Given a matrix $\mathbf{A} \in \mathbb{Z}^{n \times m}$, $m$ sufficiently large with respect to $n$ and prime $q$. The rows of $\mathbf{A}$ are linearly independent with high probability. In MR09 the authors state that the number of vectors in $\mathbb{Z}_q^m$ belonging to the $q$-ary lattice $\Lambda_q^\intercal(\mathbf{A})$ is $q^{m-n}$ and therefore it follows that $\text{det}(\Lambda_q^\intercal(\mathbf{A})) = q^n$.

I understand that the dimension of the kernel of $\mathbf{A}$ (which is equivalent to the the dimension of the dual lattice) is $m-n$. However, I don't understand how the volume immediatley follows and would be thankful for an explanation.


1 Answer 1


Note that for a lattice $L\subseteq\mathbb{R}^n$, $\det(L)$ is the volume of a fundamental domain. There are often many of these objects, but there are two that are typically of primary interest:

  1. The Voronoi Cell $\mathcal{V}(L) = \{x\in\mathbb{R}^n\mid \forall \ell\in L\setminus\{0\}, \lVert x\rVert_2\leq \lVert x-\ell\rVert_2\}$, e.g. it is the points in $\mathbb{R}^n$ that are closer to 0 than any other lattice point.
  2. The Fundamental Parallelpiped --- for a basis $\mathbf{B}$ of the lattice, this is the set $\mathbf{B}[0,1)^n$ (or sometimes $\mathbf{B}[-1/2,1/2)^n$.

Up to some issues on the boundary, a fundamental domain "tiles space", meaning that the sum

$$L + D = \mathbb{R}^n$$

is a partition. If we assume the lattice is $q$-ary, we can reduce everything mod $q$ to get that $(L\bmod q) + (D\bmod q) = \mathbb{R}/q\mathbb{R}^n$ is a partition as well [1]. Taking volumes, we get that $$|L\bmod q||D\bmod q| = q^n\implies |L\bmod q| = \frac{q^n}{|D|} = \frac{q^n}{\det L}.$$ What you want then follows from using that the lattice is $m$-dimensional, and has $|L\bmod q| = q^{m-n}$ points, so the determinant must be $q^n$.

[1] There may be some issues with particularly irregular fundamental domains $D$ here (in particular fundamental domains that are not contained in $[-q/2, q/2)^n$, but if you let $D$ be the Voronoi cell everything seems fine, and I am not even sure if this worry I mention is for a particular reason.

  • $\begingroup$ Thanks that helps a lot! Not sure if I got the worry you brought up, but the rest makes sense. $\endgroup$
    – MrCrab
    Commented Oct 1, 2021 at 8:16
  • $\begingroup$ You need that the modular reduction preserves being a partition. This is clear to me provided that $D\subseteq [-q/2, q/2)^n$, as then modular reduction "just" squishes all of these regions together (and is the identity on the copy of $D$ at the origin). The voronoi cell of a $q$-ary lattice always satisfies this, which is why I mentioned it in particular. It is plausible that modular reduction preserves being a partition even for more general $D$, but then modular reduction is no longer the identity on the copy of $D$ at the origin, and I haven't thought through what happens then. $\endgroup$
    – Mark Schultz-Wu
    Commented Oct 1, 2021 at 8:46

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