I'm reading Eisenbud's book The geometry of syzygies and I'm quite struck undestanding the argument proposed in Chapter 5, in the section named "Fitting ideals". Remember that a coherent sheaf $\mathscr{F}$ over a scheme $X$ is called $m$*-regular* if $$H^p(X,\mathscr{F}(m-p))=0$$ for every $p>0$. The minimum $m$ among integers, if it exists, such that $\mathscr{}$ is $m$-regolar is the Castelnuovo-Mumford regularity $\mathrm{reg}(\mathscr{F})$. There is an algebraic definition for module that involves (among other things) local cohomology.
Let be $X$ an irreducible and smooth projective curve over an algebraically closed field $k$. If we call $I(X)$ the homogeneous saturated ideal of $X$, id est
$$ I(X):=\bigoplus_{n\geq 0} \mathscr{I}_X(n) $$
where $\mathscr{I}_X$ is the ideal sheaf of $X$, we know thath $\mathrm{reg}(I(X))=\mathrm{reg}(\mathscr{I}_X)$.
Now, let suppose that $X$ comes with a very ample line bundle, i.e. lets assume that $X\subseteq \mathbf{P}^r_k$ for some $r\geq 2$; in this case we have $I(X)\subseteq k[x_0,\ldots,x_r]=:S$ as homogeneous ideal. Let be $\mathscr{L}\in\mathrm{Pic}(X)$, i.e. an invertibile sheaf over $X$ and let be $$F:=\bigoplus_{n\geq 0} H^0(X,\mathscr{L}(n))$$ the generated cone; it's well known that $F$ is a graded $S$-module and has a free minimal presentation $$L_1\overset{\psi}{\longrightarrow} L_0\longrightarrow F\longrightarrow 0$$
Sheafifying this sequence, we obtain a sequence
$$\bigoplus_{j=1}^s \mathscr{O}_{\mathbf{P}^r_k}(-h_j)\overset{\Psi}{\longrightarrow} \bigoplus_{l=1}^t \mathscr{O}_{\mathbf{P}^r_k}(-l_i)\longrightarrow \mathscr{L}\longrightarrow 0$$
In both cases we can define
the 0-th Fitting ideal $I(\psi)$ of $\psi$, that is the ideal generated by $\psi$'s maximal minors;
the sheaf $\mathscr{I}(\Psi)$ of Fitting ideals of $\Psi$.
The problems that arise are the following:
how can we relate the sheaf $\widetilde{I(\psi)}$ to the sheaf $\mathscr{I}(\psi)$? Since Fitting ideals commute with localizations, I find quite reasonable that the two sheafs are isomorphic, but Eisenbud don't even give an hint of a formal way to prove it;
What can we say about $\mathrm{reg}(\mathscr{I}(\Psi))$ and $\mathrm{reg}(\mathscr{I}_X)=\mathrm{reg}(I(X))$? Following Eisenbud considerations (which don't assume $X$ smooth), there should be inequality $\mathrm{reg}(\mathscr{I}_X)\leq\mathrm{reg}(\mathscr{I}(\Psi))$, but following the cumbersome proof of this fact it seems that, if $X$ is smooth, then we have $\mathscr{I}_X=\mathscr{I}(\Psi)$.
Anyone can help me finding the way to cleat out these facts?