用户: Fyx1123581347/类域论/整体类域论/理元群上同调

命题 0.1. 设 $\mathfrak{P}|\mathfrak{p}$, 则$$H^q(G,{\mathbb{I}}_L^{\mathfrak{p}})\simeq H^q(G_{\mathfrak{P}},L_{\mathfrak{P}}^{\times })$$如果 $\mathfrak{p}$ 是有限非分歧的素数, 则$$H^q(G,U_L^{\mathfrak{p}})=1.$$

证明. $${\mathbb{I}}_L^{\mathfrak{p}}=\prod _{\sigma \in G/G_{\mathfrak{P}}}{L}_{\sigma \mathfrak{P}}^{\times }=\prod _{\sigma \in G/G_{\mathfrak{P}}}\sigma L_{\mathfrak{P}}^{\times }$$$$U_L^{\mathfrak{p}}=\prod _{\sigma \in G/G_{\mathfrak{P}}}{U}_{\sigma \mathfrak{P}}=\prod _{\sigma \in G/G_{\mathfrak{P}}}\sigma U_{\mathfrak{P}}$$因此 ${\mathbb{I}}_L^{\mathfrak{p}}$, $U_L^{\mathfrak{p}}$ 为 $G/G_{\mathfrak{P}}$ 诱模. 故$$H^q(G,{\mathbb{I}}_L^{\mathfrak{p}})\simeq H^q(G_{\mathfrak{P}},L_{\mathfrak{P}}^{\times })$$$$H^q(G,U_L^{\mathfrak{p}})\simeq H^q(G_{\mathfrak{P}},U_{\mathfrak{P}}).$$结合局部的情形, 结论得证.

$\square$

因此有$$H^q(G,{\mathbb{I}}_L^S)\simeq \prod _{\mathfrak{p}\in S}{H}^q(G,{\mathbb{I}}_L^{\mathfrak{p}})\times \prod _{\mathfrak{p}\notin S}{H}^q(G,U_L^{\mathfrak{p}})$$取 $S$ 含所有分歧的素数, 则 ${\mathbb{I}}_L={\cup }_S{\mathbb{I}}_L^S$$$H^q(G,{\mathbb{I}}_L)\simeq \underset{\begin{array}{c}⟶\\ S\end{array}}{\lim }{H}^q(G,{\mathbb{I}}_L^S)\simeq \prod _{\mathfrak{p}\in S}{H}^q(G_{\mathfrak{P}},L_{\mathfrak{P}}^{\times })\simeq \underset{\mathfrak{p}}{⨁}{H}^q(G_{\mathfrak{P}},L_{\mathfrak{P}}^{\times }).$$并且, 投影映射就是 $H^q(G,{\mathbb{I}}_L)$ 到 $H^q(G_{\mathfrak{P}},L_{\mathfrak{P}}^{\times })$ 的自然映射.