Comments 1 to 20 out of 6320 in reverse chronological order.


On left comment #6783 on Section 10.63 in Commutative Algebra

Lemma 10.63.18. All right I make a mistake, only "R is a local ring" is superfluous.

On left comment #6782 on Section 10.63 in Commutative Algebra

Lemma 10.63.18. I think the condition "R is a local ring" and "M is finite over R" is superfluous.

On left comment #6781 on Section 10.63 in Commutative Algebra

I think the condition "R is a local ring" and "M is finite over R" is superfluous.

On Mark left comment #6780 on Section 4.31 in Categories

In the sentence "The 2-category of 2-commutative diagrams is the 2-category defined as follows...", it might be better to put in the dependency of $f$ and $g$ somewhere (e.g. use instead "The 2-category of 2-commutative diagrams over $f$ and $g$ is the 2-category...")

On Bogdan left comment #6779 on Lemma 14.32.3 in Simplicial Methods

Shouldn't A and B be interchanged in the diagram?

On Bogdan left comment #6778 on Lemma 25.4.2 in Hypercoverings

I think the reference in "Thus it is a homotopy equivalence (Simplicial, Lemma 14.32.3)" should refer to some other lemma.

On left comment #6777 on Lemma 43.9.1 in Intersection Theory

Ha! Yes, this could happen. Of course, in that case $[W'_0]_k - [W'_\infty]_k$ is the zero cycle. I will fix this the next time I go through all the comments.

On Morten left comment #6776 on Lemma 43.9.1 in Intersection Theory

Why is the map $W'\rightarrow W$ generically finite? Could $W'$ not be a trivial family like $X\times \mathbb{P}^1$?

On Yuto Masamura left comment #6774 on Lemma 10.62.4 in Commutative Algebra

This also follows from Lemma 10.40.5.

Indeed, $I^nM=0$ is equivalent to $I^n\subset\operatorname{Ann}(M)$, hence to $I\subset\sqrt{\operatorname{Ann}(M)}$ since $R$ is Noetherian. This is equivalent to $V(I)\supset V(\operatorname{Ann}(M))$, thus by Lemma 10.40.5 $V(I)\supset\operatorname{Supp}(M)$.

On left comment #6773 on Lemma 5.19.8 in Topology

Once the sets $\{u_i\}$ and $\{u'_j\}$ have one element in comment they are the same.

On left comment #6772 on Remark 5.20.5 in Topology

Please feel free to ignore this remark; it isn't used in what follows. The phrase "the obstruction ... element of" is not a precise mathematical statement; it is aimed at people who have heard similar phrases in the past. The statement means that given a catenary, locally Noetherian, sober topological space $X$ there is an element $o_X \in H^1(X, \mathbf{Z})$ which is zero if and only if $X$ has a dimension function. This is not a terribly useful statement in and of itself: the only thing you can deduce from this is that if $H^1(X, \mathbf{Z}) = 0$ then $X$ has a dimension function. It gets more interesting if you know how to construct $o_X$ and how it behaves: for example, given an open $U \subset X$ the image of $o_X$ in $H^1(U, \mathbf{Z})$ by the restriction map is equal to $o_U$. And so on.

On Alejandro González Nevado left comment #6771 on Lemma 5.19.8 in Topology

At the end, shouldn't it be say that there is not only a sequence but several sequences (one for each $u_{i}$ beginning on such $u_{i}$)? If there was only a sequence, then we would be saying that all elements are in such sequence and thence proving that $u_{i}=u_{j}=u'{k}=u'_{l}$, i.e., that all the elements at play are equal and not the sets $\{u_{1},\dots,u_{n}\}$ and $\{u'_{1},\dots,u'_{m}\}$ themselves.

On Alejandro González Nevado left comment #6770 on Remark 5.20.5 in Topology

Is there a typo in the wording "the obstruction to having a dimension function is an element of"? Is the obstruction itself an element of such set. If so, what is exaclty an "obstruction"? Shouldn't it be defined somewhere?

On Oliver Roendigs left comment #6769 on Lemma 15.22.11 in More on Algebra

The first sentence after "Proof of (1)." indicates that a previous version of the Lemma addressed only PIDs. It can be removed.

On left comment #6768 on Section 15.26 in More on Algebra

@#6767: Well, if $f \in R$ and $M$ is an $R$-module with such that $M/M[f]$ is generated by $k$ elements, then of course $M/(f\text{-power torsion})$ is generated by $k$ elements too. Here $M[f] = \text{Ker}(f : M \to M)$. Or did I misunderstand your comment?

On jok left comment #6767 on Section 15.26 in More on Algebra

In the proof of Lemma 0CZM, the reference to Lemma 080Z uses variant statement. What is needed in Lemma 0CZM is a quotient by a-power torsion while in Lemma 080Z only the quotient by the torsion is proven.

On left comment #6766 on Lemma 57.9.1 in Fundamental Groups of Schemes

Yes, the statement you suggest sounds correct. Fun! I guess the proof as you sketched it should work too. The proof will use Lemmas 15.50.14 and 15.50.15 and (as you say) the extension of Theorem 15.50.8 to henselian pairs, namely Lemma 16.14.1. I will add this the next time I go through all the comments. Thanks very much!

On Thomas left comment #6765 on Lemma 10.62.1 in Commutative Algebra

In Lemma 00L0, could you write:

$aR/(J\cap aR)$ is a quotient of $R/(J+bR)$ and therefore isomorphic to $R/J'$ for some ideal $J'$ containing $J+bR,$ or even:

$aR/(J\cap aR)$ is isomorphic to $R/(J:a)$ and $J+bR \subset (J:a).$

If we start with a filtration $0=N_0\subset N_1\subset\dots \subset N_s = R/(J+bR)$ with quotients of the form $R/P$ for some prime $P,$ and then project this filtration to the quotient $aR/(J\cap aR),$ we would not necessarily get a filtration of $aR/(J\cap aR)$ with quotients of the form $R/P$ for some prime $P.$ But the text might indicate that this is the case.

On Laurent Moret-Bailly left comment #6764 on Lemma 57.9.1 in Fundamental Groups of Schemes

This should extend to henselian pairs, right? This would involve extending theorem 07QY (approximation) to henselian pairs $(R,I)$ where $R$ is a noetherian G-ring, and extending (part of) Lemma 07QR to show that the henselization of a noetherian G-ring along any ideal is a G-ring. Am I missing something?

On left comment #6762 on Lemma 39.6.3 in Groupoid Schemes

To deduce smoothness you'll need characteristic $0$ (or some other assumption), compare with Lemma 39.8.2.