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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:

  1. 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 $\mathbb C$?

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

  3. 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.

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You're not embedding ${\mathbf C}(x_1,\dots,x_n)$ into ${\mathbf C}$. Rather, since your variety is cut out by a finite number of equations, which each have a finite number of coefficients, the coefficients live in a field that is finitely generated over ${\mathbf Q}$. Such a (smaller) field can be embedded into ${\mathbf C}$. That's basically the point, provided your result behaves well under field extensions. –  KCd Apr 17 '12 at 17:15
    
Finally I see that the point is to go down to $\mathbb Q$ and to go up again via algebraic closure. Thank you, now it's clear. –  Brenin Apr 17 '12 at 22:12
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2 Answers

up vote 2 down vote accepted

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.

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I see. That's what I was looking for. Bases have to be chosen, as Martin pointed out, but I just wanted to see that such a (non canonical) embedding could really be constructed. Thank you! –  Brenin Apr 17 '12 at 21:43
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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:

http://mathoverflow.net/questions/90551/what-does-the-lefschetz-principle-in-algebraic-geometry-mean-exactly

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So if I have an extension $k/\mathbb C$, of transcendence degree at most countable - and if I choose transcendence bases - I can get an embedding. I would have never expected that such an embedding be canonical, however. Thank you, Martin. –  Brenin Apr 17 '12 at 22:08
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