Does $\lim_←$ and composite of field commutes?

Question

Does inverse limit and composite of field commutes?

For example, let $K_1$ and $K_2$ be fields. We take $\lim$ with respect to $p$-th Frobenius (not surjective), then, $\lim K_1K_2＝(\lim K_1)(\lim K_2)$ ?

I guess this is true, but I couldn't find any reference for this. Reference (book, pdf, etc..) is also appreciated.

Answer 1

Your question is very ambiguous.

Notations, assuming characteristic $p$:

• $K^{p^\infty}= \bigcap_{n\ge 1} K^{p^n}$

• $K^{1/p^\infty}= \{ a\in \overline{K}, \exists n, a^{p^n}\in K\}$.

• Let $\varprojlim K^{p^n} $ be the set/ring/field of sequences $(a_n)$ with $a_n\in K, a_n=a_{n+1}^p$. The map $(a_n)\to a_1$ is an isomorphism $\varprojlim_{n\ge 1}K^{p^n} \to K^{p^\infty}$.

Then look at $$K_1 = \Bbb{F}_p((x_0/x_1)^{1/p},(x_0/x_2)^{1/p^2},(x_0/x_3)^{1/p^3},\ldots )$$ $$K_2 = \Bbb{F}_p(x_1,x_2,x_3,\ldots)^{1/p^\infty}$$ we get that $$ x_0\in (K_1K_2)^{p^\infty}, \qquad K_1^{p^\infty}=\Bbb{F}_p \implies x_0\not \in K_1^{p^\infty}K_2^{p^\infty}$$