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Suppose that $F$ is a finite field of odd characteristic.

Suppose that $q(X_1,X_2,...,X_n)$ is a quartic (homogeneous) form with coefficients in $F$ such that

  • $q$ is irreducible over $F$
  • $q$ does not have any non-trivial zeros in $F^n$ (hence $n \leq 4$)
  • over $\overline{F}$, $q$ can be factored as the product of four linear forms

What can one say about $q$? Can we say that $q$ must factor over the degree 4 extension of $F$? What else?

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  • $\begingroup$ I don't know what sort of answer you're after, but it seems to me that what you have is some sort of combinatorial object (a configuration of 4 points/lines/planes) with an action of a (cyclic) Galois group. That might be a good point of view to start from. $\endgroup$
    – Martin Bright
    Commented Sep 2, 2010 at 10:04
  • $\begingroup$ Why a cyclic Galois group? $\endgroup$
    – Wanderer
    Commented Sep 2, 2010 at 10:20
  • $\begingroup$ The galois group of a finite extension of finite fields is always cyclic, generated by the frobenious. $\endgroup$
    – Daniel Loughran
    Commented Sep 2, 2010 at 11:04
  • $\begingroup$ Any polynomial of degree d will factor over its splitting field, which in general can be an extension of degree as large as d factorial. Perhaps it might be smaller in this specific finite field case though. $\endgroup$
    – Daniel Loughran
    Commented Sep 2, 2010 at 11:48
  • $\begingroup$ These are multivariable polynomials! So many polynomials will not factor at all. $\endgroup$
    – Wanderer
    Commented Sep 2, 2010 at 12:27

2 Answers 2

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A bit more is true when $n=4$. Suppose $F$ is any perfect field, not necessarily finite. Then, for $n=4$, the form $q$ is, up to a scalar multiple, the norm form of a quartic extension of $F$. (Proof: Suppose that $V$ is the projective $F$-scheme defined by $X_1X_2X_3X_4=0$ and that $\overline F$ is an alg. closure of $F$. Then the projective automorphism group of $V$ is a split extension of $T=\mathbb G_m^4$ by the symmetric group $S_4$. Since $H^1(F,T)=0$, the set of isomorphism classes that we seek is $H^1(F,S_4) =Hom(Gal_F,S_4)$, as stated.) For finite $F$ there is only one such extension, of course.

For $n=3$ the configuration of $4$ lines might have triple, or quadruple, points, but if not then again there is only one isomorphism class over $\overline F$. Its automorphism group is $S_4$, and the same argument shows that $q$ is unique up to scalars (you can describe it as a linear section of the quartic norm form.) For $n=2$ I have nothing to add.

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Yes. The absolute Galois group permutes the linear factors so it acts via a subgroup of $S_4$. As the Galois group of a finite field is cyclic, you have a cyclic subgroup of $S_4$ which thus has order $2,3,4$. In the case of order $3$, one of the linear factors will be fixed by the action, so the form factors over $F$. So you are left with the other two possibilities and they both factor in the unique quartic extension of $F$.

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