# Transforming Quadrics in Characteristic 2

I’m trying to solve the following problem given in a textbook:

Let $$k$$ be an algebraically closed field and $$Q=V(F)$$ a quadric in $$\mathbb{P}^3(k)$$, where $$F$$ is an irreducible polynomial in $$X,Y,Z,T$$, and hence gives rise to a quadratic form on $$k^4$$ which we assume is non-degenerate.

Show that after some change of coordinates we can write $$Q=V(XT-YZ)$$.

I’ve solved the case where $$\text{char}(k)\neq2$$ by diagonalising the quadratic form and then making a suitable change of coordinates. However this process involves using the correspondence between quadratic forms and symmetric bilinear forms which is not valid in characteristic $$2$$ (and if we could diagonalise $$F$$ then it would become reducible).

My first issue is that I can’t find a reference defining what it means for a quadratic form to be non-degenerate in characteristic $$2$$, so if I tried to come up with a counterexample I wouldn’t know if it was valid or not.

Beyond this, I’m not even sure that the result is true, I can’t seem to find anywhere claiming that it is. Has the textbook simply forgotten to specify that $$\text{char}(k)\neq2$$, or am I missing something?

Any help would be much appreciated.

Arf defines non-singularity for quadratic forms of characteristic $$2$$ here, explained in English here. From these papers, we see that if $$Q$$ is non-singular over a field of characteristic $$2$$, we can write $$Q=(aX^2+XT+bT^2)+(cY^2+YZ+dZ^2)$$ for some $$a,b,c,d\in k$$ and an appropriate choice of coordinates. Let’s consider $$aX^2+XT+bT^2$$.
If $$a=b=0$$ then we have $$XT$$ already, if say $$b=0$$ then we have $$X(aX+T)$$ so taking the inverse of the transformation sending $$X\mapsto X$$ and $$T\mapsto aX+T$$ we have $$XT$$.
Then we assume $$a,b\neq0$$. Sending $$X\mapsto\frac{1}{\sqrt{a}}X$$ and $$T\mapsto\frac{1}{\sqrt{b}}T$$ we have $$X^2+\alpha XT+T^2$$ for $$\alpha=\frac{1}{\sqrt{ab}}$$. All square roots exist since $$k$$ is algebraically closed, and for the same reason we can also find a root $$\beta$$ of the polynomial $$x^2+\alpha x+1$$. Then sending $$X\mapsto X+\frac{1}{\alpha^2}T$$ and $$T\mapsto\beta X+\frac{\alpha+\beta}{\alpha^2}T$$ we have $$XT$$. This transformation is invertible since $$\begin{vmatrix}1 & \beta \\\frac{1}{\alpha^2} & \frac{\alpha+\beta}{\alpha^2}\end{vmatrix}=\frac{1}{\alpha}\neq0$$
We can repeat the same process for $$Y$$ and $$Z$$, and so we can write $$Q=XT+YZ=XT-YZ$$.