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Comments 1 to 20 out of 7530 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 Andrea Panontin left comment #8092 on Section 7.50 in Sites and Sheaves

In the proof of theorem 00ZP you write "the sheafification functor is the right adjoint of the inclusion functor", it should be the left adjoint.


On left comment #8091 on Section 14.34 in Simplicial Methods

Thanks and fixed here.


On left comment #8090 on Section 10.25 in Commutative Algebra

Dear Miles, what about ? This corresponds to the function which is when and when , so it corresponds to in . OK?


On Steven Sam left comment #8089 on Lemma 10.90.3 in Commutative Algebra

4th paragraph of proof says: "Let \overline{x_i} \in coker(\phi)\overline{x_1},\dots,\overline{x_n} \in \coker(\phi)$ be a finite set of generators."


On Laurent Moret-Bailly left comment #8088 on Lemma 37.26.6 in More on Morphisms

Not sure this is useful, but the lemma works for any valuation ring : instead of Krull's intersection theorem, use the fact that has a content ideal (Lemma 0ASX) which is principal by Comment 8087. (I do realize that Lemma 0ASX comes a bit later).


On Laurent Moret-Bailly left comment #8087 on Lemma 38.19.6 in More on Flatness

It follows from Comment 8086 that the content ideal is principal.


On Laurent Moret-Bailly left comment #8086 on Lemma 10.89.6 in Commutative Algebra

It may be worth pointing out that is of finite type. Namely, there is of finite type such that , and since $F' is minimal.


On Yassin Mousa left comment #8085 on Section 26.12 in Schemes

Maybe one should mention oi Lemma 26.12.7, that the factorisation is unique. This is given by Lemma 26.4.6.


On left comment #8084 on Section 10.25 in Commutative Algebra

https://stacks.math.columbia.edu/tag/02LV

I find your Lemma 02LX fishy. Suppose R = k[x,y]/(x*y). The zero divisors of R are the two minimal prime ideals (x) and (y). The multiplicative set S consists of anything not in (x) or (y) -- so anything with a constant term, or anything like x^n + y^m. The total ring of fractions is certainly a subring of the Cartesian product of two fields k(x) x k(y). I believe that it consists of pairs f(x), g(y) such that both are regular functions at x=0 and y=0, and take the same value f(0)=g(0). In particular the elements (1,0) and (0,1) of the Cartesian product are not in S^-1R.

Matsumura's 2 books and the Wikipedia page on total ring of fractions just say the obvious definition without discussing what it means or any serious examples. In your treatment, I think Q(R) should be a subring of the Cartesian product

A slightly more complicated example would be k[x,y]/(x^n,y^m), where the total ring of fractions is a subring of k(x)[y]/y^m x k(y)[x]/x^n with the first and second factors equal modulo smaller powers of x,y. Tricky to state correctly, although the idea is clear.

I'm trying to read Mumford, Curves on a surface, Lecture 9: Cartier divisors.

Best, Miles


On left comment #8083 on Lemma 68.12.3 in Cohomology of Algebraic Spaces

Dear Anonymous, this is a rare instance where I slightly disagree with you. I think that this lemma and its proof never refers to modules in the Zariski topology. However, in translating to the case of schemes (at the very end), one just needs to know that the categories of (quasi-)coherent modules on a locally Noetherian scheme are the same as the corresponding categories of (quasi-)coherent modules on viewed as an algebraic space to be able to conclude. This was discussed for quasi-coherent modules much earlier, but was somehow missing in this section. So I've added that in this commit and I hope that this will help the future readers.


On left comment #8082 on Section 10.12 in Commutative Algebra

Thanks and fixed here.


On left comment #8081 on Lemma 39.9.10 in Groupoid Schemes

Thanks and fixed here.


On left comment #8080 on Section 48.2 in Duality for Schemes

Please always comment on the page of the tag you are commenting on. Thanks. Fixed here.


On left comment #8079 on Section 12.15 in Homological Algebra

Not going to do this. The thing with the short exact sequence is explained in Remark 13.12.4; as you can see we just use for the thing that fits into the ses.

Other people have mentioned they prefer to use superscripts, so use and when using truncation functors for complexes with upper numbering (so for cochain complexes). If you read this and have an opinion, please leave a comment.


On left comment #8078 on Lemma 15.41.4 in More on Algebra

Thanks and fixed here.


On left comment #8077 on Section 44.6 in Picard Schemes of Curves

Thanks and fixed here.


On Laurent Moret-Bailly left comment #8076 on Lemma 37.26.6 in More on Morphisms

First line of proof: should be . But in fact it is not very clear whether one is proving that is reduced or that is. I propose to change the first 3 lines as follows:

Assume the special fibre is reduced. Let be any point, and let us show that
is reduced. (This will prove that and are reduced). Let be a specialization with in the special fibre; such a specialization exists as a proper morphism is closed. Consider the local ring . Then is a localization of , so it suffices to show that is reduced. Let be a uniformizer. (The rest is unchanged)


On left comment #8075 on Lemma 100.45.7 in Morphisms of Algebraic Stacks

THanks and fixed here.


On left comment #8074 on Lemma 15.124.1 in More on Algebra

Thanks and fixed here.


On left comment #8073 on Section 12.19 in Homological Algebra

Please comment on each of the items separately (on the page of the tag itself) and tell me exactly what to fix.