Group structure of local residue ring For example, let $A=\mathbb Z[\sqrt2]$ and consider the residue ring $A/8A$.
The elements prime to $8A$ (hence, prime to $(\sqrt2)$) form a group $(A/8A)^*$ by multiplication. By overall research, there are 32 elements, the largest order is 8, and there are 8 elements of order 2. So I got $(A/8A)^* = \mathbb Z/2\mathbb Z×\mathbb Z/2\mathbb Z×\mathbb Z/8\mathbb Z$.
How about $(B/8B)^*$ where $B=\mathbb Z[(-1+\sqrt{-3})/2]$? Is there any good way without overall research?
EDIT(2022/5)
I learned some useful facts about this topic.
"Local Fields and Their Extensions" was a nice reference.
Let $e$ be ramification index, $f$ be residue index, and $r$ be the maximal integer such that K has primitive $p^r$-th root of $1$.
The structure of the multiplicative group of principal units is
$\mathbb{Z}/p^r\mathbb{Z} \times (\mathbb{Z}_p)^{ef}$
Let $\pi$ be prime element, $U_k = \{x|\ x-1 \in (\pi)^k \}$, $V_k = U_k-U_{k+1}$, and $x \in V_k$.
(1) If $k<\frac{e}{p-1}$, then $x^p \in V_{pk}$.
(2) If $k>\frac{e}{p-1}$, then $x^p \in V_{k+e}$.
(3) If $k=\frac{e}{p-1}$, then $x^p \in U_{k+e}$.
I could obtain the structure of $(A/8A)^*$ using these facts.
For example I can know the order of $a \in V_1$ is $8$ because in this case $e=2, f=1$ and $a^2 \in V_2, a^4 \in V_4$ are not in $(8)=U_6$ and $a^8 \in V_8 \subset U_6$.
However, I still cannot obtain that of $(B/8B)^*$ theoretically. I cannot evaluate the order of $a \in V_1$ because this time $e=1$ so it is the third case above, where I cannot evaluate the least $k$ such that $a^2 \in U_k$.
 A: The unramified case is relatively easy.
Let $K$ be a number field, $\mathfrak{p}$ a prime above $p$ of $O_K$. We assume that $p\not \in \mathfrak{p}^2$.
We consider the $\mathfrak{p}$-adic completion $R = \varprojlim O_K/\mathfrak{p}^n$,
as $Frac(R)/\Bbb{Q}_p$ is unramified then $(p)$ is prime in $R$.
$(1+p R)_{tors} = 1$ or $\pm 1$ depending on if $p=2$.
$G=(1+p R)/(1+p R)_{tors}$ is a finitely generated torsion-free $\Bbb{Z}_p$-module, so it is a free $\Bbb{Z}_p$-module, whence $G\cong (\Bbb{Z}_p)^f$ where $f=[Frac(R):\Bbb{Q}_p]$ (ie. $O_K/\mathfrak{p}\cong \Bbb{F}_{p^f}$)
Any $\Bbb{F}_p$-basis $\beta_1,\ldots,\beta_f$ of $G/G^p$ gives a $\Bbb{Z}_p$-basis of $G$ and $1+pR = (1+p R)_{tors} \times \prod_{j=1} ^f \beta_j^{\Bbb{Z}_p}$.
Write $R/(p) = \sum_{j=1}^f \alpha_j \Bbb{F}_p$ where $\alpha_1=1$ (we can take $\alpha_j = \zeta_{p^f-1}^{j-1}$)

*

*If $p\ne 2$ then $$1+p R= \prod_{j=1}^f (1+\alpha_j p)^{\Bbb{Z}_p}, R^\times = \langle \zeta_{p^f-1}\rangle \times 1+p R$$ $$O_K/(\mathfrak{p}^m)^\times=R/(p^m)^\times=
\langle \zeta_{p^f-1}\rangle \times \prod_{j=1}^f (1+\alpha_j p)^{\Bbb{Z}_p/(p^{m-1})}$$


*If $p=2$ then
$$1+p R=\pm 5^{\Bbb{Z}_p}\prod_{j=2}^f (1+\alpha_j p)^{\Bbb{Z}_p}$$
$O_K/\mathfrak{p}^\times=\langle \zeta_{p^f-1}\rangle$ and for $m\ge 2$
$$O_K/(\mathfrak{p}^m)^\times=R/(p^m)^\times=
\pm \langle \zeta_{p^f-1}\rangle\times 5^{\Bbb{Z}_p/p^{m-2}}\times \prod_{j=2}^f (1+\alpha_j p)^{\Bbb{Z}_p/(p^{m-1})}$$

If $\mathfrak{p}$ is ramified then it is much trickier. Fortunately, for $K$ fixed, this holds for only finitely many $p$.

For any ideal $I=\prod_i \mathfrak{p}_i^{d_i}$, $O_K/I^\times \cong \prod_i O_K/(\mathfrak{p}_i^{d_i})^\times$.
