# Ideals of affine space curves

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.

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How are you given the curve? –  Qiaochu Yuan Sep 25 '12 at 20:17
Let us say that I have the ideal $I$ given by generators (3 or more), and I can compute the quotient of $\mathbb{C}[X,Y,Z]/I$ and find that it is a reduced irreducible curve. How to know if it is possible to have only $2$ generators? I have seen some algorithms on homogeneous ideals, but nothing on non-homogeneous ones. Also, I do not know any example of curve where we can prove that the ideal is not generated by $2$ elements. –  Jérémy Blanc Sep 26 '12 at 10:29

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.

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Nice! I like this. However, if $I/I^2$ is a free sheaf of rank $2$ on the curve (in fact a free $\mathbb{C}[X,Y,Z]/I$-module of rank $2$) how can we decide whether the ideal is generated by $2$ elements? Taking the preimages of the generators of $I/I^2$ does not seem to work... –  Jérémy Blanc Sep 26 '12 at 17:39
I just updated my answer to address this. –  Jason Starr Sep 26 '12 at 19:08
Yes. Thanks. I have to read the notes to be sure to understand it fully but it seems to be exactly the answer I wanted. –  Jérémy Blanc Sep 27 '12 at 6:39
The same result holds for any $n$-space, not just the 3-space. In other words, for any (smooth) curve $C$ in affine space, it is a complete intersection if and only if the canonical bundle $K_C$ is trivial. For $n>3$, Serre construction no longer works, but the result is still true. –  Mohan Sep 27 '12 at 15:50