Let $k=\Bbb{F}_p(u,v)$ or any field of the same kind, let $k_s$ be the largest separable algebraic extension of $k$.

Claim $\overline{k} =(k_s)^{1/p^\infty}= \bigcup_n (k_s)^{1/p^n}$

For $a \in \overline{k}$ let $a^{1/p^n}$ be the unique root of $x^{p^n}-a =(x-a^{1/p^n})^{p^n}$ then $(k_s)^{1/p^n}$ is the field containing $k_s$ and the $p^n$-th root of all the elements of $k_s$.

Assume $(k_s)^{1/p^\infty}$ is not algebraically closed, take $c \in \overline{k}, \not \in (k_s)^{1/p^\infty}$ and $f \in k_s[x]$ its $k_s$-minimal polynomial. Let $p^r$ be the largest power such that $f(x) = g(x^{p^r})$ for some polynomial $g \in k_s[x]$. Then $g$ is irreducible and it is not of the form $g(x)=h(x^p)$ thus $g' \ne 0$. If $g$ is not separable then $\gcd(g,g')$ divides $g$ which can't be irreducible. Thus $g$ is separable and $c^{p^r}$ being one of its roots, we obtain $c^{p^r} \in k_s$ and $c \in (k_s)^{1/p^r}$.

Let $(k^{1/p^\infty})_s$ be the separable closure of $k^{1/p^\infty}$.

Claim $\overline{k}= (k^{1/p^\infty})_s=\Bbb{F}_p(u^{1/p^\infty},v^{1/p^\infty})_s$

Since $(s^{1/p^r}+t^{1/p^r})^{p^r} = s+t$ we have $(s+t)^{1/p^r} = s^{1/p^r}+t^{1/p^r}$ and the map $s \mapsto s^{1/p^r}$ is an automorphism of $\overline{k}$.

As $(\Bbb{F}_p(u^{1/p^n},v^{1/p^n}))^{p^n} = (\Bbb{F}_p)^{p^n}((u^{1/p^n})^{p^n},(v^{1/p^n})^{p^n})=\Bbb{F}_p(u,v)$ we get $k^{1/p^\infty} = \Bbb{F}_p(u^{1/p^\infty},v^{1/p^\infty})$.

For any $a \in (k^{1/p^\infty})_s$ take its $k^{1/p^\infty}$-minimal polynomial $g=\sum_{j=0}^d s_j x^j \in k^{1/p^\infty}[x]$ which is separable, then $f=\sum_{j=0}^ds_j^{1/p^r} x^j\in k^{1/p^\infty}[x]$ is separable and $f(a^{1/p^r})^{p^r} = g(a) = 0$, thus $a^{1/p^r} \in (k^{1/p^\infty})_s$ and $(k^{1/p^\infty})_s = ((k^{1/p^\infty})_s)^{1/p^\infty}$ contains $(k_s)^{1/p^\infty} = \overline{k}$.

Is it correct ?

Is there a way to understand those separable / $p$-th roots things in greater generality ?


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