[Math] About Frobenius of Witt vectors

nt.number-theorywitt-vectors

Let $k$ be a characteristic $p$ alg. closed field, Let $W(k)$ be the Witt vectors, Let $\sigma$ be the Frobenius, then we also have $\sigma: W(k)^{\times} \to W(k)^{\times}$, where $W(k)^{\times}$ are the units in $W(k)$. Thus we can define a map $f: W(k)^{\times} \to W(k)^{\times}$, $f(x) = \frac{\sigma(x)}{x}$. My question is, is $f$ surjective?

Here is what I think is a proof.
Suppose $a \in W(k)^{\times}$, write $a$ as $(a_0, a_1, \ldots)$, suppose $x =(x_0, x_1, \ldots)$, then we are looking for $x$ such that $\sigma(x)=x\cdot a$, which means $x_0^p =x_0a_0$ and $x_1^p =x_1 a_0^p + x_0^pa_1$, etc, and clearly, we can solve $x_0$ in the first equation, then solve $x_1$, etc since $k$ is alg. closed.

Is the proof correct? And is there any other proof?
Also, is the alg. closedness necessary? Of course, if $k= \mathbb{F}_p$, $f$ is identity map,but what about $k$ other than $\mathbb{F}_p$$?
Thank you!

Best Answer

You may find the following more transparent, since it uses only the fact that the Witt vectors are a complete DVR with residue field $k$. Call the Witt vectors $R$, and let $y$ be a unit for which you want to find $z$ with $z^\sigma=yz$. First do it mod $p$, by solving $\zeta^p=\eta\zeta$ for $\zeta$ in $k$, where $\eta$ is the image of $y$ in $k$. Now you can assume that you have $z\in R$ satisfying $z^\sigma\equiv yz \mod{(p^m)}$, in other words $z^\sigma \equiv yz + p^m\delta \mod{(p^{m+1})}$. Now you want to adjust $z$ to $z'=z+p^m x$ so that $z'$ satisfies your congruence modulo $(p^{m+1})$. This boils down to solving $\xi^p - \xi \eta + \delta = 0$ in $k$, which you can do. So you see that you don't need $k$ to be algebraically closed, just separably closed.

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