Question related to a determinant of lattice There is this equation related to the determinant of lattice and I 
have been stuck on it for a little while. I would greatly appreciate if someone could explain to me how to prove it!
Let $\Lambda$ be a lattice of full rank in $\mathbb{R}^l$
and let $\Pi$ be its dual lattice.
Suppose $\mathbf{p}$ is a primitive point in $\Pi$.
Denote $V = \mathbf{p}^{\perp}$ and $\Lambda' = \Lambda \cap V$.
How do you prove that $\det (\Lambda') = |\mathbf{p}|\det (\Lambda)$?
Thank you!
 A: We note that $\Lambda' = \{\textbf{x} \in \Lambda : \textbf{p} \cdot \textbf{x} = 0\}.$ By the theory of Geometry of Numbers, we can choose a basis for $\Lambda$, say $\textbf{b}_1, \cdots, \textbf{b}_n$ and a corresponding basis $\textbf{b}_1^\ast, \cdots, \textbf{b}_n^\ast$ of the polar lattice such that $\textbf{b}_i \cdot \textbf{b}_j^\ast = \delta_{ij}$, where $\delta_{ij}$ is the Kronecker delta. We may suppose, without loss of generality, that $\textbf{p} = \textbf{b}_n^\ast$ say. Let us write $\textbf{a} = \textbf{p}/|\textbf{p}|$. Then consider the matrices $(\textbf{b}_1, \cdots, \textbf{b}_n)$ and $(\textbf{b}_1, \cdots, \textbf{b}_{n-1}, \textbf{b}_n^\ast)$. Multiply the first row of each matrix by $a_1$, the second row by $a_2$ etc. (in effect taking the dot product of $\textbf{b}_i$ with $\textbf{a}$) to obtain that
$$\displaystyle a_1 \cdots a_n \det(\Lambda) = a_1 \cdots a_n\det(\textbf{b}_1, \cdots, \textbf{b}_n) = (a_1 \cdots a_n)(a_1 b_{n,1} + \cdots + a_n b_{n,n}) \det(C),$$
where $C$ is the lower left hand $(n-1) \times (n-1)$ minor resulting from above, and that
$$\displaystyle a_1 \cdots a_n \det(\Lambda') = a_1 \cdots a_n |\textbf{p}| \det(C).$$
Equating the two and noting that by construction we have $a_1 b_{n,1} + \cdots + a_n b_{n,n} = 1$, we get the desired result.
