Is it possible to embed $\Bbb{C}(x)$ (the field of rational functions over the complex numbers) in $\Bbb{C}$ ?
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Embed as via a field homomorphism? Or as a set (i.e. just an injection)? – Daniel Fischer Aug 12 '13 at 08:37
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It is explicitly noted that $\mathbb{C}(x)$ is considered as a field. – Martin Brandenburg Aug 12 '13 at 08:52
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Related: http://math.stackexchange.com/questions/1598358 – Watson Aug 17 '16 at 13:10
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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 $\bf{C}$.
BBBB
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Is it furthermore true that all fields of characteristic zero can be embedded in $\Bbb C$? That's a statement I heard, but from what you said it should be wrong if field extensions of $\Bbb Q$ of transcendence degree greater than the continuum exist. – Rodrigo Mar 05 '14 at 16:23
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3Assuming choice, there are algebraically closed fields of arbitrarily large cardinality (take the algebraic closure of the field of rational functions over $\mathbb{Q}$ with in many variables), so they cannot be embedded into $\mathbb{C}$. But every finitely generated field of characteristic zero can be embedded in the complex numbers, and this fact is very useful. – BBBB Mar 18 '14 at 23:51
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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 $\overline{\mathbb{C}(X)}\xrightarrow{\approx}\mathbb{C}$.
Alex Youcis
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