Volume of an NTRU lattice

Question

Let $K$ be a number field of degree $n$ and $\Lambda^q_h=\{(f,g)\in\mathcal{O}_K\text{ : }fh-g = 0\bmod q\mathcal{O}_K\}$, where $h$ is an NTRU public key. Then $\{(1,h),(0,q)\}$ generates a lattice. I've found it stated in the literature that $Vol(\Lambda^q_h) = Vol(\mathcal{O}_K)^2q^n$ (e.g. here), but how does the proof of this statement run? Or where can I find a proof?

Answer 1

This is a standard computation in number theory. The idea behind it is that the matrix you have written down is a basis of the lattice as an $\mathcal{O}_K$-module, but to find the volume you first find a $\mathbb{Z}$-basis for the lattice, and then do "standard" computations with this. If $B$ is a $\mathbb{Z}$-basis of $\mathcal{O}_K$, then one has that:

$$B' = \begin{pmatrix}B & hB\\ 0 & qB\end{pmatrix} = \begin{pmatrix}1 & h\\ 0 & q\end{pmatrix}\otimes B$$

is a $\mathbb{Z}$-basis for your lattice. You can then compute the volume of this in the "standard" way, e.g. taking determinants, to get that:

$$\det B' = q^{\deg \mathcal{O}_K}(\det B)^2$$

which is precisely the expression you have written.

You can probably find this in many (if not all) textbooks on algebraic number theory. For example, I believe this is a corollary of lemma 2.23 of Milne's notes, but there are a number of abstractions one would have to unwind to verify this.