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We know that (for an algebraically closed field $k$) there is an equivalence between algebraic curves over $k$ (up to birational equivalence) and fields of transcendence degree $1$ over $k$.

Is there something similar for higher dimensional algebraic varieties?

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  • $\begingroup$ For algebraic varieties the transcendence degree of the function field is equal to the dimension, so the answer is yes: just replace $1$ with $n := \dim X$ $\endgroup$
    – Francesco Polizzi
    Commented Jun 19, 2012 at 16:14
  • $\begingroup$ Is this a functorial equivalence? Where can I read about this? $\endgroup$
    – expmat
    Commented Jun 19, 2012 at 16:15
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    $\begingroup$ Hartshorne, Chapter I, Section 4 $\endgroup$
    – Francesco Polizzi
    Commented Jun 19, 2012 at 16:21
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    $\begingroup$ In your "we know that" sentence, you really need the fields of transcendence degree 1 to be finitely generated. $\endgroup$
    – S. Carnahan ♦
    Commented Jun 20, 2012 at 8:30

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For $X,Y$ integral $k$-varieties, there is a bijection {$f: X \to Y$ dominant rational} <-> $\mathrm{Hom}_k(K(Y),K(X))$.

Also, there is an equivalence of categories {integral $k$-varieties with morphisms as above} and {finitely generated field extensions of $k$}.

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