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!


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.