Is every algebraic curve birational to a planar curve 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? 
 A: 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.
A: 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. 
A: 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 $L=k(X)$.
