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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 !

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    $\begingroup$ What have you tried? $\endgroup$ – user113102 Mar 7 at 20:00
  • $\begingroup$ I've tried the following : for all coherent sheaf on $X$ there is an exact sequence $$0 \longrightarrow F_{r-i} \longrightarrow F_{r} \cdots \longrightarrow F_{0} \longrightarrow \mathcal{F} \longrightarrow 0$$ where $F_{i}$ are locally free sheaves. I can do the item (a), My problem now is the item (b). $\endgroup$ – Allan Ramos Mar 8 at 0:07
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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.

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  • $\begingroup$ thank you very much. $\endgroup$ – Allan Ramos Mar 8 at 20:35
  • $\begingroup$ @AllanRamos If you are satisfy with this answer, you should accept it so that your question won't be mark as unanswered. $\endgroup$ – Roland Mar 8 at 21:05
  • $\begingroup$ How do I do this? I'm new to the forum. Thank you. $\endgroup$ – Allan Ramos Mar 8 at 22:15
  • $\begingroup$ taking the opportunity, I just did not understand the part: If it were, there would be local retractions of $Z_{n} \rightarrow F_{n}$. Thank you. $\endgroup$ – Allan Ramos Mar 8 at 22:27
  • $\begingroup$ Left to the beginning of the answer there should be a zero (the current note), an arrow up (to upvote an answer) an arrow down (to say that an answer was not good) and below the acceptation mark. As for the part you didn't understand, I will edit my answer, tell me if it is clearer that way. $\endgroup$ – Roland Mar 8 at 23:12

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