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I do not see the following statement:

Suppose $\mathcal{F}$ is a locally free coherent sheaf on a smooth affine curve $X=\text{Spec }A$. Then for any coherent sheaf $\mathcal{G}$, there is an exact sequence $$ \mathcal{F}^m\to\mathcal{F}^n\to\mathcal{G}\to 0.$$

Clearly it would be enough to have a surjection $\mathcal{F}^n\to \mathcal{G}$.I think I have to use projectivity of the module $M$ associated to $\mathcal{F}$ here. Any ideas?

Thanks.

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    I removed the algebra tag, as it is a deprecated tag.2012-01-01

1 Answers 1

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Let $\mathcal{F}$ be a locally free coherent sheaf on $X$ and $\mathcal{G}$ be any coherent sheaf on $X$. Let $\text{rank}(\mathcal{F})=r$. Suppose $r>1$. We have $\mathcal{F}=\tilde{M}$ for some projective $A$-module $M$. As $X$ is integral, $X$ is connected. Let $\eta$ be the generic point of $X$. Set $\mathcal{O}_{X,\eta}=K$. Then the rank of $\mathcal{F}$ is the dimension of the $K$-vectorspace $M\otimes_AK$ and the latter is the rank of the projective $A$-module $M$. As $A$ is a Dedekind ring, $M\cong A^{r-1}\oplus I$ for some ideal $I$ (see the structure theorem in http://www.math.uchicago.edu/~may/MISC/Dedekind.pdf ) and hence for any finitely generated $A$-module $N$ there is an exact sequence $$ M^n\to M^m\to N\to 0.$$ Thus for any coherent sheaf $\mathcal{G}$ there is an exact sequence $$\mathcal{F}^n\to\mathcal{F}^m\to\mathcal{G}\to 0.$$

If $r=1$ $\mathcal{F}$ is an invertible sheaf. As $X$ is affine, for any coherent sheaf $\mathcal{G}$, $\mathcal{G}\otimes\mathcal{F}^{-1}$ is globally generated. So there is an exact sequence $$\mathcal{O}_X^m\to\mathcal{G}\otimes\mathcal{F}^{-1}\to 0.$$ Tensoring with $\mathcal{F}$ yields an exact sequence $$\mathcal{F}^m\to\mathcal{G}\to 0$$ and again we obtain an exact sequence

$$\mathcal{F}^n\to\mathcal{F}^m\to\mathcal{G}\to 0.$$

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    The $F$ in the question is only locally free, not invertible.2012-01-14
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    Thanks! I deleted the statement about invertible sheafs.2012-01-14
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    «As A is noetherian, X is connected» is not quite right.2012-01-15
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    Thank you, I replaced it by "as $X$ is integral, $X$ is connected".2012-01-15