What is the difference between the Grothendieck and the Čech cohomology?

Question

Grothendieck has proven that whenever $X\longrightarrow\operatorname{Spec}(A)$ is a proper morphism of Noetherian schemes, $F$ is coherent over $X$ and flat over $\operatorname{Spec}(A)$, then there exists a finite complex of finitely generated projective modules over $A$ \begin{equation*} 0\longrightarrow K^{0}\longrightarrow...\longrightarrow K^{n}\longrightarrow 0 \end{equation*} such that for any $A$-module $M$ there exists an isomorphism of $A$-modules \begin{equation*} H^{p}(X,F\otimes_{A}M)\cong H^{p}(K\otimes_{A}M)\text{.} \end{equation*} So far, so good. Is not the Čech complex precisely one example of such a Grothendieck complex? By construction, the Čech complex consists of projective modules, and it satisfies the isomorphism condition.

Answer 1

The answer is that the groups of the Cech complex aren’t finitely generated except when $X$ is finite over $A$. I’m studying the cohomology of $\mathcal{O}_X$.

Indeed, let $X=\cup_i{U_i}$ be a finite affine cover. Thus every finite intersection of the $U_i$ is affine and the Cech complex computes the sheaf cohomology.

$\bigoplus_i{\mathcal{O}_X(U_i)}$ is the first nonzero group in the Cech complex, so if it is finitely generated, then all the $\mathcal{O}_X(U_i)$ are finitely generated $A$-modules, thus they are finite (hence proper) over $A$, and thus are closed in $X$, so that $X$ is the disjoint reunion of the $U_i$ which are finite.

Answer 2

The argument becomes easy if $A$ is assumed hereditary. In the construction given in [https://amathew.wordpress.com/2011/01/09/the-grothendieck-complex], all $K^{i}$ are free for $i\geq 1$ and therefore projective. The module $K^{0}$ is a quotient of a free module $K'^{0}=F'\oplus F$, and we have maps $\alpha:F'\longrightarrow K^{1}$ and $\beta:F\longrightarrow K^{1}$. Since $C^{0}$ and $K^{1}$ are projective, so are the images $\operatorname{im}(\alpha)$ and $\operatorname{im}(\beta)$ (by hereditarity). Therefore, there exist splitting maps $\operatorname{im}(\alpha)\longrightarrow F$ and $\operatorname{im}(\beta)\longrightarrow F'$. The maps $K^{0}\longrightarrow C^{0}$ and $K^{0}\longrightarrow K^{1}$ factor as $K^{0}\longrightarrow\operatorname{im}(\alpha)\longrightarrow C^{0}$ and $K^{0}\longrightarrow\operatorname{im}(\beta)\longrightarrow K^{1}$. Let $\gamma$ denote the composite map $K^{0}\longrightarrow\operatorname{im}(\beta)\longrightarrow F\longrightarrow K'^{0}$ and $\delta$ the composite map $K^{0}\longrightarrow\operatorname{im}(\beta)\longrightarrow F'\longrightarrow K'^{0}$, and let $\zeta:K'^{0}\longrightarrow K^{0}$ be the quotient map. Then $\zeta\circ(\gamma\oplus\delta)=\operatorname{id}_{K^{0}}$, hence $K^{0}$ is a direct summand of $K'^{0}$ and therefore projective.

All in all, only the projectivity of $C^{0}$ was needed in the entire proof, hence the statement remains true if one demands projectivity only for $C^{0}$. Very intriguing and enlightening indeed!

This does however not prove the general (non-hereditary) case...