Is there such a basis for a $p$-adic lattice? I wonder where I can find the following theorem? Is it right?
Let $\mathcal L$ be a full rank lattice in $K$.
Then there exists a $\mathbb Z_p$-basis $(e_1, \ldots, e_n)$ of 
$\mathcal L$ and some real numbers $(r_1, \ldots, r_n)$ such that the 
norm on $K$ is given by:
$$\Vert \alpha_1 e_1 + \cdots + \alpha_n e_n \Vert =
 \max \big(|\alpha_1|_p r_1, \ldots, |\alpha_n|_p r_n\big)$$
for $\alpha_1, \ldots, \alpha_n \in \mathbb Q_p$.
 A: *

*The OP meant $K/\Bbb{Q}_p$ a finite extension of degree $n$ and  $$a\in K, \qquad \|a\| = |a|_p\overset{def}= |N_{K/\Bbb{Q}_p}(a)|^{1/n}_p$$ where $K,\|.\|$ is seen as a $n$-dimensional normed $\Bbb{Q}_p$-vector space.


Let $\pi\in O_K$ be an uniformizer (an element of minimal non-zero valuation) and a basis of the residue field  $O_K/(\pi)= \sum_{j=1}^f b_j\Bbb{F}_p,b_j \in O_K^\times$ then $|\pi|_p^e = |p|_p$ and $$K = \sum_{j=1}^f \sum_{m=0}^{e-1} b_j \pi^m \Bbb{Q}_p, \qquad \|\sum_{j=1}^f \sum_{m=0}^{e-1} b_j \pi^m x_{j,m}\| = \sup_{j,m} |b_j \pi^m|_p | x_{j,m}|_p, x_{j,m} \in \Bbb{Q}_p$$


*

*If $K/\Bbb{Q}_p$ is unramified that is $e=1, \pi = p$ we are in the following situation : 
Let the vector space $\Bbb{Q}_p^n$ with the norm $$\|x\|= \sup_j |x_j|_p$$
This norm is the most natural because it is $GL_n(\Bbb{Z}_p)$ invariant : $\|M x\| \le \|x\| = \|M^{-1}(Mx)\|\le \|Mx\|$ where $GL_n(\Bbb{Z}_p)$ is a maximal compact subgroup of $GL_n(\Bbb{Q}_p)$ (in the same way that the euclidean norm on $\Bbb{R}^n$ is invariant for a maximal compact subgroup $O_n(\Bbb{R})$ of $GL_n(\Bbb{R})$)

*A full rank $\Bbb{Z}_p$-lattice is of the form $$L = A \Bbb{Z}_p^n, \qquad A \in GL_n(\Bbb{Q}_p)$$
The smith normal form gives $$A = U D V, \qquad U,V \in GL_n(\Bbb{Z}_p),D\in diag(\Bbb{Q}_p^{*n}),  \qquad L =UD \Bbb{Z}_p^n$$ 
Whence $$ \|UDy\| = \|Dy\| = \sup_j D_j |y_j|_p$$

  
*
  
*If $K/\Bbb{Q}_p$ is ramified it doesn't work anymore because there is no isomorphism $K,\|.\| \to \Bbb{Q}_p^n, \|.\|$. 
  

For example $$K = \Bbb{Q}_p(p^{1/2}) , \qquad L=(1+p^{1/2}) \Bbb{Z}_p + p \Bbb{Z}_p$$ any basis of $L$ is of the form $$L = (1+p^{1/2}+pu)\Bbb{Z}_p+ p(v+p^{1/2}w)\Bbb{Z}_p, \qquad u,w\in  \Bbb{Z}_p[p^{1/2}],v \in \Bbb{Z}_p^\times$$ and $$\| (1+p^{1/2}+pu)p - p(v+p^{1/2}w)v^{-1}\|\qquad\qquad$$ $$\qquad\qquad \ne \ \sup| (1+p^{1/2}+pu)|_p|p|_p ,|p(v+p^{1/2}w)|_p|v^{-1}|_p$$
