My question may be not interesting or easy to answer ! but I am really not familiar with proba.
Let $p$ be an odd prime number. and let $r\geq1$ an integer. choose an element $A\in\mathrm{GL}_2(\mathbb{Z}/p^{r+1}\mathbb{Z})$
What is the probability for which $Tr(A)\equiv p^{r}\pmod{p^{r+1}}$ and
$Tr(A)\not\equiv 0\pmod{p^{r+1}}$ ?
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5$\begingroup$ Not really a probability question: it's counting solutions for some Diophantine equations mod $q$. Also $F_q$ (or ${\bf F}_q$) usually denotes a finite field of $q$ elements, which is not the same as the integers mod $q$ once $q=p^e$ with $e>1$. $\endgroup$– Noam D. ElkiesCommented Mar 14, 2017 at 18:04
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$\begingroup$ Do you want to know exact or approximate probabilities? The latter is quite easy, the former not so much. $\endgroup$– Igor RivinCommented Mar 14, 2017 at 18:23
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1$\begingroup$ @IgorRivin both if possible $\endgroup$– Zakariae.BCommented Mar 14, 2017 at 18:28
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2$\begingroup$ I removed the tags "probability" and "probabilistic number theory" and "group theory" and "finite groups", because your question did not belong to these subjects (even though it is formulated in terms of matrices, probability, and congruences). I added the tag "number theory" though, because your question is purely number theoretic. $\endgroup$– GH from MOCommented Mar 14, 2017 at 21:49
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1$\begingroup$ @GHfromMO Thanks for your understanding :) I actually added a group tag as I'm myself interested by questions pertaining in any respect to group theory. Actually for a question on groups (such as counting), it's not always clear a priori if using the group structure will help. By the way I also often remove tags when they seem off-topic or too narrow, especially when I expect the potentially interested users to be already in the scope of another tag. $\endgroup$– YCorCommented Mar 14, 2017 at 22:25
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Total number of elements in $$\bigl|\mathrm{GL}_2(\mathbb{Z}/p^{r+1}\mathbb{Z})\bigr|=p^{4r}(p^2-1)(p^2-p).$$ Let us count the number of matrices with a given trace $s$. It is $$\sum_{a\in \mathbb{Z}/p^{r+1}\mathbb{Z}} \bigl|(c,d):p\, \text{does not divide}\, a(s-a)-cd\bigr|.$$ Clearly it depends only on the remainder of $s$ modulo $p$. For $s$ divisible by $p$, we have $p^{2r}(p-1)^2$ pairs $(c,d)$ for any $a$ divisible by $p$ and $p^{2r+2}-(p^{r+1}-p^r)$ pairs for $a$ not divisible by $p$. The total number is $$p^{3r}(p-1)^2+(p^{2r+2}-(p^{r+1}-p^r))(p^{r+1}-p^r)\\=(p^{r+1}-p^r)(p^{2r}(p-1)+p^{2r+2}-p^{r+1}+p^r).$$ So, for any given remainder $s$ modulo $p^{r+1}$, which is divisible by $p$, the probability that a random matrix from $\mathrm{GL}_2(\mathbb{Z}/p^{r+1}\mathbb{Z})$ has a trace equal to $s$, equals $$\frac{p^{r}(p-1)+p^{r+2}-p+1}{p^{2r+1}(p^2-1)}.$$
answered Mar 14, 2017 at 18:33