In Groupes de Brauer II, Remark 1.11(b) Grothendieck notes that results of Mumford’s paper “The topology of normal singularities of an algebraic surface and a criterion for simplicity” gives one an example of a normal surface Y over the complex numbers such that H^2(Y, G_m) isn’t torsion and does not inject into H^2(C(Y), G_m). Grothendieck even references a page number, namely 16. To explain this in the graduate student seminar on Brauer groups this semester I came up with the following, which may be what Grothendieck had in mind.

Let E ⊂ P^2 be a smooth degree 3 curve. Let P ∈ E be a flex point. Blow up P exactly 10 times on E, i.e., blow up P in P^2, then blow up P on the strict transform of E, etc. The result is a surface X with an embedding E ⊂ X such that

1. the self square of E in X is -P, and
2. the image of the map Pic(X) —> Pic(E) is contained in ZP.

This means you can blow down E on X to get a normal projective surface Y with a unique singular point y. Part 2 implies that the local ring of O_{Y, y} is factorial (this is one of Grothendieck’s claims — in fact we won’t need it). Now look at the Leray Spectral Sequence for G_m and the morphism f : X —> Y. You get something like

Pic(X) —> H^0(Y, R^1f_*G_m) —> H^2(Y, G_m) —> H^2(X, G_m)

We have R^1f_*G_m = Pic(E) placed at y and H^2(X, G_m) = 0 as X is a smooth projective rational surface. Using 1 and 2 above we conclude that H^2(Y, G_m) = E as abelian groups. By Gabber’s result on Brauer groups of quasi-projective schemes it follows that Br(Y) = E_{tors}. Of course both H^2(Y, G_m) and Br(Y) map to zero in the Brauer group of the generic point.

This is just a quick note on the paper Brown representability does not come for free by Casacuberta and Neeman. This is going to be completely bare bones as you can read more details in the paper.

We are going to define a “big” abelian category A as follows. An object of A consists of a pair (M, α, s_β) where M is an abelian group and α is an ordinal and s_β : M —> M is a commuting family of homomorphisms parametrized by β ∈ α. A morphism (M, α s_β) —> (N, γ, t_δ) is given by a homomorphism of abelian groups f : M —> N such that f(s_β(m)) = t_β(f(m)) for any ordinal β where the rule is that we set s_β equal to zero if β is not in α and similarly we set t_β equal to zero if β is not an element of γ.

A special object is Z = (Z, 0, ∅), i.e., all the operators are zero. The observation is that computed in A the “group” Ext^1_A(Z, Z) is a proper class and not a set. Namely, for each ordinal β we can find an extension M of Z by Z whose underlying group is M = Z ⊕ Z and where s_β acts by a nonzero operator s_β, e.g. via the matrix (0, 1; 0, 0). This clearly produces a proper class of isomorphism classes of extensions.

In my world forming the category D(A) doesn’t make sense because the Hom’s aren’t sets. Another conclusion is that in K(A) the subcategory of acyclic complexes does not give rise to a Bousfield localization or colocalization.

Scarier than Halloween?