Let $X$ be an integral (and singular) curve over the complex field and let $A$ be a torsion free sheaf (not necessarily locally free) of rank $1$ on $X$. We denote by $\mathrm{Quot}^n_A$ the Quot scheme parametrizing quotient $q:A\to Q$ of length $n$. Is this Quot scheme always isomorphic to the Hilbert scheme $\mathrm{Hilb}^n(X)$? If not, can you give me a counterexample?

1$\begingroup$ Why would it be? $\endgroup$ – Mariano SuárezÁlvarez Aug 23 '13 at 19:27
That is definitely not true. Let $X$ be an integral curve that has a single ordinary double point $p$ (and no other singular points). Let $\nu:\widetilde{X}\to X$ be the normalization. The fiber of $\nu$ over $p$ consists of two closed points, $p_1$ and $p_2$.
Let $A$ be $\nu_*\mathcal{O}_{\widetilde{X}}$. Let $n$ be $1$. Then $\text{Quot}^1_A$ admits a "fundamental cycle" morphism, $$\text{FC}:\text{Quot}^1_A \to X.$$
But this is not an isomorphism. In fact $\text{Quot}^1_A$ has two irreducible components, one of which is $\widetilde{X}$, and the second of which is the projective line associated to the twodimensional vector space $\nu_*\mathcal{O}_{\widetilde{X},p}/\mathfrak{m}_p\nu_*\mathcal{O}_{\widetilde{X},p}$, which is canonically isomorphic to $$
(\mathcal{O}_{\widetilde{X},p_1}/\mathfrak{m}_{p_1} \mathcal{O}_{\widetilde{X},p_1})
\oplus
(\mathcal{O}_{\widetilde{X},p_2}/\mathfrak{m}_{p_2} \mathcal{O}_{\widetilde{X},p_2}).
$$
This $\mathbb{P}^1$ is attached to $\widetilde{X}$ at the two points $p_1$ and $p_2$, with $p_i$ being attached to the point of $\mathbb{P}^1$ determined by the corresponding summand of the twodimensional vector space. Also $\text{FC}$ is constant on the $\mathbb{P}^1$, mapping it to $p$. Allinall, $\text{Quot}^1_A$ is quite different from $X$.