Extension to an automorphism of topological fields

Question

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?

Answer 1

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 $\tilde{\tau}(\lambda)=\mu$.