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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:

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up vote 1 down vote accepted

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.

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Thanks. This looks better. I get the dual of the module you specified by transposing and then replace the $t$'s by $t^{-1}$? Then taking the determinant gives a polynomial in $t^{-1}$ with the desired pole order. So one does not need $\mathbb{Q}$-divisors at all, very good! – TonyS Aug 24 '10 at 18:44

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