Suppose X is a closed subscheme of Y, with Y locally Noetherian. Is there a locally free resolution of $i_* O_X$? Let $I : X \to Y$ be a closed subscheme of a locally Noetherian scheme. I am secretly trying to show that sheaf exts of $i_* O_X$ to coherent sheaves on Y are coherent (in order to find a dualizing sheaf on X) hence I am trying to resolve the left entry by locally frees of finite rank. 
I know how to do this if Y is projective, but otherwise I don't know if it is even possible. I know in general that locally free resolutions don't always exist, but maybe the case of an ideal sheaf is special?
I am trying to answer exercise 30.4.A in Ravi Vakils notes.
 A: Here is an example which shows you need to impose more finiteness conditions for it to possibly work.
Consider the rings
$$R_n = k[x_1,\ldots,x_n]_{(x_1,\ldots,x_n)} \ \text{for each}\ n \in \mathbf{Z}_{\ge 0}$$
formed by localizing the polynomial ring in $n$ variables at the maximal ideal $(x_1,\ldots,x_n)$. Now consider the scheme:
$$Y = \coprod_n Y_n \ \text{where}\ Y_n := \operatorname{Spec} R_n.$$
Note that the topology on $Y$ is the disjoint union topology, and so the subscheme
$$X = \coprod_n \operatorname{Spec} k$$
formed by choosing the unique closed point in each $\operatorname{Spec} R_n$ is closed.
We claim that $\mathscr{I}_X$, the sheaf of definition of $X$, is such that there does not exist a surjection
$$\mathcal{O}_Y^{\oplus r} \longrightarrow \mathscr{I}_X \longrightarrow 0$$
for any $r < \infty$, which implies $i_*\mathcal{O}_X$ does not have a resolution by locally free sheaves of finite rank. But taking stalks at the closed point in $Y_n$ gives the exact sequence
$$R_n^{\oplus r} \longrightarrow (x_1,\ldots,x_n)_{(x_1,\ldots,x_n)} \longrightarrow 0,$$
which is impossible if $n > r$.
