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Comments 1 to 20 out of 5816 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 DatPham left comment #6246 on Definition 96.11.1 in Artin's Axioms

Maybe this is trivial to ask but is there a definition for the notion (i.e. a colimit of groupoids)? My understanding is that this is just a suggestive notation (and the meaning of the equivalence is explained in the paragraph followed the above definition). Is this correct or I am missing something?


On left comment #6245 on Lemma 10.119.2 in Commutative Algebra

Yes, I agree. So the fix is to say in part (4) that is nonzero, right? In other words, this is the usual mistake of forgetting to check what happens with .


On Brendan Seamas Murphy left comment #6244 on Lemma 10.119.2 in Commutative Algebra

*that the kernel and cokernel of the map are annihilated are automatic


On Brendan Seamas Murphy left comment #6243 on Lemma 10.119.2 in Commutative Algebra

I think the lemma is slightly wrong as stated, because condition 1 implies condition 4, taking . If is a local Artinian ring then for some , so the conditions that the kernel and cokernel of the map are automatic. The map isn't an isomorphism since isn't a local ring, it's a finite map, and as is zero there are no associated primes (thus isn't an associated prime of ).


On left comment #6242 on Section 10.102 in Commutative Algebra

If I am not wrong, I think that when the equivalent conditions of the above proposition are fulfilled, there exists such that if and only if . You could perhaps mention it? This is related to the fact that if all your matrices have coefficients in , then for a given one has .


On Matthieu Romagny left comment #6241 on Lemma 5.18.7 in Topology

In the two bracketted sets of the statement, the backslashes are visible but they shouldn't be, I guess ?


On Yuto Masamura left comment #6240 on Section 4.3 in Categories

typo in the next paragraph of Exaple 4.3.4 (maybe too small one to point out): I think we should add a period after "".


On Fred Diamond left comment #6239 on Section 37.55 in More on Morphisms

X/Y in Lemma 0FK1 (3)


On Owen left comment #6238 on Lemma 15.29.4 in More on Algebra

same thing as comment #6205: rather than


On Owen left comment #6237 on Lemma 47.3.9 in Dualizing Complexes

I don't think it's necessary to take maximal among ideals of this form. Indeed, the inclusion is automatic, and the reverse inclusion is also automatic, since iff for some , but , so . Clearly , but it is just shown that . So I don't see where maximality of figures into the argument.


On left comment #6236 on Lemma 107.23.4 in Moduli of Curves

The inequality here should be . This is fixed here.


On Owen left comment #6235 on Lemma 15.92.1 in More on Algebra

I think what we want and have is that for , because we want to apply Artin-Rees for , , and (with the notation of 10.51.2).


On Yuto Masamura left comment #6234 on Lemma 10.20.1 in Commutative Algebra

just typos: in (1), (3) and (5) of the statement, "there exists a " should be "there exists an ". (already so in (7))


On 57Jimmy left comment #6233 on Lemma 12.19.13 in Homological Algebra

@Johan Sorry for making a bit of a mess with my comments. I agree with you, my point is precisely that with abelian and endowed with a functorial filtration (possibly boundedness needed, not sure), many properties follow automatically: all morphisms are strict, is exact, is exact for all . So it might be worth pointing this out somewhere. (I am totally unfamiliar with how this project works, so I don't know who should do it. I just thought it could be worth doing it. But if it doesn't fit with the philosophy of the site or of this specific page, just ignore all my comments :) ).


On left comment #6232 on Lemma 12.19.13 in Homological Algebra

What is a counterexample where you have the filtration on but morphisms aren't automatically strict? OK, maybe the category of finite modules over the polynomial ring with filtration on given by where is the torsion submodule?


On 57Jimmy left comment #6231 on Lemma 12.19.13 in Homological Algebra

Actually, I do think that, if the filtration on is assumed to be bounded from below, then isomorphisms are automatically strictly compatible with the filtration (by induction on filtration steps), and so exactness of also follows.


On 57Jimmy left comment #6230 on Lemma 12.19.13 in Homological Algebra

I think my first comment is flawed ((2) is not automatically satisfied), but it would still be interesting to write somewhere that is exact iff all morphisms are strict


On 57Jimmy left comment #6229 on Lemma 12.19.13 in Homological Algebra

(Sorry for the double comment, I don't think I can delete one). My aim was just to point out that from the lemma mentioned here one can easily deduce "global" results about abelian categories that come with a functorial filtration. It might be relevant to point this out, as there are many cases of interest (Hodge structures, motives).


On 57Jimmy left comment #6228 on Lemma 12.19.13 in Homological Algebra

Have I got this right? If is abelian and comes endowed with a functorial filtration on (in particular, all morphisms are compatible with the filtration), then condition (2) is automatically satisfied (as is abelian), hence all morphisms are actually strictly compatible with the filtration (by (1)), and moreover the functor is exact (by (6)).


On ToumaKazusa left comment #6226 on Lemma 10.127.11 in Commutative Algebra

Just a typo. The last map is from R_\lambda[x_1,...,x_n]/(f_1,...,f_m) to R[x_1,...,x_n]/(f_1,...,f_m)