Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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. –  mqx May 8 at 10:11
1  
Any direct summand of a locally free sheaf is saturated (since the quotient is the other direct summand which is automatically locally free). –  Sasha May 8 at 10:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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