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The following is the proposition (1.4) of Mumford's book Algebraic Geometry

If $\mathbb C$ has infinite transcendence degree over $k$, then every variety has a $k$-generic point.

In the proof

Then $L$ is an extension field of $k$ of finite transcendence degree. But any such field is isomorphic to a subfield of $\mathbb C$. i.e. there exists a monomorphism $\phi : L \rightarrow {\mathbb C}$.

Is any extension field of $k$ of finite transcendence degree isomorphic to a subfield of $\mathbb C$ ? If so, how do you prove it.

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Yes, it is.

Assume the transcendence degree of $L$ is equal to $n$, and choose $n$ algebraically independent elements $\xi_1,...,\xi_n \in \Bbb C \setminus k$. We have an isomorphism $$ k(X_1,...,X_n) \cong k(\xi_1,...,\xi_n) \subset \Bbb C. $$ Now by definition, $L$ is a finite algebraic extension of $k(X_1,...,X_n)$, hence isomorphic to a finite algebraic extension $L'$ of $k(\xi_1,...,\xi_n) \subset \Bbb C$. But every such $L'$ is necessarily contained in $\Bbb C$, and we are done.

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