# Extension to an automorphism of topological fields

Here is what I am trying to prove:

Let $$L$$ be an algebraically closed topological field of characteristic zero and $$K$$ be a proper algebraically closed and topologically closed subfield of $$L$$. Let $$\lambda,\mu\in L\setminus K$$ and $$\tau:K\cup\lbrace\lambda\rbrace\to L$$ be given by $$\tau(x):=x$$ for all $$x\in K$$ and $$\tau(\lambda):=\mu$$. Then there exists a topological field automorphism $$\tilde{\tau}\in\text{Aut}(L)$$ with $$\tilde{\tau}|_{K\cup\lbrace\lambda\rbrace}=\tau$$ i.e. an extension of $$\tau$$ to a field automorphism that is also a homeomorphism.

I am aware that $$L:K$$ and $$K(\lambda):K$$ are transcendental extensions. I am not very clued up on field theory or Galois theory but it seems that very little if any of ordinary field/Galois theory can work for automorphisms of topological fields. I do not need to construct one explicitly but rather just confirm whether or not there exists an extension. Any ideas on how I might go about doing this?

If I understand correctly what you are trying to prove, it is just not true: Take e.g. $$K=\mathbb{C}$$, $$L=\bigcup_{n\in\mathbb{N}}\mathbb{C}((t^{1/n}))$$ the Puiseux series together with the topology coming from the $$t$$-adic valuation $$v_t$$. Every topological automorphism of $$L$$ will preserve the valuation ring $$\bigcup_{n\in\mathbb{N}}\mathbb{C}[[t^{1/n}]]$$ of $$v_t$$, hence if you set $$\lambda=t$$ and $$\mu=t^{-1}$$ there will be no topological automorphism $$\tilde{\tau}$$ with $$\tilde{\tau}(\lambda)=\mu$$.