About quadratic forms

Question

I know that every element is finite field of odd order can be written as sum of two squares. Also I know that since $\mathbb{F}_q$/$\mathbb{F}_q^2$ has index 2; it represent 2 classes. One class represents elements which are sqaures and other class represent non-squares.

So take $\epsilon$ which can be represent as $\langle 1,1\rangle$ meaning is $\epsilon = x^2+y^2$ then how to show that $\langle 1,1\rangle$ is isometric to $\langle \epsilon,\epsilon\rangle$?

Basically I want to show $\langle 1,1\rangle$ is isometric to $\langle x^2+y^2,x^2+y^2 \rangle$? How to show this in finite field?

Answer 1

I think you need to read the earlier chapters and understand the terminology properly. I don't have a copy of the book, but here are some things.

1. By saying that $1,\epsilon$ are the elements of $\mathbb{F}_{q}^{*}/\mathbb{F}_{q}^{*2}$, this is saying that $\epsilon$ is an arbitrary nonsquare.
2. I am going to guess that when Scharlau says the quadratic form $\langle 1,1 \rangle = x^{2}+y^{2}$ represents $\epsilon$, this is defined to mean that there is some vector $v \in \mathbb{F}^{2}$ for which the quadratic form evaluated at $v$ is equal to $\epsilon$. It does not mean that $\langle 1,1 \rangle = \epsilon$.
3. As a continuation of 2., the quadratic form $\langle \epsilon, \epsilon \rangle$ would then be $\epsilon x^{2} + \epsilon y^{2}$.

Based on this, you want to show that the quadratic forms given by $$\langle 1,1 \rangle = \begin{bmatrix} 1 & 0 \\ 0&1 \end{bmatrix}, \langle \epsilon, \epsilon \rangle = \begin{bmatrix} \epsilon & 0 \\ 0 & \epsilon \end{bmatrix}$$ are isometric, where $\epsilon$ is an arbitrary nonsquare.

To see that these two matrices give equivalent forms, you want to take $a,b$ such that $a^{2}+b^{2} = \epsilon$. Then if you consider $$ \begin{bmatrix} a & b\\ -b & a \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & -b\\ b& a \end{bmatrix}$$ you should be able to see your result.