On an isomorphism between $p$-adic power series and an inverse limit

Question

Let $K$ be an extension field of $\mathbb{Q}_p$, let $O$ be the ring of integers of $K$, and let $P$ be the maximal ideal of $O$.

If $K$ is a finite extension of $\mathbb{Q}_p$, there is the well-known algebraic isomorphism

$$O[[X]]\cong\varprojlim O[X]/((1+X)^{p^n}-1).$$

A proof of this fact can be found, for example, in Lang's "Cyclotomic fields", in Washington's "Introduction to Cyclotomic Fields", and in Neukirch et al's "Cohomology of number fields". In all of this proofs, at some point one needs a generator of the maximal ideal $M$, that is, a uniformizer element of $K$. Hence, a priori, this proofs do not work if we let $K$ to be a non discretely valued extension.

Now, in page 96 of Lang's "Cyclotomic fields I and II", he claims (and uses) that this isomorphism is true when one let $K=\mathbb{C}_p$, the completion of the algebraic closure of $\mathbb{Q}_p$. My question is: where can I find a proof of this?, or, why is this obviously true?, since he only proves this isomorphism for $K=\mathbb{Q}_p$.

Note: Since I do not know too much about (the foundations of) Iwasawa theory, I do not know if this is an elementary question. If it is, please let me know and I will ask it in MSE.

Answer 1

There is unique division with remainder by a monic polynomial in $O[X]$, where $O$ is any commutative ring. When $O$ is a $p$-adically complete ring, meaning the natural ring homomorphism $O \to \varprojlim O/(p^n)$ is an isomorphism, the Weierstrass division theorem tells us there is unique division with remainder by a polynomial in $O[[X]]$ that is distinguished: monic with lower degree coefficients in the maximal ideal of $O$. Therefore, since $(1+X)^{p^n}-1$ is both monic and distinguished when $O$ is $p$-adically complete (as its non-leading coefficients are all divisible by $p$), the natural composite mapping $O[X] \rightarrow O[[X]] \rightarrow O[[X]]/((1+X)^{p^n}-1)$ is surjective with kernel the ideal $((1+X)^{p^n}-1)$ in $O[X]$, so we have a natural ring isomorphism $$ O[X]/((1+X)^{p^n}-1) \cong O[[X]]/((1+X)^{p^n}-1). $$ This is compatible with the inverse system built on both sides from the ideals $((1+X)^{p^n}-1)$ for $n = 0, 1, 2, \ldots$, so we get an isomorphism of inverse limits $$ \varprojlim O[X]/((1+X)^{p^n}-1) \cong \varprojlim O[[X]]/((1+X)^{p^n}-1). $$ The ring $O[[X]]$ is complete with respect to the $(p,X)$-adic topology and let's show $(1+X)^{p^n}-1 \subset (p,X)^n$. The ideal is $(p^n,p^{n-1}X,\ldots,pX^{n-1},X^n)$, the polynomial is $\sum_{k=1}^{p^n} \binom{p^n}{k}X^k$, and ${\rm ord}_p\binom{p^n}{k} = n - {\rm ord}_p(k) \geq n - k$, so for $1 \leq k \leq n$ we have $\binom{p^n}{k}X^k \in (p^{n-k}X^k)$ and for $k > n$ we have $\binom{p^n}{k}X^k \in (X^n)$, so $(1+X)^{p^n} - 1 \in (p,X)^n$. Thus $(1+X)^{p^n} - 1$ tends to $0$ in the $(p,X)$-adic topology on $O[[X]]$, so $\varprojlim O[[X]]/((1+X)^{p^n}-1) \cong O[[X]]$ for the same reason $\varprojlim \mathbf Z_p/(p^{n_i}) \cong \mathbf Z_p$ for any increasing sequence $n_i$. (Note: the polynomials $(1+X)^{p^n} - 1$ do not tend to $0$ in the $X$-adic topology or in the $p$-adic topology.)