Leaving comments

If you want to leave a comment about a lemma, definition, remark, etc, please click through to the page of the lemma, definition, remark, etc and then leave a comment. Only leave a comment on a section if your comment is about the whole section or about text which is not in any environment, e.g., it is about the introduction to the section. This should be pretty rare.

If you have a longish comment involving mathematics and especially latex code, it is helpful to send a copy of your comment to the email address for the Stacks project (stacks.project..at..gmail.com). This will make it much faster for me to incorporate your suggested changes.

Technologically advanced people can use github and do a pull request.

Finally, if you don’t want to comment in public, then you can email the stacks project at the address mentioned above. Of course, I will be the one dealing with your email, since I am currently the maintainer of the project. So you could also decide to email to my personal email address. But if you do so, then there is a chance that I will overlook your email due to the large volume of incoming email in my personal email account. So I strongly prefer emails addressing issues with the Stacks project to be sent to the Stacks project email address.

Thanks for all your comments!


One of the original goals of the Stacks project was to work through most of the “preliminary” material in the paper of Deligne and Mumford. Here I mean the material on algebraic stacks and on moduli stacks of curves, before one actually gets to the “interesting” part, namely, why the moduli stack of curves of a given genus is irreducible. This is now done. Currently the last theorem of the Stacks project is about how the moduli stack Mgbar is a proper and smooth Deligne-Mumford stack over Z for g >= 2.


PS: I will make an effort to write more frequently here about what is going on with the Stacks project. In particular, I should write about the very successful Stacks project workshop which we just had, about what is next in line to be put in the Stacks project, about the wonderful people who help out with the Stacks project, and about how we’d like more people to help Pieter Belmans to code up parts of the new Stacks project web site!

6000 pages

OK, as promised here this means a party! It will be at my house on Saturday evening. Drop me an email if you are in town and I’ll give you directions, etc. See you there!

Yet another update and …

OK, I’ve worked through your comments and made changes in the LaTex. Later today I will update the Stacks project so the changes will be reflected online too. Thanks to everybody.

Also, we’ve reached 6000 pages which means we’re going to have a party soon (similarly to when we reached 5000 pages). Keep watching this blog to learn more.

Updated again

Thanks for all your comments on the Stacks project. I especially enjoy the comments pointing out actual mathematics errors and even more those that calmly explain what went wrong and how to fix it. But all comments are good. We now have more than 250 people who have contributed a bit (and some contributed a lot). Thanks to all.

Challenge Accepted

Remember this challenge? Probably not. But wait, don’t click! Namely, I will do something more general in this post.

Suppose we have a ring A and a contravariant functor F on (Sch/A) with the following properties:

  1. F satisfies the sheaf property for fpqc coverings
  2. the value of F on a scheme is either a singleton or empty
  3. for every quasi-compact scheme T/A such that F(T) is nonempty, there is an ideal I of A such that F(Spec(A/I)) is nonempty and such that T —> Spec(A) factors through Spec(A/I).

Example: A = k[x, y] for a field k and F(T) is nonempty if and only if the generic point of Spec(A) is not in the image of T —> Spec(A). Here F is not a representable functor.

I’d like to add some conditions that guarantee that F is representable by a closed subscheme of Spec(A). Here is what I just came up with; I think it is obviously correct and the right thing to do. If A is Noetherian we add the following two conditions

  1. If s_1, s_2, s_3, … is an infinite sequence of points of Spec(A) such that F(s_i) is nonempty and s is a limit point of the sequence, then F(s) is nonempty.
  2. If A —> …. —> A_n —> A_{n – 1} —> … —> A_1 are surjections such that the kernels A_n —> A_{n – 1} are locally nilpotent ideals and F(Spec(A_i)) is nonempty, then F(Spec(A_∞)) is nonempty where A_∞ = lim A_n.

I leave it as an exercise to show that 1 — 5 imply F is representable in the desired manner. If A is not Noetherian, then somehow these should still be enough although maybe you need to replace the natural numbers by a bigger directed set.

Why am I excited by this observation? It is because I want to apply this to the situation of the other blog post mentioned above: X is an algebraic space of finite presentation over A, u : H —> G is a map between quasi-coherent O_X-modules. We assume G is flat over A, of finite presentation, and universally pure relative to A (this is a technical condition which is satisfied if the support of G is proper over A). The functor F is defined by F(T) is nonempty if and only if the base change u_T of u is zero.

Properties 1, 2, 3 hold for F and are easy to prove. The proof of property 4 still doesn’t use purity of G relative to A (I think because we already have 3 it follows from an argument using generic freeness, but I also have an argument using \’etale localization). The key is to prove property 5.

To see 5 is true, I argue as follows. Suppose that the base change u_∞ to A_∞ is nonzero. Choose a weakly associated point ξ of the image of u_∞. This is also a weakly associated point of G_∞. The image t’ of ξ in Spec(A_∞) specializes to a point t in V(I_1) = Spec(A_1) because I_1 is contained in the radical of A_∞. Because G is universally pure relative to A, there is a specialization θ of ξ which lies over t (indeed this is the definition of being pure relative to the base). Then since u_∞ is zero at θ (in a suitable \’etale neighbourhood Edit: Argh… I just discovered this doesn’t work!) it is zero at ξ, a contradiction.


PS: A finitely presented module G on X flat and pure over A is universally pure relative to A. However, this is harder to prove than the above and it is easy to see that support proper over A implies universal purity.

Limit preserving and finite presentation

Let A —> B be a ring map. Let R = colim R_i be a filtered colimit of A-algebras. Then there is a canonical map

colim Hom_A(B, R_i) ——> Hom_A(B, R)

By Tag 00QO the following are equivalent

1. A —> B is of finite presentation,
2. the map above is bijective for all R = colim R_i
3. the map above is surjective for all R = colim R_i

Let S be a scheme. Let X be a scheme over S. Let T = lim T_i be a directed limit of affine schemes over S. Then there is a canonical map

colim Mor_S(T_i, X) ——> Mor_S(T, X)

By Tag 01ZC and Tag 0CM0 the following are equivalent

1. X —> S is locally of finite presentation,
2. the map above is bijective for all T = lim T_i
3. the map above is surjective for all T = lim T_i

The same thing is true if X and S are algebraic spaces (Tag 04AK and Tag 0CM6).

I didn’t know you could replace bijectivity by surjectivity in the criterion. But somewhere in the Stacks project we used this fact without proof, so it had better be true, right?

A related result is that to check a morphism f of algebraic stacks is locally of finite presentation, you need only check f is limit preserving on objects (this is the analogue of the above and it says that certain functors are essentially surjective). You can find this in Tag 0CMQ.

Caveat: as this only applies to situations where you already know your functors (or stacks in groupoids) are algebraic spaces (or stacks), it probably won’t be that useful. Often when we try to show a stack is limit preserving, it is part of applying Artin’s criteria and then we don’t yet know our stack is algebraic of course.

Thanks for reading!

[Edit on 6/30/2016: Matthew Emerton just pointed out that this observation was already in Lemma 2.3.15 of his paper with Toby Gee. I must have read it and then forgotten that I had. Apologies to everybody.]