How to prove that two $p$-adic lattices are isomorphic? Let $\mathbb{Z}_{p}$ be the ring of p-adic integers.
A pair $(L,<>)$ is called lattice if $L$ be a free $\mathbb{Z}_{p}$ module of finite rank and $<>:L×L \to \mathbb{Z}_{p}$be a nondegenerate symmetric bilinear
form on $\mathbb{Z}_{p}$.
Two lattices $L_1,L_2$ is called isomorphic if there exist  isomorphism of $\mathbb{Z}_{p}$ module $L_{1} \to L_{2}$ preserving $<>$.
Let $X_1,X_2$ be 2-adic lattices of rank 2 determined by matrices
$\begin{pmatrix}0&2^k&\\2^k&0&\end{pmatrix},\begin{pmatrix}2^{k+1}&2^k&\\2^k&2^{k+1}&\end{pmatrix}$.
How to prove $X_{1}\oplus X_{1} \cong X_{2}\oplus X_{2}$ and write this isomorphism explicitly ?
 A: I fiddled a little with your $\mathbb{Z}_2$-lattices, and believe that a lattice isomorphic of $X_1 \oplus X_1$ with $X_2 \oplus X_2$ is not hard to make. Clearly we may assume $k = 0$. The lattice $X_2$ may be identified with the additive group of the ring $A = \mathbb{Z}[\zeta_3]$, which is a quadratic ring extension of $\mathbb{Z}_2$---defined by$$\zeta_3^2 + \zeta_3 + 1 = 0$$---the inner product being given by$$\langle a, b\rangle = a\overline{b} + b\overline{a},$$where the overline is the automorphism of $A$ sending $\zeta_3$ to its inverse and each element of $\mathbb{Z}_2$ to itself. Hence $X_2 \oplus X_2$ can similarly be identified with $A \oplus A$. Now it is easy to see that $A$ has an element $u$ with $u\overline{u} = -1$. Then the subgroup$$H = \{(x, u.x) \mid x \in A\}$$of $A$ is totally isotropic, and so is$$I = \{(x, u.x.\zeta_3) \mid x \in A\},$$while $A\, \oplus$ is the direct sum of $H$ and $I$. The inner product is unimodular, so it identifies $H$ with the $\mathbb{Z}_2$-dual of $I$; hence one can now conclude that $A \oplus A$ is as a lattice isomorphic to $X_1 \oplus X_1$.
