammar typo at the start of the second sentence.


On Liam left comment #9035 on Lemma 10.43.6 in Commutative Algebra

In Lemma 030U. Why we can choose x1,…,xr+1∈K as in Lemma 030Q. Does separable + finitely generated property impliy separably generated ?


On left comment #9034 on Lemma 10.63.16 in Commutative Algebra

Regarding "the displayed inclusion and equality in the Noetherian case follows from Lemma 10.63.15": I think the displayed inclusion is independent of the invoked lemma (one observes that the -annihilator of in is the -annihilator of in if ), and it is only the converse inclusion what requires the mentioned lemma.


On left comment #9033 on Lemma 10.62.2 in Commutative Algebra

Just to save thinking time, here are the details: by induction in . For there's only prime , and we have , whence , by Lemma 10.40.5. Supposing the result true for , we apply Lemma 10.40.9 and obtain . By the induction hypothesis, ; on the other hand, .