The Stacks project

Comments 601 to 620 out of 9050 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 left comment #9118 on Definition 14.11.1 in Simplicial Methods

Thanks and fixed here.


On left comment #9117 on Section 10.18 in Commutative Algebra

OK, I tried to improve this section a bit. I didn't add the equivalent condition suggested by Cesar Massri but I will in the future if it turns out to help. See changes


On left comment #9116 on Lemma 27.11.2 in Constructions of Schemes

Thanks and fixed here.


On left comment #9115 on Section 10.95 in Commutative Algebra

Thanks and fixed here.


On left comment #9114 on Lemma 9.21.6 in Fields

Because the degree of is always .


On left comment #9113 on Lemma 10.37.10 in Commutative Algebra

Thanks and fixed here.


On left comment #9112 on Lemma 35.10.2 in Descent

Thanks and fixed here.


On left comment #9111 on Lemma 10.60.5 in Commutative Algebra

Well, it is finite because has already been shown to be the product of its localizations.


On left comment #9110 on Section 27.10 in Constructions of Schemes

Thanks and fixed here.


On left comment #9109 on Lemma 29.53.10 in Morphisms of Schemes

Going to leave as is.


On left comment #9108 on Section 10.9 in Commutative Algebra

Thanks and fixed here.


On left comment #9107 on Lemma 61.27.2 in Pro-étale Cohomology

Good catch! Fixed as you suggested here.


On left comment #9106 on Section 31.16 in Divisors

Thanks and fixed here.


On left comment #9105 on Lemma 10.137.12 in Commutative Algebra

@#8495: The lemma states that the category of finitely presented modules is the colimit of the categories, so if the colimit is free then it is free at some stage. So I think this is easy enough to leave as is.

@#8612: This is because of the somewhat "bad" choice of writing as a shorthand for . See Definition 10.134.1. I guess we should have used the classical and less confusion would have been caused. Sorry, but unless more people complain I am going to leave this as is.


On left comment #9104 on Lemma 12.29.5 in Homological Algebra

Yes, that was silly! Thanks! Fixed here.


On left comment #9103 on Lemma 47.16.7 in Dualizing Complexes

Yes, the zero module is Cohen-Macaulay. Thanks for pointing out the issue with the statement. I fixed the statement as you suggested here.


On left comment #9102 on Lemma 29.11.8 in Morphisms of Schemes

That would be a much harder proof.


On left comment #9101 on Section 64.6 in The Trace Formula

Thanks and fixed here.


On left comment #9100 on Theorem 97.16.1 in Criteria for Representability

Yes, I agree that is a slogan for this result. Added here.


On left comment #9099 on Lemma 29.3.1 in Morphisms of Schemes

No, because here we also get that is an immersion.