Recent comments—The Stacks project

Comments 1 to 20 out of 5091 in reverse chronological order.

On Aug 12, 2020 Sandeep left comment #5464 on Section 10.36 in Commutative Algebra

@Johan Yes, you are right, the argument is faulty.

On Aug 11, 2020 Leo left comment #5463 on Lemma 53.24.2 in Algebraic Curves

Let $P \subseteq X$ be the union of rational bridges and tails. Is the composite $X \to X_1 \to \cdots X_{n+n'}$ equal to the pushout $\pi_0 P \bigsqcup_P C$ ? ( $\pi_0 P$ is a copy of the base field for each connected component).

It seems that the iterative construction for each $X_i \to X_{i+1}$ in both cases 0E3M and 0E3H is the pushout of $X_i$ along some unstable component mapping to the base field, which one could do in one step as above. This universal property might make it easier to study the stabilization in families if it holds.

A related question is whether stabilization is functorial: If I have an automorphism $X \to X$ , does this induce an automorphism of $Y$ ?

On Aug 11, 2020 Anonymous left comment #5462 on Lemma 15.9.10 in More on Algebra

I do believe proof should instead say something like " $D(b_1)$ and $D(b_2)$ form a disjoint union because $b_1b_2=0$ and cover $\mathrm{Spec}(B)$ because $(b_1,b_2)=B$ "

On Aug 11, 2020 Johan left comment #5461 on Section 10.36 in Commutative Algebra

@Sandeep: Why is $g \cdot \frac{1}{r}R \subset \frac{1}{r}R$ ?

On Aug 11, 2020 Sandeep left comment #5460 on Section 10.36 in Commutative Algebra

I don't think you need Noetherianness for the converse part of Lemma 10.36.4. Since, $g(1/rR)\subset 1/rR$ and $1/rR$ is finitely generated, use Cayley-Hamilton to conclude that there exists some monic polynomial $p$ such that $p(g)1/rR=0$ , implying in particular that $p(g)=0$ .

On Aug 10, 2020 Pavel left comment #5459 on Lemma 10.74.8 in Commutative Algebra

what is this in the end of the proof? there is a long exact sequence of tors with three tensor products being the last three nonzero terms of it, AND if they all the tors vanish we are left with short exact sequence of products, hence M is flat, it's just simple as that. You don't even have to consider some ideals of the ring, it doesn't change anything. Link to lemma at the very end leads to some immense amount of logic manipulations as well

On Aug 10, 2020 Du left comment #5458 on Lemma 30.5.2 in Cohomology of Schemes

Lemma 02N7 requirs $g^\prime$ to be flat. So this assumption should be added.

On Aug 10, 2020 Du left comment #5457 on Lemma 20.17.1 in Cohomology of Sheaves

It looks like there is not need to assume $g$ to be flat. I went over the proof and the lemmas it uses, only Lemma 02N5 needs $g^\prime$ to be flat. $g$ shows up at the ending part of the proof, but no place indicates that $g$ needs to be flat.

On Aug 09, 2020 Brian Shih left comment #5456 on Section 10.115 in Commutative Algebra

Should it be d instead of r in the proposition 00P0?

On Aug 08, 2020 James A. Myer left comment #5455 on Section 5.10 in Topology

Oh yeah, sorry: I thought "the usual Euclidean space $\mathbb{R}^n$ " meant the scheme $\mathbb{A}^n_{\mathbb{R}}=\text{Spec}(\mathbb{R}[x_1,...,x_n])$ , in which case there is a chain of irreducible closed subsets given by $Z_0=\mathbb{V}((x_1,...,x_n))\subset Z_1=\mathbb{V}((x_1,...,x_{n-1}))\subset...\subset Z_{n-1}=\mathbb{V}((x_1))\subset Z_n=\mathbb{V}((0))\subset\mathbb{A}^n_{\mathbb{R}}$ of length $n$ , but of course it makes more sense to interpret "the usual Euclidean space $\mathbb{R}^n$ " as $\mathbb{R}^n$ equipped with the Euclidean topology.

On Aug 08, 2020 Matt Larson left comment #5454 on Lemma 33.47.1 in Varieties

There is a missing parenthesis in the second paragraph.

On Aug 06, 2020 R left comment #5453 on Lemma 42.37.7 in Chow Homology and Chern Classes

Typo: "assignes" $\to$ "assigns".

On Aug 06, 2020 Pieter Belmans left comment #5452 on Theorem 97.5.12 in Quot and Hilbert Spaces

There is an article 'a' missing in the second sentence of the statement.

On Aug 06, 2020 Pieter Belmans left comment #5451 on Lemma 97.10.1 in Quot and Hilbert Spaces

Then should be The in the statement.

On Aug 06, 2020 Laurent Moret-Bailly left comment #5450 on Lemma 33.47.3 in Varieties

Typos in proof: Line 2: "if $Z(s)$ is disconnected..." Exact sequence: ${Z(s)}$ should be ${Z(s^n)}$ . Line -3: "either $H^0(\dots)$ or $H^1(\dots)$ ".

On Aug 05, 2020 Johan left comment #5449 on Lemma 33.47.3 in Varieties

Here $s$ has a section of $\mathcal{L}$ !

On Aug 05, 2020 Noah Olander left comment #5448 on Proposition 97.11.8 in Quot and Hilbert Spaces

Ah, great! Thanks!

On Aug 05, 2020 Pieter Belmans left comment #5447 on Proposition 97.11.8 in Quot and Hilbert Spaces

This is done in Lemma 106.9.3, the section it is contained in discusses more geometric properties of the Picard functor.

On Aug 05, 2020 Noah Olander left comment #5446 on Proposition 97.11.8 in Quot and Hilbert Spaces

Should probably add that $\mathrm{Pic} _{X/B} \to B$ is locally of finite presentation in this situation. I think this follows from the proof but it's a little bit hard to tell.

On Aug 05, 2020 Hao left comment #5445 on Lemma 59.18.4 in Crystalline Cohomology

In the proof, "sheaves are presheaves" is somehow confusing. Maybe "any presheaf is a sheaf" is better.

What is the definition of "total complex" in the statement?