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Stacks project, see http://stacks.math.columbia.eduenstacks.project@gmail.com (Stacks Project)pieterbelmans@gmail.com (Pieter Belmans)http://stacks.math.columbia.edu/stacks.pngStacks project -- Comments
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#2552 on tag 051E by Johan
http://stacks.math.columbia.edu/tag/051E#comment-2552
A new comment by Johan on tag 051E.Dear Stella, Lemma \ref{051F} is not true when the module isn't flat. Please read carefully. Everybody: if you have a comment about a specific lemma, then please leave the comment on the tag page of the lemma.
]]>JohanTue, 23 May 2017 23:52:15 +0000#2551 on tag 051E by Stella Gastineau
http://stacks.math.columbia.edu/tag/051E#comment-2551
A new comment by Stella Gastineau on tag 051E.I think you can remove the condition that $M$ is flat from Lemma 10.100.1 for a much cleaner proof: If $M$ is any $R$-module and if ${\bar{x}\alpha}$ forms a basis on $M/\mathfrak{m}M$, then let $N$ be the submodule of $M$ generated by the $x\alpha$. Then the composition $N\hookrightarrow M\to M/\mathfrak{m}M$ is surjective and so $N+\mathfrak{m}M=M$ ie $\mathfrak{m}(M/N)=M/N$. Then by nilpotency of $\mathfrak{m}$, we have that $$M/N=\mathfrak{m}(M/N)=\cdots=\mathfrak{m}^n(M/N)=0$$ and so the ${x_\alpha}$ form a generating set for $M$. (These are the standard techniques for the different forms of Nakayama's lemma, but we can remove the requirement that $M$ is finitely generated because we have $\mathfrak{m}L=L\Rightarrow L=0$ for all $R$-modules when $R$ is local Artinian. You still need to use Axiom of choice to find such a $x_\alpha$, but that is used anyways in Lemma 10.100.2.)
]]>Stella GastineauTue, 23 May 2017 18:35:09 +0000#2550 on tag 0DUV by Johan
http://stacks.math.columbia.edu/tag/0DUV#comment-2550
A new comment by Johan on tag 0DUV.Exactly what you say happens a few times in this section: given a cartesian square with a uniform categorical moduli space for the downward right vertical arrow and the lower rightwards horizontal arrow is flat and representable by algebraic spaces, then the left downward vertical is a uniform categorical moduli space too. Although I think we need to leave this kind of stuff to the reader in general, I've made the change you requested in this case. See this commit.
]]>JohanMon, 22 May 2017 18:30:51 +0000#2549 on tag 0DUV by Matthew Emerton
http://stacks.math.columbia.edu/tag/0DUV#comment-2549
A new comment by Matthew Emerton on tag 0DUV.The statement of the lemma doesn't say that $\mathcal X' \rightarrow M'$ is as in the Keel--Mori theorem, although the proof starts with that set-up. Maybe this could be put into the lemma statement too? (I guess it's automatic, given the rest of the lemma, b/c the base-change of a uniform categorical moduli space is again such a thing, but might be worth making explicit?)
]]>Matthew EmertonMon, 22 May 2017 17:15:28 +0000#2548 on tag 01LG by Dario Weißmann
http://stacks.math.columbia.edu/tag/01LG#comment-2548
A new comment by Dario Weißmann on tag 01LG.In 26.2.1 and 26.2.2: I don't think the $X$ constructed is unique, just unique up to unique isomorphism.
Noah Olander's comment should also apply to 26.2.2.
]]>Dario WeißmannSun, 21 May 2017 22:39:00 +0000#2547 on tag 0BI1 by sdf
http://stacks.math.columbia.edu/tag/0BI1#comment-2547
A new comment by sdf on tag 0BI1.I think this name was first used by Grothendieck in EGA. My guess is it in honor of the strong Tokyo school, Nakayama, Takagi, Nagata and the many others there doing things around representation theory around 1920s-1960s
]]>sdfWed, 17 May 2017 18:17:29 +0000#2546 on tag 0BI1 by Ashwin Iyengar
http://stacks.math.columbia.edu/tag/0BI1#comment-2546
A new comment by Ashwin Iyengar on tag 0BI1.Where did the name "Japanese" come from?
]]>Ashwin IyengarWed, 17 May 2017 14:23:05 +0000#2545 on tag 002V by Ingo Blechschmidt
http://stacks.math.columbia.edu/tag/002V#comment-2545
A new comment by Ingo Blechschmidt on tag 002V.Is there a clash of notation here? The chapter on limits of schemes speaks of "directed limits", by which limits over directed sets are meant (and not more general limits over filtered categories). Here, however, the terms "directed" and "filtered" are used interchangeably.
]]>Ingo BlechschmidtTue, 16 May 2017 08:36:04 +0000#2544 on tag 0BQS by Pieter Belmans
http://stacks.math.columbia.edu/tag/0BQS#comment-2544
A new comment by Pieter Belmans on tag 0BQS.It should say vector bundles, not vectorbundles.
]]>Pieter BelmansTue, 16 May 2017 07:14:02 +0000#2543 on tag 033H by Grayson Jorgenson
http://stacks.math.columbia.edu/tag/033H#comment-2543
A new comment by Grayson Jorgenson on tag 033H.There appears to be a minor typo in the proof of Lemma 27.7.9: the summations are indexed $i = 1,...,d$, but should be indexed $i = 0,...,d-1$ or the exponents changed from $f^i$ to $f^{d - i}$.
]]>Grayson JorgensonMon, 15 May 2017 22:59:24 +0000