About Frobenius of Witt vectors - MathOverflow

About Frobenius of Witt vectors

2011-04-20T23:07:24Z

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!

Answer by Felipe Voloch for About Frobenius of Witt vectors

2011-04-21T00:38:13Z

I think your argument is essentially correct.

Here is a proof for the algebraic closure of a finite field. It is enough to deal with the units of the ring of truncated Witt vectors $W_n(k)$ for all $n$. But then this is an algebraic group over a finite field and a theorem of Lang (Amer J Math 1956) states that $x \mapsto \sigma(x)x^{-1}$ is surjective for any algebraic group over such a field. I think from the algebraic closure of a finite field, the result follows for any algebraically closed field of positive characteristic.

It's not going to hold for any finite field, as you'll get the elements of norm one only. For $\mathbb{F}_p$ you don't get the identity but the function identically equal to $1$.

Answer by Lubin for About Frobenius of Witt vectors

2011-04-23T05:41:13Z

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.