# isomorphisms of algebraic closures

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

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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, $\overline{\mathbb{C}(X)}\cong \mathbb{C}$.

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Transcendence degree... over what? $\mathbb{C}$ is algebraic over $\mathbb{R}$, for example. I was going to guess transcendence degree over $\mathbb{Q}$ but that doesn't make sense for fields of positive characteristic. –  Zhen Lin Jul 25 '11 at 13:05
@Zhen: The transcendence degree of a field is its transendence degree as an extensions of its prime subfield: that is, either $\mathbb{Q}$ or $\mathbb{Z}/p\mathbb{Z}$, depending on the characteristic. –  Chris Eagle Jul 25 '11 at 13:43
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 $F$.