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Let $\overline{\mathbb{F}_p}$ be the algebraic closure of $\mathbb{F}_p$. Let $W({-})$ denote the functor of taking $p$-typical Witt vectors. Then the extension $\mathbb{F}_p\rightarrow \overline{\mathbb{F}_p}$ induces a map $f: \mathbb{Z}_p=W(\mathbb{F}_p)\rightarrow W(\overline{\mathbb{F}_p})$.

Is $f$ split as a $\mathbb{Z}_p$-module map?

As far as I understand, $W(\overline{\mathbb{F}_p})$ is built out of $\mathbb{Z}_p$ by adjoining prime-to-$p$-th-power roots of unity, so it seems like you could try sending all those adjoined roots to zero. But some roots live in $\mathbb{Z}_p$, so you better not send those to zero.

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$W(\overline{\mathbb F_p})$ is the completion of the direct limit of $W(\mathbb F_{p^{n!}})$. The ring $W(\mathbb F_{p^{n!}})$ is a free module of rank $n$ over $W(\mathbb F_{p^{(n-1)!}})$, so the natural map $W(\mathbb F_{p^{(n-1)!}}) \to W(\mathbb F_{p^{n!}})$ has a splitting.

Combining all these splittings, we get a splitting of the direct limit. Since it is a $\mathbb Z_p$-module homomorphism, it extends to the completion, giving a splitting of the completion.

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  • $\begingroup$ Thanks! Does a similar result hold for F_q, q=p^k? Or even more ambitiously, for k a random perfect field of char p? $\endgroup$
    – kiran
    Commented Jan 18, 2022 at 19:58
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    $\begingroup$ @kiran For the first case, yes, by the same proof. For the second case, you have to use a more complicated proof (for example looking at the set of all finite extensions of $k$) but the answer is still yes. $\endgroup$
    – Will Sawin
    Commented Jan 18, 2022 at 20:28

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