TITLE: Is every algebraic curve birational to a planar curve
QUESTION [13 upvotes]: 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?

REPLY [8 votes]: 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.