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Below let's work over coherent sheaves on a smooth projective algebraic curve.

We call a subsheaf $\mathcal{F'}$ of $\mathcal{F}$ saturated if it $\mathcal{F/F'}$ is locally free.

We call a locally free sheaf indecomposible if it cannot be written as a direct sum of two saturated subsheaves.

Suppose $\mathcal{E=F_1 \oplus F_2\oplus\dots \oplus F_n=G_1 \oplus G_2\oplus\dots \oplus G_m}$ where the summands are locally free and indecomposible in the sense above.

Then how can we prove $m=n$ and the summands are isomorphic after a permutation?

Thanks for comments and references!

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This is proven by Atiyah (Theorem 3) in On the Krull-Schmidt theorem with application to sheaves. I think it's worth noting that techniques used are very broadly applicable; he really just uses that the algebra $A=\mathrm{End}(\mathcal{E})$ is artinian (since it is finite dimensional over a field).

EDIT: As Sasha notes below, the saturated hypothesis is a red herring. Any Remak decomposition of $\mathcal{E}$ will automatically be into saturated subsheaves.

In a more modern perspective, I think we would just note that the Remak decompositions above correspond canonically to Remak decompositions $A\cong \mathrm{Hom}(\mathcal{E},\mathcal{F}_1)\oplus \cdots\oplus\mathrm{Hom}(\mathcal{E},\mathcal{F}_n)$, etc. of $A$ as a left module over itself: such a Remak decomposition must be of the form $A\cong Ae_1\oplus \cdots \oplus Ae_n$ for idempotents $e_n$, and $\mathcal{F}_i$ is the image of $e_i$. We can then use the usual Krull-Schmidt theorem from artinian rings.

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  • $\begingroup$ Thanks for your reference! But I am a bit confusing that here indecomposible is defined by direct sum of saturated subsheaves while in the article it only asked decomposible as direct sum of subsheaves? So I think a sheaf indecomposible here may not be indecomposible in the sense in the article. $\endgroup$ – Qixiao May 8 '14 at 10:11
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    $\begingroup$ Any direct summand of a locally free sheaf is saturated (since the quotient is the other direct summand which is automatically locally free). $\endgroup$ – Sasha May 8 '14 at 10:21

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