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Comments 1 to 20 out of 4063 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 dpw left comment #4308 on Section 12.5 in Homological Algebra

In lemma 12.5.17, both the object and the morphism are denoted by . I think it looks a bit confusing.


On Rankeya left comment #4307 on Lemma 10.114.7 in Commutative Algebra

Does the proof use that S is a domain anywhere?


On correction_bot left comment #4306 on Lemma 28.45.7 in Morphisms of Schemes

The proof says we'll only prove (1) and (2) but then talks about which only plays a role in (3) and (4) (should be instead).


On left comment #4305 on Lemma 14.19.13 in Simplicial Methods

The right hand side has U,V and W mixed up.


On bogdan left comment #4304 on Proposition 10.156.15 in Commutative Algebra

Two small typos:

1) The phrase ''Let be the subring...'' should read ''Let be the subring...''

2) The sentence ''We claim that is generated by all linear forms such that in '' should read ''We claim that is generated by all linear forms such that in ''.


On David Speyer left comment #4303 on Section 104.16 in Examples

It seems to me that $\mathrm{Rad}(M) = \mathfrak{m} B + \mathfrak{n} B$ should read $\mathrm{Rad}(M) = \mathfrak{m} B \cap \mathfrak{n} B$.

On Laurent Moret-Bailly left comment #4302 on Section 16.1 in Smoothing Ring Maps

Line 11: I guess "filtered limit" should be "filtered colimit".


On Rankeya left comment #4301 on Section 16.1 in Smoothing Ring Maps

Okay, may be you are assuming is henselian in para 4 as well, in which case my question is redundant since henselian rings are excellent.


On bogdan left comment #4300 on Lemma 10.155.16 in Commutative Algebra

The proof implicitly uses the fact that is noetherian (when it cites 00PD). This follows from Tag 00PG but this result is not explicitly cited now (the proof can be significantly simplified in this situation).


On left comment #4299 on Section 5.26 in Topology

"but the converse does not holds in general" has a typo.


On left comment #4298 on Lemma 10.136.7 in Commutative Algebra

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On Bogdan left comment #4297 on Lemma 10.136.7 in Commutative Algebra

A formula should read


On Goodluckthere left comment #4296 on Section 5.8 in Topology

Possible typo: In the proof of the third part of lemma 004.W you take to be a set of irreducible sets but later you consider to be a set of indices of elements in .


On comment_bot left comment #4295 on Lemma 36.54.11 in More on Morphisms

It would be useful to strengthen this statement: it suffices to assume that is a local complete intersection (with still regular). A similar comment applies to https://stacks.math.columbia.edu/tag/0E9J


On Xuande Liu left comment #4294 on Lemma 28.12.2 in Morphisms of Schemes

Without loss of generality we can assume that is affine. Let . And hence we can factorize as . The former morphism is an open immersion and the latter is morphism between affine schemes. Therefore is separated.


On Xuande Liu left comment #4293 on Lemma 28.12.2 in Morphisms of Schemes

Without loss of generality we can assume that is affine. Let . And hence we can factorize as . The former morphism is an open immersion and the latter is morphism between affine schemes. Therefore is separated.


On Bogdan left comment #4292 on Lemma 15.45.5 in More on Algebra

Let K be the fraction field of should read .

I guess (the last line) should read .


On Laurent Moret-Bailly left comment #4291 on Lemma 7.47.11 in Sites and Sheaves

I think one can simplify the exposition by reducing (via 00Z7(1)) to the case of just one presheaf.


On Kazuki Masugi left comment #4290 on Lemma 7.47.11 in Sites and Sheaves

I can't make sure that is right inverse to the map in the statement. Would you please describe the precise way to verify?


On David Hansen left comment #4289 on Lemma 35.3.7 in Derived Categories of Schemes

In the commutative square here, A and B should be swapped.