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?

  • $\begingroup$ 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 Sep 2 '10 at 10:04
  • $\begingroup$ Why a cyclic Galois group? $\endgroup$ – Wanderer Sep 2 '10 at 10:20
  • $\begingroup$ The galois group of a finite extension of finite fields is always cyclic, generated by the frobenious. $\endgroup$ – Daniel Loughran Sep 2 '10 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 Sep 2 '10 at 11:48
  • $\begingroup$ These are multivariable polynomials! So many polynomials will not factor at all. $\endgroup$ – Wanderer Sep 2 '10 at 12:27

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.

| cite | improve this answer | |

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 $F$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.