You may find the following more transparent, since it uses only the fact that the Witt vectors are an unramified 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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2 | removed "unramified" from the description of $R$, unnecessary to proof | ||
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You may find the following more transparent, since it uses only the fact that the Witt vectors are an unramified 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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