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


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

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    $\begingroup$ A degenerate conic may also be the form $C=Q^2=0$ which is a double line, in particular $X^2=0$. $\endgroup$ – Yuchen Liu May 10 '13 at 12:13

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.

  • $\begingroup$ I do not think Reid requires a closed algebraic field, but that is a key part of your proof. Also the factoring only works over $\mathbb{C}$ also, right? $\endgroup$ – levitopher May 11 '13 at 16:19
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    $\begingroup$ Well, over $\mathbb R$ you can arrange that $A$, $B$, and $C$ are all either $0$, $1$, or $-1$. If we have one zero, then we get either $X^2+Y^2=0$, which is empty, or $X^2-Y^2=(X+Y)(X-Y)=0$, which is the intersection of lines, or $X^2=0$, which is a double-line. If we can't take square-roots in your field, then we don't quite get here :) $\endgroup$ – Ted Shifrin May 19 '13 at 18:44

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