Hartshorne Theorem III.3.5 Proof

78 Views Asked by At

Let $X=\operatorname{Spec}A$ be the spectrum of a noetherian ring $A$. I am trying to understand the proof of the following fact from Hartshorne: For all quasi-coherent sheaves $F$ on $X$, and for all $i>0$, we have $H^i(X,F)=0$.

We start by letting $M=\Gamma(X,F)$ and take an injective resolution $0\to M\to I^\cdot$. From this we get an exact sequence of sheaves $0\to F\to\tilde{I^\cdot}$ (noting that $\tilde{M}=F$). Each $\tilde{I^i}$ is flasque by an earlier result, and so we can use this resolution of $F$ to calculate cohomology.

We then apply the global sections functor to the sequence of sheaves above, and we get back $0\to M\to I_1\to I_2\to\cdots$. By definition, $H^i(F,X)=\operatorname{im}(I_i\to I_{i+1})/\operatorname{ker}(I_{i-1}\to I_i)$. My question is why is this necessarily zero for $i>0$?