The Stacks project

Comments 621 to 640 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 #9098 on Lemma 10.99.1 in Commutative Algebra

Dear Ryo Suzuki, for generalizations of this lemma, please see Section 38.25.


On left comment #9097 on Section 7.5 in Sites and Sheaves

Thanks to both of you and fixed here.


On left comment #9096 on Lemma 22.5.4 in Differential Graded Algebra

To prove that Homs from into a product is the product of the homs (in the homotopy category).


On Christian Merten left comment #9094 on Lemma 10.138.14 in Commutative Algebra

It is not clear to me why the constructed is a right inverse to the projection : To prove this, we need to check that for each . Since is a right inverse, we have and by the construction of , . But is not necessarily equal to .

This can be fixed in the following way: Since , there exist such that If we adjoin the (finitely many) coefficients of the to , the above relation also holds in and we may conclude.


On left comment #9093 on Section 20.25 in Cohomology of Sheaves

Thanks and fixed here.


On left comment #9092 on Lemma 10.99.11 in Commutative Algebra

Tried to clarify in a slightly different way. See this.


On left comment #9091 on Lemma 10.134.12 in Commutative Algebra

No, it is not a splitting.


On left comment #9090 on Lemma 10.99.9 in Commutative Algebra

OK, I think that induction is needed but it wasn't clear because of the confusing typo in the last formula in the proof which I fixed here.


On left comment #9089 on Lemma 10.99.8 in Commutative Algebra

Yes, this is much better! Changes are here.


On left comment #9088 on Lemma 6.29.1 in Sheaves on Spaces

OK, I fixed the issue in the proof of (2) but I think the proof of (4) is correct as written now. Change is here.


On left comment #9087 on Lemma 10.5.4 in Commutative Algebra

OK, I made the changes except for the very last one. Thanks and changes are here.


On left comment #9086 on Lemma 34.9.2 in Topologies on Schemes

@#8641 Thanks. I added your example here and I updated the blog post as well.


On left comment #9085 on Lemma 34.9.2 in Topologies on Schemes

@#8469 Thanks and fixed here.


On left comment #9084 on Lemma 26.21.7 in Schemes

Thanks and fixed here.


On left comment #9083 on Lemma 26.21.9 in Schemes

Thanks.


On left comment #9082 on Lemma 26.18.2 in Schemes

Going to leave as is.


On left comment #9081 on Lemma 26.17.6 in Schemes

Going to leave as is.


On left comment #9080 on Section 26.18 in Schemes

Going to leave as is.


On left comment #9079 on Section 9.8 in Fields

Thanks and fixed here.


On left comment #9078 on Section 26.10 in Schemes

Going to leave as is.