Let $A$ be a Dedekind ring, $K=\operatorname{Frac}A$, $L/K$ a finite separable extension and $B$ the integral closure of $A$ in $L$. (Perhaps we should suppose that $K$ is a global field.)
Each non-zero prime ideal $\mathfrak{p}\in \operatorname{Spec}A$ determines an absolute value $|\cdot|_{\mathfrak{p}}$ of $K$ and each prime ideal $\mathfrak{P}\in\operatorname{Spec}B$ over $\mathfrak{p}$ determines an absolute value $|\cdot|_{\mathfrak{P}}$ of $L$ which extends $|\cdot|_{\mathfrak{p}}$. We denote $\hat{K}$ and $\hat{L}$ the completions of $K$ and $L$ with respect to their respective absolute values.
We also denote $\hat{A}$ the valuation ring of $\hat{K}$ (which coincides with the completion of $A$ and the closure of $A$ in $\hat{K}$) and $\hat{B}$ the valuation ring of $\hat{L}$ (which coincides with the completion of $B$, the closure of $B$ in $\hat{L}$ and the integral closure of $\hat{A}$ in $\hat{L}$).
If $L=K(\theta)$ for some $\theta\in L$, I understand that $\hat{L}=\hat{K}(\theta)$. But I also know that $\hat{B}=\hat{A}[\alpha]$ for some $\alpha$. I wonder if we can take $\alpha=\theta$. If not, how can we calculate $\hat{B}$? I'm having trouble calculating it even in the simplest cases (for example when $A=\mathbb{Z}$ and $L=\mathbb{Q}(i)$).
Best Answer
$L=K(\theta)$ an finite extension of number field.
A completion $L_Q$ above $K_P$ gives an embedding $h:K(\theta)\to \overline{K_P}$, it maps $\theta$ to a root $h(\theta)$ of $f\in K[x]$ the minimal polynomial of $\theta$.
$h(L)=K(h(\theta))$ is dense in $K_P(h(\theta))$ and the latter is complete so $L_Q=K_P(h(\theta))$.
Then we look at $O_{L_Q}$ the closure of $O_L$ in $L_Q$. The main theorem is that it is a DVR, with uniformizer $(\pi_Q)$. Take some $b_Q\in O_L$ such that $O_L/(\pi_Q)=O_L/Q=(O_K/P)[b_Q+P]$. Let $S_Q$ be some representatives in $O_K[b_Q]$ of the residue field $O_L/Q$. By definition of a complete DVR $$O_{L_Q} = \{ \sum_{n\ge 0} s_n \pi_Q^n, s_n\in S_Q\}= \sum_{j=0}^{f(Q/P)-1} b_Q^j \sum_{m=0}^{e(Q/P)-1} \pi_Q^m O_{K_P}=O_{K_P}[b_Q,\pi_Q]$$ It works the same way when replacing $b_Q$ by $b_Q^p+\pi_Q$ and $\pi_Q$ by $(b_Q^p+\pi_Q)^{N(Q)-1}-1$ (prove that the latter is an uniformizer if $p$ isn't an uniformizer) thus $$O_{L_Q} = O_{K_P}[b_Q^p+\pi_Q]$$