Lefschetz Principle: explicit embeddings into $\mathbb C$. I am very confused about the Lefschetz Principle. I read the Tarski Principle, but I am not acquainted with logic. Is there a statement more close to the language of field theory? 
Most of all, I would like to see concretely how a field $k$ of characteristic zero, containing $\mathbb C$, can be embedded into $\mathbb C$. For example:


*

*Let $x_1,\dots,x_n$ be transcendental elements over $\mathbb C$. How to construct an embedding of $k=\mathbb C(x_1,\dots,x_n)$ into $\mathbb C$?

*Is it possible to construct a similar embedding starting with a field $k/\mathbb C$ of infinite transcendence degree over $\mathbb C$?

*If we start with $k$ algebraically closed of finite transcendence degree over $\mathbb C$, should we get an isomorphism $k\cong \mathbb C$ after the embedding constructed in 1?
In addition, I would like to know how one can use this result in Algebraic Geometry: if we work with algebraic varieties over $k$, they are all determined by finitely many coefficients in $k$. What kind of statements can we prove just "as if" they were defined over $\mathbb C$? What happens with a scheme over $k$ which cannot be covered by finitely many affine schemes? In fact, as in 2, I'm wondering if some finiteness condition is essential for performing "reduction steps" via Lefschetz Principle.
Thank you.
 A: I donot know "Lefschetz Principle". But for 1, I think you can embed $\mathbb{C}(x_1,\ldots,x_n)$ into $\mathbb{C}$.
Let $\{y_i\mid i\in I\}$ be a transcendental basis of $k$ over $Q$ and $\{z_j\mid j\in J\}$ be a transcendental basis of $\mathbb{C}$ over $Q$. We see that $I$ and $J$ have the same cardinality. Any bijection between $I$ and $J$  will enduce an isomorphism from $\mathbb{Q}(y_i\mid i\in I)$ to $\mathbb{Q}(z_j\mid j\in J)$. Since $k$ is an algebraic extension of $\mathbb{Q}(y_i\mid i\in I)$ and $\mathbb{C}$ is an algebraic closure of $\mathbb{Q}(z_j\mid j\in J)$, we may extend our former map to get an embedding from $\mathbb{C}(x_1,\ldots,x_n)$ to $\mathbb{C}$.
Similar argument for 2,3 . It depends on the transcendent degree.
And 3 is true.
A: You won't get any explicit embeddings into $\mathbb{C}$ because choices of transcendence bases (or well-orderings, or alike) are needed.
As for the Lefschetz principle, see this mathoverflow question:
https://mathoverflow.net/questions/90551/what-does-the-lefschetz-principle-in-algebraic-geometry-mean-exactly
