It is well known that if a smooth curve $C \subset \mathbb{P}^3$ has degree $ d \leq 6$. Then $ g(C) \leq 4$ (Hartshorne pg 354). I know that the case $g=4$ correspond to the complete intersection of a quadratic and a cubic surface. Therefore, the ideal of the curve is $$ I_C = (f_2(x_0, x_1,x_2, x_3),f_3(x_0, x_1,x_2, x_3)) $$ I am wondering: Is it possible to find explicitly the ideal for the other genus? For example, what are the equation defining a rational curve in $\mathbb{P}^3$ of degree $6$ ? Is there a general strategy for finding those equations? or at least some bounds? What does it happen if we drop the smooth condition for admitting some mild singularities i.e $A_1$

Thanks for any reference!

  • $\begingroup$ There is a cubic surface containing all these curves as explained by Charles Staats in mathoverflow.net/questions/131940/…. So you could start by looking for rational curves on cubic surfaces.. $\endgroup$
    – J.C. Ottem
    Jun 13, 2013 at 15:51
  • $\begingroup$ 1. The degree $6$ rational curve comes from a the Veronese embedding $\mathbb P^1 \to \mathbb P^6$ and a nontrivial projection $\mathbb P^6 \\ \mathbb P^2 \to \mathbb P^3$. Given an explicit projection (essentially 3 degree 6 polynomials in 2 variables) you could try to search for explicit equations by searching for relations among the variables. 2. If you add some singularities, you have to decide if you care about the arithmetic or geometric genus. The complete intersection family has arithmetic genus $4$, but I believe it has members of geometric genus $0$ with simple singularities. $\endgroup$
    – Will Sawin
    Jun 13, 2013 at 18:34

1 Answer 1


As explained by J.C. Ottem, the curves of degree $6$ are contained in a cubic. Let us assume that the surface is smooth, so you can see the cubic surface as the blow-up of six points in $\mathbb{P}^2$ with no $3$ collinear and not all on the same conic.

The surface $X$ has Picard group generated by $L$, the preimage of a line in $\mathbb{P}^2$, and by $E_1,\dots,E_6$, the exceptional divisors.

The trace of an hyperplane is the anti-canonical divisor, $-K_X=3L-E_1-\dots-E_6$.

$1)$ To obtain degree $6$ rational curves in $X$, you take a conic of $\mathbb{P}^2$ not passing through the points. It is equivalent to $C=2L$. You can thus see that no quadric of $\mathbb{P}^3$ contains it, because $C$ it is not equivalent to $-2K_X$. There is a cubic $X'$ which contains $C$ is and only if $-3K_X-C=7L-3E_1-\dots-3E_6$ is effective. Intersecting this system with the line $2L-E_1-\dots-E_6+E_i$, we obtain $-1$, so the system should contain the $6$ lines, which is not possible. Hence, the cubic $X$ is the unique one which contains $C$. However, there are quartics containing $C$, because $-4K_X-C=10L-4E_1-\dots-4E_6$ is effective. This can be seen by taking the union of two general quintics of $\mathbb{P}^2$ with $6$ double points. These curves are other rational sextics of $\mathbb{P}^3$. The system $-4K_X-C$ being moreover base-point free, the ideal of $C$ in $\mathbb{P}^3$ is generated by the cubic and equations of degree $4$. You can do this easily in coordinates for a given curve, but this takes time.

$2)$ To obtain degree $6$ curves of genus $1$, you can take a cubic of $\mathbb{P}^2$ passing through three of the points, i.e. $C=3L-E_1-E_2-E_3$. The same type of calculation gives you the ideal.

You can of course continue with all curves of degree $6$. If you want more details on the type of the curves you obtain on the cubic surface, see "Weak Fano threefolds obtained by blowing-up a space curve and construction of Sarkisov links" by S. Lamy and myself, especially Proposition 4.2.

  • $\begingroup$ Thanks for answering! with the notation of your article set up 4.1. It seems that for a fixed degree $d \leq 6$ and genus $g\leq 4$ I can choose an arbitrary k for generating those curves on the cubic surface. Of course, as long as I also select appropriate values of multiplicities $m_i$. Is that right? Is this a way to construct different kind of curves for a fixed genus and degree? $\endgroup$
    – eventually
    Jun 17, 2013 at 1:27
  • $\begingroup$ Since you have $d=3k-\sum m_i$ and $g=(k-1)(k-2)/2-\sum m_i(m_i-1)/2$, you only have finitely many solutions for $k$ and $m_i$. And then most of them are equivalent by changing the choice of the six smooth curves that you could contract from the cubic to $\mathbb{P}^2$. However, the changing of the cubic surface and the curves in the families gives you a moduli space of curves. For degree $\le 6$ it should be classical to compute the dimension. $\endgroup$ Jun 17, 2013 at 23:12

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