How to find unit group of $\Bbb Z[x]/(nx)$? I am trying to work out the group structure of the group of units of $$\Bbb Z[x]/(nx)$$
where $n\ge 2$ is an integer.
So finding the units isn't the hard part, they will be of the form $u+t$ where $u$ is a unit of $\Bbb Z$ and $t$ is a nilpotent element of the polynomial ring. The difficult part is trying to work out how they multiply.
As far as I can tell these groups don't seem to have any torison, which is probably why I couldn't understand their structure.
Is there any way to determine what the group structure might be?
 A: In fact the unit group is entirely torsion, and it can be decomposed as an infinite direct sum of finite groups.
Write $r$ for the radical of $n$, that is, $r$ is the product of the distinct primes dividing $n$. Consider the group homomorphism
\begin{align*}
\varphi:\{\pm 1\}\times\bigoplus_{i=1}^\infty \frac{\mathbb{Z}}{\frac{n}{r}\mathbb{Z}}&\to \big(\mathbb{Z}/(nx)\big)^\times,\\
\big(\epsilon,(a_i)\big)&\mapsto \epsilon\prod_{i=1}^\infty (1+rx^i)^{a_i}.
\end{align*}
This map is an isomorphism. I will sketch a proof.
First, the product in the definition of $\varphi$ converges as all but finitely many terms are $1$. To check that this is well-defined amounts to $(1+rx^i)^{n/r}\equiv 1\mod nx$, which is not too hard to prove by playing with binomial coefficients. The map $\varphi$ is injective, as $\epsilon$ can be read off of the constant term, and $\min\{i:a_i\neq 0\}=\min\{i:\varphi(\epsilon,(a_i))[x^i]\neq 0\}$.
For surjectivity, suppose
$$
f(x)=\sum_{i=0}^k b_i x^i\in \big(\mathbb{Z}/(nx)\big)^\times.
$$
This means $b_0\in\{\pm 1\}$ and $r|b_i$ for $i\geq 1$. I claim that for every $m$, there exists an element $A_m$ such that $\varphi(A_m)\equiv f(x)\mod r^m x$. The proof is induction on $m$. For $m=1$, we can take $A_1=(b_0,(0))$. Now assume the claim holds for $m$, and let $A_m$ be such that $\varphi(A_m)\equiv f(x)\mod r^m x$. Then
$$
\frac{f(x)}{\varphi(A_m)}=1+\sum_{i} c_i x^i,
$$
with $r^m|c_i$. It is an exercise in binomial coefficient to show that we can take
$$
A_{m+1}=A_m\cdot\left(1,\left(\frac{c_i}{r}\right)\right).
$$
There is some positive integer $m$ such that $n|r^m$, so we have $\varphi(A_m)=f(x)\in\big(\mathbb{Z}[x]/(nx)\big)^\times$. This proves that $\varphi$ is surjective.
