Suppose $K=\mathbb{Q}_p(\sqrt p)$ be the quadratic extension of the p-adic field $\mathbb{Q}_p$ for $p>2$ and let $O_K$ be the ring of integers (free $\mathbb{Z}_p$-module). We know that $\mathbb{Q}_p$ has three quadratic extensions for primes $p \geq 3$ which are $\mathbb{Q}_p(\sqrt n),~ \mathbb{Q}_p(\sqrt{p}),~ \mathbb{Q}_p(\sqrt{pn})$, where $n$ is the degree of extension.
What are the integral basis of the ring of integers $O_K$ in the $3$ quadratic extensions $\mathbb{Q}_p(\sqrt n), \ \mathbb{Q}_p(\sqrt{p}), \ \mathbb{Q}_p(\sqrt{pn})$ ?
I know the case when it was rational quadratic extension. But in this case, it becomes difficult.
Best Answer
Let $K/\Bbb{Q}_p$ be a finite extension and $L$ its normal closure. $|.|_p$ extends uniquely to $L$ through the rule $|a|_p = |\sigma(a)|_p$ for all $a\in L,\sigma\in Gal(L/\Bbb{Q}_p)$. Let $O_K = \{ a \in K, |a|_p\le 1\}$ and $\pi_K \in O_K$ such that $\forall a \in O_K,|a|_p > |\pi_K|_p \implies |a|_p = 1$. Then $O_K/(\pi_K)$ is a finite field. Take $b_1,\ldots,b_m \in O_K$ whose reduction $\bmod \pi_K$ is a $\Bbb{F}_p$-basis of $O_K/(\pi_K)$. Then $$ O_K = \sum_{l=0}^{d-1} \sum_{j=1}^m \pi_K^l b_j \Bbb{Z}_p$$ where $md = [K:\Bbb{Q}], m = [O_K/(\pi_K):\Bbb{F}_p], |\pi_K|^d_p=|p|_p$.
Moreover we can take $b_j$ as $p^m-1$ roots of unity obtaining $$O_{\Bbb{Q}(\zeta_{p^m-1})} = \sum_{j=1}^m b_j \Bbb{Z}_p, \qquad O_K = \sum_{l=0}^{d-1} \pi_K^l O_{\Bbb{Q}(\zeta_{p^m-1})}$$