If $K$ is algebraically closed and $X$ is transcendental, is $\overline{K(X)}\simeq K(X)$?

Question

let $K$ be an algebraically closed field. Consider the algebraic closure $\overline{K(X)}$ of $K(X)$, with $X$ trascendent over $K$. Are there cases in which $\overline{K(X)}\cong K$? where $\cong$ is isomorphism in whatever sense u prefer. Example: if we consider $K=\mathbb{C}$ then this is true if we take $\cong$ isomorphism of $\mathbb{Q}$ vector spaces. What about field structure?

Thanks

Answer 1

Yes. All algebraically closed fields of the same characteristic and transcendence degree are isomorphic as fields. $\overline{K(X)}$ has transcendence degree one more than that of $K$, so $\overline{K(X)}\cong K$ whenever $K$ has infinite transcendence degree. For example, $\overline{\mathbb{C}(X)}\cong \mathbb{C}$.

Answer 2

The answer is yes, there are such cases. Indeed such cases are in a sense the norm.

If $\kappa$ is an uncountable cardinal, any two algebraically closed fields of cardinality $\kappa$ and characteristic $p$, where $p$ may be $0$, are isomorphic.

In model theory, the standard way to refer to this fact is by saying that the theory of algebraically closed fields of characteristic $p$ is $\kappa$-categorical for any uncountable $\kappa$.

Thus in particular, if $K$ is the field of complex numbers, then the algebraic closure of $K(X)$, where $X$ is transcendental over $K$, is isomorphic to $K$.

The algebraically closed $K$ for which the algebraic closure of $K(X)$ is not isomorphic to $K$ are the "unusual" ones! For example, if $F$ is the field of algebraic numbers, then the only extensions $K$ of $F$ with this property are, up to isomorphism, the algebraic closures of fields obtained by adding a finite number of transcendentals to $F$.