Explanation for devissage argument

Question

Let $K$ be a local field of characteristic $0$ with the ring of integers $\mathcal{O}_K$ and uniformizer $\pi$. Let $k$ be the residue field of $K$ with $\text{card}(k)=q$. Let $\mathcal{O}_\mathcal{E}$ be the $\pi$-adic completion of $\mathcal{O}_K((u))$, where $u$ is a fixed local co-ordinate. Then $\mathcal{O}_\mathcal{E}$ is complete local ring with uniformizer $\pi$ and residue field $E:=k((u))$. Let $\mathcal{E}$ be the field of fractions of $\mathcal{O}_\mathcal{E}$. Let $\widehat{\mathcal{E}^{ur}}$ be the completion of the maximal unramified extension of $\mathcal{E}$. Let $\mathcal{O}_{\widehat{\mathcal{E}^{ur}}}$ denotes the ring of integers of $\widehat{\mathcal{E}^{ur}}$. Then $\mathcal{O}_{\widehat{\mathcal{E}^{ur}}}$ is a complete local ring with uniformizer $\pi$ and residue field as $E^{sep}$. Then we have the following exact sequence \begin{equation*} 0\rightarrow k \rightarrow E^{sep}\xrightarrow{x\mapsto x^q-x}E^{sep}\rightarrow 0. \end{equation*} In other words, the sequence \begin{equation*} 0\rightarrow \mathcal{O}_K/\pi\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto x^q-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} is exact as $k$ is the residue field of $\mathcal{O}_K$ and $E^{sep}$ is the residue field of $\mathcal{O}_{\widehat{\mathcal{E}^{ur}}}$. Then by devissage the sequence \begin{equation} 0\rightarrow \mathcal{O}_K/\pi^n\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto x^q-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0, \end{equation} is exact for all $n\geq1$. I don't want to say the word "by devissage" and want to write the explicit proof. I am trying to induction on $n$. but somehow I am not able to prove that the sequence is exact. Is there any other way to prove the exactness of this sequence.

Answer 1

Let $\varphi_q: \mathcal{O}_{\widehat{\mathcal{E}^{ur}}} \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}$ be a ring homomorphism such that its reduction mod $\pi$ is the map $E^{sep}\rightarrow E^{sep}, x \mapsto x^q$. Knowing that the sequence

$$\begin{equation*} 0\rightarrow \mathcal{O}_K/\pi\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto x^q-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} $$

(the left arrow being induced by the inclusion $\mathcal{O}_K\hookrightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}$) is exact, I claim the following:

A necessary and sufficient condition for $$\begin{equation*} 0\rightarrow \mathcal{O}_K/\pi^n\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto \varphi_q(x)-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} $$ to be exact for all $n$, is that the restriction of $\varphi_q$ to $\mathcal{O}_K$ is the identity, i.e. $\varphi_{q \vert \mathcal{O}_K} = id_{\mathcal{O}_K}$.

Proof: Induction on $n$. The case $n=1$ is known as stated above. For general $n \ge 2$, put the three sequences

$$\begin{equation*} 0\rightarrow \pi\mathcal{O}_K/\pi^n\mathcal{O}_K \rightarrow \pi\mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto \varphi_q(x)-x} \pi\mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} $$ $$\begin{equation*} 0\rightarrow \mathcal{O}_K/\pi^n\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto \varphi_q(x)-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^n \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} $$

$$\begin{equation*} 0\rightarrow \mathcal{O}_K/\pi^{n-1}\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^{n-1} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto \varphi_q(x)-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^{n-1} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} $$ into a commutative diagram with the obvious maps in the columns. By induction hypothesis, the top and bottom row are exact. A generalised version of the Nine Lemma shows that the middle row is exact if and only if it is a complex, i.e. if for all $\bar x \in \mathcal{O}_K/\pi^n$, we have $\varphi_q(x) - x = 0$ mod $\pi^n$. By $\pi$-adic completeness of $\mathcal{O}_K$, this last condition holds for all $n$ if and only if $\varphi(x)=x$ for all $x \in \mathcal{O}_K$. QED.

To be clear, as soon as $\varphi_q$ does not restrict to the identity on $\mathcal{O}_K$, that means that there is some $x \in \mathcal{O}_K$ such that $\varphi_q(x) \neq x$, which means there is some $m \in \mathbb N$ such that $\varphi_q(x)-x \neq 0$ mod $\pi^m$, which means that the sequence $$\begin{equation*} 0\rightarrow \mathcal{O}_K/\pi^m\mathcal{O}_K \rightarrow \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^m \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\xrightarrow{x\mapsto \varphi_q(x)-x} \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}/\pi^m \mathcal{O}_{\widehat{\mathcal{E}^{ur}}}\rightarrow0 \end{equation*} $$ (and likewise for all $n \ge m$) is not exact because it is not even a complex: that $\bar x$ is not sent to $0$ by composition of the two middle arrows.

---

Note further that the condition $\varphi_{q \vert \mathcal{O}_K} = id_{\mathcal{O}_K}$ is equivalent to the simple condition $\varphi_q(\pi)=\pi$, by $\pi$-adic expansion and the fact it's automatically true modulo $\pi$. Now e.g. if $K$ is unramified and we can choose $\pi=p$, this is automatically true because $\phi(p)=p$ for every ring homomorphism. But for the general case, I do not see see an argument why $\varphi_q(\pi) = \pi$ should hold automatically. Actually, I see no reason why e.g. in the case that $K\vert \mathbb Q_p$ is Galois and ramified, one could not take $\varphi$ to be any non-trivial Galois automorphism of $K\vert K_u$ where $K_u$ is the maximal unramified subextension of $K \vert \mathbb Q_p$.