Let $K$ be a local field. Let $\mathcal{O}_K$ be its ring of integers. Let $f(x)\in \mathcal{O}_K[x]$ be monic and irreducible. Consider the extension $L=K[x]/ (f(x))$. Then $L$ is a local field with (unique) absolute value extending $|\cdot|_K$. Clearly $\mathcal{O}_K[x]/(f(x))\subset \mathcal{O}_L$, but I am trying to show we have equality.
Suppose $\alpha$ is a root of $f(x)$. Why can't we have something like $\bigg|\sum_{i=0}^{n-1} a_i\alpha^i\bigg|_L\leq 1$ when not all $a_i$ satisfy $|a_i|_K\leq 1$?
Best Answer
Let $O_K$ be a complete DVR with finite residue field $O_K/(\pi_K)=\Bbb{F}_q$, let $L/K$ a finite extension, $O_L/(\pi_L) = \Bbb{F}_{q^f}$. The valuation extends uniquely to $O_L$, by Hensel lemma $\zeta_{q^f-1} \in O_L$. The ramification index is $e= v(\pi_K)/v(\pi_L)$.
$O_L$ is complete, its completion being $$\overline{O_L}=\{ \sum_{n\ge 0} c_n \pi_L^n, c_n\in \{\zeta_{q^f-1}^j\}\cup 0\}= \sum_{m=0}^{e-1}\pi_L^m \sum_j \zeta_{q^f-1}^j O_K\subset O_L$$ Thus $$O_L= O_K[\zeta_{q^f-1},\pi_L]$$
Let $$\alpha = \zeta_{q^f-1} (1+\pi_L)$$ Being a finitely generated $O_K$-module $O_K[\alpha]$ is complete too, and since $v( \alpha^{q^f-1}-1) =v(\pi_L)$ we get that $$O_L = \{ \sum_{n\ge 0} c_n ( \alpha^{q^f-1}-1)^n, c_n\in \{\alpha^j\}\cup 0\}=\sum_{m=0}^{e-1} ( \alpha^{q^f-1}-1)^m\sum_j \alpha^j O_K= O_K[\alpha]$$