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!


2 Answers 2


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.

  • 2
    $\begingroup$ Looking more closely at the argument above you see that all that's necessary for the map to be onto is for $k$ to be closed under the operation of taking $n$-th root when $n|(p-1)$, and to have no cyclic extensions of degree $p$. So the smallest field in each characteristic with the property that the map is onto the units is the union of the finite fields of cardinality $p^{(p^2-p)^m}$ for all positive $m$. In particular, in characteristic two, all you need is the maximal $2$-extension of $\mathbb{F}_2$. $\endgroup$
    – Lubin
    Apr 24, 2011 at 5:45

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 $1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.