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I have a (geometrically irreducible) cubic surface defined over a finite field $F_q$ with three non-$F_q$-rational singularities (defined over the cubic extension of $F_q$).

Counting the number of $F_{q^3}$-rational points is easy by projection from a singular point.

What can be said about the number of $F_q$-rational points of the surface?

I would like upper/lower bounds that are finer than the Lang-Weil bound.

Thanks, H.

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  • $\begingroup$ The number $N_{\mathbb{F}_q}$ of $\mathbb{F}_q$-rational points of your cubic surface is congruent to $1$ modulo $q$ by the Chevalley-Warning theorem. $\endgroup$
    – Jason Starr
    Commented Nov 12, 2017 at 14:13
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    $\begingroup$ There is at least one $\mathbb{F}_q$-rational point by Chevalley-Warning. If this point is contained in no line contained in the hypersurface, the tangent planar cubic at this point contains $q$ points, resp. $q+1$ points depending on whether it is nodal, resp. cuspidal. You can iterate the construction of the tangent planar cubic with each of those points. $\endgroup$
    – Jason Starr
    Commented Nov 12, 2017 at 15:15
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    $\begingroup$ Resolving singularities gives a map $\pi \colon Y \to X$ where $Y$ is a generalised del Pezzo surface. Since your singularities aren't defined over the base field, I guess $\pi$ will give a bijection on $\mathbb{F}_q$-points. You can count points on $Y$ using the Weil conjectures, which in this case comes down to $q^2+1+q$(trace of Frobenius on $\mathrm{Pic}\ Y$). You have $\mathrm{Pic}\ Y \cong \mathbb{Z}^7$ and the Galois action depends on whether your singularities are $A_1$ or $A_2$ and the Galois action on the lines on $X$. $\endgroup$
    – Martin Bright
    Commented Nov 12, 2017 at 17:33

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