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To be specific my question is how to determine $| \mathbb{Z}^m/\Lambda_q^{\bot}(A)|$ which $$\Lambda_q^{\bot}(A) = \{y =\mathbb{Z}^m \mid Ay =0 \bmod q \}$$ for full rank $A_{q}^{n*m}$ that $m\geq n$ and $q$ a prime(a q-ary lattice). And more generally what is $| \mathbb{Z}^m/\Lambda|$ when $\Lambda$ is a additive subgroup of $\mathbb{Z}^m$(a general integer lattice)?

alfred
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  • To find the structure of ${\mathbb Z}^m/\Lambda$, you calculate the Smith Normal Form of the matrix defining $\Lambda$. – Derek Holt Jan 24 '21 at 16:14
  • Could you please explain more? What do you mean by the matrix defining $\Lambda$? – alfred Jan 24 '21 at 16:43

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The quotient $\mathbb{Z}^m/\Lambda$ is the quotient group of cosets $x + \Lambda$ with $x \in \mathbb{Z}^m$. Each $x+ \Lambda$ is a shift (translation) of $\Lambda$ by $x$. Then $|\mathbb{Z}^m/\Lambda|$ is the volume of $\Lambda$. This is more extensive on this.

For $\Lambda = \Lambda^{\perp}_q(A)$, note that $\Lambda$ is the kernel of the composition homomorphism $a:\mathbb{Z}^m \to (\mathbb{Z}/q\mathbb{Z})^m \to (\mathbb{Z}/q\mathbb{Z})^n$, where the first arrow is reduction modulo $q$ and the second arrow sends $x$ to $Ax$. Since $A$ has full rank, $a$ is surjective. Therefore, by the isomorphism theorem, there is a group isomorphism $\mathbb{Z}^m/\ker(a) \simeq (\mathbb{Z}/q\mathbb{Z})^n$. Therefore it follows $|\mathbb{Z}^m/\Lambda^\perp_q(A)| = |\mathbb{Z}^m/\ker(a)| = q^n$.

luca_no
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