2

Let $GR(p^2,m)$ be the Galois ring with $p^{2m}$ elements and characteristic $p^2$. Let $Z^m_{p^2}$ be the cross product of $m$ copies of $Z_{p^2}$ which is the set of integers from zero up to $p^2-1$.

Let $a \in GR(p^2,m)$ where $p$ is an odd prime and $m$ is a positive integer. As $a$ varies over $GR(p^2,m)$, consider the set of values $ A=\left\{\sum_{x \in GR(p^2,m)}w^{Tr(ax)} \right\}$ where $w=e^{2\pi i/p^2}$, $Tr:GR(p^2,m)\rightarrow GR(p^2,1)$ is the trace function.

Let $b \in Z^m_{p^2}$ where $p$ is an odd prime. As $b$ varies over $Z^m_{p^2}$, consider the set of values $ B=\left\{\sum_{x \in Z^m_{p^2}}w^{b \cdot x} \right\}$ where $w=e^{2\pi i/p^2}$, $b \cdot x$ is the classical dot product of $b$ and $x$. Are the sets $A$ and $B$ equal?

Considering finite fields, the answer would be yes. However, there are zero divisors here which is confusing.

Many thanks in advance.

Math_D
  • 335
  • 1
    Something is a bit wrong. If $a=0$ you get $w^{Tr(ax)}=1$ for all $x$, so the first sum is $p^{2m}$. In the second sum there are only $p^2$ terms. Is the range of summation perhaps supposed to be $\mathbb{Z}_{p^2}^m$? – Jyrki Lahtonen Mar 26 '13 at 12:23
  • What is $,GR(p^2,m),$ ? Regular matrices $,m\times m,$ over $,\Bbb F_{p^2},$ ?...Galois Rings? – DonAntonio Mar 26 '13 at 12:31
  • $GR(p^2,m)$ is the Galois ring with $p^{2m}$ elements and characteristic $p^2$. $Z^m_{p^2}$ is the cross product of $m$ copies of $Z_{p^2}$ which is the set of integers from zero up to $p^2-1$. I will add this to question. Thanks. – Math_D Mar 26 '13 at 12:36
  • 2
    @DonAntonio: I've created a tag for Galois rings and put a short explanation to the tag wiki. – azimut Mar 26 '13 at 12:48
  • 1
    The question is now ok. But the answer is trivial in the sense that the sum is equal to $p^{2m}$, if $a=0$ (resp. $b=0$), and the sum is equal to zero, if $a\neq0$ (resp. $b\neq0$). What I was implying yesterday is more accurately written as follows. Pick an automorphism of additive groups $\phi: GR(p^2,m)\to \mathbb{Z}{p^2}^m$. Then we get two sets of additive characters on $GR(p^2,m)$. One set consists of the functions $\psi_a(x)=w^{Tr(ax)}$ with $a$ ranging over $GR(p^2,m)$. The other set consists of the functions $\chi_b(x)=w^{\phi(x)\cdot b}$ with $b$ ranging over $\mathbb{Z}{p^2}^m$. – Jyrki Lahtonen Mar 26 '13 at 13:19
  • 1
    (cont'd): The exercise is to show that there is another bijection $g:GR(p^2,m)\to\mathbb{Z}{p^2}^m$ such that $\psi_a=\chi{g(a)}$. Sorry about not making this clear right away. I have done this too many times, so I tend to rush through several details :-). Have fun! – Jyrki Lahtonen Mar 26 '13 at 13:22
  • 1
    @JyrkiLahtonen: Why not make an answer out of this very nice explanation? – azimut Mar 26 '13 at 13:37

0 Answers0