Q:

Isomorphism of an extension field of a field of finite transcendence degree

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.

A:

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.

