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Let $X$ be a projective curve over a field $K$ (any characteristic). Let $\mathcal{F}$ be a coherent simple sheaf (In the sense, that $\mathcal{F}$ doesn't have non-trivial subsheaves). What is the Euler characteristic of $\mathcal{F}$ ?

I would like to have $\chi(\mathcal{F})=1$ ? But, I don't know how to prove it, or where to find a reference about simple sheaves on curves? Any help?

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    $\begingroup$ The only simple sheaves with your definition are skyscraper sheaves $K(x)$ for a closed point $x\in X$. Thus $\chi(\mathcal{F})$ for such a sheaf would be $[K(x):K]$. If you want it to be one, you must assume that $K$ is algebraically closed. $\endgroup$
    – Mohan
    Commented Nov 24, 2015 at 18:03
  • $\begingroup$ Thank you for your help. I believe I am starting to understand: the sheaf $\mathcal{F}$ in my context must be a skycraper sheaf of $X$ at a point $x$ determined by a simple module $M$. In such a case, it is known for skycraper sheaves that $H^1(X,\mathcal{F})=0$. But, since $X$ is projective $\mathcal{O}_X=K$, hence $M$ is a $K$ vector space, therefore $M\cong K^1$ because $M$ is simple. But, $H^0(X,\mathcal{F})=M$, hence $dim_K (H^0(X,\mathcal{F}))=1$. Is this correct? I am not very sure! $\endgroup$
    – Francisco Portillo
    Commented Nov 25, 2015 at 1:56
  • $\begingroup$ $M=K^1$ only if the point $x$ is $K$-rational. If $K$ is algebraically closed, all closed points are $K$-rational. $M$ is simple as an $\mathcal{O}_X$ module is not the same as being simple as a $K$-vector space. $\endgroup$
    – Mohan
    Commented Nov 25, 2015 at 15:09

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