# Volume of an NTRU lattice

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?

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