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 $\mathbb{C}$?
Algebraic closure of $\mathbb{C}(x)$ is isomorphic to $\mathbb{C}$
2 Answers
If $B$ is a transcendence basis of $\mathbb C$ over $\mathbb Q$, then $B$ is uncountable, and $\mathbb C$ is an algebraic closure of $\mathbb Q(B)$.
Now $B\cup\{x\}$ is a transcendence basis of $K$ over $\mathbb Q$. Since $B$ and $B\cup\{x\}$ have the same cardinal, $K$ and $\mathbb C$ are isomorphic: they are both algebraic closures of purely transcendental extensions of $\mathbb Q$ of the same transcendence degree.
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0No. Uncountable means only «not countable» and there are many other cardinalities! For example $\mathbb R$ and the set $\mathcal P(\mathbb R)$ of all subsets of $\mathbb R$ are uncountable of different cardinalities. – 2012-04-29
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.