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


How to prove $X_{1}\oplus X_{1} \cong X_{2}\oplus X_{2}$ and write this isomorphism explicitly ?

  • $\begingroup$ Do you not just want to find an isomorphism $X_1 \simeq X_2$? $\endgroup$ Oct 10, 2018 at 22:56
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    $\begingroup$ @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. $\endgroup$ Oct 11, 2018 at 9:03
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    $\begingroup$ @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. $\endgroup$ Oct 16, 2018 at 5:46
  • $\begingroup$ @TorstenSchoeneberg, All: In case you find the time for another look at this I will appreciate it. $\endgroup$ Dec 12, 2018 at 8:58
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    $\begingroup$ @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 $n\equiv1\pmod 8$. $\endgroup$ Dec 14, 2018 at 13:47

1 Answer 1


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 $X_1 \oplus X_1$.

  • $\begingroup$ 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! $\endgroup$ Dec 14, 2018 at 20:02
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    $\begingroup$ Anyway, I will let the bounty sit for a few more days at least. Might as well maximize the extra exposure! $\endgroup$ Dec 14, 2018 at 20:03
  • $\begingroup$ 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. $\endgroup$ Dec 15, 2018 at 9:42
  • $\begingroup$ 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 $\Bbb{Q}_2$ :-) $\endgroup$ Dec 15, 2018 at 9:45
  • $\begingroup$ 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 $0$? $\endgroup$
    – MAS
    Nov 8, 2020 at 18:01

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