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Say $L_1,L_2$ are contained in $\mathbb Z^r$ with

\begin{gather*} \operatorname{rank}(L_1) = \operatorname{rank}(L_2) = r, \\ \gcd(\det(L_1), \det(L_2)) = 1. \end{gather*}

How do I prove $\operatorname{det}(L_1 \cap L_2) = \det(L_1)\cdot \det(L_2)$?

It's easy to show that $\det(L_1 \cap L_2)$ is divisible by $\det(L_1) \cdot \det(L_2)$. My question is how to prove $\det(L_1 \cap L_2) \leq \det(L_1) \cdot \det(L_2)$?

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I know this is really old, but it's still an interesting question that is semi-related to something I am working on. I will refer to the lattices as $L_0$ and $L_1$, so I can write $L_i$ and $L_{1 - i}$ to refer to both of them without making a particular choices of indices. I will use $L \leq L'$ to denote sublattice containment.

Let the operation $L_0 + L_1$ be the minkowski (not direct) sum of lattices. If $L_i = \mathcal{L}(M_i)$ is the lattice generated by the matrix $M_i$, then $L_0 + L_1 = \mathcal{L}([M_0 | M_1])$ is the lattice generated by the concatenation of the matrices. You can equivalently view this as the sum of all pairs of lattice points, i.e. $L_0 + L_1 = \{l_0 + l_1 | l_0\in L_0, l_1\in L_1\}$. This operation can be seen as "dual" to the intersection in a formal sense [1], which I will exploit below.

Note that $L_0\cap L_1\leq L_i$, so the quotient $L_i / (L_0\cap L_1)$ is well-defined. Moreover note that $|L_i / (L_0\cap L_1)| = [L_i : L_0 \cap L_1] = \det(L_0\cap L_1)/\det(L_i)$. By the second isomorphism theorem (viewing lattices as $\mathbb{Z}$-modules), we have that: \begin{align*} L_i / (L_0\cap L_1) &\cong (L_0 + L_1) / L_{1-i}\\ \implies \det(L_0\cap L_1)/\det(L_i) &= \det(L_{1-i})/\det(L_0 + L_1)\\ \implies \det(L_0 \cap L_1) &= \frac{\det(L_0)\det(L_{1})}{\det(L_0 + L_1)} \end{align*} This holds for any pair of lattices $L_0, L_1$ such that $L_0 \cap L_1$ and $L_0 + L_1$ are lattices (interestingly, we have not assumed the lattices are integral yet). This suggests that all we have to do is prove that $\det(L_0 + L_1) = 1$, and we have:

  • $L_i$ are integral (i.e. $L_i \leq \mathbb{Z}^r$)
  • The GCD condition

To work with. Note that integral lattices satisfy $\det(L)\mathbb{Z}^r \leq L \leq \mathbb{Z}^r$. One can show that $L_i \leq L_i'$ implies that $L_0 + L_1 \leq L_0' + L_1'$. Applying this to the above, we get that: $$\det(L_0)\mathbb{Z}^r + \det(L_1)\mathbb{Z}^r \leq L_0 + L_1 \leq \mathbb{Z}^r$$ It is not hard to show that $a\mathbb{Z}^r + b\mathbb{Z}^r = \mathsf{gcd}(a, b)\mathbb{Z}^r$, so upon using our GCD condition we get that $\mathbb{Z}^r \leq L_0 + L_1 \leq \mathbb{Z}^r$, and therefore $\det(L_0 + L_1) = 1$.


[1] The submodules of a module form a modular lattice in the sense of order theory, where $\cap$ is the meet, and $+$ is the join respectively. This is to say that $+$ and $\cap$ are "dual" in the same sense that the set union and intersection are "dual".

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