2
$\begingroup$

Let $\mathbb{F}_p$ be a prime finite field for $p > 2$. Consider two ternary quadratic forms $$Q_1\!: x^2 - a_1(t)y^2 - b_1(t)z^2,\\ Q_2\!: x^2 - a_2(t)y^2 - b_2(t)z^2$$ over the field $\mathbb{F}_p(t)$ of rational functions with coefficients from $\mathbb{F}_p$. For simplicity let $a_1, a_2, b_1, b_2 \in \mathbb{F}_p[t]$ are polynomials without multiple roots and $a_1$, $b_1$ (respectively $a_2$, $b_2$) have no common roots.

Is there an efficient algorithm to find a linear transformation (over $\mathbb{F}_p(t)$) between $Q_1$ and $Q_2$ if it exists?

There is the theory that relates such forms and quaternion algebras (see, for example, $\S$1.4 in Book of Gille, Szamuely - Central Simple Algebras and Galois Cohomology). For example, for any non-zero polynomial $f \in \mathbb{F}_p[t]$ and $p > 2$ the following quadratic forms are isomorphic: $$ Q_1\!: x^2 - y^2 - f(t)z^2,\\ Q_2\!: x^2 - y^2 - z^2 $$ This is true, because $Q_1$ can be reduced to the quadratic form $Q_3\!: x^\prime y^\prime-(z^\prime)^2$ by the transformation $$ x := x^\prime+\frac{y^\prime}{4f},\qquad y := x^\prime-\frac{y^\prime}{4f},\qquad z := \frac{z^\prime}{f}. $$ It is well known that any two conics (including $Q_2$, $Q_3$) over a finite field are isomorphic.

$\endgroup$
  • $\begingroup$ What do you mean by a linear transformation? Is it $x'=c_0x+c_1$, $y'=c_2x+c_3$, $z'=c_4z+c_5$? If so it seems easy to prove that there can never be any non-trivial linear transformation, since the coefficients of $x,y,z,1$ are zero. $\endgroup$ – D.W. Jan 21 at 16:51
  • $\begingroup$ I mean a non-degenerate projective transformation: $x := c_1x + c_2y + c_3z$, $y := d_1x + d_2y + d_3z$, $z := e_1x + e_2y + e_3z$, where coefficients from $\mathbb{F}_p(t)$. $\endgroup$ – Dima Koshelev Jan 21 at 17:15
  • $\begingroup$ Still seems impossible for the same reasons. Can you edit the question to give an example of two such quadratic forms where a linear transformation does exist? $\endgroup$ – D.W. Jan 21 at 17:18
  • $\begingroup$ I added the comments. $\endgroup$ – Dima Koshelev Jan 21 at 17:59
  • 1
    $\begingroup$ You claim that they are isomorphic in the question; how can you know that, if you can't find such a transformation? I think you should put some more effort into your question first. I suggest trying an example and try proving whether such a transformation exists. You should be able to write down a system of 8 equations on the 9 unknowns $c_1,c_2,c_3,d_1,d_2,d_3,e_1,e_2,e_3$ and then see if any solution exists, and thus whether any such linear transformation exists. I think you should also edit the question to show your definition of "linear transformation" in the question. $\endgroup$ – D.W. Jan 21 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.