3
$\begingroup$

Let $p>2$ be a prime number, $V=\left(\mathbb{Z}/p^n\mathbb{Z}\right)^{2k+1}$. The bilinear form $$B:V\times V \rightarrow \mathbb{Z}/p^n\mathbb{Z}$$ is a perfect pairing. That is, mapping $x\in V$ to $B(x,-)\in V^*$ is an isomorphism between $V$ and $V^*$.

Is it true that the number of solutions to $B(x,x)=0$ does not depend on $B$. Also, what is intuition that this is happening for odd rank $V$, but not even ones? How about over general rings?

Edit: It has been pointed out that this won't hold for general $B(x,x)=c$ (which was the original version of this question). Looks like it is true for $c=0$, though. Still want to ask the intuition behind.

I believe this invariance can lead to some interesting facts. Like this MO post, tries to count the number of solutions to the quadratic equation $$𝑥^2_1+⋯+𝑥^2_m=0.$$ If $m=2k+1$ is odd, This is indeed our case when $B$ is the identity matrix. Using the invariance, one can compute it by counting the number of solutions to $$x_1(x_2+\ldots+x_{2k+1})+x_2(x_3+\ldots+x_{2k+1})+\ldots + x_{2k}x_{2k+1}=0,$$ which comes from the case $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.

$\endgroup$
2
$\begingroup$

This is already false for $k=0$ (and $n$ arbitrary)

Le $B_0(x,y)=xy$ and $B_1(x,y)=-xy$, and $c=-1$. We then have two equations $x^2=-1$ and $-x^2=-1$. The second one always has at least two solutions (maybe more), while the first have no solutions if $p\equiv 3 \mod 4$.

$\endgroup$
1
  • $\begingroup$ Thank you. It looks like it is only true for c=0. Any intuition why it is happening? $\endgroup$
    – Ted Mao
    Jul 9 at 18:31
0
$\begingroup$

Your claim is never true for $n=1$, assuming nondegeneracy. This gives also many counterexamples for $n>1$ using Hensel's lemma.

In Lidl and Niederreiter's book "Finite Fields", 2nd edition, Chapter 6, section 2 (quadratic forms) you'll find plenty of information on the $n=1$ case. In particular, from their Theorem 6.27 it follows that for fixed $k$ and nondegenerate $B\colon (\mathbb{Z}/p\mathbb{Z})^{2k+1}\times (\mathbb{Z}/p\mathbb{Z})^{2k+1} \to \mathbb{Z}/p\mathbb{Z} $, the number of solutions to $B(x,x)=c$ is a non-constant (explicit) function of the Legendre symbol of $c$ mod $p$.

The reason for the parity difference is essentially their Lemma 6.24, showing that quadratic forms in two variables are well-behaved, namely the function $b \mapsto \#\{(x_1,x_2): a_1 x_1^2 + a_2 x_2^2 =b\}$ is essentially constant (depends only on whether $b=0$ or not), and then some linear algebra allows you to reduce the study of $B(x,x)=c$ when $x \in (\mathbb{Z}/p\mathbb{Z})^{2k}$ to $k=1$ (which behaves almost like a constant) and the study of $B(x,x)=c$ when $x \in (\mathbb{Z}/p\mathbb{Z})^{2k-1}$ to $k=1$, that is, counting $b$ with $ax^2 = b$, which clearly depends on the Legendre symbol of $b$.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.