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.


1 Answer 1


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.

  • $\begingroup$ It looks it works prefectly! $\endgroup$ Dec 24, 2009 at 12:48
  • 3
    $\begingroup$ Thank you, Maharana! I think that your solution is not exactly correct as written (despite all the upvotes), but it is easy to fix. Your map in Step 2 need not be a morphism. To fix this, I would first modify Step 1 in order to assume not only that the singularities are on the $y$-axis but also that $(0,0)$ is not on the curve. Next, rewrite Step 2 as follows: if after Step 1 the curve is $f(x,y)=0$, then the new curve $f(x,xy)=0$ has no points with $x=0$, so it is isomorphic to the part of $f(x,y)=0$ where $x \ne 0$, and in particular it is smooth and birational to the original curve. $\endgroup$ Dec 24, 2009 at 23:43
  • $\begingroup$ You are right! Thanks for the correction. I meant to avoid the origin but forgot to mention. That it is possible to do so is also a tiny observation. Cheers! $\endgroup$
    – Maharana
    Dec 25, 2009 at 6:04
  • $\begingroup$ Yes. And my reason for rewriting Step 2 was that after Step 1 there might also be nonsingular points on the $y$-axis where your map would not be defined. So we are led to a variant of the original question: Is it nevertheless true, as claimed in your original solution, that the smooth locus of any singular curve can be embedded as smooth affine plane curve? More generally, is every smooth affine curve isomorphic to some smooth affine plane curve? Maybe we should ask a new MO question about this. $\endgroup$ Dec 25, 2009 at 18:39
  • $\begingroup$ Hello Poonen, I understand that and in fact I was tempted to say initially that we just need to sacrifice a few more points from our curve $X$ to make my original map a morphism without changing any birational property. However, I have a feeling that it is too good to be true that any smooth affine curve can be embedded in the plane in a closed manner. In particular, I am also asking myself if any Zariski open set of a smooth affine plane curve can again be given a closed embedding in the plane! I am going to ask a new question about it. Will be glad to know your thoughts on this. Cheers! $\endgroup$
    – Maharana
    Dec 25, 2009 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.