TITLE: Algebraic closure of $\mathbb{C}(x)$ is isomorphic to $\mathbb{C}$
QUESTION [8 upvotes]: Let $x$ be transcendental over $\mathbb{C}$. Let $K$ be the algebraic closure of $\mathbb{C}(x)$. How to show that $K$ is isomorphic to $\mathbb{C}$?

REPLY [5 votes]: This is an immediate consequence of a famous theorem of Steinitz that uncountable algebraically closed fields are isomorphic iff they have equal characteristic and equal cardinality (a prototypical example of Morley's categoricity theorem). 
This fails in the countable case, e.g. if $\mathbb A = $ algebraic numbers then, though countable, $\overline{\mathbb A(x)}\not\cong \mathbb A$ since they have unequal transcendence degree over $\mathbb Q,\:$ viz. $1$ vs. $0$, resp.