Comments 1 to 20 out of 5816 in reverse chronological order.


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 $\mathrm{colim}\; \mathcal{X}_{T_i}$ (i.e. a colimit of groupoids)? My understanding is that this is just a suggestive notation (and the meaning of the equivalence $\mathrm{colim}\;\mathcal{X}_{T_i}\to \mathcal{X}_T$ 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 $R'$ is nonzero, right? In other words, this is the usual mistake of forgetting to check what happens with $0$.

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 $R' = 0$. If $(R, \mathfrak{m})$ is a local Artinian ring then $\mathfrak{m}^r = 0$ for some $r$, so the conditions that the kernel and cokernel of the map are automatic. The map $R \to R'$ isn't an isomorphism since $0$ isn't a local ring, it's a finite map, and as $R'$ is zero there are no associated primes (thus $\mathfrak{m}$ isn't an associated prime of $R'$).

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 $j$ such that $I(\varphi_i)=R$ if and only if $i\geqslant j$. You could perhaps mention it? This is related to the fact that if all your matrices have coefficients in $\mathfrak m$, then for a given $i$ one has $I(\varphi_i)=R\iff \varphi_i=0$ .

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 "$h:\mathcal C\to\mathit{PSh}(\mathcal C)$".

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 $J$ maximal among ideals of this form. Indeed, the inclusion $J:=\{a\in R|ax\in E\}\subset\{a\in R|ax'\in E\}$ is automatic, and the reverse inclusion is also automatic, since $ax'\in E$ iff $f^max'=0$ for some $m$, but $f^max'=f^{m+n}ax$, so $ax\in E$. Clearly $\{a\in R|ax'\in E\}\supset\operatorname{Ann}(x')$, but it is just shown that $J\subset\operatorname{Ann}(x')$. So I don't see where maximality of $J$ figures into the argument.

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

The inequality here should be $g \geq 2$. 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 $J^t\supset (f_1^{m-n},\ldots,f_r^{m-n})$ for $m-n\geq tn$, because we want to apply Artin-Rees for $I=J$, $M=K_n^p$, and $N=\operatorname{Ker}(K_n^p\to K_n^{p+1})$ (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 $f$" should be "there exists an $f$". (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 $\mathcal{A}$ abelian and endowed with a functorial filtration (possibly boundedness needed, not sure), many properties follow automatically: all morphisms are strict, $\mathrm{gr}_*(-)$ is exact, $F^k$ is exact for all $k$. 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 $\text{id}_\mathcal{A}$ but morphisms aren't automatically strict? OK, maybe the category of finite modules $M$ over the polynomial ring $k[x]$ with filtration on $M$ given by $M_{tors} \subset M$ where $M_{tors}$ 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 $\mathcal{A}$ 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 $\mathrm{gr}_*(-)$ 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 $\mathrm{gr}_*(-)$ 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 $\mathcal{A}$ is abelian and comes endowed with a functorial filtration on $\mathrm{Id}_\mathcal{A}$ (in particular, all morphisms are compatible with the filtration), then condition (2) is automatically satisfied (as $\mathcal{A}$ is abelian), hence all morphisms are actually strictly compatible with the filtration (by (1)), and moreover the functor $\mathrm{Grad}_*(-)$ 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)