Let $R$ be a graded commutative ring such that $X=\operatorname{Proj}(R)$ is a smooth projective variety. Let $M$ be a finite graded module over $R$ and let $\mathcal{F}=\widetilde{M}$ be the associated coherent sheaf on $X$.

If $M$ is indecomposable (i.e. can not be presented as a direct sum of two nonzero graded modules) is it true that $\mathcal{F}$ is also indecomposable?


1 Answer 1


Let $R=\mathbb{C}[x,y]$ (with $x$ and $y$ in degree one), so we're considering sheaves on the projective line.

Let $M=\frac{\mathbb{C}[x,y]}{(xy)}$. Then $M$ is indecomposable.

$M$ also has a graded submodule $xM\oplus yM$, such that the quotient $M/(xM\oplus yM)$ is one dimensional and so has zero associated coherent sheaf.

So the inclusion $xM\oplus yM\to M$ induces an isomorphism of associated coherent sheaves $\widetilde{xM}\oplus\widetilde{yM}\cong\widetilde{M}$.

  • $\begingroup$ Thanks! Do you know examples with vector bundles? $\endgroup$
    – Alex
    Commented Oct 1, 2015 at 18:42
  • $\begingroup$ @Alex I think something similar will work, "gluing together" two modules associated to vector bundles. For example, let $R=\mathbb{C}[x,y]$, and take the graded ideals $M=(x,y^2)$ and $N=(x^2,y)$, so that both are zero in degrees $\leq0$ and one-dimensional in degree one, and $\widetilde{M}=\widetilde{N}=\mathcal{O}$. Let $\alpha:M\oplus N\to\mathbb{C}$ be a map that is non-zero on both factors, where $\mathbb{C}$ is in degree one. Then $\widetilde{\ker(\alpha)}\cong\mathcal{O}\oplus\mathcal{O}$, and I think $\ker(\alpha)$ is indecomposable. $\endgroup$ Commented Oct 5, 2015 at 8:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .