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Let $R$ be a perfect ring of characteristic $p$, $W(R)$ its Witt vectors. We know that any element $x\in W(R)$ has a unique Teichmuller expansion:

$$x=[c_0]+[c_1]p+[c_2]p^2+\cdots$$ here all $c_i$ are in $R$. I want to know whether the following assertion is true:

$x$ is invertible $\Leftrightarrow$ $c_0$ is invertible.

The "$\Rightarrow$" direction is trivial. But can we prove the inverse?

Any help is appreciated. Thanks!

Richard
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    Your hunch is correct. If you can show that $[c_0]$ is invertible, then you are reduced to the case of $[c_0]=1$, when your series has the form $1-n$ such that $1+n+n^2+n^3+\cdots$ converges. In the cases I'm familiar with the set of Teichmüller elements is closed under multiplication, taking care of $[c_0]^{-1}=[c_0^{-1}]$. I'm a bit rusty to be sure that this is always the case (may depend on your notation). – Jyrki Lahtonen Feb 10 '22 at 13:42

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To expand on Jyrki Lahtonen's comment, we can remove the perfect assumption. For simplicity, let us still suppose that $R$ is a commutative ring of characteristic $p$. Any element $x\in W\left(R\right)$ can be uniquely written as: \begin{equation*} x=\left[c_{0}\right]+V\left(x'\right)\text{,} \end{equation*} where $V$ is the Verschiebung operator, and $x'\in W\left(R\right)$. In the setting of your question, you can define $x'=F\left(\sum_{u\in\mathbb{N}^{*}}p^{u-1}F^{-1}\left(\left[c_{u}\right]\right)\right)$ (because you assumed that $R$ is perfect, but we shall not).

As you said, it is clear that when $x$ is invertible, so is $c_{0}$. Conversely, as the Teichmüller operator is multiplicative, we have $\left[c_{0}\right]\left[{c_{0}}^{-1}\right]=1$, and in particular: \begin{equation*} x\left[{c_{0}}^{-1}\right]=\left[c_{0}\right]\left[{c_{0}}^{-1}\right]+V\left(x'\right)\left[{c_{0}}^{-1}\right]=1+V\left(x'F\left(\left[{c_{0}}^{-1}\right]\right)\right)\text{.} \end{equation*}

So, as Jyrki Lahtonen wrote, we can just suppose without loss of generality that $c_{0}=1$.

Now, the series $\sum_{u\in\mathbb{N}}\left(-V\left(x'\right)\right)^{u}$ converges for the $V$-adic topology, because we have assumed that $R$ has characteristic $p$, so that $\left(-V\left(x'\right)\right)^{u}\in V^{u}\left(W\left(R\right)\right)$. And a direct computation yields: \begin{equation*} \left(1+V\left(x'\right)\right)\times\sum_{u\in\mathbb{N}}\left(-V\left(x'\right)\right)^{u}=1 \end{equation*}

user242429
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