Intersection of two lattices

Question

Suppose that $\Lambda_1, \Lambda_2$ are two sub-lattices of $\mathbb{Z}^n$ of full rank, defined by congruence modulo a prime $p$. That is, there exist two vectors with integer entries $\mathbf{a}, \mathbf{b}$ such that

$$\displaystyle \Lambda_1 = \{\mathbf{x} \in \mathbb{Z}^n : \mathbf{a} \cdot \mathbf{x} \equiv 0 \pmod{p} \}$$

and

$$\displaystyle \Lambda_2 = \{\mathbf{x} \in \mathbb{Z}^n : \mathbf{b} \cdot \mathbf{x} \equiv 0 \pmod{p}\}.$$

Further, suppose that $\mathbf{a} \not \equiv \mathbf{b} \pmod{p}$. It is easy to see that $\det(\Lambda_1) = \det(\Lambda_2) = p$.

Now let $\Lambda = \Lambda_1 \cap \Lambda_2$. It is clear that $\Lambda$ is a lattice. If it is a lattice of full rank, then $\det(\Lambda)$ is a multiple of $p$. Under what circumstances can we ensure that $\det(\Lambda)$ is a multiple of $p^2$?

Answer 1

You can define two homomorphisms $\phi_1\colon\mathbb Z^n\to \mathbb Z/p\mathbb Z$ and $\phi_2 \colon\mathbb Z^n\to \mathbb Z/p\mathbb Z$ given by $\phi_1(\mathbf x)=\mathbf a\cdot \mathbf x\bmod p$ and $\phi_2(\mathbf x)=\mathbf b\cdot \mathbf x\bmod p$. Your two lattices (necessarily full-dimensional; and also of co-volume $p$ in case $a\not\equiv 0\pmod p$ and $b\not\equiv 0\pmod p$) are the kernels of $\phi_1$ and $\phi_2$ respectively. If you define $\Phi(\mathbf x)=(\phi_1(\mathbf x),\phi_2(\mathbf x))$, you obtain a map $\mathbb Z^n\to (\mathbb Z/p\mathbb Z)^2$. The range of $\Phi$ is a subgroup of $(\mathbb Z/p\mathbb Z)^2$. The subgroups of $(\mathbb Z/p\mathbb Z)^2$ are vector subspaces, so are the trivial subgroup, the entire group, or a line.

The case of a line is precisely the situation where $\mathbf a\bmod p$ and $\mathbf b\bmod p$ are linearly dependent in $\mathbb Z^n/p\mathbb Z^n$.

Answer 2

Shouldn't this happen as long as $\mathbf{a}$ and $\mathbf{b}$ are linearly dependent mod $p$? For if so, you are talking about points in $\mathbb{Z}^n$ which reduce mod $p$ in two some $n-2$ dimensional subspace of $\mathbb{F}_p^n$. After choosing a basis, this amounts to choosing $p^{n-2}$ coefficients. Thus the lattice contains $p^{n-2}$ points in any box $p\times\stackrel{n}{\cdots}\times p$ box.