Is every curve over $\mathbf{C}$ birational to a smooth affine plane curve?
Bonnie Huggins asked me this question back in 2003, but neither I nor the few people I passed it on to were able to answer it. It is true at least up to genus 5.
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Sign up to join this communityIs every curve over $\mathbf{C}$ birational to a smooth affine plane curve?
Bonnie Huggins asked me this question back in 2003, but neither I nor the few people I passed it on to were able to answer it. It is true at least up to genus 5.
Yes. Here is a proof.
It is classical that every curve is birational to a smooth one which in turn is birational to a closed curve $X$ in $\mathbb{C}^2$ with atmost double points. Now my strategy is to choose coordinates such that by an automorphism of $\mathbb{C}^2$ all the singular points lie on the $y$-axis avoiding the origin. Now the map $(x,y)\rightarrow(x,xy)$ from $\mathbb{C}^2$ to itself will do the trick of embedding the smooth part of $X$ in a closed manner. Below are the details.
The only thing we need to show is that the smooth locus of a closed curve $X\in\mathbb{C}^2$ with only double points can again be embedded in the plane as a closed curve.
Step 1. Let $S$ be the set of singular points of $X$. Choose coordinates on $\mathbb{C}^2$ such that the projection of $X$ onto both the axes gives embeddings of $S$. Call the projection of $S$ on the $y$-axis as $S'$. By sliding the $x$-axis a little bit we can make sure that $S'$ doesn't contain the origin of the plane. Now I claim that there is an automorphism of $\mathbb{C}^2$ which takes $S$ to $S'$. This is easy to construct by a Chinese remainder kind of argument: There is an isomorphism of the coordinate rings of $S$ and $S'$ and we need to lift this to an isomorphism of $\mathbb{C}[x,y]$. I will illustrate with an example where #$\{S\}=3$. Let $(a_i,b_i)$ be the points in $S$. Then there exists a function $h(y)$ such that $h|S'=x|S$ as functions restricted to the sets $S$ and $S'$. Here is one recipe: $h(y)=c_1(\frac{y}{b_2}-b_2)(\frac{y}{b_3}-b_3)(y-b_1+1)+\dots$ where $c_1=a_1(\frac{b_1}{b_2}-b_2)^{-1}(\frac{b_1}{b_3}-b_3)^{-1}$ etc.
Look at the map $\phi:(x,y)\rightarrow(x-h(y),y)$ on $\mathbb{C}^2$. It is clearly an automorphism and takes the set $S$ to $S'$.
Step 2. Now consider the map $\psi:(x,y)\rightarrow(x,xy)$ from the affine plane to itself. It is an easy check that $\psi^{-1}\circ\phi:X-S\rightarrow\mathbb{C}^2$ is a closed embedding.