# is generically split Azumaya algebra locally split?

Let $A$ be an Azumaya algebra over a scheme $X$ (or maybe more specifically a scheme of finite type over a field). Suppose that the restriction of $A$ to $U=X\setminus Z$ (where $Z$ is a closed set) is split, i.e. isomorphic to $M_n(\mathcal{O}_U)$.

Let $x \in Z$. Is it necessarily true that there exists a Zariski open neighbourhood $U' \ni x$, such that the restriction of $A$ to $U'$ is split?

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No, this is not true; the earliest counterexample I know of is described by Auslander and Goldman, "The Brauer group of a commutative ring", Trans. AMS 97 (1960), pp. 367–409, available freely online from the AMS. Given that I think they were the first to define the Brauer group of a ring, there is unlikely to be an earlier example. They prove (Theorem 7.2 on page 388) that, whenever $R$ is a regular ring, the Brauer group of $R$ injects into the Brauer group of the function field (and so your statement is true if $X$ is regular). They then give the following counterexample when $R$ is not regular.

Take $X$ to be the scheme $x^2+y^2=0$ over the real numbers, and the Azumaya algebra to be simply the usual quaternions, considered as an Azumaya algebra on $X$. Take $Z$ to be the origin. On $U$, the function $y$ is invertible, so in the coordinate ring of $U$ we have $(x/y)^2 = -1$ and the algebra splits. But the algebra does not split in any neighbourhood of $(0,0)$, since evaluating it at $(0,0)$ gives a non-split algebra.

If you don't like the fact that $X$ is geometrically reducible, there are plenty of other examples of singularities where the Brauer group of the local ring fails to inject into the Brauer group of the function field. In his articles on the Brauer group, Grothendieck refers to an example of Mumford where the Brauer group of the local ring is not even torsion, in contrast to the Brauer group of the function field which must always be torsion.

Edit:

You asked whether it is enough that $X$ be factorial. Grothendieck showed, in the second of his three articles on the Brauer group, that what you want is true if all the étale local rings of $X$ are factorial. Briefly, this is because you get an exact sequence

$0 \to H^1(X,\mathcal{D}iv) \to H^2(X,\mathbb{G}_m) \to H^2(k(X),\mathbb{G}_m)$

where $\mathcal{D}iv$ is the sheaf of Cartier divisors on $X$; the hypothesis means that this is the same as the sheaf of Weil divisors, whose cohomology vanishes.

I have been trying to think of an example where $X$ is factorial but not étale-locally factorial, giving non-trivial Brauer group. I'm sure such things are well known to the geometers, but I wonder whether something like this works: take an affine nodal cubic curve $y^2 = x^2(x+1)$ and rotate it about the $x$-axis. It feels like you should get a singularity which is a simple double point, but where the non-Cartier divisor class only reveals itself after an étale localisation (adjoining a square root of $x+1$). This would then lead to a class of order 2 in $\mathrm{Br} X$ which is trivial in $\mathrm{Br} k(X)$.

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Do I understand it correctly that a sufficient condition for the indjectivity is that every Weil divisor is a Cartier divisor? –  Dima Sustretov Mar 13 '12 at 10:37
No, I'm pretty sure that condition is not sufficient, but I don't have a counterexample off the top of my head. It is closely related to this question: <mathoverflow.net/questions/14961/…;. (Not sure how to put links in comments.) –  Martin Bright Mar 13 '12 at 13:24
Of course it is true if $X$ is regular, by the Grothendieck-Serre conjecture for $\textbf{PGL}_n$. Also you should check out the "purity theorem" in Grothendieck's "Groupes de Brauer III": quite often the result is true if $Z$ has codimension at least $2$. –  Jason Starr Mar 14 '12 at 18:09
Thank you, @Martin and @Jason. By the way (I might consider posting it as a separate question), are there any results similar to Auslander-Goldman, along the lines "restriction to the generic points induces injective map on Brauer groups" for complex analytic spaces? Of course there is no such thing as "generic point" for complex spaces, so the precise statement is also a part of the question. –  Dima Sustretov Mar 14 '12 at 21:13
@Jason - out of interest, are the situations where X is not regular, but some version of the purity theorem is known (or conjectured) to hold? –  Martin Bright Mar 15 '12 at 7:41