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The definition for "birational map" I was given was as follows:

Let $V$ and $W$ be irreducible projective varieties. If $\phi$ is a dominant rational map from $V$ to $W$ and $\psi $ is a dominant rational map from $W$ to $V$ and $\phi \circ \psi$ and $\psi \circ \phi$ are identity maps (on the domains they can be defined), the $\phi$ and $\psi$ are called birational maps.

$\phi$ being dominant means that $\phi(\text{dom}\phi)$ is dense in $W$.

Later in the lectures I think it was assumed that birational equivalence between projective varieties was an equivalence relation, but I don't immediately see why--are the composite of dominant rational maps dominant?

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Yes. It suffices to show the image of the composite is dense - try to play with continuity to get that. –  user27126 Feb 17 '13 at 19:41
    
But what are the continuous maps? $\phi$ and $\psi$ are not even everywhere defined? –  Montez Feb 17 '13 at 19:46
    
they are continuous on the open set they are defined. –  user27126 Feb 17 '13 at 19:49
    
I got it, thank you. –  Montez Feb 17 '13 at 20:14

1 Answer 1

You can show that for projective varities, birationnality is equivalent to say that the function fields are isomorphic.

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