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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.

Katie Dobbs
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    You probably should assume some separability hypothesis. But here's a hint: take a transcendence basis for the function field... – Zhen Lin Jun 13 '13 at 18:44
  • @ZhenLin Okay, say I take $k(X)$ to have a transcendence basis $f_1, \cdots, f_n$ over $k.$ Then $k(X)$ is algebraic over $k(f_1,\cdots, f_n).$ Say maybe I take $g\in k(X).$ We know there is some polynomial with coefficients in $k$ such that $p_g(f_1,\cdots, f_n)=g.$ Does this do anything? – Katie Dobbs Jun 13 '13 at 19:04
  • That's where the separability hypothesis comes in – you need to be able to know that $k (X)$ is generated by one element over your transcendence basis. – Zhen Lin Jun 13 '13 at 19:12
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    @ZhenLin Ahh Ok if I assume separability, then by the primitive element theorem there is a single element $t$ such that $k(X) = k(f_1,\cdots, f_n, t).$ Let $I$ be the kernel of the natural map from $k[X_1,\cdots, X_{n+1} ] $ onto $k[f_1,\cdots, f_n, t].$ Then $I$ is prime and $V:= V(I)$ is a variety in $\mathbb{A}^{n+1}.$ The coordinate ring $k[X_1,\cdots, X_{n+1}]/I$ is isomorphic to $k[f_1,\cdots, f_n, t]$, and taking the fields of fractions of both sides shows that the function field of $V$ is isomorphic to $k(X),$ which gives the result. Is this reasoning correct? – Katie Dobbs Jun 13 '13 at 19:23
  • Ahh I've forgotten to show that $I$ is principle. – Katie Dobbs Jun 13 '13 at 19:25
  • @ZhenLin I'm not sure how to show $I$ is principle (assuming my reasoning up to that point is fine). – Katie Dobbs Jun 13 '13 at 19:38

1 Answers1

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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 $\{ f (y) = 0 \} \subset \mathbb{A}^{n+1}$.

Zhen Lin
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  • Thank you, I understand now. – Katie Dobbs Jun 14 '13 at 04:40
  • Is there any counterexample to the conclusion when $K(X)$ is inseparable over any transcendence basis? – Yuchen Liu Jun 16 '13 at 14:03
  • If the conclusion holds, then $K (X)$ must admit a transcendence basis over which it is simply generated – separability is sufficient but not necessary. So for a counterexample, I suppose one has to find a field that is not simply generated over any transcendence basis. Perhaps something like the variety ${ x^p - z, y^p - w } \subset \mathbb{A}^4$ over $\overline{\mathbb{F}_p}$? – Zhen Lin Jun 16 '13 at 20:47
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    Sorry I thought I had understood but after some more thinking I really do not. I had thought this answer shows that $f(t)$ is the minimal polynomial over $k[x_1,\cdots, x_n]$ as well, and therefore generates the kernel of the natural map $k[X_1,\cdots, X_{n+1}]\to k[x_1,\cdots, x_n,y],$ but after clearing denominators $f$ may not be monic anymore so I don't know if it is. I may be interpreting your answer incorrectly. – Katie Dobbs Jun 20 '13 at 15:44
  • It may not be monic but I think it doesn't matter. You can also change variables to make it monic, as in the proof of Noether normalisation. – Zhen Lin Jun 20 '13 at 15:49
  • It turns out that we can always choose a transcendence basis satisfying the separability condition, as stated in Shafarevich's Basic Algebraic Geometry I, Proposition A.7. – Ken Nov 20 '20 at 06:49