Nondegenerate Conics

Question

I am having trouble seeing why the set of all nondegenerate conics is Zariski open in the parameter space $\mathbb{P}^5$.This is an exercise in Smith 5.2.2 that I am trying to do. I will be grateful to receive any help on this.

Answer 1

Scaling the l.h.s. of the (homogeneous) conic equation $$A x^2 + 2 B x y + C y^2 + 2 D x z + 2 E y z + F z^2 = 0$$ preserves the locus, but other transformations of the l.h.s. do not, so we can regard $$[A : B : C : D : E : F]$$ as homogeneous coordinates on the space of conics, which we identify with $\mathbb{P}^5$. A conic is degenerate iff the discriminant of this form, namely, $$\det \begin{pmatrix} A & B & D \\ B & C & E \\ D & E & F\end{pmatrix} = ACF + 2 BDE - AE^2 - FB^2 - CD^2$$ is zero.

In particular, the set of degenerate conics is a projective subvariety of $\mathbb{P}^5$, and hence is Zariski closed, and so its complement, the set of nondegenerate conics, is Zariski open.