Problem: Show every $n$-dimensional variety is birationally equivalent to a hypersurface in $\mathbb{A}^{n+1}.$
Thoughts: For a (quasi-projective) variety $X,$ the function field $k(X)$ is a finitely generated extension of $k.$ The dimension of $X$ has been defined as the transcedence degree of $k(X)$ over $k.$
Two varieties $X$, $Y$ are birationally equivalent if and only if their function fields $k(X)$ and $k(Y)$ are isomorphic.
Any help is greatly appreciated. Thank you.
Best Answer
Let $X$ be an $n$-dimensional variety (irreducible, over an algebraically closed field $k$, etc.). Choose a transcendence basis $x_1, \ldots, x_n$ for the function field $K (X)$, and assume that $K (X)$ is separable over that transcendence basis. By the primitive element theorem, there exists an element $y$ in $K (X)$ that generates $K (X)$ over $k (x_1, \ldots, x_n)$, and this has a minimal polynomial over $k (x_1, \ldots, x_n)$, say $f (t)$.
Clearing denominators, we may assume that $f (t)$ has coefficients in $k [x_1, \ldots, x_n]$, and by Gauss's lemma, $f (t)$ remains irreducible over $k [x_1, \ldots, x_n]$. It follows that $X$ is birationally isomorphic to the hypersurface $\{ f (y) = 0 \} \subset \mathbb{A}^{n+1}$.