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I am a little confused about the log structure of Witt ring and its Frobenius map. Let $k=\bar{\mathbb{F}}_p$, $W:=W(k)$ the Witt ring. We know that to deal with the semistable reduction, i.e. scheme over $W$ etale locally of the form $X=Spec(W[x_1,\cdots, x_d]/(x_1\cdots x_d-p))$, if we want to say that $X$ is log smooth over $W$, we need the following log structure $(Spec(W),M,\alpha)$, $\alpha : M=W\setminus\{0\}\hookrightarrow W$. However, we know that $W$ has the Frobenius map $\sigma$, the corresponding map on the monoid $M$ is the restrction of $\sigma$ on $W\setminus \{0\}$. If we consider the closed point of $Spec(W)$, such that $(Spec(k),N, \beta)\hookrightarrow (Spec(W),M,\alpha)$ is exact closed immersion, then the log structure $N$, is the following $N=(k\setminus\{0\})\oplus \mathbb{N}$, and $\beta: N\to k$, is defined as follows; $(b,0) \to b$ and $(b,n)\to 0$, for $n>0$, $b\in k\setminus\{0\}$. We then see that the Frobenius map $\sigma$ induced a map on $(Spec(k),N, \beta)$, which on the underying scheme is just the usual power $p$ map on $k$. But the map on the monoid is the direct sum of power $p$ map and identity map , i.e. $N\to N$, $(b,n)\to (b^p,n)$, which is different from the absolute Frobenius map. Because we know that for any log scheme $(Y,L)$, the absolute Frobenius map on the monid $L$ is just the power $p$ map, i.e. $l\to p\cdot l$. On the other hand , when deal with absoulte (relative )Frobenius map, people ususally consider the trivial log structure on $W$ and $k$.

So does it means that one can not deal with semistable reduction and Frobenius at the same time ? For example, if one want to generalized the calssical Cartier Descend theorem over smooth curve (which said that a vector bundle with connection is come from Frobenius pull back iff its $p$- curvature vanishing ) to node curves, one will have problem?

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  • $\begingroup$ Is this the absolute Frobenius given in Definition 4.7 of Kato? $\endgroup$
    – S. Carnahan
    Commented May 18, 2014 at 0:03
  • $\begingroup$ @Carnahan, Yes. $\endgroup$
    – Lan
    Commented May 18, 2014 at 10:43

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