# 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 ?

• Do you not just want to find an isomorphism $X_1 \simeq X_2$? Oct 10, 2018 at 22:56
• @Torsten: $X_1$ and $X_2$ are not isomorphic because there are no isotropic vectors in $X_2$. Basically because $-3$ is not a square in $\Bbb{Z}_2$. It is easy to find several 2-dimensional subspaces of $X_2\oplus X_2$. such that the restriction of the bilinear form to such a subspace is zero. I used the existence of $\sqrt{-7}\in\Bbb{Z}_2$. But I couldn't quite get an isometry for I don't remember this piece of theory (and didn't have the time to look it up from O'Meara or another tome). I'm thinking about placing a bounty here. If you see a way forward, I will do it. Oct 11, 2018 at 9:03
• @JyrkiLahtonen: Interesting. I follow all your steps, and agree there are what one might call totally isotropic rank 2 sublattices in $X_2 \oplus X_2$. If we had vector spaces, that would mean we're abstractly done; the question is if the usual trick of splitting off hyperbolic planes works in the $\mathbb{Z}_2$-integral setting here as well, maybe with minor adjustments. I am actually quite confident it would, but I have not had time to write down the steps explcitly. Oct 16, 2018 at 5:46
• @TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it. Dec 12, 2018 at 8:58
• @san I think there are many. Let $k=0$ for simplicity. Then the vector $(2,1)$ in $X_2$ has squared length $14$. Therefore $((2,1),(\sqrt{-7},0))\in X_2\oplus X_2$ is isotropic. Recall that $\sqrt n\in\Bbb{Z}_2$, $n$ a square-free integer, if and only if $n\equiv1\pmod 8$. Dec 14, 2018 at 13:47

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 $$X_1 \oplus X_1$$.
• Looks good. I will, of course, double check everything later :-). The first time I got the off-diagonal entries of your inner product on $A$ with opposite signs (to that of $X_2$). That is, of course, trivial to fix - even I can do it! Dec 14, 2018 at 20:02
• Question: Are you sure that $A\oplus A$ is the direct sum $H\oplus I$? The matrix $$\pmatrix{1&u\cr 1&u\zeta_3\cr}$$ has determinant $u(\zeta_3-1)$. A problem with this is that $u=(1\pm\sqrt{-7})/2$ is not a 2-adic unit. Dec 15, 2018 at 9:42
• Never mind! One of them actually is a unit - for an appropriate choice of $\sqrt{-7}$. I was caught thinking that the choice of sign doesn't affect the value, "conjugates" you see, and their product is two. But they aren't conjugates over $\Bbb{Q}_2$ :-) Dec 15, 2018 at 9:45
• I didn't understand the calculation how $H$ is isotropic ? We have,$\left\langle x,ux \right\rangle=x \overline{ux}+ux \bar x$. But how to show this is $0$?