By Grothendieck's vanishing theorem, every variety $ X $ over a field $ k $ has a finite cohomology sequence; i.e. we have an exact sequence
$$ 0 \to H^0(X, \mathcal{O}) \to \cdots \to H^r(X,\mathcal{O}) \to 0 \to 0 \to \cdots $$
where each cohomology group $ H^i $ is a module over $ k $. Set $ h^i = \dim H^i(X, \mathcal{O}) $ for each $ i $.
Now by rank-nullity, the alternating sum $ h^0 - h^1 + \cdots \pm h^r $ is just $ 0 $.
But this final alternating sum is just the Euler characteristic of $ X $ - which is obviously not zero for every variety.
We are having some trouble finding the mistake (which is probably really obvious) - any hint would be appreciated.
Yes the mistake turned out to be really obvious...
We thought that there was a long exact sequence like the one in the question that we could obtain from the cohomology groups, but this was a mistake (obviously there is a complex involved, namely the one induced from an injective resolution $ 0 \to \mathcal{O} \to \mathcal{O}^0 \to \mathcal{O}^1 \to \cdots $, i.e. $ 0 \to \Gamma(X,\mathcal{O}^0) \to \Gamma(X, \mathcal{O}^1) \to \cdots $, and our mistake was thinking that this set up a natural exact sequence of the quotients $ H^i $, which it does not).