# Are the sets $\left\{\sum_{x \in \operatorname{GR}(p^2,m)}w^{Tr(ax)} \right\}$ and $\left\{\sum_{x \in Z^m_{p^2}}w^{b \cdot x} \right\}$ equal?

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.

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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
@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
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
(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
@JyrkiLahtonen: Why not make an answer out of this very nice explanation? –  azimut Mar 26 '13 at 13:37
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