Sunday, 13 July 2008

ra.rings and algebras - Quartic form which is irreducible but not geometrically irreducible

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 X1X2X3X4=0 and that overlineF is an alg. closure of F. Then the projective automorphism group of V is a split extension of T=mathbbG4m by the symmetric group S4. Since H1(F,T)=0, the set of isomorphism classes that we seek is H1(F,S4)=Hom(GalF,S4), 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 overlineF. Its automorphism group is S4, 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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