matrix interesting problem about determinant

Question

Hi guys i am a new member and i need help about this problem . Let $A=(a_{ij}) \in M_{n,n}(\mathbb{Z})$ , $p$ a prime, and $\tilde{A}\in M_{n,n}(\mathbb{F}_p)$ be the matrix obtained by reducing every $a_{ij}$ modulo $p$ . If $rank(\tilde{A})=r$ , how can I show that $\det(A)$ is divisible by $p^{n−r}$ ?

Answer 1

$\tilde{A}$ has rank $r$ if and only if there exists invertible matrices $g,h\in M_{n,n}(\mathbb{F}_p)$ such that $$ g\tilde{A}h=\begin{bmatrix} I_r & 0\\ 0 & 0 \end{bmatrix} $$ where $I_r$ denotes the $r\times r$-identity matrix and $0$'s denote $0$ matrices of correct sizes. This matrix is called the Smith Normal Form of $\tilde{A}$.

Pick $G,H\in M_{n,n}(\mathbb{Z})$ such that $\tilde{G}=g$ and $\tilde{H}=h$. Then, $\tilde{GAH}=g\tilde{A}h$, so each entry of $GAH$ is divisible by $p$, except the first $r$ diagonal entries. In particular, $n-r$ columns of $GAH$ is divisible by $p$ and the multilinearity of $\det$ implies that $\det(GAH)$ is divisible by $p^{n-r}$. On the other hand, $\det(GAH)=\det(G)\det(A)\det(H)$ and $\det(G),\det(H)$ are not divisible by $p$ since $g,h$ are invertible. Thus, $p^{n-r}$ divides $\det(A)$.