# Higher direct image of the inclusion of the generic point

$\newcommand{\Spec}{\operatorname{Spec}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Gal}{\operatorname{Gal}}$

I am trying to understand the proof of the following statement:

Let $X=\Spec(D)$ where $D$ is the ring of intergers of a number field K. Let I be the ideal class group of $K$. Then the etale cohomology $H^1(X,Z/n) = \Hom(I,Z/n)\\$ The proof goes as follow :

Consider the Leray spectral sequence for $j : \Spec(K)\rightarrow X$ we have $$H^r(X,R^sj_{*}Z/n) \Rightarrow H^{r+s}(G_K,Z/n)\\$$ The first few terms of the spectral sequence gives the exact sequence $\\$ $$0\rightarrow H^1(X,Z/n)\rightarrow H^1(G_K,Z/n)\rightarrow H^0(X,R^1j_{*}Z/n) \rightarrow \ker(H^2(G_K,Z/n)\rightarrow H^0(X,R^2j_{*}Z/n))$$

Then as $H^0(X,R^rj_{*}Z/n)=H^r(G_L,Z/n)$ where L is the Hilbert Class Field of K and $I=\Gal(L/K)=G_K/G_L$ we get

$$H^1(X,Z/n)=\ker(H^1(G_K,Z/n)\rightarrow H^1(G_L,Z/n))=H^1(I,Z/n)=\Hom(I,Z/n)$$

My question is that why do we have $$H^0(X,R^rj_{*}Z/n)=H^r(G_L,Z/n)$$

I know that $R^rj_{*}F$ is the sheaf associated to the presheaf $$U\rightarrow H^r(G_{K(U)},F)$$ where $K(U)$ is the function field of U. Any help would be much appreciated. Thank you.

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