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Let $R$ be an arbitrary ring with maximal ideal $I$, such that the quotient ring $\frac{R}{I}$ is finite. For each integer $i$, we define the quotient ring number as below: $$N_i(R,I)=\sum_{a\in \frac{R}{I}}{a^i}.$$

It is easy to see that if $\frac{R}{I}=\mathbb{F}_q$ be a finite field with $q$ elements, then we have two case for $N_i(R,I)$. Actually, if $i$ is not a multiple of $q-1$, this summation is zero and if $i$ is a multiple of $q-1$, this summation is $-1$. So, for commutative ring with maximal ideal $I$, we have: $$N_i(R,I)\in \{0,-1\}.$$

Now, my questions is:

Is there a non-commutative ring $R$ with maximal ideal $I$ such that we have $N_i(R,I)\in \{0,-1\}$?

Also, for arbitrary non-commutative ring $R$ with maximal ideal $I$, what can be the maximum number of elements of the set $\{N_i(R,I); i\in \mathbb{N}\}$?

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    $\begingroup$ $R/I$ is a finite simple ring and hence isomorphic to $M_n(\mathbb{F}_q)$ for some $n \geq 1$ and some finite field $\mathbb{F}_q$. Now one can probably make a direct calculation. It seems to be always $\{0,-1\}$. $\endgroup$
    – HeinrichD
    Nov 23, 2016 at 13:23
  • $\begingroup$ Does $N_i(R,I)$ depends on $R$ and $I$, or only on the quotient $R/I$? $\endgroup$
    – Ehud Meir
    Nov 23, 2016 at 14:49
  • $\begingroup$ @EhudMeir for the commutative ring it does not depend on $R$ and $I$. But, if HeinrichD's answer be true, $N_i(R,I)$ does not depend on $R$ and $I$. I do not know the answer now. $\endgroup$
    – Shahrooz
    Nov 24, 2016 at 15:34
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    $\begingroup$ In any case: the definition of $N_i(R,I)$ seems to depend only on the quotient, regardless of the commutativity of $R$. $\endgroup$
    – Ehud Meir
    Nov 25, 2016 at 12:11

1 Answer 1

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The definition of $N_i(R,I)$ depends only on $R/I$, which is a ring isomorphic to $M_n(q)$ as HeinrichD has already observed. Thus, the question is: what is the sum $$ S = \sum_{A\in M_n(q)} A^i \quad? $$ The answer that has been proposed is: either $0$ or $-I$.

This answer is correct. It is proved in

Brawley, J. V.; Carlitz, L.; Levine, J. Power sums of matrices over a finite field. Duke Math. J. 41 (1974), 9-24.

that, if $n>1$, the sum $S$ is $0$ unless $n=q=2$, $i>1$, and $i\equiv -1, 0, 1\pmod{6}$, in which case $S=-I=I$ is the $2\times 2$ identity matrix over the $2$-element field.

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  • $\begingroup$ Thanks for your good answer and the reference. I am reading the paper and it is so nice dear Keames. $\endgroup$
    – Shahrooz
    Nov 27, 2016 at 9:30

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