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Let $\mathbb F _q$ be finite field of odd characteristic and consider the polynomials $$ \xi_i = x_1^{q^i+1} - x_2^{q^i+1} + x_3^{q^i+1} - x_4^{q^i+1} \in \mathbb F_q[x_1,x_2,x_3,x_4].$$ I'm interessted in $\xi_0,\xi_1,\xi_2$ and relations between them. Using a computer algebra system I checked that $\xi_2$ is in $(\xi_1,\xi_0)$, the ideal generated by $\xi_0,\xi_1$, for $3 \leq q \leq 37$. Therefore I have the impression that $\xi_2 \in (\xi_0,\xi_1)$ for all $q$. Unfortunately I was not able to prove it.

What I have tried: For $3 \leq q \leq 37$ I (my computer) computed polynomials $f,g \in \mathbb F_q[x_1,x_2,x_3,x_4]$ satisfying $\xi_2 = f \xi_1 + g \xi_0$. Since they are too complicated I can not recognize any pattern or general form for arbitrary $q$.

Motivation: The polynomials $\xi_0,\xi_1,\xi_2$ are part of a generating system of the polynomial invariants of the four-dimensional orthogonal group of plus type:

Chu , H., 2001, Polynomial invariants of four-dimensional orthogonal groups. Comm. Algebra 29, 1153–1164,

Long story short: Does anyone know how to show $$x_1^{q^2+1} - x_2^{q^2+1} + x_3^{q^2+1} - x_4^{q^2+1} \in ( x_1^{2} - x_2^{2} + x_3^{2} - x_4^{2}, x_1^{q+1} - x_2^{q+1} + x_3^{q+1} - x_4^{q+1})$$ over a finite field with $q$ elements? Or maybe it is just wrong for arbitrary $q$? Thank you in advance.

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Do a change of variables $$2y_1=x_1+x_2, \quad 2y_2=x_1-x_2, \quad 2y_3=x_3+x_4,\quad 2y_4=x_3-x_4.$$ The ideal $I$ in question is now generated by $$ y_1y_2+y_3y_4, \qquad y_1y_2(y_1^{q-1}+y_2^{q-1}-y_3^{q-1}-y_4^{q-1}).$$ Now it is a question of solving equations. For example, if $y_1y_2\neq 0$, then
$$\begin{cases} y_1y_2=-y_3y_4 \\ y_1^{q-1}+y_2^{q-1}-y_3^{q-1}=y_4^{q-1}\end{cases}.$$

Take the $(q-1)$-th power of both sides of the first equation and eliminate $y_4$ using the second, you get $$ (y_3^{q-1}-y_1^{q-1})(y_3^{q-1}-y_2^{q-1})=0, $$ which factors into a product of linear terms. Everything else should be straight forward from there.

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Thanks. I will try this. – Hans Giebenrath Apr 15 '13 at 13:56
I followed your steps and arrived at the factorization into linear factors. But how to proceed? And why does $y_1y_2 \neq 0$ imply $y_1^{q-1}+y_2^{q-1}-y_3^{q-1}-y_4^{q-1} eq 0$, i.e. why does $y_1y_2 \not\in I$ imply $y_1^{q-1}+y_2^{q-1}-y_3^{q-1}-y_4^{q-1} \in I$? – Hans Giebenrath Apr 16 '13 at 14:35

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