I need to describe the kernel of the canonical group homomorphism $GL(n, \mathbb{Z}) \to GL(n, \mathbb{Z}_{m})$.
Generally speaking, the kernel would be the set of all matrices in $GL(n,\mathbb{Z})$ whose entries along the main diagonal modulo $m$ would equal 1, with all other entries modulo $m$ equaling zero. For example, in the case where $n = 2$, the kernel of this homomorphism would be the set of all $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ such that $\begin{pmatrix} a_{11} \mod m = 1 && a_{12} \mod m = 0 \\ a_{21} \mod m = 0 & &a_{22}\mod m = 1 \end{pmatrix}$.
Is this all there is to this problem? Or is there something additional that I need to prove? It is an exercise in a section on Ideals and Isomorphism Theorems for Rings, so I thought it was kind of weird that I was able to come up with a solution to this problem that has nothing to do with ideals or ring isomorphisms. Therefore, I feel like I'm missing something...
The kernel of this homomorphism consists of
$ \begin{align} \begin{pmatrix} a_{11} \dots a_{1n} \\ \vdots \, \, \ddots \, \, \vdots \\ a_{n1} \, \dots \, a_{nn} \end{pmatrix} \in GL(n,\mathbb{Z}) \end{align}$
such that
$ \begin{cases} a_{ij} \equiv 0 \, mod \, m \, \, if \, \, i \neq j \\ a_{ij} \equiv 1 \, mod \, m \, \, if \, \, i = j \end{cases} $
By the first isomorphism theorem for groups, the kernel of this homomorphism is a normal subgroup of $GL(n, \mathbb{Z})$.