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I cannot find what I am missing in the following computation. Let $K=\mathbb{Q}_p(p^{1/{(p-1)p^{\infty}}})$ and $L=\mathbb{Q}_p(\zeta_{p^{\infty}})$, where $\zeta_{p^n}$ is a primitive $p^n$-th root of unity. I think I have computed that $\mathcal{O}_K/p\cong \mathcal{O}_L/p\cong \mathbb{F}_p[t^{1/p^{\infty}}]/t^{p-1}:=R$, which is perfect (I can provide more details for this computation if needed). We should then have that for the $p$-adic completions $\widehat{\mathbb{Z}_p[p^{1/{(p-1)p^{\infty}}}]}\cong \widehat{\mathbb{Z}_p[\zeta_{p^{\infty}}]}\cong W(R)$ since there is a unique strict $p$-ring having a given perfect ring $R$ of characteristic $p$ as its residue ring (Theorem 5, pg.39 Local Fields Serre). However, I know this cannot be true. Maybe there is an easier way to see it but here is an argument: By Corollary 5.10 in Kuhlmann - The algebra and model theory of tame valued fields, we get that $L$ is relatively algebraically closed in its completion. In particular, if the above isomorphism is true, this would give us that $p^{1/p}\in L$. But since $L/\mathbb{Q}_p$ is abelian, this would imply that $\mathbb{Q}_p(p^{1/p})$ is Galois, which is not true. p.s. This is related to a question that I posted on mathstack: https://math.stackexchange.com/questions/3662451/what-is-wrong-with-this-application-of-ax-kochen

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    $\begingroup$ For a ramified extension $\mathbf Z_p \subseteq \mathcal O_K$, the ring $\mathcal O_K/p$ is never perfect, because an element $x$ with valuation $0 < v(x) < v(p)$ satisfies $x^{p^r} = 0$ in $\mathcal O_K/p$ for $r \gg 0$, but $x$ is not zero in $\mathcal O_K/p$. $\endgroup$ May 8, 2020 at 0:00
  • $\begingroup$ Perfect here means that the Frobenius is surjective, not bijective. $\endgroup$ May 8, 2020 at 0:02
  • $\begingroup$ Edit: But you might be right; in Serre's book pg.39 he defines a ring to be perfect if the Frobenius is actually an automorphism. So probably the theorem that I used doesn't apply. $\endgroup$ May 8, 2020 at 0:08
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    $\begingroup$ As far as I know, the usual definition of perfect ring is that the Frobenius is an isomorphism, but I agree that the wording in Serre is confusing. $\endgroup$ May 8, 2020 at 0:10
  • $\begingroup$ A characteristic $p$ ring in which Frobenius is merely surjective is apparently sometimes called semiperfect. (also, I believe the word "perfect" also meant "bijective" in the lecture we were attending this week :) ) $\endgroup$
    – Wojowu
    May 8, 2020 at 9:39

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