I have some questions about a proof given in my class (unfortunately I not found the original source) of following statement:
Let $Y$ a Noetherian scheme and $\mathcal{F},\mathcal{G}$ two coherent sheaves on $Y$ then $\mathscr{Ext}^i _{O_Y}(\mathcal{F}, \mathcal{G})$ is also coherent for $i \ge 0$.
The PROOF works as follows:
The problem is in following way local: Since $\mathcal{F}$ is coherent locally for an appropriaty small open $U \subset Y$ there exist a free resolution
(*)$$... \to \mathcal{E}_1 \to \mathcal{E}_0 \to \mathcal{F} \vert _U \to 0 $$
where the $\mathcal{E}_j$ are free $O_U$ modules of finite rank (works since $\mathcal{F}$ coherent).
Since the Ext-sheaves bahave well under localizations there is an identity
$${Ext}^i _{O_Y}(\mathcal{F}, \mathcal{G}) \vert _U ={Ext}^i _{O_U}(\mathcal{F} \vert _U, \mathcal{G} \vert _U)$$
So we can assume $U=Y$ and $\mathcal{F} \vert _U=\mathcal{F} $ has free resolution from above.
Appling Hom to the exact free resolution (*) we obtain the the complex $\mathscr{Hom}_{O_Y}(\mathcal{E}_{\bullet}, \mathcal{G})$ of $O_Y$-modules.
Taking it's cohomology we obtain the sheaves $\mathscr{Ext}^i (\mathcal{F}, \mathcal{G})$.
Now comes first inderstanding problem: the author of the proof wants to compare them with the functor "$i$-th comology functor" $\mathcal{E}^i(\mathcal{G})$ of $\mathscr{Hom}_{O_Y}(\mathcal{E}_{\bullet}, \mathcal{G})$.
Local calculations show that $\mathcal{E}^i(\mathcal{G})$ is a so called $\delta$-functor.
Making a comparison we conclude
$$ \mathscr{Ext}^i (\mathcal{F}, \mathcal{G})=\mathcal{E}^i(\mathcal{G})$$
because:
The booth coinside in $i=$ since $\mathscr{Hom}(-, \mathcal{G})$ is left exact and both vanish for injective $\mathcal{G}$. Therefore both are Universal $δ$-functors (by Grothendieck's criterion for universal $\delta$-functors (Hartshorne's AG Theorem 1.3A, page 206). Since agin both coinside in $i=0$ and and are characterized by a universal property both sides must couside in all $i$.
But $\mathcal{E}^i(\mathcal{G})$ is "obvious" (WHY?) coherent. Fini.
Following two questions:
Why is $\mathcal{E}^i(\mathcal{G})$ is "obvious" coherent?
Why do we need to make this tour around the theory of (universal) $\delta$-functors? Can we just observe that all $\mathscr{Hom}_{O_Y}(\mathcal{E}_{j}, \mathcal{G})$ are coherent and coclude directly that their cohomology is also coherent?
Or is the story a bit more difficult and I don't see the advantange of the delta functor $\mathcal{E}^i(\mathcal{G})$?
Aren't $\mathscr{Ext}^i (\mathcal{F}, \mathcal{G})$ and $\mathcal{E}^i(\mathcal{G})$ both the same by definition of cohomology based on freeness of choice of projective resolution? (free resolutions are projective.
From naive point of view for me the proof seems to work like:
"we want show that an object is coherent, we take this object, give it another name and then - after giving this new name to it - it is obviously coherent.
Which detail I oversee?
Remark N# 1: Regarding the coherence of $\mathscr{Hom}_{O_Y}(\mathcal{E}_{j}, \mathcal{G})$.
I think that it is obvious since we are work locally!. so
- $\mathcal{E}_{j}= \mathcal{O}_{Y}^n$
- $\mathscr{Hom}_{O_Y}(\mathcal{O}_{Y}, \mathcal{G}) \cong \mathcal{G}$
Combining 1. and 2. we obtain $\mathscr{Hom}_{O_Y}(\mathcal{E}_{i}, \mathcal{G})\cong \mathcal{G}^n $ locally and coherence is a local property. Is the argument ok?
But if we know that $\mathscr{Hom}_{O_Y}(\mathcal{E}_{i}, \mathcal{G})$ are coherent why it is obvious that $\mathcal{E}^i(\mathcal{G})$ are also coherent?
Remark N# 2: I know that there are maybe more ways to proof the statement but the main point of this thread is to understand the logic behind this proof.