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Feb 20, 2010 at 20:42 comment added BCnrd The etale property in this rank-1 case mod p is indeed the same as phi being nonzero on some basis vector. Concerning the 2nd part of your comment, nothing is being missed; I was just expressing (perhaps in an unclear way) that there's a lot of nontrivial action on the coefficients. When reading the question I misread it as not accounting for that (but I see it was implicit when you wrote "(on some basis element)"). Anyway, see the Example 13.6.6, which I hope will clear things up.
Feb 20, 2010 at 20:29 comment added sibilant Yes. I certainly meant to include etale. In this case (i.e. 1-dimensional in char p), this is simply asserting that $\phi$ is non-zero, yes? Regarding the trivial representation, its associate $(\phi,\Gamma)$-module must just be $F_p((T))$ with the standard $\phi$ and $\Gamma$ actions. Since $\phi(1)=1$ and $\gamma(1)=1$ for all $\gamma \in \Gamma$, with respect to the basis ``1" all of these operators have matrix 1 (certainly in $F_p^\times$). What am I missing?
Feb 20, 2010 at 17:55 history answered BCnrd CC BY-SA 2.5