Suppose I have a $n\times n$ matrix $A$ with integer entries. Suppose $\det(A)=1$.
I was thinking how determinant will change if I reduce the entries of $A$ modulo $k$ where $k$ is a positive integer.
What will be the determinant of this reduced matrix?
I think $\det$ of the reduced matrix will again be $1$.
This is what I have tried so far.
I took a arbitrary $2\times 2$ matrix $$A=\begin{pmatrix}a &b\\c&d\end{pmatrix}$$then we have $ad-bc=1$. Consider the reduced matrix(after going modulo each entry by say $k$)
$$\bar{A}=\begin{pmatrix}\bar{a} &\bar{b}\\\bar{c}& \bar{d}\end{pmatrix}$$ where $\bar{*}:=* \mod k$
Then we are interested in $\bar{a}\bar{d}-\bar{b}\bar{c}$
Now $$ab \mod n = (a \mod n) · (b \mod n)$$ and $$(a + b) \mod n = (a \mod n + b \mod n) \mod n$$
Using these two results we conclude that $\det(\bar{A})=1$
Since $\det$ is a polynomial expression in entries of matrix. We can generalize the process for any $n\times n$ matrix.
Hence in conclusion I propose this lemma.
Let $A=(a_{ij})$ be a $n\times n$ integer matrix with $\det(A)=1$ then consider the matrix $\bar{A}$ whose $ij$ th entry is $a_{ij}\mod k$ where $k$ is a positive integer then we can conclude that $\det(\bar{A})=1$
Am I correct?
You have shown $\overline{ab} = \overline{\overline a \overline b}$ and $\overline{a+b} = \overline{\overline a + \overline b}$. Then you can express $\det(A)$ as in the big formula and obtain $$\overline{\det(\overline A)} = \overline{\sum_{\tau \in S_n}\prod_{i=1}^n\overline{A_{i,\tau(i)}}} = \overline{\sum_{\tau \in S_n}\prod_{i=1}^nA_{i,\tau(i)}} = \overline{\det(A)}$$ with the special case $\overline{\det(A)} = 1 \implies \det(\overline A) = 1$.