About Frobenius of Witt vectors - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:19:58Z http://mathoverflow.net/feeds/question/62468 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62468/about-frobenius-of-witt-vectors About Frobenius of Witt vectors asker 2011-04-20T23:07:24Z 2011-04-23T07:02:27Z <p>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?</p> <p>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.</p> <p>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!</p> http://mathoverflow.net/questions/62468/about-frobenius-of-witt-vectors/62476#62476 Answer by Felipe Voloch for About Frobenius of Witt vectors Felipe Voloch 2011-04-21T00:38:13Z 2011-04-21T01:35:05Z <p>I think your argument is essentially correct.</p> <p>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.</p> <p>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$1$.</p> http://mathoverflow.net/questions/62468/about-frobenius-of-witt-vectors/62706#62706 Answer by Lubin for About Frobenius of Witt vectors Lubin 2011-04-23T05:41:13Z 2011-04-23T07:02:27Z <p>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.</p>