# Non-separatedness of moduli space of sheaves

The moduli space of sheaves over a smooth variety is in general not separated. That is, there exists a flat family of coherent sheaves over a punctured disk which extends to a flat family of coherent sheaves over a disk in several ways. Could anyone give me easy examples?

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Take sheaves of ideals of the origin with different multiplicities. – Alex Degtyarev Mar 1 '14 at 9:17
It is better to write "the moduli STACK of sheaves" here, because the moduli space implicitly means the moduli space of semistable sheaves, and the latter is separated. – Sasha Mar 1 '14 at 10:20

There is a vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{A}^1$ whose restriction to $\mathbb{P}^1 \times (\mathbb{A}^1 - \{ 0\})$ is isomorphic to $\mathcal{O}^2$, but whose restriction to $\mathbb{P}^1 \times 0$ is $\mathcal{O}(-1)\oplus \mathcal{O}(1)$.
One can take $E$ to be the cokernel of the map $$(x, y, t) : \mathcal{O}(-1) \to \mathcal{O}\oplus \mathcal{O} \oplus\mathcal{O}(-1),$$ where $x$ and $y$ are the coordinates on $\mathbb{P}^1$ and $t$ is the coordinate on $\mathbb{A}^1$. Indeed, for nonzero $t$ the map has a section, and for $t=0$ we get $\mathcal{O}(-1)\oplus\mathcal{O}(1)$.
The whole point of course is that $\mathcal{O}(-1)\oplus \mathcal{O}(1)$ is not semistable.
It seems to me that graded quotients (with obvious filtration) of the two bundles are different. One consists of two $\mathcal{O}$ and the other consists of $\mathcal{O}(-1)$ and $\mathcal{O}(1)$. – user2013 Mar 1 '14 at 23:43