Determine if a conic is degenerate with the determinant.

There is a natural bijection between conics (written as homogeneous quadratic) and $3 \times 3$ matrices:

$$C=aX^2+2bXY+cY^2+2dXZ+2eYZ+fZ^2\Leftrightarrow \left(\begin{array}{ccc} a&b&d\\ b&c&e\\ d&e&f\end{array}\right)$$

I'm reading Reid's Undergraduate Algebraic Geometry, and on page 21 he says that a conic $C$ is degenerate iff the determinant of the matrix vanishes.

So I think of degenerate conics as $XY=0$, decomposed as the set $X=0$ and $Y=0$. So if the determinant vanishes we have a condition on the $a,b,c,d,e,f$. Should I believe that if I put that condition into my conic $C$, I will find that the conic decomposes like

$$C=Q_1Q_2=0?$$

I don't see how one can guarantee $C$ is degenerate from the vanishing of the determinant.

Solution 1:

Working over $\mathbb C$, yes. By a change of basis, the quadratic form can be written as $AX^2 + BY^2 + CZ^2$, where $A,B,C$ are either $0$ or $1$. The determinant condition says you have at least one zero. And note either $X^2+Y^2$ or $X^2$ factors as a product of linear polynomials.