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In the book GTM 52 by R.Hartshone, there is a theorem as following :

Every variety of dimension $r$ is birational to a hypersurface in $\mathbb{P}^{r+1}$

Could you please tell me, who is the author of this theorem ?

I have read somewhere that birational equivalence was considered in Italian school, so could you please show me what exactly they did found relating to classification of variety(w.r.t birational equivalence).

Thank!

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See mathoverflow.net/questions/4324/…, which is quite related. –  Mariano Suárez-Alvarez Jan 30 '13 at 7:15

1 Answer 1

That theorem is probably folkloric, in that there is no name attached to it.

The proof is very simple (let me work over an algebraically closed field $k$ of characteristic zero) If $K(X)$ is the field of rational functions on a variety $X$ of dimension $r$, then the trascendence degree of $K(X)$ over $k$ is $r$ and there are functions $x_1,\dots,x_r\in K(X)$ algebraically independent over $k$ such that the extension $K(X)\supset k(x_1,\dots,x_r)$ is finite and separable. It is therefore simple, so there exists an $x_{r+1}\in K(X)$ such that $K(X)=k(x_1,\dots,x_{r+1})$. Moreover, since the extension is finite, there is a polynomial $f$ such that $f(x_1,\dots,x_{r+1})=0$.

Now $f$ defines a hypersurface in the affine space $\mathbb A^{r+1}$, and its projective closure is a hypersurface in $\mathbb P^{r+1}$. It is easy to see that the field of rational functions on the latter is isomorphic to $K(X)$.

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Could you please explain for me why the field of rational function of the latter is isomorphic to $K(X)$ ? –  Arsenaler Jan 30 '13 at 16:22

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