# determinant of intersection of two lattices

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 $\det(L_1 \cap L_2) \leq \det(L_1) \cdot \det(L_2)$?

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 , 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 $$\det(L_0 + L_1) = 1$$.

 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".