Comments 1 to 20 out of 4766 in reverse chronological order.


On Matthieu Romagny left comment #5126 on Section 99.26 in Morphisms of Algebraic Stacks

Typo in the section introduction : replace "spaces" by stacks" in the sentence "we may define what it means for a morphism of algebraic spaces..."

On Peng DU left comment #5125 on Lemma 10.22.2 in Commutative Algebra

May need add comma before "in other words".

On left comment #5124 on Lemma 56.12.6 in Derived Categories of Varieties

Yes, this is very confusing! Thanks and fixed here.

On left comment #5120 on Lemma 56.12.6 in Derived Categories of Varieties

The categories in the statement are claimed to be triangulated, but it needs abelian categories (and this is what it used in the proof anyway).

On slogan_bot left comment #5119 on Lemma 37.38.2 in More on Morphisms

Suggested slogan: Quasi-finite, separated morphisms are quasi-affine

Also, why is there an "(!)" in the proof?

On Weixiao Lu left comment #5118 on Section 21.2 in Cohomology on Sites

If $f: Sh(\mathcal C) \to Sh(\mathcal D)$ is a morphism of topoi, then $f_\ast(\mathcal F)$ might just be a sheaf of sets, not a sheaf of abelian groups, even if $\mathcal F$ is. Then how do we define $Rf_\ast(\mathcal F)$?

On J left comment #5117 on Section 56.2 in Derived Categories of Varieties

Typo: last paragraph $M\in D(\mathcal O_Y)$ not $\mathcal O_U$

On Laurent Moret-Bailly left comment #5116 on Lemma 62.3.2 in The Trace Formula

The meaning of the statement is not completely formal. To make sense of "the identity on cohomology" we need to show that we can identify $\mathcal{F}$ with $g_*(\mathcal{F})$ and/or $g^{-1}(\mathcal{F})$, for any $\mathcal{F}$. This is of course the case in subsequent lemmas where $\mathcal{F}$ is a constant sheaf.

On Tongmu He left comment #5115 on Lemma 53.20.12 in Algebraic Curves

It seems that in the statement of 53.20.12, we could add "the image of $a$ vanishes at the base point $v$ of $V$, and the base point $u$ of $U$ maps to the node of the fiber $W_v$", right?

On left comment #5114 on Section 5.10 in Topology

@James A. Myer: No. Every closed set with $\geq 2$ elements is reducible.

On pippo left comment #5113 on Section 62.3 in The Trace Formula

Typo: in (03SX) the first $\pi_X$ should be $\pi_x$.

On Mingchen left comment #5112 on Equation 90.6.0.2 in The Cotangent Complex

g should be from Sh(C) to Sh(C')

On left comment #5110 on Lemma 66.9.1 in Decent Algebraic Spaces

Dear Shiji, thanks for finding this error. There is a way of fixing this using limit arguments, but that would mean pushing this much later in the project. Another fix is the following.

We have to show: Given a quasi-compact open immersion $j : V \to Y$ of a quasi-separated and quasi-compact algebraic spaces and an integral morphism $\pi : Z \to V$ we can extend $\pi$ to an integral morphism $\pi' : Z' \to Y$ in the sense that $Z$ is isomorphic to the inverse image of $V$ in $Z'$. To do this, let $\mathcal{A}$ be the quasi-coherent sheaf of $\mathcal{O}_V$-algebras on $V$ corresponding to $\pi : Z \to V$, in other words, $\mathcal{A} = \pi_*\mathcal{O}_Z$. Pushforward along $j$ preserves quasi-coherence. Hence $j_*\mathcal{A}$ is a quasi-coherent $\mathcal{O}_Y$-algebra. Now we let $\mathcal{A}' \subset j_*\mathcal{A}'$ be the integral closure of $\mathcal{O}_Y$. By Lemma 29.51.1 (translated over to the category of algebraic spaces; details omitted) we see that $\mathcal{A}'$ is a quasi-coherent $\mathcal{O}_Y$-algebra and we see that $\mathcal{A}'|_V \cong \mathcal{A}$ because the integral closure of $\mathcal{O}_V$ in $mathcal{A}$ is $\mathcal{A}$ as $\pi$ is integral! Thus $Z' = \underline{\text{Spec}}_Y(\mathcal{A}') \to Y$ is the answer to our problem.

On Shiji Lyu left comment #5109 on Lemma 66.9.1 in Decent Algebraic Spaces

In the last paragraph we applied Lemma 0ABS to the morphism $Z \to Y$. However, it seems that this morphism is not necessarily quasi-finite since the morphism $Z \to V$ is just integral, not finite. Is there a way to resolve this?

On Tongmu He left comment #5108 on Lemma 53.19.13 in Algebraic Curves

Typo: in the proof of 53.19.13, the map $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}$ should be $\mathcal{O}_{X, x}^\wedge \to \mathcal{O}_{Y, y}^\wedge$.

On typo_bot left comment #5107 on Lemma 15.8.6 in More on Algebra

In (3), element' should beelements'. In (2) it would be good to parenthesize the argument of dim, even though one could argue that no confusion is possible and that this is a matter of taste.

On anon left comment #5106 on Lemma 10.119.7 in Commutative Algebra

If $x/1\in A[S^{-1}]$ is prime then doesn't ideal correspondence for localizations imply $x \in A$ is prime?

On Jordan Levin left comment #5105 on Section 10.130 in Commutative Algebra

There is a nice discussion of the exact sequences using only the universal properties in a down-to-Earth way in the book "Commutative Algebra" by Singh on page 249.

On Noah Olander left comment #5104 on Lemma 90.13.2 in The Cotangent Complex

I think the method of proof here works immediately for Koszul regular sequences (and is even a little simpler since you don't need the induction and you don't have to argue flat locally in the next lemma). Just compute $L_{B/A}$ for $A = \mathbf{Z} [x_1, \dots , x_r]$, and then if $(f_1, \dots , f_r)$ is a Koszul regular sequence on $C$, then $C / (f_1, \dots , f_r) = B \otimes _A ^{\mathbf{L}} C$.

On left comment #5103 on Lemma 48.18.3 in Duality for Schemes

Argh! It seems you are correct. This just is a terrible lemma. Luckily it seems we only use it once!