Stacks project -- Comments https://stacks.math.columbia.edu/recent-comments.xml Stacks project, see https://stacks.math.columbia.edu en stacks.project@gmail.com (The Stacks project) pieterbelmans@gmail.com (Pieter Belmans) https://stacks.math.columbia.edu/static/stacks.png Stacks project -- Comments https://stacks.math.columbia.edu/recent-comments.rss #5890 on tag 015E by Johan https://stacks.math.columbia.edu/tag/015E#comment-5890 A new comment by Johan on tag 015E. 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.

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Johan Mon, 18 Jan 2021 02:44:19 GMT
#5889 on tag 015E by Ingo Blechschmidt https://stacks.math.columbia.edu/tag/015E#comment-5889 A new comment by Ingo Blechschmidt on tag 015E. Can we remove the hypothesis that $RF$ is everywhere defined? At least in the bounded case it's obvious that we can.

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Ingo Blechschmidt Mon, 18 Jan 2021 01:02:27 GMT
#5888 on tag 01J6 by Zhenhua Wu https://stacks.math.columbia.edu/tag/01J6#comment-5888 A new comment by Zhenhua Wu on tag 01J6. @#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.

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Zhenhua Wu Sun, 17 Jan 2021 09:11:09 GMT
#5887 on tag 00DY by Abhiram Natarajan https://stacks.math.columbia.edu/tag/00DY#comment-5887 A new comment by Abhiram Natarajan on tag 00DY. Slight abuse of notation in point 11 of lemma 00E0. It should be V((f)).

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Abhiram Natarajan Fri, 15 Jan 2021 05:03:55 GMT
#5886 on tag 05QC by Anna Cadoret https://stacks.math.columbia.edu/tag/05QC#comment-5886 A new comment by Anna Cadoret on tag 05QC. 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.

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Anna Cadoret Thu, 14 Jan 2021 03:52:27 GMT
#5884 on tag 077N by Laurent Moret-Bailly https://stacks.math.columbia.edu/tag/077N#comment-5884 A new comment by Laurent Moret-Bailly on tag 077N. 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 $(i,k,m)$.

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Laurent Moret-Bailly Thu, 14 Jan 2021 03:11:19 GMT
#5883 on tag 0AK9 by Johan https://stacks.math.columbia.edu/tag/0AK9#comment-5883 A new comment by Johan on tag 0AK9. The proof of this lemma is wrong unfortunately. Please refrain from using this lemma for now.

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Johan Wed, 13 Jan 2021 06:31:49 GMT
#5882 on tag 01J6 by Johan https://stacks.math.columbia.edu/tag/01J6#comment-5882 A new comment by Johan on tag 01J6. @#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.

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Johan Mon, 11 Jan 2021 08:30:19 GMT
#5880 on tag 01J6 by Zhenhua Wu https://stacks.math.columbia.edu/tag/01J6#comment-5880 A new comment by Zhenhua Wu on tag 01J6. 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.

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Zhenhua Wu Mon, 11 Jan 2021 12:29:59 GMT
#5879 on tag 01J6 by Zhenhua Wu https://stacks.math.columbia.edu/tag/01J6#comment-5879 A new comment by Zhenhua Wu on tag 01J6. 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 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 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 $g=h$.

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