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I was reading the beautiful Brinon and Conrad's introduction to $p$-adic Hodge theory (link) and I came to the introduction of de rham period ring. During the book, they always mention that the quotient of the ring of integers of the completion of the algebraic closure of $\mathbb{Q}_p$ modulo the ideal generated by $p$ is not perfect. How to prove it? Thank you for any suggestion.

Torsten Schoeneberg
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Being perfect would mean that the map $x\mapsto x^p$ is bijective, or in other words, every element would have a unique $p$-th root and unique $p$-th power. Raising to the $p$-th power also raises the (multiplicative) valuation $|x|$ of an element $x$ to the $p$-th power. Now with $0 < c := |p| < 1$, can you see a whole bunch of elements in $O_{\mathbb{C}_p}$ which are nontrivial mod $p$, but whose $p$-th powers become $0$ mod $p$?

Torsten Schoeneberg
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  • +1 for this, but I would have thought that the right definition of perfectness of a ring like this would be that the multiplicative group was $p$-divisible. And I can’t offhand see that this condition fails. – Lubin Jan 17 '17 at 13:44
  • @Lubin: I think the multiplicative group satisfies $G^p = G$ since $\mathbb{C}p$ is algebraically closed; and indeed $O{\mathbb{C}p}/\mathfrak{m}$ is perfect for the maximal ideal $\mathfrak{m} = \lbrace x \in \mathbb{C} p : |x| < 1\rbrace$. But the OP refers to Brinon/Conrad notes, section 4.2 I presume, where the issue is that one cannot apply the Witt vector formalism to lift homomorphisms. The problem is that w.r.t. $\mathfrak{m}$, $O_{\mathbb{C}_p}$ is not a $p$-ring in their setting (specifically, not $\mathfrak{m}$-adically separated because $\mathfrak{m} = \mathfrak{m}^2$), ... – Torsten Schoeneberg Jan 17 '17 at 19:54
  • [cont'd] ... whereas w.r.t. to any ideal $\mathfrak{b} = \lbrace x \in \mathbb{C}p: |x| \le b \rbrace$ ($|p| < b < 1$) (which they might mean when they write "proper ideal" in remark 4.2.4), the ring $O{\mathbb{C}_p}/\mathfrak{b}$ is a variant of the non-domain with non-injective $p$-powers in the question, which makes lifting to Witt vectors non-unique. – Torsten Schoeneberg Jan 17 '17 at 19:54