Is every algebraic curve birational to a planar curve

Solution 1:

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.

Solution 2:

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 $L=k(X)$.