用户: Fyx1123581347/平展上同调/平展上同调/平展上同调群的计算

There is an exact sequence of sheaves, $$0\to {\mathbb{G}}_{m,X}\to g_{\ast }{\mathbb{G}}_{m,K}\to D_X\to 0$$where $g:\mathrm{Spec}(K)\to X$ is the inclusion of generic point of $X$ into $X$ and $D_X={⨁}_{v\in X_1}{i}_{v\ast }\mathbb{Z}$ where $X_1$ is the set of points of $X$ of codimension $1$ and $i_v:\mathrm{Spec}(k(v))\to X$. This exact sequence is equivalent to $A={\bigcap }_{\mathrm{ht}(\mathfrak{p})=1}{A}_{\mathfrak{p}}$ for a normal domain $A$. There is the corresponding long exact sequence $$\cdots \to H^1(X,D_X)\to H^{r+1}(X,{\mathbb{G}}_{m,X})\to H^{r+1}(X,g_{\ast }{\mathbb{G}}_{m,K})\to \cdots .$$The Leray spectral sequence for $i_v$ is $$H^p(X,R^q{i}_{v\ast }\mathbb{Z})\Rightarrow H^{p+q}(v,\mathbb{Z}).$$Note that $H^0(v,\mathbb{Z})=\mathbb{Z},H^1(v,\mathbb{Z})=0$ and $H^2(v,\mathbb{Z})=\mathrm{Hom}(G_v,\mathbb{Q}/\mathbb{Z})$. The same reasoning that showed $H^1(v,\mathbb{Z})=0$ shows that $R^1{i}_{v\ast }\mathbb{Z}=0$, or by checking stalks. Thus $$H^0(X,i_{v\ast }\mathbb{Z})=\mathbb{Z},H^1(X,i_{v\ast }\mathbb{Z})=0,H^2(X,i_{v\ast \mathbb{Z}})\hookrightarrow \mathrm{Hom}(G_v,\mathbb{Q}/\mathbb{Z}),$$and so $$H^0(X,D_X)=\underset{v\in X_1}{⨁}\mathbb{Z},H^1(X,D_X)=0,H^2(X,D_X)\hookrightarrow \underset{v}{⨁}\mathrm{Hom}(G_v,\mathbb{Q}/\mathbb{Z}).$$The Leray spectral sequence of $g:\mathrm{Spec}(K)\to X$ is $$H^p(X,R^q{g}_{\ast }{\mathbb{G}}_{m,K}\Rightarrow H^{p+q}(\mathrm{Spec}(K),{\mathbb{G}}_m).$$The stalk of $R^q{g}_{\ast }{\mathbb{G}}_{m,K}$ at a geometric point $\bar{x}$ of $X$ is $H^q(K_{\bar{x}},{\mathbb{G}}_m)$ where $K_{\bar{x}}$ is the field of fractions of the strict Henseliaztion of ${\mathcal{O}}_{X,x}$. Thus $R^1{g}_{\ast }{\mathbb{G}}_{m,K}=0$ by Hilbert’s Theorem 90. It follows that $$H^0(X,g_{\ast }{\mathbb{G}}_{m,K})=K^{\times },H^1(X,g_{\ast }{\mathbb{G}}_{m,K})=H^1(K,{\mathbb{G}}_m)=0,H^2(X,g_{\ast }{\mathbb{G}}_{m,K})\hookrightarrow H^2(K,{\mathbb{G}}_m).$$

Putting these together, we get exact sequences $$0\to \Gamma (X,{\mathcal{O}}_X{)}^{\times }\to K^{\times }\to \underset{v}{⨁}\mathbb{Z}\to H^1(X,{\mathbb{G}}_m)\to 0,$$$$0\to H^2(X,{\mathbb{G}}_m)\to H^2(K,{\mathbb{G}}_m).$$The first exact sequence recovers the fact that $$H^1(X,{\mathbb{G}}_m)=\text{(Divisors)/(Principal Divisors)}=\mathrm{Pic}(X).$$