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Let $f: X \rightarrow Y$ be a finite, flat morphism of curves of degree $n$. The direct image of the structure sheaf $f_* O_X$ is a locally free $O_Y$-module.

Given a local section $s$ of $f_* O_X$ on a trivializing subset, this induces by multiplication an automorphism of $f_* O_X$ which is represented by a $n \times n$ matrix with entries in $O_Y$. This operation gives locally a morphism $f_* O_X \rightarrow Mat(n, O_Y)$ whose image is a commutative subalgebra of $Mat(n, O_Y)$.

How does this algebra looks like? Is it possible that it is made of diagonalizable matrices?

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  • $\begingroup$ Only sections which are units give an invertible matrix, so you do not have such a map to $GL$. $\endgroup$ – Mohan Sep 4 '18 at 12:30
  • $\begingroup$ Right, I fixed my question $\endgroup$ – Ramac Sep 4 '18 at 12:35
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    $\begingroup$ If you restrict over an open subset of $Y$ where $f$ is unramified, then the algebra is semisimple. If $f$ is everywhere ramified, e.g., if $f$ is a morphism of integral curves and the extension of fraction fields is inseparable, then typically the algebra is nowhere semisimple. $\endgroup$ – Jason Starr Sep 4 '18 at 12:43
  • $\begingroup$ Thanks, this is interesting! Do you know where I can find a reference for this? $\endgroup$ – Ramac Sep 4 '18 at 13:18
  • $\begingroup$ Or how do you see this? :D $\endgroup$ – Ramac Sep 6 '18 at 9:15
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Assume $s$ locally generates $O_X$ over $O_Y$. Then $$ O_X = O_Y[s]/(s^n - a_1s^{n-1} - \dots - a_n) $$ for some $a_i \in O_X$. Then $1,s,\dots,s^{n-1}$ form a basis of $O_X$ over $O_Y$, and in this basis $s$ is given by the matrix $$ \left( \begin{array}{cccccc} 0 & 0 & 0 & \dots & 0 & a_n \\ 1 & 0 & 0 & \dots & 0 & a_{n-1} \\ 0 & 1 & 0 & \dots & 0 & a_{n-2} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 0 & a_2 \\ 0 & 0 & 0 & \dots & 1 & a_1 \\ \end{array} \right). $$

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  • $\begingroup$ Why is it always possible to find such a generator? $\endgroup$ – Ramac Sep 6 '18 at 9:14

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