the integral basis of the ring of integers $O_K$ in the $3$ quadratic extensions of p-adic field $ \mathbb{Q}_p$

p-adic numbers (algebraic number theory):

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 $3$ quadratic extension for $p>2$ which are $\mathbb{Q}_p(\sqrt n), \ \mathbb{Q}_p(\sqrt{p}), \ \mathbb{Q}_p(\sqrt{pn})$, where $n$ is the degree of extension.

My question is-

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 tough.

Solution 1:

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 $$O_{\Bbb{Q}(\zeta_{p^m-1})} = \sum_{j=1}^m b_j \Bbb{Z}_p, \qquad O_K = \sum_{l=0}^{d-1} \pi_K^l O_{\Bbb{Q}(\zeta_{p^m-1})}$$

Solution 2:

You can compute directly as in the case of quadratic extensions of $\mathbb{Q}$. In fact, it is simpler than the rational case. This is because $\frac 12 \in \mathbb Z_p =\{ x \in \mathbb{Q}_p : v_p(x) \geq 0\}$. Concretely, $$ \frac 12 = \frac{p+1}{2} + \sum_{n=1}^{\infty} \frac{p-1}{2} p^n$$

Such phenomenon does not occurs for the quadratic extensions of $\mathbb{Q}$.

Let $d \in \{n, p, pn \}$ where is $n \in \mathbb{Z}$ is not a square modulo $p$ and suppose $a + b \sqrt{d}$ is integral over $\mathbb{Z}_p$. Consider the norm and trace of $\mathbb{Q}_p(\sqrt{d}) / \mathbb{Q}_p$. Then both $\mbox{Tr}(a+b\sqrt{d})=2a$ and $\mbox{N}(a+b\sqrt{d})=a^2-b^2d$ lie in $\mathbb{Z}_p$. It is immediate that $a = \frac12 \cdot 2a \in \mathbb{Z}_p$.

On the other hand $4b^2d = (4b^2d - 4a^2) + (2a)^2 \in \mathbb{Z}_p$. It follows that $v_p(4b^2d) = v_p(4) + 2v_p(b) + v_p(d) = 2v_p(b)+1 \geq 0$. Hence $v_p(b) \geq 0$. This shows that the ring of integers in $\mathbb{Q}_p(\sqrt{d})$ equals to $\mathbb{Z}_p[\sqrt{d}]$. Hence $\{1, \sqrt{d} \}$ is an integral basis.