Algebraically closed field

Question

Is there an example of a field $E$ which is not algebraically closed but each finite simple field extension $K$ of $E$, is isomorphic (not necessarily is identity over $E$) as a field to $E$?

Answer 1

Let $K=\mathbb{C}((x))$. Let $L/K$ be a finite extension. Since $K$ is a non-Archimedian complete discretely valued field, there is a unique extension of the absolute value on $K$ to $L$.

Since the residue field of $K$ is algebraically closed, $L/K$ is totally ramified.

Let $\varpi$ be a uniformizer of $L$, then $\varpi^e \in xO_L^{\times}$, where $e=[L:K]$.

By Hensel’s lemma, $u\in O_L^{\times} \longmapsto u^e \in O_L^{\times}$ is onto, so we can choose $\varpi$ such that $\varpi^e=x$.

It’s easy to show that $O_L=O_K[\varpi]$, so the above shows that $L=\mathbb{C}((x^{1/e})) \cong \mathbb{C}((x)) = K$.