Consider a quotient ring $\dfrac{F_p[x]}{(f)}$, where $p$ is prime. I want to find all the possible orders in this ring.
I know that with given factorization of $f(x) = f_1(x)^{k_1} * ... * f_n(x)^{k_n}$ we are able to represent this ring as a direct product $\dfrac{F_p[x]}{(f)}^{\times} = \dfrac{F_p[x]}{(f_1^{k_1})}^\times \times\cdots\times \dfrac{F_p[x]}{(f_n^{k_n})}^\times$. Then I came up with the result that the orders I am interested in are LCM's of all the possible orders in these "little" quotient rings.
When $k_i = 1$ it's trivial, since the order is $p^{deg(f_i)} - 1$ and for all the divisors of that number there's an element of this order. However, when I tested some cases with $k_i \ne 1$ I totally lost it. I figured out(with simple observation) that $|\dfrac{F_p[x]}{(f_n^{k_n})}^{\times}| = p^{deg(f_i) * (k_n-1)} * (p^{deg(f_i)} - 1)$ (also I didn't find any confirmation of that fact so I might be wrong). But in that very case not all the divisors have the element of a particular order. I guess something like the cyclic groups modulo $p^k$ is going on here, but still I was not able to find any info.
Also I'm interested in the case when $p = 2$. There's always some unusual behavior in binary fields, so it would be nice if you can help me with an answer or maybe with some links to the literature.
Edit 1
Well, I was playing with it for several hours and now I have a feeling that the orders of all elements in the case $k_i=2$ are the divisors of square freed order of the ring. It would be nice to know why it happens.
Well, after two days of research I have finally found the answer. Here's the related question and absolutely fascinating answer Multiplicative group modulo polynomials.
We got
$\dfrac{F_p[x]}{(f)}^{\times} \cong \dfrac{F_p[x]}{(f_1^{k_1})}^\times \times\cdots\times \dfrac{F_p[x]}{(f_m^{k_m})}^\times$.
Let $deg(f_i) = n_i$.
According to the question above, each $K_i = \dfrac{F_p[x]}{(f_i^{k_i})}^\times \cong Z_{p^{n_i} - 1}\times \displaystyle\prod_{i \ge 0}^{k_i \gt p^i}Z_{p^{i+1}}^{n_i(\lceil \dfrac{k_i}{p^i} \rceil - 2 \lceil \dfrac{k_i}{p^{i + 1}} \rceil + \lceil \dfrac{k_i}{p^{i+2}} \rceil)} $
In my very question I do not need all the corresponding powers of $Z_{p^i}$, but I need the highest nonzero of them.
So the elements in $K_i$ may be represented as vector $(a_0, a_1, a_2, ..., a_{r+1})$, where r is the highest power of $p$ s.t. $k_i > p^r$
Note that some powers may be zero, as in the case:
$\dfrac{F[x]}{(x+1)^6} \cong Z_1 \times Z_2^2 \times Z_8$ and the elements will be of the form $(a_0, (a_{1_1}, a_{1_2}), a_3)$
However it doesn't really matter, since the last term will always be with power $\ge n_i$:
Since $p^r \lt k_i \le p^{r + 1} \implies 1 \lt \dfrac{k_i}{p^r} \le p \implies 2 \le \lceil \dfrac{k_i}{p^r} \rceil \le p,\ a = \lceil \dfrac{k_i}{p^r} \rceil$
Hence
$\lceil \dfrac{k_i}{p^r} \rceil - 2 \lceil \dfrac{k_i}{p^{r + 1}} \rceil + \lceil \dfrac{k_i}{p^{r + 2}} \rceil = a - 2 + 1 = a - 1 \ge 1$
So all the possible orders of element $(a_0, a_1, ..., a_r)$ are the divisors of $LCM(p^{n_i} - 1, p, p^2,..., p^{r+1}) = LCM(p^{n_i} -1, p^{r+1})$, where $r = \lfloor log_p(k_i) \rfloor$(well sometimes when r is the power of p then we have to choose r - 1)
The rest is clear too: all the possible orders in $\dfrac{F_p[x]}{(f)}^{\times}$ are all the divisors of $LCM(LCM(p^{n_1} -1, p^{r_1+1}),...,LCM(p^{n_m} -1, p^{r_m+1}))$
and as far as I'm sure it's the same as
$LCM(p^{n_1} -1, ..., p^{n_m} -1, p^{r_1+1},..., p^{r_m+1})$, where $r_i = \lfloor log_p(k_i) \rfloor$
And even nicer:
$LCM(p^{n_1} -1, ..., p^{n_m} -1, p^{R+1}) = p^{R+1}LCM(p^{n_1} -1, ..., p^{n_m} -1)$, since $gcd(p, p^k - 1) = 1$. where $R = Max_i(\lfloor log_p(k_i) \rfloor)$