TITLE: Is the Quot-scheme over non-singular curve reduced
QUESTION [6 upvotes]: Let $k$ be an algebraically closed field, $C$ a non-singular projective curve over $k$ of genus at least $2$ and $\mathcal{F}$ a locally free sheaf on $C$. Let $r,d$ be two integers satisfying $\mathrm{gcd}(r,d)=1$ and $r>0$. Denote by $Q$ the Quot-scheme parametrizing quotients of $\mathcal{F}$ of rank $r$ and degree $d$. Is every irreducible component of $Q$ (generically) reduced i.e., for a general point $x$ of the irreducible component of $Q$, the local ring $\mathcal{O}_{Q,x}$ is reduced?

REPLY [3 votes]: This is false, at least for special $C$ (perhaps it is true for $C$ that are sufficiently general in moduli).  The simplest counterexample I know of is a genus $4$ curve $C$ that is non-hyperelliptic and whose canonical image is contained in a singular quadric hypersurface in $\mathbb{P}^3$.  In this case, the scheme $G^1_3$ equals $W^1_3$ inside $\text{Pic}^3_C$, and this is one nonreduced point.  For the invertible sheaf $\mathcal{L}$ on $C$ parameterized by this point, $\mathcal{L}$ is globally generated.  Thus for $\mathcal{F}$ defined to be $H^0(C,\mathcal{L})\otimes_k \mathcal{O}_C$, the natural homomorphism, $$\phi:\mathcal{F} \to \mathcal{L},$$ is surjective.  
Since being invertible is an open condition on flat families of coherent $\mathcal{O}_C$-modules, every small deformation of the quotient $\mathcal{L}$ is an invertible $\mathcal{O}_C$-module of degree $3$ that has a $2$-dimensional vector space of global sections, and conversely.  Thus, there is a smooth morphism from a Zariski open neighborhood $U$ of $[\phi]$ in the Quot scheme to $W^1_3$.  Since $W^1_3$ is nonreduced with nilradical a rank $1$ vector space over the residue field, also $U$ is nonreduced with nilradical an invertible sheaf on the reduced scheme of $U$ (which is smooth).