Stacks project -- Comments

#5890 on tag 015E by Johan

Johan — Mon, 18 Jan 2021 02:44:19 GMT

Let $A^\bullet$ be a bounded below complex of acyclic objects. I think we can replace the assumption that $RF$ is everywhere defined by the assumption that $RF$ is defined at $A^\bullet$ . I will make this change the next time I go through all the comments.

I do not see is how to prove that $RF$ is defined at $A^\bullet$ . The problem, I think, is the following. Let $s : A^\bullet \to M^\bullet$ be a quasi-isomorphism. Then I think the arguments in the proof and Lemma 13.5.8 allow us to conclude that given an integer $n$ there is a quasi-isomorphism $s' : M^\bullet \to N^\bullet$ such that the composition $F(A^\bullet) \to F(M^\bullet) \to F(N^\bullet)$ is an isomorphism on cohomology in degrees $\leq n$ . But in order to show that $RF$ is defined at $A^\bullet$ with value $F(A^\bullet)$ we would need to show that we can do this in all degrees simultaneously!

Thanks for the interesting question.

#5889 on tag 015E by Ingo Blechschmidt

Ingo Blechschmidt — Mon, 18 Jan 2021 01:02:27 GMT

Can we remove the hypothesis that $RF$ is everywhere defined? At least in the bounded case it's obvious that we can.

#5888 on tag 01J6 by Zhenhua Wu

Zhenhua Wu — Sun, 17 Jan 2021 09:11:09 GMT

@#5882, denote the functor as $c$ , for me the problem of confusion is that the relation between $c$ restricted to open affine $U$ and $c$ on $X$ is not clear. For example, before we prove that $U\hookrightarrow X$ is a monomorphism in the category of schemes in tag 01L7 (so for this tag not proved yet), we cannot use the fact $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ injects into $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$ . So in this case I prefer we emphasis the difference between the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ and the set $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$ . Actually I would drow a square for it if I know how to draw it in this editor.

And we don't really need the injection in this proof. For the shortest suggestion on the proof, I would add something like:

We proved every element of $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,X)$ can be lifted to an element in $\mathop{\mathrm{Hom}}(\mathop{\mathrm{Spec}}R,U)$ for some open affine $U$ . Providing $c$ is functorial. Then the injectivity of $c$ on $X$ reduces to the injectivity of $c$ on $U$ , which is already proved.

#5887 on tag 00DY by Abhiram Natarajan

Abhiram Natarajan — Fri, 15 Jan 2021 05:03:55 GMT

Slight abuse of notation in point 11 of lemma 00E0. It should be V((f)).

#5886 on tag 05QC by Anna Cadoret

Anna Cadoret — Thu, 14 Jan 2021 03:52:27 GMT

This is a stupid comment but the notion of kernel in 10.12.5 is only introduced for an additive functor between Abelian categories while, here, it is used in the more general context of an additive functor between additive categories.

#5884 on tag 077N by Laurent Moret-Bailly

Laurent Moret-Bailly — Thu, 14 Jan 2021 03:11:19 GMT

In the statement, delete the second (or the third) occurrence of "quasi-coherent". Also, I had some trouble understanding the definition of $S'_j$ . For more clarity, I think the union should run over triples $(i,k,m)$ .

#5883 on tag 0AK9 by Johan

Johan — Wed, 13 Jan 2021 06:31:49 GMT

The proof of this lemma is wrong unfortunately. Please refrain from using this lemma for now.

#5882 on tag 01J6 by Johan

Johan — Mon, 11 Jan 2021 08:30:19 GMT

@#5879: In stead of giving your own proof, could you comment on what sentence(s) of the current proof should be clarified? I'm a bit surprised you are complaining about the deduction of the general case from the affine case, as that seems to me is explained quite well.

@#5880. Yes. I'm sorry about the mismatch between the editor and what readers get to see. As you say, this has been a problem for a while. We don't quite know how to fix this.

#5880 on tag 01J6 by Zhenhua Wu

Zhenhua Wu — Mon, 11 Jan 2021 12:29:59 GMT

The typo above is an example of this issue \ref{https://stackoverflow.com/questions/29014573/mathjax-how-to-deal-with-this-strange-behavior/29040570#29040570}. Note that adding underscore after right curly bracket instantly turned into italc form (you can see the italic button on the tool row becomes highlighted). I added backslash sign before underscore sign to make it looks right in preview. But now in real output it screws up. Let me test again. One backslash: $\mathcal{O}\_{X,x}, \mathcal{O}\_{X,x}$ . Correct in preview, probably wrong in real output. No backslash: $\mathcal{O}_{X,x}, \mathcal{O}_{X,x}$ . Wrong in preview, probably correct in real output.

#5879 on tag 01J6 by Zhenhua Wu

Zhenhua Wu — Mon, 11 Jan 2021 12:03:54 GMT

I don't see how bijection is proved in the general scheme case. It's not that the proof is wrong, but morelike it's not enough. The rest of the proof still needs some non-trivial logic. $\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Spec}{Spec}$ Define $c:\Hom(\Spec R,X)\to \\{ (x,\varphi),x\in X,\varphi\in \Hom_{local}(\mathcal{O}\_{X, x},R)\\},f\mapsto (f(\mathfrak{m}),f^\sharp)$ as above. We first show it's surjective. Given any $(x,\varphi)$ , there exists an open affine $U=\Spec A$ containing $x$ , and a map $A \to \mathcal{O}\_{X, x}$ , by composing with $\varphi:\mathcal{O}\_{X, x}\to R$ , we have $f:\Spec R\to \Spec A\to X$ . It's easy to see $c(f)=(x,\varphi)$ . Next we show it's injective. Given any two maps $g,h:\Spec R\to X$ , s.t. $c(g)=c(h)=(x,\varphi)$ . We know $g(\mathfrak{m})=h(\mathfrak{m})=x$ . Pick an open affine $U=\Spec A$ containing $x$ we know both $g$ and $h$ factor through $U$ using the orginal argument. So we have $g_0,h_0:\Spec R\to U$ . It suffices to show $g_0=h_0$ . Clearly $c$ is functorial w.r.t. the open immersion $U\to X$ , and the canonical map $\\{ (x,\varphi),x\in U,\varphi\in \Hom_{local}(\mathcal{O}\_{U, x},R)\\}\to \\{ (x,\varphi),x\in X,\varphi\in \Hom_{local}(\mathcal{O}\_{X, x},R)\\}$ is an identity map. We have shown $c_U$ is a bijection (defined as $c$ with $U$ replacing $X$ ). So $c\_U (g\_0)=c\_U(h\_0)$ implies that $g\_0=h\_0$ , which in turn implies $g=h$ .