I really hope the question's title isn't misleading, but unfortunately no better one came to my mind (EDIT: I adjusted the title, but i'm still not happy with it). I'm trying to understand the "lectures on hilbert schemes" by M. Lehn (available e.g. at www.mathematik.uni-mainz.de/Members/lehn/ar/montreal.ps) and i got stuck at the proof of Proposition 3.8 -- i'll try to sketch the setup:
Let $X$ be a smooth, projective surface (over the complex numbers $\mathbb{C}$, but i'd prefer to consider any algebraically closed field $k$ here). Then the Hilbert Scheme $X^{[n]}$ of $n$ points on $X$ is again a smooth projective variety, of dimension $2n$. Consider the universal family $\Xi_n\subset X\times X^{[n]}$ and the corresponding sheaf of ideals $I_{\Xi_n}\subset \mathcal{O}_{X\times X^{[n]}}$. Now there shall be constructed a locally free resolution of $I_{\Xi_n}$ (of length $2$) $$0\rightarrow A\rightarrow B\rightarrow I_{\Xi_n}\rightarrow 0$$ with $\text{rank}(B)=\text{rank}(A)+1$. Now my actual question consists of two parts (sorry for the lengthy prologue):
What i don't understand in the proof cited above is the statement in the very beginning: Choose an ample divisor $H$ on $X$. Then $B:= p^{\ast}\left( p_{\ast}(\mathcal{O}_{\Xi_n}(mH))\right)(-mH) $ is locally free for $m$ large enough (where $p$ is the map $\Xi_n\hookrightarrow X\times X^{[n]}\rightarrow X^{[n]}$ -- at least it seems to me that this must be meant). By what i read so far it seems to me that twisting by a divisor means tensoring with the corresponding invertible sheaf. But why is it obvious that the above is locally free? And what is the "evaluation" $B\rightarrow I_{\Xi_n}$ and why is it surjective? And finally, having this and restricting to $\{\xi\}\times X$ (for a closed point $\xi $ of $X^{[n]}$), why does "the global dimension of $I_{\xi}$ (the ideal defining $\xi $ in $X$) is less or equal to $1$" imply that $A\vert_{\{\xi\}\times X}$ is locally free? (EDIT: I forgot to say that $A$ is choosen as the kernel of the evaluation .) Any hint would help me quite a lot! As it seems to me that i'm missing some essentials needed for this i'd besides be glad about any references approaching those as well.
I wondered if one could find a different argument for the existence of a locally free resolution of length $2$ the following way: Exercise III.6.5 (c) of Hartshorne's Algebraic Geometry tell's me that the homological dimension (which shall be proven to be $2$) of $I_{\Xi_n}$ equals $\sup_x\text{pd}_{\mathcal{O}_x}(I_{\Xi_n,x})$ where $x$ runs over the points of $X\times X^{[n]}$. Using the Auslander-Buchsbaum-formula one knows that the projective dimension is given by $\text{pd}(I_{\Xi_n,x})=2n+2-\text{depth}(I_{\Xi_n,x})$. Shouldn't it be possible to show $\text{hd}(I_{\Xi_n})=2$ this way? Could anyone give me a hint on how to do so?
Finally in both cases: Why is it obvious that, having such a resolution, it is $\text{rank}(B)=\text{rank}(A)+1$?
Thanks in advance for any hint! As i'm quite new to the topic i hope these weren't stupid questions...
Kind regards!
Thank you very much for your answer! This explanation indeed clarifys the situation excellently... As this question was my first post and i signed up a little bit later, it is unfortunately not connetcted to my account -- so i can't write a comment to my question because i'm missing the necessary reputation and needed to write an answer instead. Especially, though i know this shouldn't be done, i would -- despite of my guilty conscience -- like to use this "answer" to ask for another clarification according to the rest of question(s) 1.:
Is it possible that there is another typing mistake/ambiguous use of notation and it is meant to be "homological dimension of $I_{\xi}$ is $\leq 1$ instead of "global dimension of ..." (By the way, what is the global dimension of a module defined to be? I only know the global dimension of a ring...)? In this case the implication "$A\vert_{\{\xi\}\times X}$ is locally free" would seem rather plausible to me, though one would still need a result which assures one does indeed construct a locally free resolution by taking $A$ (resp. $A\vert_{\{\xi\}\times X}$) as the kernel, described in the corresponding EDIT of the question.
Finally, to make this some kind of answer after all, i think i found an answer to question 2.:
First, i have to admit i made another mistake when asking the question: When formulating the statement, i said $I_{\Xi_n}$ should be proven to have a locally free resolution of length 2, by what i meant -- as written there -- that there is a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow I_{\Xi_n}\rightarrow 0$$ with $A$ and $B$ locally free. Of course this amounts to showing that $\text{hd}(I_{\Xi_n})=1$ (instead of "...$=2$") as i accidently wrote), as counting starts at $0$ here. Showing $\text{hd}(I_{\Xi_n})\leq 1$ would actually be fine as well, and to do so, i thought one could do the following:
As is described in the question, one has to show $\text{depth}(I_{\Xi_n,x})\geq 2n+1$ for all points $x$ of $X\times X^{[n]}$. Now consider the short exact sequence $$0\rightarrow I_{\Xi_n}\rightarrow \mathcal{O}_{X\times X^{[n]}}\rightarrow i_{\ast}\mathcal{O}_{\Xi_n}\rightarrow 0$$ on stalks at $x$, where $x\in\Xi_n\subset X\times X^{[n]}$, so that $(i_{\ast}\mathcal{O}_{\Xi_n})_x=\mathcal{O}_{\Xi_n,x}\neq 0$, identifying $x$ with $i^{-1}(x)$, where $i$ is the closed embedding $\Xi_n\hookrightarrow X\times X^{[n]}$.
Then $\text{depth}(I_{\Xi_n,x})\geq \min (\text{depth}(\mathcal{O}_{X\times X^{[n]},x}),\text{depth}((i_{\ast}\mathcal{O}_{\Xi_n})_x)+1)$ (Corolarry 18.6 of Eisenbud's Commutative Algebra).
(For $x$ not in $\Xi_n$, it is of course $\text{depth}(I_{\Xi_n,x})=\text{depth}(\mathcal{O}_{X\times X^{[n]},x})=2n+2$, as $\mathcal{O}_{X\times X^{[n]},x}$ is regular of dimension $2n+2$.)
As furthermore $\Xi_n$ is flat over $X^{[n]}$ by definition, and $X^{[n]}$ is smooth of dimension $2n$, it is $\text{depth}(\mathcal{O}_{\Xi_n,x})=\dim (\mathcal{O}_{X^{[n]},\text{pr}_2(x)})=2n$, by "flattness and depth" - Theorem 18.16 of Eisenbud's Commutative Algebra ($\text{pr}_2$ is meant to be the projection $X\times X^{[n]}\rightarrow X^{[n]})$.
So $\text{depth}(I_{\Xi_n,x})\geq 2n+1$ for all points $x$ of $X\times X^{[n]}$ is shown, as deisred -- at least if i haven't done any mistakes...
At last, as $X$ is connected, $X^{[n]}$ and $X\times X^{[n]}$ are as well, so by considering the locally free resolution $0\rightarrow A\rightarrow B\rightarrow I_{\Xi_n}\rightarrow 0$ on stalks a point $x$ outside of $\Xi_n$ (so that $I_{\Xi_n,x}\cong\mathcal{O}_{X\times X^{[n]},x}$, one gets $\text{rank}(B)=\text{rank}(A)+1$.
Kind regards!