Hartshorne Exercise III.5.1.
Let $X$ be a projective scheme over a field $k$, and let $\mathcal F$ be a coherent sheaf on $X$. We define the Euler Characteristic of $\mathcal F$ by
$$\chi(\mathcal F)=\sum_i(-1)^i\dim _kH^i(X,\mathcal F).$$
If
$$0\ra \mathcal F’\ra\mathcal F\ra\mathcal F”\ra 0$$
is a short exact sequence of coherent sheaves on $X$, show that $\chi (\mathcal F)=\chi(\mathcal F’)+\chi(\mathcal F”).$
First note that this sum is finite because $X$ is noetherian and separated. We instead prove a slightly stronger fact about exact complexes of modules over a PID $R$. Indeed, let
$$0\ra M_1\xrightarrow{f_1}M_1\xrightarrow{f_2}\cdots \xrightarrow{f_{n-1}}M_n\ra 0$$
be a long exact sequence of free $R$-modules with $n$ a multiple of 3, and let $r_i:=\op{rank}M_i$. For each $1<i<n$, we have a split SES,
$$0\ra \op{im}f_{i-1}\ra M_i\ra \op{im}f_i\ra 0.$$
Note that these SES split because everything is a free module (sub-module of free is free, over a PID). I claim that $\sum_{i=1,4,7,…,n}(-1)^{i}(-r_i+r_{i+1}-r_{i+2})=0$. This will be the desired result. We proceed by induction on $n$. When $n=3$ the result is clear. Suppose the result holds for long exact sequences of length $n=3m$ (with $m<\dim X+1$). Then, with $n=3(m+1)$ we have a LES,
$$\begin{equation} 0\ra M_1\ra\cdots\ra M_{n-4}\ra\op{im}f_{n-4}\ra 0, \end{equation}$$
as well as several SES,
$$0\ra\op{im}f_{n-2}\ra M_{n-1}\ra M_n\ra 0 $$
and
$$0\ra\op{im}f_{n-3}\ra M_{n-2}\ra \op{im}f_{n-2}\ra 0$$
and
$$0\ra\op{im}f_{n-4}\ra M_{n-3}\ra \op{im}f_{n-3}\ra 0.$$
By the inductive hypothesis,
$$\sum_{i=1,4,…,n-8}(-1)^{i}(-r_i+r_{i+1}-r_{i+2})-r_{n-5}+r_{n-4}-\op{rank}(\op{im}f_{n-4})=0.$$
Moreover, we see that
$$\op{rank}(\op{im}f_{n-4}) =r_{n-3}-\op{rank}(\op{im}f_{n-3}) = r_{n-3}-r_{n-2}+\op{rank}(\op{im}f_{n-2})=r_{n-3}-r_{n-2}+r_{n-1}-r_n.$$
This concludes the proof.
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