Primal attack On Kyber algorithm specifications

Question

In the Kyber Algorithm specifications document, chapter 5.1.2 primal attack, it says that:

Given the matrix LWE instance $(A,b=As+e)$, one builds the lattice $\Lambda = \{x\in \mathbb{Z}^{m+kn+1}:(A\vert I_m\vert -b)x=0 \,mod \,q\}$ of dimension $d=m+kn+1$, volume $q^m$, and with a unique-SVP solution $v=(s,e,1)$ of norm $\lambda\approx\varsigma\sqrt{kn+m}$, where $\varsigma$ is the standard deviation of the individual secret/error coefficients.

To my mind, $\Lambda = q\,L(A\vert I_m\vert -b)^{\lor}$, where $L^{\lor}$ represent the dual of lattice $L$.and $L(A\vert I_m\vert -b)^{\lor}=L(b_1, \dots, b_{m})$, where $b_i$ is the column vector with $b_i^{(j)}$ represtion its j-th component(start from 0), satisfy $$b_i^{(nk+i-1)}=1, b_i^{(nk+j)}=0, j\ne i-1, i=1, \dots, m.$$ So $\Lambda =L(q\,b_1,\dots,q\,b_{m})$...

Is that correct?

If it is right, then how can $v=(s,e,1)$ be linearly represented by $q\,b_1,\dots,q\,b_{m}$?

If this is not right, how can I figure out

What is the basis of $\Lambda$ ?

and

Why is the volume $q^m$?

Or, are there any references?

Answer 1

Whereas it is true that each of the $b_i$ vectors lies in the dual to the lattice generated by the rows of $A|I_m|b$, there are vectors in the dual that do not lie in the span of the $b_i$. For instance, let $\mathbf w\in \mathbb Z^{kn}$ be any vector in the row span of $A^{-1}$ then $(\mathbf w|0|0)$ is also in the dual. Likewise the vector $\frac1q(s,e,1)$ lies in the dual, but not in the span of the $b_i$.