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 $p^2$?



Intersection of two lattices
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 $\mathbb Z^n/p\mathbb Z^n$.


