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Let $K$ be a complete valued field extension of $\mathbb{Q}_p$ with possibly imperfect residual field $k$. Assume that there exists a ring endomorphism $\sigma: K\to K$ lifting the $p$-th power map of $k$.

The question is : given an (arbitrary) finite extension $L/K$, is it true that there exists a finite extension $M/L$ together with a Frobenius endomorphism lifting $\sigma$, or a power of $\sigma$ ?

The answer is yes if $K/\mathbb{Q}_p$ is finite (and hence $\sigma$ is bijective). In fact (up to my mistakes) by Galois theory arguments any Galois extension has a Frobenius automorphism lifting $\sigma$.

The answer is also positive if $L/K$ is unramified. In other words this means (by definition) that $L = K \otimes_{C(k)} C(k')$, where $k'/k$ is any field extension and $C(k)$, $C(k')$ are Cohen rings of $k$, and $k'$ respectively (Cohen rings are the analogous of Witt Vectors for imperfect fields). In fact maps between fields of characteristic $p$ lifts (not uniquely, hence not functorially) to their Cohen rings.

Do you have any information about the existence of such a Frobenius in the ramified case ? I need a Frobenius up to possibly extend my field $L$.

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    $\begingroup$ 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. $\endgroup$
    – user29720
    Commented Dec 21, 2012 at 4:03
  • $\begingroup$ What is the Galois theory argument? $\endgroup$
    – Will Sawin
    Commented Dec 21, 2012 at 4:05
  • $\begingroup$ 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 $k_M$. $\endgroup$
    – PULITA ANDREA
    Commented Dec 21, 2012 at 11:32
  • $\begingroup$ 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. $\endgroup$
    – PULITA ANDREA
    Commented Dec 22, 2012 at 16:57

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