It is an exercise in Hartshorne to classify nonsingular quartic curves in projective 3-space. I am interested in what happens when we allow singularities. In particular, I am looking for an explanation or source for how to exclude the possibility of a space curve of arithmetic genus 1 and a single node or cusp.
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2 Answers
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Singular nondegenerate irreducible degree $4$ curves certainly exist. They can be obtained as complete intersections of two quadric surfaces which are tangent at some point.
Thinking differently, any curve of class $(2,2)$ on a nonsingular quadric $Q$ is a degree $4$ space curve. The series $|\mathcal{O}_Q(2)|$ is $8$-dimensional, and the singular curves in this family form a $7$-dimensional subvariety. Cuspidal members of this family form a smaller $6$-dimensional locus.
answered Jan 22, 2012 at 22:30
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Take a rational normal quartic curve $C$ in ${\mathbb P}^4$ and project it to ${\mathbb P}^3$ from a point $P\notin C$. The image $X$ of $C$ is a quartic. Moreover $X$ is smooth if $P$ is general, it has a node if $P$ lies on a secant line of $C$ and a cusp if $P$ lies on a tangent line to $C$. All rational irreducible non degenerate quartics in ${\mathbb P}^3$ arise in this way.
