Let $K$ be a non-Archimedean local field, so either a finite extension of $\mathbb{Q}_p$ or a finite extension of $\mathbb{F}_q((t))$. Let $\mathcal{O}$ denote its ring of integers and $\pi$ a uniformizer.
Is there a simple description of the finite ring $\mathcal{O}/ \pi^k \mathcal{O}$, for example in terms of the degree of the finite extension?
The case of $\mathbb{Q}_p$ giving $\mathbb{Z}/p^k\mathbb{Z}$ is what I would call very simple, and the case of $\mathbb{F}_q((t))$ giving $\mathbb{F}_q[t] / t^n$ is what I would call simple. But I have no intuition about how the case of finite extensions of these two fields could look like.
Best Answer
$F=Q_p$ or $F_p((t))$, $$O_K/(\pi_K^k) = \{ \sum_{m=0}^{k-1} \pi_K^m s_m, s_m \in S\}$$ where $S$ are representatives of the residue field, from which the $O_F/(\pi_F^{\lceil k/e \rceil})$-module structure follows easily (free module if $e | k$, non-free otherwise).
The ring structure is less obvious, it depends on the minimal polynomial of $\pi_K$. $$O_K = O_F[\zeta_{q-1}][\pi_K]=O_F[x]/\psi(x)[y]/(f(x,y))$$ For $k$ large enough then $$O_K/(\pi_K^k)=O_F/(\pi_F^{\lceil k/e \rceil})[x]/\psi(x)[y]/(f(x,y))$$
When $p\nmid e$ (tamely ramified) then $f(x,y) = y^e-\pi_F y^r$ for some $r$.
(If $e\nmid k$ then quotient by $y^k$)