Suppose that $S$ is a perfect ring of char $p$ and is t-adically complete for some non-zero-divisor $t$. Is the ring of Witt vectors $W(S)$ $[t]$-adically complete? If so can i get a proof please?
1 Answer
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This is actually true for any ring $S$ of characteristic $p$, and the argument I have in mind requires considering Witt vectors of non-perfect rings(such as $S/t^n$), so let me do it in full generality.
Fix an integer $k\geq 0$. For any ring $S$ with an ideal $I\subset S$ the canonical map induces an isomorphism $W_k(S)/([I]+V[I]+\dots+V^{k-1}[I])\to W_k(S/I)$.
In our situation we have surjections $W_k(S)/[t]^n\to W_k(S/t^n)$ with kernel generated by $V[t^n],\dots V^{k-1}[t^n]$. Denote this kernel by $K_n$. For any $m\geq 0$ the reduction map $W_k(S)/[t]^{p^km}\to W_k(S)/[t]^{m}$ sends $V^i[t^{p^km}]=V^iF^k[t^m]=p^i[t^{p^{k-i}m}]$ to zero, hence kills $K_{p^km}$. It means that these maps induce an injections on inverse limits(the above computation shows that the $m$-th component of an element of the kernel is zero) $$\lim\limits_{n}W_k(S)/[t]^n\hookrightarrow \lim\limits_{n}W_k(S/t^n)$$
Next, the map $W_k(S)\to\lim\limits_{n}W_k(S/t^n)$ is an isomorphism, because on the level of sets $W_k(R)$ is canonically isomorphic to $\prod\limits_{i=0}^{k-1} R$ for any ring $R$ and the map $S\to\lim S/t^n$ is an isomorphism by assumption. It implies that the two maps $$W_k(S)\to \lim\limits_{n} W_k(S)/[t]^n\to\lim\limits_{n}W_k(S/t^n)$$ are both isomorphisms.
It remains to pass to the inverse limit over $k$. Since $t$ is not a zero-divisor in $S$, so is any power $[t^n]$ in $W_k(S)$ and this gives an exact sequence of inverse systems $$0\to \{W_k(S)\}_k\to \{W_k(S)\}_k\to \{W_k(S)/[t]^n\}_k\to 0$$ The exactness of the induced sequence of inverse limits implies that $W(S)/[t]^n$ is isomorphic to $\lim\limits_{k}{W_k(S)/[t^n]}$.
It follows that the map $W(S)=\lim\limits_k W_k(S)\to \lim\limits_k\lim\limits_n W_k(S)/[t]^n=\lim\limits_{n}W(S)/[t]^n$ is an isomorphism.
