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I am looking at Miles Reid's UAG book. There he claims that every projective conic is projectively equivalent to $XZ = Y^2$. He asks to show that $Q$ a non-degenerate quadratic form is such that $Q(e_1) = 0$ then $V$ has a basis $e_1,e_2,e_3$ such that $Q(x_1e_1 +x_2e_2 + x_3e_3) = x_1x_3 + ax_2^2$.

First I confusd by hint in exercise 1.5. He say if $\dim_k V = 3$ and $e_1$ in $V$ satisfies $Q(e_1) = 0$ then if $\varphi$ is symmetric bilinear form stuck to $Q$ then there is $e_3$ such that $\varphi(e_1,e_3) = 1$ (okay?????? ) Then he asks to find suitable $e_2$.

I confuse because actually we want to find a basis for $V$ so that $\varphi$ in that basis

$$\varphi = \left[\begin{array}{ccc} 0 & 0 & \frac{1}{2} \\ 0 & a & 0 \\ \frac{1}{2}& 0 & 0\end{array}\right]$$

no? So how does hint help?

Please help I confuse.

1 Answers1

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We have $Q(e_1)=\varphi(e_1,e_1)=0$. Since $\varphi$ is non-degenerated, the functional $v\mapsto \varphi(e_1,v)$ is not $0$, i.e. we can fix a $v$ such that $\varphi(e_1,v)\ne 0$. Then, let $u:=v/\varphi(e_1,v)$, this will satisfy $\varphi(e_1,u)=1$. What we still need for an $e_3$ is $\varphi(e_3,e_3)=0$. For this, observe that for $u-\lambda e_1$, we have $$Q(u-\lambda e_1)=\varphi(u-\lambda e_1,u-\lambda e_1)=\varphi(u,u)-2\lambda\,\varphi(e_1,u)=Q(u)-2\lambda\,.$$ So, let's define $e_3:=u-\frac{Q(u)}2\,e_1$.

Now we need an $e_2$ which is $\varphi$-orthogonal to both $e_1$ and $e_3$. This can be done similarly: first extend $e_1,e_3$ to a basis arbitrarily by a vector $w$. Then search $e_2$ in the form $$e_2=w-\alpha\,e_1-\gamma\,e_3$$ that satisfy $\varphi(e_1,e_2)=\varphi(e_3,e_2)=0$.

Berci
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