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Let $R$ be an $\mathbf{F}_p$-algebra. Kunz's theorem says that if $R$ is Noetherian, then the Frobenius of $R$ is flat iff $R$ is regular. Following the philosophy that valuation rings often behave like regular Noetherian rings, it is shown in Datta and Smith - Valuations and Frobenius that if $R$ is a valuation ring, then the Frobenius of $R$ is flat.

I'm wondering about a mixed characteristic variant of this. Let $R$ be a $p$-complete valuation ring (in which $p$ is nonzero and not invertible) with maximal ideal $\mathfrak{m}$, and let $\pi \in \mathfrak{m}$ be a regular element such that $\pi^p$ divides $p$. Is the Frobenius map $R/\pi \to R/\pi^p$ flat?

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  • $\begingroup$ I think you could choose an absolute integral closure $R \to R^+$, which is necessarily faithfully flat, and then compare your question for $R$ and $R^+$ to get the desired flatness since $\text{Frob}:R^+/\pi \to R^+/\pi^p$ is an isomorphism. $\endgroup$
    – Anonymous
    Dec 4, 2022 at 13:44

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