ag.algebraic geometry - Is every curve birational to a smooth affine plane curve?

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 $Xinmathbb{C}^2$ 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 $psi^{-1}circphi:X-Srightarrowmathbb{C}^2$ is a closed embedding.