Given some algebraic closed curve in the affine space $\mathbb{A}^3\_\mathbb{C}$, is there a way to decide whether its ideal (polynomials in $\mathbb{C}[X,Y,Z]$ vanishing on the curve) is generated by two elements?


I am mostly interested in the case where the curve is smooth and irreducible.



Ideals of affine space curves
There is a necessary condition: the element in the divisor class group of the affine curve
coming from the cotangent sheaf of the curve must be trivial. If $I$ is generated by two elements, then $I/I^2$ is a free sheaf of rank $2$ on the curve. By adjunction, this forces the cotangent sheaf of the curve to be trivial. This immediately suggests how to make examples where $I$ is not generated by two elements. Begin with a smooth, projective curve $X$ of genus $g>1$. Let $x$ be a closed point such that $\omega\_X(-(2g-2)\underline{x})$ is nontorsion. For integers $N\gg 0$, the divisor $N\underline{x}$ is very ample. For three general sections $f\_1,f\_2,f\_3$ of $\Gamma(X,\mathcal{O}\_X(N\underline{x}))$, form the morphism $(f\_1,f\_2,f\_3):X\setminus\{x\} \to \mathbb{A}^3$. If $f\_1,f\_2,f\_3$ are sufficiently general, this will be a closed immersion. The image is a smooth curve whose ideal $I$ is not generated by two elements.


$\textbf{Edit.}$ According to my computations, if $X$ is a genus $2$ curve and $x$ is a generic (hence non-hyperelliptic) point, then the linear system $|5\underline{x}|$ is sufficient for this argument. This will embed $X$ in $\mathbb{P}^3$ as a degree $5$ curve whose "osculating 2-plane" $H$ at $x$ has contact of order $5$. Thus the complement of $H$ will be an affine space $\mathbb{A}^3$ and $X\setminus (X\cap H)$ equals $X\setminus\{x\}$ is an affine curve whose ideal is not generated by two elements.


$\textbf{Second edit.}$ Using "Serre's Construction" and the Quillen-Suslin theorem (perhaps avoidable), you can show that the necessary condition above is also sufficient. Let $R$ denote $k[x\_1,x\_2,x\_3]$, and let $I$ denote the defining ideal of the affine curve $C$. Then $\text{Ext}^1\_R(I,R)$ is annihilated by $I$, i.e., equivalent to an $R/I$-module. As an $R/I$-module, it is isomorphic to the dual of $\bigwedge^2(I/I^2)$, which is isomorphic to $R/I$ by hypothesis. Choose an element that generates this $R/I$-module, i.e., $0\to R \to F \to I \to 0$. Since $I$ is locally generated by 2 elements, it is not hard to check that $F$ is a locally free $R$-module of rank 2. By the Quillen-Suslin theorem (or perhaps something weaker), $F$ is a free $R$-module of rank $2$.


$\textbf{Third edit.}$ I just noticed that there are some nice notes on "Serre's construction" (used above) written by my colleague, Christian Schnell. Here is the URL:[<http://www.math.sunysb.edu/~cschnell/pdf/notes/serre.pdf>](http://www.math.sunysb.edu/~cschnell/pdf/notes/serre.pdf).


