Timeline for Existence of a Frobenius endomorphism of ramified extensions with unperfect residual field

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Dec 22, 2012 at 16:57	comment	added	PULITA ANDREA		Thanks kreck, could you provide some detail please ? I do not see why a linear combination $\sum_n\alpha_n\sigma^n(a)^{1/p}=0$, with the $\alpha_n\in K$ is impossible.
Dec 21, 2012 at 11:32	comment	added	PULITA ANDREA		The Galiois theory argument is the following. Let $M/L$ be any extension such that $M/\mathbb{Q}_p$ is Galois. Let $M_0$ be its absolutely unramified extension, i.e. the Fraction field of the Witt vectors $C(k_M)$ of the residual field $k_M$ of $M$. There is a Frobenius on $M_0$. This Frobenius extends to the algebraic closure of $M$, and hence to an automorphism of $M$ inducing the $p$-th power map on $k_M$.
Dec 21, 2012 at 4:05	comment	added	Will Sawin		What is the Galois theory argument?
Dec 21, 2012 at 4:03	comment	added	user29720		It seems to fail for $L/K$ totally ramified and perfect $k$ not algebraic over $\mathbf{F}_p$. Let $K = W(k)[1/p][\zeta_p]$ for perfect $k$ that contains $t$ not algebraic over $\mathbf{F}_p$. Let $\sigma:K\simeq K$ fixing $\zeta_p$ lift a positive power of Frobenius on $k$. The $t^{p^n}$'s are linearly independent over $\mathbf{F}_p$, so for $a = 1 + p[t] \in W(k)^{\times}$ the $\sigma^n(a)$'s are multiplicatively independent in $K^{\times}/(K^{\times})^p$. Thus, the elements $\sigma^n(a)^{1/p}$ generate an infinite extension of $K$, so $L=K(a^{1/p})$ seems to be a counterexample.
Dec 20, 2012 at 21:31	history	asked	PULITA ANDREA	CC BY-SA 3.0