I'm trying to prove the following:

Let $X$ be a non-singular scheme then it is known that there exists a finite free resolution (resolution by locally free sheaves) for any coherent sheaves on $X$. Given a short exact sequence of coherent sheaves on $X$: \begin{equation} 0 \rightarrow \mathcal{F}' \rightarrow \mathcal{F} \rightarrow \mathcal{F}'' \rightarrow 0. \end{equation} If we had finite free resolutions $E'_* \rightarrow \mathcal{F}' \rightarrow 0, E_* \rightarrow \mathcal{F} \rightarrow 0$ and $E''_* \rightarrow \mathcal{F}'' \rightarrow 0$ then we have the following equality in $K(X)$ (the Grothendieck group of locally free sheaves on $X$): \begin{equation} \sum_{i=0}^\infty (-1)^i[E_i] = \sum_{i=0}^\infty (-1)^i[E'_i] + \sum_{i=0}^\infty (-1)^i[E''_i] \end{equation} (these are finite sums, of course).

My first thought/attempt on this:

This statement would be easy to proof if we had a Horseshoe lemma for sheaves. But as far as I know, a locally free sheaf $E$ doesn't have to be a projective object in the category of coherent sheaves (unless $X$ is affine or something, I think, please correct me if I'm wrong here).

Trying to read B.8.3 in Fulton (intersection theory):

Fulton did exactly this proof in B.8.3. I was trying to understand that, but the proof is too brief for me. First step is to construct this diagram (the rightmost vertical map is missing in the book, I think this is by design) \begin{array}{llllllllllllll} 0 & \rightarrow & E_n & \rightarrow & ... & \rightarrow & E_0 & \rightarrow & \mathcal{F} & \rightarrow & 0\\ & & \downarrow & & & & \downarrow & & & & \\ 0 & \rightarrow & E''_n & \rightarrow & ... & \rightarrow & E''_0 & \rightarrow & \mathcal{F}'' & \rightarrow & 0\\ & & \downarrow & & & & \downarrow & & & & \\ & & 0 & & ... & & 0 & & & & \\ \end{array} where $E_0$ is mapped surjectively into the kernel of the canonical map $E''_0 \oplus \mathcal{F} \rightarrow \mathcal{F}''$. I'm not sure what this mean, but I will assume the map is by projection then the horizontal map $E''_0 \oplus \mathcal{F} \rightarrow E''_0 \rightarrow \mathcal{F}''$. Then I don't see why $E_0 \rightarrow E''_0$ should be surjective. If I looked at stalks, at any $x \in X$, shouldn't the kernel of $E''_{0,x} \rightarrow \mathcal{F}''_x$ be a submodule of $E''_{0,x}$, hence not surjective? The book repeated the similar construction to get the rest of resolution $E_* \rightarrow \mathcal{F}$, so I didn't understand them.

Could anyone please guide me through this proof or the alternative one please? Thank you.


First prove that if there are two locally free resolutions of the same sheaf and a morphism between these resolutions, inducing the identity on the sheaf, then the alternating sums for the two resolutions coincide.

Second, prove the same without an assumption about a morphism (for this construct a third resolution with maps to the first two).

Third, construct a compatible system of locally free resolutions for the exact sequence. This will give an equality for these resolutions, and then by the previous observation also for the original resolutions.

EDIT (some details on the first part). Assume $\{E'_k\}$ and $\{E''_k\}$ are the resolutions and $f_k \colon E'_k \to E''_k$ is a morphism of resolutions. Consider it as a bicomplex $\{E_{k,l}\}$ with $E_{k,0} = E'_k$, $E_{k,1} = E''_k$, and with the other terms being zero (just two lines). Then the totalization $Tot(E)_n = \oplus_{k+l = n}E_{k,l}$ of this bicomplex is acyclic, hence $\sum (-1)^n[Tot(E)_n] = 0$. This means that $\sum (-1)^k[E'_k] = \sum (-1)^k[E''_k]$.

  • $\begingroup$ This is exactly what I don't know how to do. Could you provide me more detail please? $\endgroup$ – user113988 Dec 4 '16 at 1:56
  • $\begingroup$ Which of the parts? $\endgroup$ – Sasha Dec 4 '16 at 12:59
  • $\begingroup$ I incline to say every parts, but I guess the first part and I might get an idea how to do the rest. Thank you. $\endgroup$ – user113988 Dec 4 '16 at 14:35
  • $\begingroup$ OK, I edited the answer. $\endgroup$ – Sasha Dec 4 '16 at 15:32
  • $\begingroup$ Thank you for your answer. I will have time to look through it more carefully in a few days and might ask you more question. $\endgroup$ – user113988 Dec 5 '16 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.