# Fields of characteristics and affine conics

Let $k$ be any field of characteristic not equal to $2$, and $V$ a $3$-dimensional $k$-vector space. Let $Q: V \to k$, be a non degenerate quadratic form on $V$. How can we show that if $\,0\neq e_1\in V\,$ satisfies $Q(e_1) = 0$, then $V$ has a basis $e_1,e_2,e_3$ such that $Q(x_1 e_1 + x_2 e_2 + x_3 e_3) = x_1 x_3 + ax_2^2$.

And how to deduce that a nonempty nondenerate conic $C \subseteq \mathbb{P}_k^2$, is projectively equivalent to $XZ = Y^2$?

What I am able to note is I need to work the symmetric bilinear form $\phi$ associated to $Q$; and since $\phi$ is non-degenerate, there is a vector $e_3$ such that $\phi(e_1,e_3) = 1$.

#### Solutions Collecting From Web of "Fields of characteristics and affine conics"

You are off to a good start with considering $\phi$ and noting that you can choose $e_3$ so that $\phi(e_1, e_3) = 1$.

Now let $c = \phi(e_3, e_3)$, which may or may not be $0$. But we can replace $e_3$ by $e’_3 = \frac{c}{2} e_1 – e_3$ and compute that $\phi(e’_3, e’_3) = \frac{c^2}{4}\phi(e_1,e_1) – 2\frac{c}{2}\phi(e_1,e_3) + \phi(e_3,e_3) = 0$. So, replacing $e_3$ with $e’_3$, we may assume $\phi(e_1,e_1) = \phi(e_3,e_3) = 0$ while $\phi(e_1,e_3) = \phi(e_3,e_1) = 1$.

Let $e_2$ be a third vector to complete the basis. Set $b = \phi(e_1,e_2)$. Setting $e’_2 = e_2 – b e_3$, we have that $\phi(e_1,e’_2) = 0$, so replacing $e_2$ by $e’_2$, we may assume $\phi(e_1,e_2) = \phi(e_2, e_1) = 0$. Now let $c = \phi(e_2,e_3)$ and set $e’_2 = e_2 – c e_1$. The $\phi(e’_2,e_3) = 0$ and $\phi(e’_2,e_1) = 0$, so replacing $e_2$ by $e’_2$ again, we get that the matrix for $\phi$ has the form
$$\begin{pmatrix} 0 & 0 & 1 \\ 0 & a & 0 \\ 1 & 0 & 0 \end{pmatrix}$$
for some $a$. Because $\phi$ is non-degenerate, $a \neq 0$. This yields the desired $Q$.

If $-a$ has a square root in $k$ (always the case if $k$ is algebraically closed), then replacing $e_2$ by $e’_2 = \frac{1}{\sqrt{-a}} e_2$, we may assume that $Q(e_2) = \phi(e_2,e_2) = -1$.

For the second part, let $C \subset \mathbb{P}^2_k$ be a non-degenerate conic defined by a non-degenerate quadratic form $Q$. Note that the existence of the vector $e_1 \neq 0$ such that $Q(e_1) \neq 0$ is equivalent to the fact that the conic $C$ is not empty.

Applying a projective equivalence corresponds to making a change of basis in $V$. And we just showed that we could change our basis so that $Q$ has the form
$$\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix}.$$
But this $Q$ defines the conic $C$ given by the equation $XZ – Y^2 = 0$, or $XZ = Y^2$. (Edit: Actually, depending on conventions, you are more likely to interpret this $Q$ as defining the conic $2 XZ = Y^2$, but see the remark below.)

Edit: As Georges Elencwajg pointed out in the comments, you can use $Z \mapsto Z/a$ to get $XZ = aY^2$ projectively equivalent to $XZ = Y^2$ for any $a \neq 0$.