Assume we are given a Brauer class $\xi\in Br(k(\mathbb{P}^n))$ ramified at some divisor $D\subset \mathbb{P}^n$, here $k=\mathbb{C}$.

If $f: \mathbb{P}^n\mathrel{-\,}\rightarrow \mathbb{P}^n$ is a birational morphism, what can we say about the ramification of $f^{*}(\xi)\in Br(k(\mathbb{P}^n))$?

Is this still $D$ or can it be smaller, bigger or totally different from $D$? Is it birational to $D$? Do we have any control over it?

Maybe one can use the Weak Factorization Theorem, which states that $f$ can be written as a sequence of blow-ups and blow-downs with smooth centers. So one must just understand what happens to the ramification under smooth blow-ups or smooth blow-downs. For example if we blow up a point, can(must ?) the exceptional divisor be part of the ramification divisor? Or can one of the blow-downs reduces part of $D$ to a point? Is everything is possible here?

Is this known or written down somewhere in the literature?

  • 1
    $\begingroup$ By purity, the ramification locus of $f^*(\xi)$ is a divisor. The irreducible components of this divisor are mapped under $f$ into $D$. An irreducible component of $f^*D$ is in the ramification of $f^*(\xi)$ if $f$ is locally isomorphic at the generic point of the irreducible component. However, $f^*(\xi)$ may be unramified at other components. For instance, if an associated Severi-Brauer variety is regular over the generic point of $Z\subset D$, and if the restriction of the Severi-Brauer over $Z$ has a rational section, then the class is unramified at the exceptional divisor over $Z$. $\endgroup$ Nov 16, 2015 at 18:26
  • 1
    $\begingroup$ I should have said above that the class is unramified over the exceptional divisor if there exists a rational section over $Z$ with image in the smooth locus of the projection morphism. For the class $\xi$ over $\mathbb{P}^2\setminus \text{Zero}(x_0x_1x_2)$ with Severi-Brauer $\{([x_0,x_1,x_2],[y_0,y_1,y_2]) \in \mathbb{P}^2\times \mathbb{P}^2 | x_0y_0^2 + x_1y_1^2 + x_2y_2^2 = 0\}$, for the blowing up at the closed point $[x_0,x_1,x_2] = [1,1,0]$, the pullback of $\xi$ is unramified over the exceptional divisor. On the other hand, the blowing up over $[1,0,0]$ is ramified. $\endgroup$ Nov 16, 2015 at 18:54
  • $\begingroup$ @Jason Starr: Thanks. I think i can mainly follow. But what associated Severi-Brauer variety do you mean? Shall we pick an order $O$ (maximal?) which has $\xi$ as its Brauer class at the generic point of $\mathbb{P}^n$ and look at $SB(O)\rightarrow \mathbb{P}^n$? The existence of a rational section implies that the Sveri Brauer variety is the projectivization of a locally free sheaf, so the algebra must be unramified? $\endgroup$
    – Bernie
    Nov 17, 2015 at 12:53
  • $\begingroup$ There is typically not a unique Severi-Brauer variety, just as there is not a unique Azumaya algebra representing a given Brauer class. There are uniqueness results in codimension 1, but you are asking about higher codimension behavior. For a finitely presented, flat morphism $\pi:P\to V$ that is a Severi-Brauer over a dense open $U$, for $Z\subset V$ an integral closed subscheme, if $s:Z\to P$ is a morphism whose image is contained the smooth locus of $\pi$, then the associated class is unramified over every prime exceptional divisor with center $Z$. $\endgroup$ Nov 17, 2015 at 13:03
  • $\begingroup$ @Jason : Do you know any references dealing with such kind of questions? I could not find any. I'm still not getting the whole picture, I think. For example I would be happy to understand what happens if instead of a birational map, I just look at blow ups. If the class $\xi$ ramifies in just one irreducible divisor $D$ and one blows up a smooth $Z\subset D$, the ramification of $f^{*}(\xi)$ lies in the union of the strict transform of $D$ and the exceptional divisor. When is the exceptional divisor ramified? Can this be understood in terms of $Z$? Can the strict transform be unramified? $\endgroup$
    – Bernie
    Jan 13, 2016 at 16:38


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.