Given a smooth projective surface $X$, let $A$ be a sheaf of maximal orders in a division ring.
Let us for simplicity assume $A$ ramifies in one curve $C$ with ramification index $e$. Let $A^\*$ be the dual sheaf.


How can I see that the determinant map is a map from $A^\*$ to $O\_X((1-\frac{1}{e})C)$?
And how to understand the invertible sheaf $O\_X((1-\frac{1}{e})C)$? How to handle the rational coefficients?


Since $A$ only ramifies in C, we have that $A$ is Azumaya on $U:=X\backslash C$. So on 
$U$ we have $A^\*\cong A$. There the determinant map induces a map $A^{\\*} \rightarrow O\_U$ since $A$ is just a matrix algbera etale locally. So I see that we have a map $A^\* \rightarrow O\_X(rC)$. But how to find $r=1-\frac{1}{e}$? How can i determine $A^\*$ on $C$?


The question arose reading Theorem 7.1.4. on page 157 of this script: <http://www.math.lsa.umich.edu/courses/711/ordersms-num.pdf>



Q-Divisor and Determinant Map on a Maximal Order
The determinant should map to ${\cal A}^\* \rightarrow {\cal O}\_X(e(1-1/e)$. You can see this along $C$ in codimension one since if you \'etale localize at the generic point of $C$ then the structure Theorem for maximal orders says that ${\cal A}$ localizes to something Morita
equivalent to


R tR ... tR 


R R ... tR


... ... ...


R ... R tR 


R ... ... R


with equal size blocks. Where $R$ is the strictly henselian d.v.r at the generic point of $C$. Now if you if you dualize and count $t^{-1}$, you can have at most a pole of order $e-1$
in the determinant.


