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Non-hyperelliptic curves of genus three are smooth quartics. Is the moduli space of such curves affine?

I think this follows from a more general result on smooth complete intersections, but I'm looking for a simple proof.

One idea would be to use a "good" compactification, i.e., such that the boundary divisor is ample. This can be done by using a suitable Grassmannian containing the Hilbert scheme.

I'd like to avoid something like this and give a more elementary argument. Is that possible?

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    $\begingroup$ Every nonempty divisor in $\mathbb{P}^{14}$ is ample and its complement is affine. Is that what you are looking for? $\endgroup$
    – Jason Starr
    Commented Dec 23, 2012 at 13:53
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    $\begingroup$ Beware that analogous results for smooth complete intersections do not hold in general. For instance, the moduli space of non-hyperelliptic genus $4$ curves, that is the moduli space of smooth complete intersections of degrees $(2,3)$ in $\mathbb{P}^4$ is not affine, beacuse it contains complete curves. $\endgroup$
    – Olivier Benoist
    Commented Dec 23, 2012 at 14:24
  • $\begingroup$ @Olivier. You're right. I didn't mean to say that. Rather, more generally, the moduli space of smooth hypersurfaces of degree $d$ in $N$-projective space is affine if $d>N+1$. $\endgroup$
    – Masse
    Commented Dec 23, 2012 at 16:27
  • $\begingroup$ @Jason. That's what I was looking for indeed. Any nonempty effective divisor on $\mathbf P^{14}$ is ample. $\endgroup$
    – Masse
    Commented Dec 23, 2012 at 16:29

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