There is an example in 3.1.10 $\mathbb{A}^1$-homotopy theory of schemes demonstrating that the Čech cohomology on an affine scheme with Zariski topology can be different from the derived cohomology.

Let $x_1,x_2 \in \mathbb{A}^2$ be two distinct closed points and let $S$ be the semilocal ring at $x_1$ and $x_2$. Let $C_1,C_2 \subset \mathbb{A}^2$ be two irreducible curves intersecting transversely at $x_1,x_2$. Let $U= \mathrm{Spec}\, S\backslash(C_1\cup C_2), V = \mathrm{Spec}\, S \backslash\{x_1,x_2\}$. Let $j$ be the inclusion $U \rightarrow \mathrm{Spec}\, S$.

We have the decomposition $$V= (V\backslash(V\cap C_1))\cup (V\backslash(V\cap C_2))$$ whose Mayer-Vietoris sequence give $$0=H^0(V\backslash(V\cap C_1),j_!\mathbb{Z})\oplus H^0(V\backslash(V\cap C_2),j_!\mathbb{Z})\rightarrow H^0(U,j_!\mathbb{Z}) = H^0(U,\mathbb{Z})=\mathbb{Z}\rightarrow H^1(V,j_!\mathbb{Z}).$$ Therefore, $H^1(V,j_!\mathbb{Z})\neq 0$. The decomposition $$\mathrm{Spec}\, S = (\mathrm{Spec}\, S\backslash x_1 )\cup (\mathrm{Spec}\, S\backslash x_2 )$$ gives a Mayer-Vietoris sequence $$H^1(\mathrm{Spec}\, S\backslash x_1,j_!\mathbb{Z})\oplus H^1(\mathrm{Spec}\, S\backslash x_2,j_!\mathbb{Z})\rightarrow H^1(V,j_!\mathbb{Z}) \rightarrow H^2(S,j_!\mathbb{Z}).$$

Then he conclude that $H^2(S,j_!\mathbb{Z})\neq 0$. But how can he says that the image of the nonzero element in $H^1(V,j_!\mathbb{Z})\neq 0$ is not zero in $H^2(S,j_!\mathbb{Z})$? (He says that this is because $C_1$ and $C_2$ are irreducible. But I don't know how this can leads to the conclusion.)


Write $S$ for the spectrum of the semi-local ring our considering, $S_i := S \setminus \{x_i\}$, and $D_i := S_i \cap (C_1\cup C_2) = S_i \setminus U$. Then, I think $H^1(S_i,j_!\mathbb{Z}) = 0$. So the morphism $H^1(V,j_!\mathbb{Z}) \to H^2(S,j_!\mathbb{Z})$ is injective.

I would suggest you to consider the following exact sequence on $S_i$: $$ 0 \to j_! \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}|_{D_i} \to 0. $$ Since $\mathbb{Z}$ is a constant sheaf, and any open subset of $S_i$ is irreducible (in particular, connected), $\mathbb{Z}$ is flasque. Hence $H^l(S_i,\mathbb{Z}) = 0$ for any $l \geq 1$, and $H^0(S_i,\mathbb{Z}) = \mathbb{Z}$. Moreover, since $D_i$ is connected, $H^0(S_i,\mathbb{Z}|_{D_i}) = H^0(D_i,\mathbb{Z}) = \mathbb{Z}$. These imply that $H^1(S_i,j_!\mathbb{Z}) = 0$ (note that the morphism $\mathbb{Z} = H^0(S_i,\mathbb{Z}) \to H^0(S_i,\mathbb{Z}|_{D_i}) = \mathbb{Z}$ is induced by $\mathbb{Z} \to \mathbb{Z}|_{D_i}$, hence this is an isomorphism): $$ [H^0(S_i,\mathbb{Z}) = \mathbb{Z}] \xrightarrow{1\mapsto 1} [H^0(S_i,\mathbb{Z}|_{D_i}) = \mathbb{Z}] \to H^1(S_i,j_!\mathbb{Z}) \to [H^1(S_i,\mathbb{Z}) = 0]. $$

  • $\begingroup$ また別のいい解決策だよ。お見事、よくやった! $\endgroup$ Nov 1 '21 at 13:04
  • $\begingroup$ I'm glad your compliment and your excellent Japanese!! 😉 $\endgroup$
    – YJ_cat
    Nov 2 '21 at 0:44

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