Non singular degenerate conics in the projective plane

Question

I was trying to count degenerate conics in the projective plane $\mathbb{P}^2(\mathbb{F}_q)$ and I discovered what ooks like a "paradox" (I know it's not but I can't solve it).

Given a point (let's say (0:0:1)) in $\mathbb{P}^2(\mathbb{F}_q)$ the space of singular conics passing throught it is nothing but the spaace of solutions of $ax_0+bx_0x_1+cx_1^2$ that has cardinality $q^2+q+1$.

On the other hand I thought that degenerate conics were of the form $L_1*L_2$ (the two lines can also be the same), there are q+1 lines through any point and so there should be $(q+1)^2=q^2+2q+1$ degenerate conics.

Why are the two numbers different? Are there q degenerate conics that are not singular?

EDIT: The first formula, that is indeed correct, comes from the fact that a point is said to be "smooth", and a point s that is not smooth is singular, if the conic equation vanishes only to order 1 at s, or equivalently, if at least one of the partial derivatives of it is nonzero at s.

Let's now consider the generic conic equation $ax_0^2+bx_1^2+cx_2^2+dx_0x_1+ex_0x_2+fx_1x_2=0$, if we evaluate it at (0:0:1) we get $c=0$ and so the equation becomes $ax_0^2+bx_1^2+dx_0x_1+ex_0x_2+fx_1x_2=0$, we can now compute the partial derivative and find they are $P_0=2ax_0+dx_1+ex_2$, $P_1=2bx_1+dx_0+gx_2$, $P_2=ex_0+fx_1$, for the point to be singular we need all of these to be 0 if valuated at (0:0:1), now $P_0(0:0:1)=e\Rightarrow e=0$, $P_1(0:0:1)=g\Rightarrow g=0$, $P_2(0:0:1)=0$ and so the right equation we need to look for solutions to is indeed $ax_0^2+dx_0x_1+ex_1^2$.

My question is why doesn't the second way work too

Answer 1

To count the conics in $\mathrm{PG}(2,q)$, I will consider the set of homogeneous degree 2 polynomials $$c_{0}X_{0}^{2} + c_{1}X_{1}^{2} + c_{2}X_{2}^{2} + c_{3}X_{0}X_{1} + c_{4}X_{0}X_{2} + c_{5}X_{1}X_{2}$$ there are $q^{6}-1$ such nonzero polynomials, and since we are associating a conic with the zero locus of such a polynomial, we get that there are $\frac{q^{6}-1}{q-1}$ conics in the plane (in the sense of their algebraic equation; it is possible that some of these equations define the same point set of $\mathrm{PG}(2,q)$).

If we want to count the number of degenerate conics, it is worth noticing that these come in three forms (I borrowed the example from Stanley Payne's extensive Topics In Finite Geometry, which sadly seems to no longer be available online):

1. a line (counted twice); e.g. $X_{0}^{2}$
2. a pair of distinct lines (contrary to what you state in the question body, order does not matter); e.g. $X_{0}X_{1}$
3. a single point (this is a possibility since, unlike $\mathbb{C}$, $\mathbb{F}_{q}$ is not algebraically closed; this case corresponds to a pair of conjugate lines in $\mathrm{PG}(2,q^{2})$); e.g. $X_{0}^{2}+aX_{0}X_{1}+bX_{1}^{2}$ with $a$ and $b$ chosen so this polynomial is irreducible over $\mathbb{F}_{q}$.

There are $q^{2}+q+1$ degenerate conics of the first form, and ${q^{2}+q+1 \choose 2}$ of the second form.

The number of conics of the third form is given by $\frac{1}{2}(q^{2}+q+1)(q^{2}-q)$; I'm being a little hand-wavy here; there are of course only $q^{2}+q+1$ point sets of $\mathrm{PG}(2,q)$ of the third type but this counts the number of these degenerate conics considered over $\mathrm{PG}(2,q^{2})$. This count comes from noticing that each of the $q^{2}+q+1$ points of $\mathrm{PG}(2,q) \subset \mathrm{PG}(2,q^{2})$ is incident with $q^{2}+1$ lines, $q+1$ of which intersect $\mathrm{PG}(2,q)$ in $q$ other points, and the remaining $q^{2}-q$ containing no other points of $\mathrm{PG}(2,q)$; we cut this number in half because we are counting the conjugate pairs of lines. Of course, if you are defining a conic as a point set of $\mathrm{PG}(2,q)$ then there are only $q^{2}+q+1$ degenerate conics of this type.