# Algebraic closure of $\Bbb{F}_p(u,v)$

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 ?