# Is every smooth affine curve isomorphic to a smooth affine plane curve?

As suggested by Poonen in a comment to an answer of his question about the birationality of any curve with a smooth affine plane curve we ask the following questions:

Q) Is it true that every smooth affine curve is isomorphic to a smooth affine plane curve?

(a) In particular, given a smooth affine plane curve $X$ with an arbitrary Zariski open set $U$ in it, can one give a closed embedding of $U$ in the plane again?

(b) An extremely interesting special case of (Q) above: Suppose $X$ is a singular plane algebraic curve with $X_{sm}$ the smooth locus. Can one give a closed embedding of $X_{sm}$ in the plane?

All varieties in question are over $\mathbb{C}$.

UPDATE:
Bloch, Murthy and Szpiro have already proven in their paper "Zero cycles and the number of generators of an ideal" , a much more general result (see Theorem 5.7, op.cit), that every reduced and irreducible prjective variety has an affine open set which is a hypersurface. This settles the above question birationally, in particular.
The authors give a very short and beautiful alternate proof of their result by M.V. Nori which I include here for its brevity and for anyone who may not have access to the paper:
Proof: Suppose $X$ is an integral projective variety of dimension $d$. By a generic projection, easily reduce to the case of a (possibly singular) integral hypersurface $X$ of $\mathbb{A}^{d+1}$. Suppose the coordinate ring of $X$ is $A=\mathbb{C}[x_1,\dots,x_{d+1}]$ and its defining equation is $F=\Sigma_0^m{f_i}x_{d+1}^{i}=0$ with $f_0\neq{0}$. For some element $h$ in $J\cap\mathbb{C}[x_1,\dots,x_d]$, where $J$ defines the singular locus of $X$, put $x_{d+1}'=x_{d+1}/(hf_0^2)$ in $F=0$ to observe that $1/(hf_0)\in\mathbb{C}[x_1,\dots,x_{d+1}']$ and $A_{hf_0}=\mathbb{C}[x_1,\dots,x_{d+1}']$. Clearly $\rm{Spec}\ {A_{hf_0}}$ admits a closed immersion in $\mathbb{A}^{d+1}$.

However, the above authors also prove in their Theorem 5.8 that there exist affine varieties of any dimension, which are not hypersurfaces. This answers our question in negative. This was also known to Sathaye for curves, see On planar curves. He gives a nice example of a double cover of a punctured elliptic curve, ramified at 9 points and also at the point at infinity. This curve cannot be embedded in $\mathbb{A}^2$. Sathaye uses the value semigroup at the only point at infinity to prove this. His example has trivial canonical divisor. So it answers Poonen's question in the comments below, negatively.
In short, $K=0$ for an affine curve is necessary but not sufficient for the curve to be planar, however one should note that $K=0$ is necessary and sufficient for an affine curve to be a complete intersection.

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Re your comment below: send me an email to hdao@math.ku.edu, I will send you the paper. –  Hailong Dao Dec 30 '09 at 7:24
Your link to a paper by Bloch, Murthy, Szpiro leads to another paper by Murthy alone. –  Matthieu Romagny Aug 25 '11 at 20:52

You can try this:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.52.6348

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The article cited gives a negative answer. The argument is as follows: A smooth affine plane curve has trivial canonical bundle. But if you start with a smooth projective curve X of genus greater than 1 and remove a finite number of sufficiently general points, then the subgroup of Pic(X) generated by the classes of the removed points does not contain the class of the canonical bundle of X, so the resulting affine curve has nonzero canonical class. –  Bjorn Poonen Dec 25 '09 at 21:25
Is the canonical class the only obstruction to embedding a smooth affine curve as a closed subscheme of A^2? –  Bjorn Poonen Dec 26 '09 at 21:18
I would expect so but not sure. This is definitely worthwhile thinking about. My only worry is a paper of the above cited author which says that given any set of points on a projective curve, he can find nearby points such that their complement has no closed embedding in the plane. Unfortunately I am unable to download any of these papers, so details are not available. Relying just on Mathscinet review so far! –  Maharana Dec 30 '09 at 6:48