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Is it true that any affine variety is birational to affine space of the proper dimension? For example, say I have an affine curve in $\mathbb{A}^n$, it makes sense that this curve should be birational to $\mathbb{A}^1$.

Trying to prove this statement I tried to use the classification of birational varieties, saying that they have isomorphic function fields, however, if all affine varieties of dimension $d$ have isomorphic function fields to $k(x_1,...,x_d)$, then this would imply that all function fields over an algebraically closed field of transcendence degree $d$ are isomorphic, which strikes me as a rather suspicious statement.

Thanks in advance.

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    $\begingroup$ No, this is very far from true; if it were then algebraic geometry would be much more boring. Take, for example, any punctured elliptic curve. $\endgroup$ Feb 19, 2019 at 20:45

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Firstly, there is no need to focus on affine varieties here. Any affine variety is birationally equivalent to its projective closure. So you may as well ask the question about projective varieties.

As for the question itself, the statement isn't even true for curves. In fact, two smooth projective curves are birationally equivalent if and only if they are isomorphic. So for example, if you take two smooth projective curves with different genuses, then they cannot be birationally equivalent.

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  • $\begingroup$ This is very eye opening. $\endgroup$
    – kindasorta
    Feb 19, 2019 at 20:55

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