Can I embed $\Bbb{C}(x)$ into $\Bbb{C}$?

Solution 1:

Yes, it is possible; although the solution relies heavily on the axiom of choice, and is not constructive or geometric in any way. If you believe the axiom, then every pair of algebraically closed fields of the same transcendence degree over the base field and of the same characteristic are isomorphic - simply pick a bijection between transcendence bases for both, and extend it to an isomorphism of fields.

Observe now that the algebraic closure of $\bf{C} (x)$ has transcendence degree continuum, just like $\bf{C}$. They are therefore isomorphic and this isomorphism induces an embedding of $\bf{C} (x)$ into $\bf{C}$.

Solution 2:

Yes. Note that $\overline{\mathbb{C}(X)}$ is of characteristic zero, and still of the same cardinality as $\mathbb{C}$. So, there is an isomorphism $\overline{\mathbb{C}(X)}\xrightarrow{\approx}\mathbb{C}$.