Multiplicative group modulo polynomials When working over $\mathbb{Z}$, it is well known what the structure of the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^{\times}$ exactly is:
If $n=p$ for a prime $p$, then $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is the $p-1$ cyclic group $C_{p-1}$.
If $n=p^e$ for an odd prime $p$, then $(\mathbb{Z}/p^e\mathbb{Z})^{\times}$ is again a cyclic group $C_{\phi(p^e)}=C_{p^e-p^{e-1}}$.
If $n=2^e$ for $e \geq 2$, then $(\mathbb{Z}/2^e\mathbb{Z})^{\times}$ is no longer a cyclic group, but a product of two cyclic groups. It is of the form $C_{2} \times C_{2^{e-2}}$.
For a more general integer $n$, we can simply use the Chinese remainder theorem to obtain the precise structure of $(\mathbb{Z}/n\mathbb{Z})^{\times}$ using the above results.
My question is: 

Is there a similar result for the multiplicative group of a quotient ring of $\mathbb{F}_{p}[X]$ where $\mathbb{F}_{p}$ is the finite field of $p$ elements (where $p$ is a prime power)? That is, what is the structure of 
  $$(\mathbb{F}_{p}[X]/\left<f(X)\right>)^{\times}$$
  for a polynomial $f(X) \in \mathbb{F}_{p}[X]$? 

If $f(X)$ is irreducible, then this group is again cyclic (since we know that the multiplicative group of a finite field is cyclic). But what about the more general case? Again, we can use Chinese remainder theorem to reduce the general question to the case of $f(X)$ being a power of an irreducible. So what do we know about the structure of
$$(\mathbb{F}_{p}[X]/\left<f(X)^e\right>)^{\times}$$
for an irreducible $f(X) \in \mathbb{F}_{p}[X]$ and $e \geq 2$?
 A: what do we know about the structure of
$({\mathbb F}_p[X]/\langle f(X)^e\rangle)^{\times}$
for an irreducible $f(X)\in \mathbb F_p[X]$ and $e\geq 2$?
Assuming that $\deg(f) = n$, 
I claim that
$({\mathbb F}_p[X]/\langle f(X)^e\rangle)^{\times}$
is isomorphic to
$$
\mathbb Z_{p^n-1}\times \left(
\mathbb Z_{p}^{n(e-2\lceil\frac{e}{p}\rceil+
\lceil\frac{e}{p^2}\rceil)}\times 
\mathbb Z_{p^2}^{n(\lceil\frac{e}{p}\rceil
-2\lceil\frac{e}{p^2}\rceil+
\lceil\frac{e}{p^3}\rceil)}\times 
\mathbb Z_{p^3}^{n(\lceil\frac{e}{p^2}\rceil
-2\lceil\frac{e}{p^3}\rceil+
\lceil\frac{e}{p^4}\rceil)}\times \cdots\right).
$$
The product looks infinite, but the exponents on the
factors eventually become zero.
For example, the group
$\left(\mathbb F_3[x]/\langle(x^2+1)^{13}\rangle\right)^{\times}$ is
isomorphic to $\mathbb Z_8\times \left(\mathbb Z_3^{10}\times
\mathbb Z_9^4\times \mathbb Z_{27}^2\right)$.

Let $R = ({\mathbb F}_p[X]/\langle f(X)^e\rangle)$ and
$J = \langle f(X)\rangle$.
Reduction modulo $J$ yields a homomorphism of $R^{\times}$
onto $(R/J)^{\times}$ with kernel $1+J$.
The image $(R/J)^{\times}$ is isomorphic to
$\mathbb F_{p^n}^{\times}\cong \mathbb Z_{p^n-1}$,
which has order that is relatively prime to the order
of the kernel $|1+J| = |J| = p^{n(e-1)}$.
Thus
$$R^{\times} \cong (R/J)^{\times}\times (1+J)\cong
\mathbb Z_{p^n-1}\times (1+J).$$
It remains to determine the structure of the $p$-group
$(1+J)$.
The structure of a finite abelian $p$-group $A$ is determined
by the orders of the annihilators
$$
A[p^k]:= \{a\in A\;|\;p^ka = 0\}.
$$
These orders are not so hard to determine
when $A = 1+J$.
That is, for $\alpha\in J$, the element
$1+\alpha\in 1+J$ is annihilated by $p^k$ exactly
when $1=(1+\alpha)^{p^k} = 1+\alpha^{p^k}$, which happens
exactly when $f^e$ divides $\alpha^{p^k}$, which
happens exactly when $\alpha\in \langle f^{\lceil\frac{e}{p^k}\rceil}\rangle$.
Thus
$$|A[p^k]| = |\langle f^{\lceil\frac{e}{p^k}\rceil}\rangle|
= p^{n\left(e-\lceil\frac{e}{p^k}\rceil\right)}.$$
What remains is to determine the $m_i$'s in the expression
$$
B = \mathbb Z_p^{m_1}\times 
\mathbb Z_p^{m_2}\times 
\mathbb Z_p^{m_3}\times \cdots
$$
if this general abelian $p$-group has the same size annihilators as $A=1+J$;
i.e., if $|B[p^k]| = |A[p^k]|$ for all $k$.
One computes that
$$
\begin{array}{rl}
|B[p]| &= p^{m_1+m_2+m_3+m_4+\cdots}\\ 
|B[p^2]| &= p^{m_1+2m_2+2m_3+2m_4+\cdots}\\ 
|B[p^3]| &= p^{m_1+2m_2+3m_3+3m_4+\cdots}\\
&\textrm{ETC.}
\end{array}$$ 
Thus we need to solve
the equations
$$
\begin{array}{rl}
m_1+m_2+m_3+m_4+\cdots&=n\left(e-\lceil\frac{e}{p}\rceil\right)\\
m_1+2m_2+2m_3+2m_4+\cdots&=n\left(e-\lceil\frac{e}{p^2}\rceil\right)\\
m_1+2m_2+3m_3+3m_4+\cdots&=n\left(e-\lceil\frac{e}{p^3}\rceil\right)\\
\textrm{ETC.}&
\end{array}
$$
The solution is $m_i =
n\left(
\lceil\frac{e}{p^{i-1}}\rceil-2\lceil\frac{e}{p^i}\rceil
+\lceil\frac{e}{p^{i+1}}\rceil
\right)$. Therefore
$$
\begin{array}{rl}
({\mathbb F}_p[X]/\langle f(X)^e\rangle)^{\times}&=R^{\times}\\
&\cong (R/J)^{\times}\times (1+J)\\
&\cong \mathbb Z_{p^n-1}\times \left(
\mathbb Z_{p}^{n(e-2\lceil\frac{e}{p}\rceil+
\lceil\frac{e}{p^2}\rceil)}\times 
\mathbb Z_{p^2}^{n(\lceil\frac{e}{p}\rceil
-2\lceil\frac{e}{p^2}\rceil+
\lceil\frac{e}{p^3}\rceil)}\times 
\mathbb Z_{p^3}^{n(\lceil\frac{e}{p^2}\rceil
-2\lceil\frac{e}{p^3}\rceil+
\lceil\frac{e}{p^4}\rceil)}\times \cdots\right).
\end{array}
$$
