A universally closed, universally injective, and unramified morphism is a closed immersion.

Here are some references. The result itself is here

SCHEMES: Lemma Tag 04XV
SPACES: Lemma Tag 05W8

The definition of an unramified morphisms is here

RINGS: Definition Tag 00UT
SCHEMES: Definition Tag 02G4
SPACES: Definition Tag 03ZH

Formally unramified morphisms are defined here

RINGS: Definition Tag 00UN
SCHEMES: Definition Tag 02H8
SPACES: Definition Tag 04G7

and a morphism which is formally unramified and locally of finite type is unramified, see here

SCHEMES: Lemma Tag 02HE
SPACES: Lemma Tag 04GA

Enjoy!

This post explains some algebra related to the remarks made in the previous post. I am confident that you can either find this in the literature (and that it has a name), or that it is completely wrong. Caveat emptor.

Let k be a field of characteristic p. Let G be a finite p-group. Let R = k[G]. Then R is a finite dimensional (possibly noncommutative) local k-algebra. Recall that a left R-module is the same thing as a k-vector space with a left G-action. Set ω_R = Hom_k(R, k) and think of it as a right R-module via the left multiplication of R on R. It is an injective right R-module. Note that both R^G  and ω_R^G are 1-dimensional k-vector spaces. Choose a nondegenerate pairing

< , > : R^G x ω_R^G —> k

This is the only choice we will have to make (the rest will be independent of choices I think). Let F be a finite free left R-module. Set I = Hom(R, k) which is an injective right R-module. Then the pairing above define a canonical pairing

< , > : F^G x I^G —> k

by choosing a basis of F and using < , > for the free rank 1 case and then proving that the resulting pairing is independent of choice of basis. Moreover, this pairing is suitably functorial.

Choose a resolution by finite free left modules

… —> F_1 —> F_0 —> k —> 0

This gives an injective resolution

0 —> k —> Hom(F_0, k) —> Hom(F_1, k) —>

by right modules. We denote I^n = Hom(F_n, k). By the above we have canonical pairings

< , > : F_n^G x (I^n)^G —> k

Hence a perfect pairing between H^i(G, k) and H^{-i}((F_*)^G). This is just an instance of the following more general procedure.

If M, N are right R-modules, then M ⊗_k N is a right R-module by using the diagonal right G-action.  Similarly for left modules. Note that the functor (M, N) —> M ⊗_k N is exact in both variable (so there won’t be any derived functors). For every finite right R-module M the tensor products M ⊗_k I_n are injective right R-modules, and the tensor products Hom(M, k) ⊗_k F_n are finite free left R-modules. (Prove by induction on the length of M). OK, so we see that

A^*  = (M ⊗_k I^0 —> M ⊗_k I^1 —> M ⊗_k I^2 —> …)

is an injective resolution of M. On the other hand

B_* = (… —> Hom(M, k) ⊗_k F_1 —> Hom(M, k) ⊗_k F_0)

is a free resolution of Hom(M, k). Using that Hom(Hom(M, k) ⊗_k F_n, k) = M ⊗_k I^n we see that there are nondegenerate pairings

< , > : (B_n)^G \times (A^n)^G —> k

which are compatible with the differentials in the complexes (B_*)^G and (A^*)^G. This means that H^i(G, M) is canonically dual to H_i((Hom(M, k) ⊗_k F_*)^G) which is “in some sense” equal to H^{-i}(G, Hom(M, k) ⊗_k F_*).

I think this means that morally speaking the complex F_* is a dualizing complex in this situation. Except that you cannot think of it as an object in the derived category (since it would just be k in degree 0), but you have to think of it as an actual complex, and you have to compute the cohomology of Hom(M, k) ⊗_k F_* in the manner described above. Another way to say this is that you might want to think of F_* as the system of stupid truncations

F(n) = (F_n —> F_{n – 1} —> … —> F_0)

sitting in cohomological degrees [-n, 0], The duality above gives canonical pairings between H^i(G, M) and colim_n H^{-i}(G, Hom(M, k) ⊗_k F(n)). This is I think a slightly improved version of what I said in the previous post.

This post is about Serre duality for smooth, proper, Deligne-Mumford stacks \cX over a field k, which came up recently in a phone conversation with Max Lieblich (but please don’t blame him for what I am writing here). Disclaimer: I haven’t yet had time to think carefully about cohomology on algebraic stacks, so what I say in this post may be completely wrong or besides the point! Moreover, it is also very likely that you (= the reader) have told me a vastly more general and superior theorem and I am repeating it here in some kind of warped way. Please let me know if so.

What I want — and it is quite possible that I shouldn’t want this — is for every locally free sheaf E with dual E^* on a smooth proper \cX over k and every integer i a nondegerate k-bilinear pairing

• H^i(\cX, E) x H^{-i}(\cX, \omega^*_\cX \otimes_{O_\cX} E^*) —> k

Here \omega^*_\cX is (maybe, see below) an object of the derived category D(\cX) of \cX and the pairing should come from a map

• Tr_\cX : H^0(\cX, \omega^*_\cX) —> k

via the cuproduct as usual. The complex \omega^*_\cX and the pairings should have more properties, but let’s ignore this for now.

Here is an example: Consider \cX = B(G) over a field k of characteristic p where G is a cyclic group of order p. Then we see that H^i(B(G), O_{B(G)}) = H^i(G, k) is zero for i < 0, but nonzero in all degrees i = 0, 1, 2, … Thus we see that the complex \omega^*_{B(G)} cannot be contained in D^{+}(X) since if it were then its cohomology groups H^{-i}(B(G), \omega^*_{B(G)}) would be zero for all sufficiently positive i! This is really the main point I wanted to make, and maybe you should stop reading now and have a beer instead (or tea).

Let me explain what I think \omega^*_{B(G)} is in case G = Z/2Z and the characteristic of k is equal to 2. In this case k[G] = k[e] with e^2 = 0. In this case the category of quasi-coherent O_{B(G)}-modules is equivalent to the category of k[e]-modules, the tensor product of O_{B(g)}-modules corresponds to tensoring over k(!), and H^0 corresponds to taking the kernel of e. An injective resolution of k is the complex

• k[e] — e –> k[e] — e –> k[e] — e –> …

and it is clear that if you take the kernel of e, then you get k in each nonnegative degree with zero maps. I think that \omega^* is the “k-linear dual” of this complex. But we have to be careful when we do this because we are working with unbounded complexes. Since my brain doesn’t appear to be functioning very well right now, let me just try to say what I am thinking (and you can leave a comment if you think this is wrong). I want to think of the infinite complex above as the limit of the complexes L_n^* which are the stupid truncations of the complex above in degrees [0, n]. Then I say that

• \omega^* = colim_n Hom_k(L_n^*, k)

for some notion of colimit of complexes. Why does this work? Well, I’m not sure it does, but I checked that it works for the two interesting modules I can compute the result for in this case, namely E = k and E = k[e]. Note that both modules are selfdual so it is easy to see what you get on both sides.

Presumably, the correct thing to do is to take the homotopy colimit or something in the formula for \omega^* above. But I think a nice way to think about it is that \omega^* simply isn’t a complex, but a system of complexes. The next thing to try would be to look at a case where \cX is a global quotient \cX = [X/G] for some smooth proper X over k. Note that \cX —> B(G) is a smooth proper morphisms. Hence in this case we can presumably let \omega^*_\cX be the tensor product of the pullback of the system \omega^*_{B(G)} just constructed and the usual dualizing sheaf of X placed in degree -dim(X). Right?

[Edit 21:15: Replaced limit by colimit and vice versa, as per the comment of Bhargav below.]

[Edit March 10, 2011: See next post for a bit more of the underlying algebra.]

It turns out that the result mentioned in this post is especially useful in theoretical considerations. For example consider the following statement

Given an 1-morphism \cX —> \cY of stacks in groupoids which is representable by algebraic spaces such that \cY is an algebraic stack, then \cX is an algebraic stack.

Proof: pick a scheme Y and a surjective smooth morphism Y —> \cY. By assumption the 2-fibre product Y x_{\cY} \cX is representable by an algebraic space X. The projection map X —> \cX is surjective and smooth as a base change and we win by the result mentioned above.

Of course the result can be proved in other ways as well, but it is quite pleasing how short the argument above is. This kind of thing is especially helpful because we intend to prove many results of this kind!

[Edit March 08: Here are some links to the result mentioned above and its improvement suggested by David Rydh in the comment below.]

Let f : X —> Y be a morphism over a base B. Let P be a property of morphisms. We often want to know that

1. there is a maximal open W of B such that the restriction f_W : X_W —> Y_W of f has property P, and
2. formation of W commutes with arbitrary base change B’ —> B.

Of course this usually isn’t the case without further assumptions on X,Y,f, and B. One of the reasons this type of result is useful, is that you can check whether a point b of B is in W by looking at the base change of the morphism f to a morphism f_b : X_b —> Y_b of schemes (or algebraic spaces or algebraic stacks) over the point b.

A well known and useful case is the following result

If X is proper, flat, of finite presentation over B, Y is proper over B, and P = “being an isomorphism”, then 1 and 2 hold. 

I recently added this to the stacks project for relative maps of algebraic spaces, see Lemma Tag 05XD. When you analyze the proof you find two more basic results that lead to the above. The first is that

If X is proper over B, Y is separated over B, and P = “being a closed immersion”, then 1 and 2 hold. 

see Lemma Tag 05XA. This first result is in some sense elementary (although its proof in the current exposition is not). The second is that

If X is proper, flat, of finite presentation over B, Y is locally of finite type over B, and P = “being flat”, then 1 and 2 hold. 

see Lemma Tag 05XB. The current proof of this second result uses the “critère de platitude par fibres” which is nontrivial. Does anybody know how to prove the result on the locus where f is an isomorphism without appealing to this criterion?

Let S be a scheme. There are many ways to turn the category of schemes over S into a site, but some of the things you can do lead to the same topos, i.e., the category of sheaves are identical. In this case we say these sites define the same topology. Here are some examples of comparisons of topologies:

1. The smooth topology and the etale topology are the same. See Lemma Tag 055V.
2. The fppf topology is the same as the one you get by considering fppf coverings {T_i —> T} such that each T_i —> T is locally quasi-finite. See Lemma Tag 0572.
3. The topology generated by Zariski coverings and {f : T —> S} with f surjective finite locally free is finer than the etale topoloy, see Lemma Tag 02LH and Remark Tag 02LI.
4. The fppf topology is the same as the one generated by Zariski coverings and finite surjective locally free morphisms. See Lemma Tag 05WN.

Somehow I hadn’t realized 4 earlier. What made me think of it today was this comment by David Rydh.