Let $X$ be an algebraic curve over an algebraically closed field $k$.

Does there exist a polynomial $f\in k[x,y]$ such that $X$ is birational to the curve $\{f(x,y)=0\}$?

I think I can prove this using Noether Normalization Lemma.

Is this correct? If yes, is it too much? That is, is there an easier argument?


Two curves are birational if and only if their function fields are isomorphic. $k(X)$ has transcendence degree $1$, so pick a transcendental element $x \in X$. Then $k(X)$ is a finite extension of $k(x)$. By the primitive element theorem (in this context a birational version of Noether normalization), there exists a primitive element $y \in k(X)$ such that $k(X) = k(x)[y]$; $y$ satisfies a minimal polynomial $f(x, y) = 0$ over $k[x]$, and the conclusion follows.

Edit: Above I implicitly assumed that the extension $k(x) \to k(X)$ is separable. This is automatic if $k$ has characteristic $0$. If $k$ has characteristic $p$ we need to choose $x$ more carefully and I am not sure how to do this without going through the characteristic-$p$ proof of Noether normalization.

  • 1
    $\begingroup$ The result holds true over any perfect field because $k(X)$ is then separable (in the sense of separable algebra) over $k$. $\endgroup$ – user18119 Jul 14 '12 at 19:46

To complete Qiaochu's proof in the postive characteristic case:

Let $k$ be perfect of characteristic $p>0$. Let $L\subseteq k(X)$ be a subextension such that $L$ is a finite separable extension of some purely transcendental extension $k(x)$ of $k$. We can suppose $L$ is the biggest possible (for various $x$). Let's prove that $L=k(X)$.

By the maximality of $L$, $k(X)/L$ is purely inseparable (any finite extension is purely inseparable over a separable extension). Suppose $L\neq k(X)$. Then there exists $f\in k(X)\setminus L$ such that $f^p\in L$. Let $$Y^{n}+g_{n-1}(x)Y^{n-1}+\cdots + g_0(x)\in k(x)[Y]\quad (*)$$
be the (separable) irreducible polynomial of $f^p$ over $k(x)$. If $g_i(x)\in k(x^p)$ for all $i$, then $g_i(x)=h_i(x)^p$ because $k$ is perfect. Thus $f$ is a zero of the separable polynomial $$Y^{n}+h_{n-1}(x)Y^{n-1}+\cdots + h_0(x)\in k(x)[T]$$ (note that separable <=> $\notin k(x)[T^p]$), hence $f$ is separable over $k(x)$ and $f\in L$. Contradiction. So at least one $g_i(x) \notin k(x^p)$. Using (*) with $Y=f^p$, we see that $x$ is algebraic and separable over $k(f)$. As $L[f]$ is finite separable over $k(x,f)$ which is finite separable over (the purely transcendental extension) $k(f)$, again by the maximality of $L$, we have $L=L[f]$. Therefore $f\in L$, still contradiction. So we proved $L=k(X)$.


The result holds over an arbitrary ground field $k$ if we assume that the curve is geometrically regular: i.e., if $X \otimes_k \overline{k}$ is regular, or equivalently if the extension $k(X)/k$ is separable, as in QiL's comment above. I believe that QiL's proof works here verbatim. Also Qiaochu Yuan's argument works, as there is a separable Noether normalization theorem: the fraction field of a geometrically regular variety can be written as a finite separable extension of a rational function field. (See Corollary 16.18 in Eisenbud's text on commutative algebra.)

Note that it is often possible to prove more: in $\S 5$ of these notes, I give a (detailed) sketch of a proof that any geometrically regular curve $C$ over an infinite field is birational to a plane curve with only ordinary double points as singularities. Essentially I follow Hartshorne's proof and explain why the hypothesis therein that "$k$ is algebraically closed" can be replaced by "$k$ is infinite".

This stronger conclusion however need not hold over a finite field: if a curve $C$ can be "immersed" in $\mathbb{P}^2$ with only double point singularities, then $\# C(\mathbb{F}_q) \leq 2 \# \mathbb{P}^2(\mathbb{F}_q)$. But a curve over a finite field can have arbitrarily many $\mathbb{F}_q$-rational points.

  • $\begingroup$ I just see this nice answer. Over an arbitrary field $k$, $k(X)/k$ is separable iff $X$ is geometrically reduced (equivalently, $k(X)\otimes_k \bar{k}$ is reduced). As for the birational morphism to $\mathbb P^2$ over a finite field, your arguments show that the same conclusion holds for any projective space $\mathbb P^n$ instead of $\mathbb P^2$. $\endgroup$ – user18119 Oct 12 '12 at 20:53

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