Euler's phi function for polynomials over a finite field. I have recently learned about $\mathbb{\Phi}_q$ an analog of Euler's  phi function for polynomials $f \in \mathbb{F}_q[x]$.  I know that for $f \in \mathbb{F}_q[x]$  the function $\mathbb{\Phi}_q(f)$ is defined to be the number of polynomials in $\mathbb{F}_q[x]$ of degree less than that of $f$ and relatively prime to $f$.  I also know how to compute this value.
If I look at the set of polynomials that $\mathbb{\Phi}_q(f)$ counts, then I should have an abelian group of order $\mathbb{\Phi}_q(f)$ where the operation is polynomial multiplication modulo $f$.  Is there a way to know the structure of this group analogous to the way we know the structure of the multiplicative group of units of an integer $n \in \mathbb{Z}$.  In particular, can anyone tell me the structure of the group of order 24 consisiting of the polynomials of degree less than 6 and relatively prime to $x^6 - 1 \in \mathbb{F}_2[x]$. 
 A: The group you are refering to is nothing but the group of units in the quotient ring $R = \Bbb F_q[x]/(f)$, i.e. $G = \left(\Bbb F_q[x]\right/(f))^\times$. This is similar to the case of integers, i.e. you may compare $R$ to $\Bbb Z/n\Bbb Z$ and $G$ to $\left(\Bbb Z/n\Bbb Z\right)^\times$.
To understand the group $G$, you should first understand the structure of the quotient ring $R$. Let $f = f_1^{r_1}\dotsc f_k^{r_k}$ be the decomposition of $f$ into irreducible polynomials. By Chinese remainder theorem, we have $R = \prod_{i = 1}^k \Bbb F_q[x]/(f_i^{r_i})$. Hence it suffices to consider the case $R = \Bbb F_q[x]/(f^r)$, where $f$ is irreducible.
To study the ring $R$ in this case, note that there is a canonical surjective ring homomorphism $\rho: R \rightarrow \Bbb F_q[x]/(f)$, whose kernel $K$ is a nilpotent ideal of $R$. The quotient $\Bbb F_q[x]/(f)$ is, of course, just the finite field $\Bbb F_{q^d}$, where $d = \deg f$.
Since $K$ is nilpotent, we know that $R^\times$ is simply $\rho^{-1}(\Bbb F_{q^d}^\times)$ (for $x\in R$ such that $\rho(x) \neq 0$, there is a $y\in R$ such that $\rho(y) = \rho(x)^{-1}$, hence $\rho(xy) = 1$ and $xy = 1 + k$ for some $k \in K$. Therefore $xy \in R^\times$ because $k$ is nilpotent, and we have $x \in R^\times$).
Now $\rho$ induces a surjective homomorphism of groups $\tau: R^\times \rightarrow \Bbb F_{q^d}^\times$, with kernel $\ker(\tau) = 1 + K$. Since $K = f\Bbb F_q[x]/(f^r)$ is a vector space of dimension $r - 1$ over $\Bbb F_{q^d}$, we see that $1 + K$ is a group of $p$-power order, where $p$ is the characteristic of $\Bbb F_q$.
On the other hand, $\Bbb F_{q^d}^\times$ is a cyclic group of order $q^d - 1$, which is prime to $p$. We conclude that $R^\times$ is isomorphic to the direct product of $\Bbb F_{q^d}^\times$ and $1 + K$.
It thus only remains to study the structure of the $p$-group $1 + K$. For this, it suffices to look at the subgroups $(1 + K)^{p^m}$ for all $m$.
Since we are in characteristic $p$, we have $(1 + K)^{p^m} = 1 + K^{p^m} = 1 + f^{p^m}\Bbb F_q[x]/(f^r)$. This has dimension $r - p^m$ over $\Bbb F_{q^d}$, and hence has dimension $ds(r - p^m)$ over $\Bbb F_p$, if we write $q = p^s$.
These informations totally determine the structure of $1 + K$ as an abelian group.

To work out your example of $x^6 - 1$ in $\Bbb F_2[x]$:
We first factorize $x^6 - 1 = (x + 1)^2(x^2 + x + 1)^2$.
For $\Bbb F_2[x]/((x+1)^2)$: the quotient $\Bbb F_2[x]/(x + 1)$ is just $\Bbb F_2$, so its group of units is trivial; the kernel $1 + K$ has $2$ elements, hence is isomorphic to $\Bbb Z/2\Bbb Z$.
For $\Bbb F_2[x]/((x^2 + x + 1)^2)$: the quotient $\Bbb F_2[x]/(x^2 + x + 1)$ is $\Bbb F_4$, so its group of units is isomorphic to $\Bbb Z/3\Bbb Z$; the kernel $1 + K$ has $4$ elements, and $(1 + K)^2$ is trivial, so $1 + K$ is isomorphic to $(\Bbb Z/2\Bbb Z)^2$.
In summary, we have $(\Bbb F_2[x]/(x^6 - 1))^\times \simeq (\Bbb Z/ 2\Bbb Z)^3 \times \Bbb Z / 3\Bbb Z$.
