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Let $K = \mathbb{F}_p$ be a field of positive characteristic $p > 0$. Consider a surface in $\mathbb{A}^3_K$ of the following form $$ S = \{f_1(x_0)y_0^2+f_2(x_0)y_0y_1+f_3(x_0)y_0+f_4(x_0)y_1^2+f_5(x_0)y_1+f_6(x_0) = 0\} $$ where $f_i$ is a polynomial with coefficients in $K$ of degree $d_i$.

Assume that the $f_i$ are general. Is then $S$ smooth and if so is there any general result implying this?

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  • $\begingroup$ What precisely are you asking? That surface is smooth for some particular choices of coefficients in your finite field. When you write “general” do you mean “at the generic point of the affine space parameterizing coefficients “? $\endgroup$ Commented Feb 1, 2022 at 12:05
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    $\begingroup$ It seems that the condition "$f_i$ general" only covers a finite number of possibilities, since $K$ is finite (and $d_i$ is fixed). So it is indeed not clear what you are asking. Do you mean "are there any polynomials $f_i$ with degrees $d_i$ such that $S$ is smooth?". $\endgroup$ Commented Feb 1, 2022 at 12:08
  • $\begingroup$ I think the problem is that I do not know if this make much sense over a finite field. However, homogeneizing $f_i$ we can consider the projective space $\mathbb{P}_i$ parametrizing homogeneous polynomials of the given degree $d_i$. I would like to know whether $S$ is smooth for a random choice of $(f_1,\dots,f_6)\in \mathbb{P}_1\times\dots \mathbb{P}_6$. I think the issue consists in formally defining "random". In characteristic zero I would define "random" as "there is a Zariski open subset of $\mathbb{P}_1\times\dots \mathbb{P}_6$" but I do not know whether this make much sense for $p >0$. $\endgroup$
    – user114666
    Commented Feb 1, 2022 at 12:42
  • $\begingroup$ Yes, I can see that for suitable choices of the $f_i$ the surface $S$ is smooth. $\endgroup$
    – user114666
    Commented Feb 1, 2022 at 12:44

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For infinite fields of characteristic $p$, the Zariski open definition makes pretty much as much sense as it does in characteristic zero. Often, but not always, the same argument works.

Over finite fields, A good definition of "random" was provided by Bjorn Poonen in his paper Bertini Theorems over Finite Fields. Specialized to your case, it says that the proportion of tuples of polynomials $(f_1,\dots, f_6)$ over $\mathbb F_q$, of degrees $(d_1,\dots, d_6)$, which satisfy your condition, has positive lim inf as $d_1,\dots, d_6$ tend to $\infty$.

Note that it is too much to ask for in this setting that the probability tends to $1$ - for smoothness results, it almost never does, since one can force a singularity with a congruence condition mod $x_0^2$, which has a positive probability of being satisfied.

In this case, a sufficient condition for smoothness is that the discriminant $\det \begin{pmatrix} f_1 & f_2 & f_3 \\ f_2 & f_4 & f_5 \\ f_3 & f_5 & f_6 \end{pmatrix}$ is a squarefree polynomial. One can lower-bound the probability of this using the main Theorem 2.2 of On square-free values of large polynomials over the rational function field by Dan Carmon and viewing this discriminant as a polynomial in whichever of the $f_i$ has the largest degree, treating the other $f_i$ as fixed.

If the $d_i$ are close together, one could also use classical analytic number theory methods such as the circle method to get an estimate.

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