A question regarding conic sections

Question

I was reading Miles Reid's book titled "Undergraduate Algebraic Geometry" and I stumbled on the following in Page 106.

Write a cubic $f = AX^2 + BXY + CY^2 + DX + EY + F$, where $A, B, C, D, E, F \in k[Z, T]$. (Note that the cubic is given to be nonsingular to start with). It is claimed that if $f$ is considered to be a quadric in variables X and Y then it is singular iff $\Delta (Z, T) = 0$, where $\Delta$ is 4 times the determinant of the matrix associated to the conic in variables $X$ and $Y$. I am looking forward to see some explanation of this statement.

Can it be more generally claimed that, $f$ is irreducible $\implies$ $\Delta (Z, T) \neq 0$?

Thanks in advance!

Answer 1

The Wikipedia article Degenerate conic gives the matrix $Q$ in $A,B,C,D,E,F$ and states "The conic is degenerate if and only if the determinant of this matrix equals zero". I am guessing that degenerate means singular in your case, but I am not sure what irreducibility means as well.

Answer 2

You might want to get the updated 2013 edition of this book, from the author's web site, https://homepages.warwick.ac.uk/staff/Miles.Reid/MA4A5/UAG.pdf. In this edition, the proof you are working through is for Proposition 7.3, and the step you are asking about is on page 113. There are some details added there, so this might be helpful to you.

Now if $a,b,c,d,e,f \in k$ are scalars and we consider the (affine) plane conic $C$ defined by $aX^2 + 2bXY + cY^2 + 2dX + 2eY + f = 0$, then you can show that the following are equivalent (assuming that $(a,b,c) \neq (0,0,0)$, i.e., the degree didn't drop to $1$ or less):

1. $C$ is singular,
2. $C$ is a union of two distinct lines, or a double line,
3. $\det\begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} = 0$,
4. the defining equation of $C$ is irreducible in $\overline{k}[X,Y]$.

What Reid is doing is assuming that $\ell$ is a line in the cubic surface $S$ defined by the cubic equation $AX^2 + \dotsb + F = 0$, where $A,\dotsc,F \in k[Z,T]$. And he also assumes that $\ell$ is the line $Z=T=0$ in $\mathbb{P}^3$. Now any plane containing $\ell$ is defined by $\mu Z = \lambda T$ for some $[\lambda:\mu] \in \mathbb{P}^1$. Take such a plane $\Pi$, containing $\ell$, defined by $1Z = \lambda T$ for some $(\lambda,1)$. Then $S \cap \Pi$ is a plane cubic curve (because it's an intersection of a plane and cubic surface). It's not an arbitrary plane cubic: it always contains the line $\ell$! So it can be written as $\ell$ plus a plane conic.

Reid explains that the conic is defined by the equation $A(\lambda,1) X^2 + 2B(\lambda,1)XY + C(\lambda,1)Y^2 + 2D(\lambda,1)XT + 2E(\lambda,1)YT + F(\lambda,1)T^2 = 0$. Here the $A(\lambda,1),\dotsc,F(\lambda,1)$ are scalars again (up to a common scalar factor). So we can use the above criterion, in terms of the determinant, to see whether the conic is reducible.

We are looking at the determinant $\det\begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix}$, where the $a,\dotsc,f$ are the values of $A(\lambda,1)$, etc. Hence $A(\lambda,1),\dotsc,C(\lambda,1)$ each have degree $\leq 1$ in $\lambda$; $D(\lambda,1),E(\lambda,1)$ each have degree $\leq 2$ in $\lambda$; and $F(\lambda,1)$ has degree $\leq 3$ in $\lambda$. This shows that the determinant has degree $\leq 5$ in $\lambda$. So there are at most $5$ values $\lambda$ for which the corresponding plane through $\ell$ cuts $S$ in a reducible conic. Reid goes on to prove that there are no fewer than $5$ such values.