Let $X$ be a noetherian scheme and $\mathcal{F}$ a coherent sheaf in $X$. We define the homological dimension of $\mathcal{F}$, denoted $hd(\mathcal{F})$, to be the least lenght of a locally free resolution of $\mathcal{F}$. Then,
a) $\mathcal{F}$ is locally free $\Leftrightarrow hd(\mathcal{F}) = 0$
b) $hd(\mathcal{F}) \leq n$ $\Leftrightarrow ext^{p}(\mathcal{F}, \mathcal{G}) = 0 $ for all $p > n$ and all $\mathcal{G} \in \mathfrak{Mod}(X)$.
Can anyone help me with this? Thank you !
Best Answer
Remember that you can compute the $ext$ sheaves using a resolution of the first argument by locally free sheaves. Using this, the direct implication in (b) is obvious.
The only non trivial part is then the converse. So let's assume that $hd(\mathcal{F})>n$ and consider a resolution $$F_{n+1}\to F_n\to F_{n-1}\to...\to F_0\to\mathcal{F}\to 0$$ with the $F_i$ being locally free.
One of the most useful trick in homological algebra is to split this sequence into short exact ones : $$0\to Z_n\to F_n\to Z_{n-1}\to 0$$ $$0\to Z_{n-1}\to F_{n-1}\to Z_{n-2}\to 0$$ $$\vdots$$ $$0\to Z_0\to F_0\to \mathcal{F}\to 0$$ With $Z_i=\ker(F_i\to F_{i-1})=\operatorname{im}(F_{i+1}\to F_i)$. (And $Z_{-1}=\mathcal{F}$ so that the proof also work if $n=0$).
From the long exact sequence associated to each of these short exact sequences, we get $ext^{n+1}(\mathcal{F,G})=ext^n(Z_0,\mathcal{G})=ext^{n-1}(Z_1,\mathcal{G})=...=ext^1(Z_{n-1},\mathcal{G})$. So to conclude the proof, it is enough to find $\mathcal{G}$ such that $ext^1(Z_{n-1},\mathcal{G})\neq 0$.
By assumption $Z_{n-1}=\ker(F_{n-1}\to F_{n-2})$ is not locally free (otherwise we would be able to construct a locally free resolution of length $n$). So in the short exact sequence $$0\to Z_n\to F_n\to Z_{n-1}\to 0$$ we have $Z_{n-1}$ not locally free and $F_n$ locally free (and thus $Z_n\neq 0$).
Take $\mathcal{G}=Z_n$, then we have a short exact sequence $$0\to\mathcal{H}om(Z_{n-1},Z_n)\to\mathcal{H}om(F_n,Z_n)\to\mathcal{H}om(Z_n,Z_n)$$ I claim that the last map in this sequence is not onto. If it were, then $\operatorname{id}_{Z_n}\in\operatorname{Hom}(Z_n,Z_n)=\mathcal{H}om(Z_n,Z_n)(X)$ would have local preimage in $\mathcal{H}om(F_n,Z_n)$. This mean that there would be local retractions of $Z_n\to F_n$, so the sequence $0t\to Z_n\to F_n\to Z_{n-1}\to 0$ would be locally split. But this would mean $Z_{n-1}$ is locally free (it would be locally a direct factor of a locally free module). Since this is not the case, $\mathcal{H}om(F_n,Z_n)\to\mathcal{H}om(Z_n,Z_n)$ is not onto.
This implies that $ext^1(Z_{n-1},Z_n)\neq 0$ and that concludes the proof.