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Consider a fixed smooth algebraic curve $C$ over $\mathbb C$. It is well-known that $\mathbb CP^3$ contains curves that are abstractly isomorphic to $C$. What is the minimal degree of a curve in $\mathbb CP^3$, abstractly isomorphic to $C$, which is not on any cubic surface?

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The minimum degree of a smooth rational curve that is not contained in any cubic surface is 7. – Charles Staats May 26 '13 at 19:22
and how to get it? It seems that the claim is that on a cubic there is no a one-dimensional family of rational curves of degree 7... – Nikita Kalinin May 26 '13 at 20:03
Not exactly. The claim is that if $C$ is a general rational curve of degree 7, then there is no cubic hypersurface containing $C$. There are some rather involved ways to see this by "pure thought," but one can also simply choose a degree 7 rational curve with random coefficients and compute (using Macaulay2, for instance) its homogeneous ideal. If you do this, you will see that the homogeneous ideal contains no polynomials of degree $\leq 3$. – Charles Staats May 26 '13 at 20:12
to Charles Staats: Thank you! I just had in mind dimension computations: there is 20-dimensional space of rational curves degree 7 and 19-dimensional space of cubic surfaces. – Nikita Kalinin May 27 '13 at 18:30
Nikita: I have edited your question in an attempt to clarify your intent, as I now understand it. Obviously, you should feel free to modify/undo any or all of my edits. – Charles Staats May 27 '13 at 20:27

NOTE: The following addressed an earlier version of the question in which it was not clear that the abstract isomorphism type of the curve $C$ was supposed to be fixed. Thus, the hypothesis "if I have correctly understood the question" is false.

The answer is seven, if I have correctly understood the question.

Claim 1: If $C$ is an irreducible curve in $\mathbb P^3$ of degree $d \leq 6$, then there exists a cubic surface containing $C$.

Proof: First, note that it suffices to prove that there exists a cubic surface containing $3d+1$ distinct points on $C$. This is because a degree $3$ surface that does not contain $C$ can intersect $C$ in at most $3\cdot\deg C$ distinct points.

The complete linear system of all cubic surfaces in $\mathbb P^3$ has dimension $\binom{3+3}{3}-1 = 19$. If $V$ is any linear system and $p$ is a point, then the subsystem of $V$ consisting of divisors containing $p$ has dimension either $\dim V$ or $\dim V - 1$. Thus, if we choose $3d+1$ points on $C$, the space of all cubic surfaces containing these $3d+1$ points has dimension greater than or equal to $$19-(3d+1) = 18-3d.$$ If $d \leq 6$, then this dimension is nonnegative, showing that the space of cubics containing these points is nonempty. $\square$

Claim 2: If $C$ is a general rational curve in $\mathbb P^3$ of degree $d \geq 7$, then $C$ is not contained in any cubic surface.

Proof: First, by semicontinuity, it suffices to show that there exists a (possibly reducible) genus zero curve of degree $d$ that is not contained in any cubic surface. If $d=7$, such a curve may be exhibited as the image of the morphism $$[t,s] \mapsto [t^{7}, \,s t^{6}+s^{2} t^5,\,s^{5} t^{2}+s^{6} t,\,s^{2} t^{5}-s^{7}]$$ from $\mathbb P^1 \to \mathbb P^3$. Computing the homogeneous ideal of polynomials that vanish on its image (e.g., using Macaulay2) will reveal that no homogeneous polynomials of degree $\leq 3$ vanish on this rational curve.

If $d > 7$, take a union $C = C'' \cup C'$, where $C''$ is a rational curve of degree $7$ that does not lie on any cubic surface and $C'$ is a rational curve of degree $d-7$ such that the two curves intersect in exactly one point (which is a node). Clearly, $C$ is not contained in any cubic surface; a generic deformation of $C$ will be a smooth rational curve of degree $d$, and by semicontinuity, will not be contained in any cubic surface. $\square$

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For a more general solution to this sort of problem (that does not rely on a computer), see Hirschowitz, "Sur la postulation generique des courbes rationelles." – Charles Staats May 26 '13 at 21:39
Note that a general curve of degree $7$ and genus $g$ is not contained in a cubic, for $0\le g\le 2$ but is contained in a cubic if $g\ge 3$ (and even on a quadric if $g\ge 6$). See for example – Jérémy Blanc May 26 '13 at 21:52
Thank you! but question is about any curve $C$, i.e. we fix a complex structure on a curve of genus $g$ and we are looking for minimal degree map of such a curve to $\mathbb C^3$. (sure, it is not known does there exist such a map) – Nikita Kalinin May 27 '13 at 18:32
Here is some partial answer: Lemma 2.3 in : Let $C$ be a smooth curve of genus $g$ and degree $d$ in $\mathbb{P}^3$. If $(2g-2)/3 < d < (19+g)/3$ then $C$ is contained in a cubic surface. – Jérémy Blanc Jun 6 '13 at 15:34

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