The Stacks project

Comments 1521 to 1540 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 Laurent Moret-Bailly left comment #8104 on Lemma 58.31.3 in Fundamental Groups of Schemes

The assumptions on are very restrictive. A more general (and natural) statement, with essentially the same proof, would be:

Let be a locally Noetherian scheme. Let be open and dense. Assume that . (This holds in particular if is the support of an effective Cartier divisor which is regular, by Algebra, Lemma 10.106.7.) Let be a finite étale morphism, unramified over in codimension . Then (...)


On Mike left comment #8103 on Section 58.31 in Fundamental Groups of Schemes

In the paragraph right above Lemma 0EYD, it's written "In particular, we see that Y×USpec(Kξ) is the spectrum of a finite separable algebra Lξ/K.". I wonder if you mean Lξ/Kξ here.


On Et left comment #8102 on Lemma 13.5.10 in Derived Categories

In the last paragraph, of the 2 compositions that need to be equal one should have X_2 in it instead of X_1


On Ryo Suzuki left comment #8100 on Remark 42.32.8 in Chow Homology and Chern Classes

"... that is an isomorphism." I think it should be "... that is an isomorphism."


On Shizhang left comment #8099 on Lemma 101.38.2 in Morphisms of Algebraic Stacks

Line 3 of the proof: should it be `` is in fact in the fibre category...''?


On Laurent Moret-Bailly left comment #8098 on Lemma 10.39.15 in Commutative Algebra

Typo one line 2 of proof: "imlpications".


On teovvv left comment #8097 on Lemma 10.39.15 in Commutative Algebra

To prove 1 implies 2 wouldn't it also be correct the following argument? exact exact To me it feels very slightly cleaner. Or am I missing something?


On Rubén Muñoz--Bertrand left comment #8096 on Lemma 10.165.4 in Commutative Algebra

Okay, I feel this proof can be slightly upgraded. First, lemma 10.44.2 only works in characteristic . Of course, the characteristic zero case is obvious, but why not just use 10.43.6 instead?

Now for the tricky part. The term universal homeomorphism is not yet defined, it only appears in definition 29.45.1 and in the context of schemes, despite its use in the title of section 10.46 (maybe that's a bit confusing, but this is not the main issue here).

It seems that throughout the Stacks Project, a great care is taken to not use the term universal homeomorphism for morphisms of rings. Instead, it is written in terms of morphisms on spectra (for instance in the proof of lemma 58.14.2), or with a sentence such as "induces a universal homeomorphism on spectra" (see for instance lemma 29.46.11).

So if we want to keep this consistency, I propose to first say that by lemma 10.45.4 (or by definition) is purely inseparable, so we can apply lemma 10.46.10 to find that has a unique prime ideal.

If this is not nitpicking for you, then I would suggest to also edit the very few remaining cases of universal homeomorphisms of rings in the Stacks Project: tags 29.47.6, 37.24.7, 110.40 (twice) and maybe 29.46.8.


On Andrea Panontin left comment #8093 on Section 7.50 in Sites and Sheaves

In Lemma 00ZO I think that should be .


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 69.12.3 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.