Extension of complete discrete valuation fields and coefficient field

Let $L$ and $K$ be two complete discrete valuation fields of equal characteristic $0$. Assume that an embedding $K\subset L$ is fixed and let $F$ be the algebraic closure of $K$ inside $L$. Can you provide a proof or at least an hint for the following proposition?

Proposition: $F$ is a finite extension of $K$ and $L\cong F((t))$.

In particular this means that $F$ is actually the residue field of $L$. why is this true?

Edit: On $L$ we have an additional hypothesis. We know that its residue field $\overline L$ is again a complete discrete valuation field (of characteristic $0$). In other words $L$ has "dimension $2$''. For example $L=\mathbb Q_p((t))$.

• Are you sure this is true as stated? What if $K=L$? – Torsten Schoeneberg Jul 20 '18 at 23:32
• Seconding Torsten's concern. Consider the case of $K=\Bbb{Q}_p$ and $L$ any finite extension. Then $L$ is complete as a finite dimensional vector space over a complete field, and $F=L$. – Jyrki Lahtonen Jul 21 '18 at 10:19
• I forgot to add that $L$ is supposed to be two dimensional. It means that the residue field $\overline L$ is again a complete discrete valuation field. For example $L=\mathbb Q_p((t))$ – manifold Jul 21 '18 at 12:51
• I think there are several more conditions given in the paper you cite (I found the statement in the middle of p.7 of the current arxiv version arxiv.org/pdf/1203.6712.pdf). E.g. what you call $K$ (called $k$ in loc.cit.) is a local field of char. 0 with finite residue field, i.e. some finite extension of some $\Bbb Q_p$. – Torsten Schoeneberg Jul 21 '18 at 20:17