# Birational transformation of conic to line (Cremona inversion)

I am following Discriminants, resultants, and multidimensional determinants and I am having trouble following the geometric argument in Lemma 2.11. I will outline the lemma and the proof here:

Lemma 2.11. Let $q_1,\dots, q_4$ be four points in $\Bbb P^2$ of which no three are colinear and let $l \subset \Bbb P^2$ be a line that does not meet any of the $q_i$. Then there are distinct conics through $q_1,\dots q_4$ tangent to $l$.

Proof. By choosing the appropriate homogeneous coordinates $x_0, x_1, x_2$ in $\Bbb P^2$, we ca assume that $q_1=(1:0:0),q_2=(0 :1:0),q_3=(0:0:1),q_4=(1:1:1)$. Consider the transformation (Cremona inversion) $$\Psi:\Bbb P^2 \to \Bbb P^2,\qquad (x_0:x_1:x_2)\mapsto\left(\frac1{x_0}:\frac1{x_1}:\frac1{x_2}\right)$$ This transformation takes the system of conics through $q_1,\dots,q_4$ into the system of straight lines through $q_4$. The line $l$ is taken into a conic $\Psi(l)$ not containing $q_4$. Clearly there are two tangent lines to $\Psi(l)$ through $q_4$. Applying the inverse transformation $\Psi^{-1}=\Psi$, we get two conics through $q_1, \dots, q_4$ tangent to $l$.

I am following every part of the proof except for the claim that the transformation takes lines to conics and conics to lines. I have written out equations for these and I just cannot see it happen. (It is possible for the line to conic however). If somebody could show that birational transformation in more light I would be grateful.

## 1 Answer

Here is a direct computation. A conic $C$ in $\mathbb{P}^2$ is of the form $$C: a_0 x^2 + a_1 xy + a_2 xz + a_3 y^2 + a_4 yz + a_5 z^2 = 0 \, .$$ Since $q_1 = (1:0:0)$ lies on $C$, substituting $x = 1, y = 0, z = 0$ into the equation for $C$ we find $a_0 = 0$. Doing the same for $q_2$ and $q_3$ shows that $a_3 = a_5 = 0$. So the conic is of the form $$C: a_1 xy + a_2 xz + + a_4 yz = 0 \, .$$ Since $q_4$ lies on $C$, then $a_1 + a_2 + a_4 = 0$. Applying the Cremona involution transforms $C$ into $$a_1 \frac{1}{x} \frac{1}{y} + a_2 \frac{1}{x} \frac{1}{z} + a_4 \frac{1}{y} \frac{1}{z} = 0$$ and multiplying through by $xyz$ yields $$a_4 x + a_2 y + a_1 z = 0$$ which is a line through $q_4$.

Here is a way to proceed without doing the above computation. Note that the Cremona map is an involution, so $\Psi \circ \Psi = \text{id}$. So if you show that it maps such a line $\ell$ to such a conic $C$, then it must map the conic back to the line, since $$\Psi(C) = \Psi(\Psi(\ell)) = \operatorname{id}(\ell) = \ell \, .$$ Also note that the Cremona involution can also be written $$(x : y : z) \mapsto \left(\frac{1}{x} : \frac{1}{y} : \frac{1}{z} \right) = (yz : xz : xy)$$ by multiplying through by $xyz$. This makes the line to conic part a bit clearer. (Also this expression is valid when any two of $x,y,z$ are nonzero, whereas your expression requires all three to be nonzero.)