The proof of Corollary Sample Clauses

The proof of Corollary.Β 4.1.5 We begin this section by proving the following claim: suppose to be given πœ‚ e( RES𝐾 and 𝐸 e Ω𝐾 . . To do this we write 𝐸 for the fixed field of πœ’ and calculate 𝐾,𝑆 (𝐸) Then for every ramified character πœ’ e ξˆ³Λ†πΈ for which π‘’πœ’ (πœ‚πΈ ) β‰  0 one has that 𝐿 π‘Ÿ) ( πœ’, 0) β‰  0. πœ’ 𝐸/πΈπœ’ [𝐸 : πΈπœ’ ] Β· π‘’πœ’ (πœ‚πΈ ) = π‘’πœ’ 𝑁 π‘Ÿ (πœ‚πΈ ) = π‘’πœ’ (𝑃𝐸j/𝐸 ) Β· π‘’πœ’ (πœ‚πΈπœ’ πœ’ ) = 𝑣 e𝑆 (𝐸)\𝑆 (Xx) 1 – πœ’ (πΉπ‘Ÿπ‘£)–1 Β· π‘’πœ’ (πœ‚πΈ ) In particular, if π‘’πœ’ (πœ‚πΈ ) β‰  0 then one has πœ’ (Fr–𝑣 1) β‰  1 for all 𝑣 e 𝑆 (𝐸) \ 𝑆 (πΈπœ’ ). On the other hand, . since 𝐿𝐾,𝑆 (πΈπœ’ ) ( πœ’, 𝑠) = 𝐿𝐾 ( πœ’, 𝑠) we have 𝐿𝐾,𝑆 (𝐸) ( πœ’, 𝑠) = 𝑣 e𝑆 (𝐸)\𝑆 (πΈπœ’ ) 1 – πœ’ (Fr𝑣)𝑁 𝑣–𝑠 Β· 𝐿𝐾 ( πœ’, 𝑠), (4.3) (+ Λ† whilst (2.41) implies, with π‘ˆ denoting either 𝑆 (𝐸) or π‘†βˆž (𝐾 ), that for all πœ“ e 𝐸 we have ord𝑠=0 𝐿𝐿,π‘ˆ (πœ“, 𝑠) = dimQ (π‘’πœ“ Β· (Q βŠ—Z 𝑋𝐸,π‘ˆ )) = π‘Ÿ dimQ (π‘’πœ“ Β· (Q βŠ—Z 𝑋𝐸,𝑆ram (𝐸/𝐾 ) )) if π‘ˆ = 𝑆 (𝐸) 0 if π‘ˆ = π‘†βˆž (𝐾 ), πœ“ β‰  1
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